In Young's experiment a mixture of orange light (611 nm) and blue light (471 nm) shines on the double slit. The centers of the first-order bright blue fringes lie at the outer edges of a screen that is located 0.497 m away from the slits. However, the first-order bright orange fringes fall off the screen. By how much and in what direction (toward or away from the slits) should the screen be moved, so that the centers of the first-order bright orange fringes just appear on the screen

Answers

Answer 1

Answer:

0.5639m

Explanation:

For a young double slit experiment the expression below gives the angular separation for m dark fringe having slit width d and wavelength λ

=sin⁻¹(mλ/d)

mλ /d =y/L

for the first order,

y= mλL/d

For ratio separation y₀/yD=1 and d= 1

y₀/yD= [mλ ₀L₀/d]/[mλD.LD./d]

1=λ ₀L₀/λD.LD.

λD.LD= λ ₀L₀

L₀= λD.LD/ λ ₀..............(1)

Then substitute the given values into (1) we have

L₀=471 *0.497/611

= 0.3831m

Distance by which the screen has to be moved towards the slit is

LD- Lo

0.947-0.3831= 0.5639m


Related Questions

Which is produced around a wire when an electrical current is in the wire? magnetic field solenoid electron flow electromagnet

Answers

Answer:

A. magnetic field

Explanation:

The magnetic field is produced around a wire when an electrical current is in the wire because of the magnetic effect of the electric current therefore the correct answer is option A .

What is a magnetic field ?

A magnetic field could be understood as an area around a magnet, magnetic material, or an electric charge in which magnetic force is exerted.

As given in the problem statement we have to find out what is produced around a wire when an electrical current is in the wire.

The magnetic field is produced as a result when an electrical current is passed through the conducting wire .

Option A is the appropriate response because a wire's magnetic field is created when an electrical current flows through it due to the magnetic influence of the electric current .

Learn more about the magnetic fields here, refer to the link given below;

brainly.com/question/23096032

#SPJ6

A merry-go-round spins freely when Diego moves quickly to the center along a radius of the merry-go-round. As he does this, it is true to say that

Answers

Answer:

A) the moment of inertia of the system decreases and the angular speed increases.

Explanation:

The complete question is

A merry-go-round spins freely when Diego moves quickly to the center along a radius of the  merry-go-round. As he does this, It is true to say that

A) the moment of inertia of the system decreases and the angular speed increases.

B) the moment of inertia of the system decreases and the angular speed decreases.

C) the moment of inertia of the system decreases and the angular speed remains the same.

D) the moment of inertia of the system increases and the angular speed increases.

E) the moment of inertia of the system increases and the angular speed decreases

In angular momentum conservation, the initial angular momentum of the system is conserved, and is equal to the final angular momentum of the system. The equation of this angular momentum conservation is given as

[tex]I_{1} w_{1} = I_{2} w_{2}[/tex]    ....1

where [tex]I_{1}[/tex] and [tex]I_{2}[/tex] are the initial and final moment of inertia respectively.

and [tex]w_{1}[/tex] and [tex]w_{2}[/tex] are the initial and final angular speed respectively.

Also, we know that the moment of inertia of a rotating body is given as

[tex]I = mr^{2}[/tex]    ....2

where [tex]m[/tex] is the mass of the rotating body,

and [tex]r[/tex] is the radius of the rotating body from its center.

We can see from equation 2 that decreasing the radius of rotation of the body will decrease the moment of inertia of the body.

From equation 1, we see that in order for the angular momentum to be conserved, the decrease from [tex]I_{1}[/tex] to [tex]I_{2}[/tex] will cause the angular speed of the system to increase from [tex]w_{1}[/tex] to [tex]w_{2}[/tex] .

From this we can clearly see that reducing the radius of rotation will decrease the moment of inertia, and increase the angular speed.

Two protons moving with same speed in same direction repel each other but what about two protons moving with different speed in the same direction?

Answers

Answer:In the case of two proton beams the protons repel one another because they have the same sign of electrical charge. There is also an attractive magnetic force between the protons, but in the proton frame of reference this force must be zero! Clearly then the attractive magnetic force that reduces the net force between protons in the two beams as seen in our frame of reference is relativistic. In particular the apparent magnetic forces or fields are relativistic modifications of the electrical forces or fields. As such modifications, they cannot be stronger than the electrical forces and fields that produce them. This follows from the fact that switching frames of reference can reduce forces, but it can’t turn what is attractive in one frame into a repulsive force in another frame.

In the case of wires the net charges in two wires are zero everywhere along the wires. That makes the net electrical forces between the wires very nearly zero. Yet the relativistic magnetic forces and fields will be of the same sort as in the case of two beams of charges of a single sign. This is true even in the frame of reference of what we think as the moving charges, that is, the electrons. In the frame of reference moving at the drift velocity of these current-carrying electrons, it is the protons or positively charged ions that are moving in the other direction. Consequently in any frame of reference for current-carrying wires in parallel, the net electrical force will be essentially zero, and there will be a net attractive magnetic force

Explanation:                                                                              

Explanation:

Particles with similar charges (both positive or both negative) will always repel each other, regardless of their speed or direction.

A small omnidirectional stereo speaker produces waves in all directions that have an intensity of 8.00 at a distance of 4.00 from the speaker.

At what rate does this speaker produce energy?

What is the intensity of this sound 9.50 from the speaker?

What is the total amount of energy received each second by the walls (including windows and doors) of the room in which this speaker is located?

Answers

Answer:

A. We have that radius r = 4.00m intensity I = 8.00 W/m^

total power = power/ Area ( 4πr2)= 8.00 w/m^2( 4π ( 4.00 m)2=1607.68 W

b) I = total power/ 4πr2= 8.00 W/m2 ( 4.00 m/ 9.5 m)2= 1.418 W/m2

c) E = total power x time= 1607 . 68 W x 1s= 1607.68 J

Two beams of coherent light start out at the same point in phase and travel different paths to arrive at point P. If the maximum destructive interference is to occur at point P, the two beams must travel paths that differ by:_____

a. a whole number of half-wavelengths.
b. a whole number of wavelengths.
c. an odd number of half-wavelengths.

Answers

Answer:

(B) a whole number of wavelengths.

Explanation:

Two beams of coherent light start out at the same point in phase and travel different paths to arrive at point P. If the maximum destructive interference is to occur at point P, the two beams must travel paths that differ by a whole number of wavelengths.

When the resultant effect of the combination of two identical waves result in their annihilation or complete cancellation of the effect of each other, destructive interference takes place. Hence to have two wave sources producing waves that have the same frequency wavelength and amplitude and which are always in phase with each other or have a constant phase difference are said to be Coherent source

A 0.2-stone is attached to a string and swung in a circle of radius 0.6 m on a horizontal and frictionless surface. If the stone makes 150 revolutions per minute, the tension force of the string on the stone is:

Answers

Answer:

2960 N

Explanation:

Convert rev/min to rad/s:

150 rev/min × (2π rad/rev) × (1 min / 60 s) = 50π rad/s

Sum of forces in the centripetal direction:

∑F = ma

T = m v² / r

T = m ω² r

T = (0.2 kg) (50π rad/s)² (0.6 m)

T = 2960 N

A worker wants to load a 12 kg crate into a truck by sliding the crate up a straight ramp which is 2.5 m long and which makes an angle of 30 degrees with the horizontal. The worker believes that he can get the crate to the very top of the ramp by launching it at 5 m/s at the bottom and letting go. But friction is not neglible; the crate slides 1.6 m upthe ramp, stops, and slides back down.

Required:
a. Assuming that the friction force actingon the crate is constant, find its magnitude.
b. How fast is teh crate moving when it reachesthe bottom of the ramp?

Answers

Answer:

a) The magnitude of the friction force is 55.851 newtons, b) The speed of the crate when it reaches the bottom of the ramp is 2.526 meters per second.

Explanation:

a) This situation can be modelled by the Principle of Energy Conservation and the Work-Energy Theorem, where friction represents the only non-conservative force exerting on the crate in motion. Let consider the bottom of the straight ramp as the zero point. The energy equation for the crate is:

[tex]U_{g,1}+K_{1} = U_{g,2}+K_{2}+ W_{fr}[/tex]

Where:

[tex]U_{g,1}[/tex], [tex]U_{g,2}[/tex] - Initial and final gravitational potential energy, measured in joules.

[tex]K_{1}[/tex], [tex]K_{2}[/tex] - Initial and final translational kinetic energy, measured in joules.

[tex]W_{fr}[/tex] - Work losses due to friction, measured in joules.

By applying the defintions of translational kinetic and gravitational potential energies and work, this expression is now expanded:

[tex]m\cdot g \cdot y_{1} + \frac{1}{2}\cdot m\cdot v_{1}^{2} = m\cdot g \cdot y_{2} + \frac{1}{2}\cdot m\cdot v_{2}^{2} + \mu_{k}\cdot m \cdot g \cdot \cos \theta[/tex]

Where:

[tex]m[/tex] - Mass of the crate, measured in kilograms.

[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.

[tex]y_{1}[/tex], [tex]y_{2}[/tex] - Initial and final height of the crate, measured in meters.

[tex]v_{1}[/tex], [tex]v_{2}[/tex] - Initial and final speeds of the crate, measured in meters per second.

[tex]\mu_{k}[/tex] - Kinetic coefficient of friction, dimensionless.

[tex]\theta[/tex] - Ramp inclination, measured in sexagesimal degrees.

The equation is now simplified and the coefficient of friction is consequently cleared:

[tex]y_{1}-y_{2}+\frac{1}{2\cdot g}\cdot (v_{1}^{2}-v_{2}^{2}) = \mu_{k}\cdot \cos \theta[/tex]

[tex]\mu_{k} = \frac{1}{\cos \theta} \cdot \left[y_{1}-y_{2}+\frac{1}{2\cdot g}\cdot (v_{1}^{2}-v_{2}^{2}) \right][/tex]

The final height of the crate is:

[tex]y_{2} = (1.6\,m)\cdot \sin 30^{\circ}[/tex]

[tex]y_{2} = 0.8\,m[/tex]

If [tex]\theta = 30^{\circ}[/tex], [tex]y_{1} = 0\,m[/tex], [tex]y_{2} = 0.8\,m[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]v_{1} = 5\,\frac{m}{s}[/tex] and [tex]v_{2} = 0\,\frac{m}{s}[/tex], the coefficient of friction is:

[tex]\mu_{k} = \frac{1}{\cos 30^{\circ}}\cdot \left\{0\,m-0.8\,m+\frac{1}{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)}\cdot \left[\left(5\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right] \right\}[/tex]

[tex]\mu_{k} \approx 0.548[/tex]

Then, the magnitude of the friction force is:

[tex]f =\mu_{k}\cdot m\cdot g \cdot \cos \theta[/tex]

If [tex]\mu_{k} \approx 0.548[/tex], [tex]m = 12\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]\theta = 30^{\circ}[/tex], the magnitude of the force of friction is:

[tex]f = (0.548)\cdot (12\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot \cos 30^{\circ}[/tex]

[tex]f = 55.851\,N[/tex]

The magnitude of the force of friction is 55.851 newtons.

b) The energy equation of the situation is:

[tex]m\cdot g \cdot y_{1} + \frac{1}{2}\cdot m\cdot v_{1}^{2} = m\cdot g \cdot y_{2} + \frac{1}{2}\cdot m\cdot v_{2}^{2} + \mu_{k}\cdot m \cdot g \cdot \cos \theta[/tex]

[tex]y_{1}+\frac{1}{2\cdot g}\cdot v_{1}^{2} =y_{2} + \frac{1}{2\cdot g}\cdot v_{2}^{2} + \mu_{k}\cdot \cos \theta[/tex]

Now, the final speed is cleared:

[tex]y_{1}-y_{2}+ \frac{1}{2\cdot g}\cdot v_{1}^{2} -\mu_{k}\cdot \cos \theta= \frac{1}{2\cdot g}\cdot v_{2}^{2}[/tex]

[tex]2\cdot g \cdot (y_{1}-y_{2}-\mu_{k}\cdot \cos \theta) + v_{1}^{2} = v_{2}^{2}[/tex]

[tex]v_{2} = \sqrt{2\cdot g \cdot (y_{1}-y_{2}-\mu_{k}\cdot \cos \theta)+v_{1}^{2}}[/tex]

Given that [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{1} = 0.8\,m[/tex], [tex]y_{2} = 0\,m[/tex], [tex]\mu_{k} \approx 0.548[/tex], [tex]\theta = 30^{\circ}[/tex] and [tex]v_{1} = 0\,\frac{m}{s}[/tex], the speed of the crate at the bottom of the ramp is:

[tex]v_{2}=\sqrt{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot [0.8\,m-0\,m-(0.548)\cdot \cos 30^{\circ}]+\left(0\,\frac{m}{s} \right)^{2}}[/tex]

[tex]v_{2}\approx 2.526\,\frac{m}{s}[/tex]

The speed of the crate when it reaches the bottom of the ramp is 2.526 meters per second.

During the spin cycle of your clothes washer, the tub rotates at a steady angular velocity of 31.7 rad/s. Find the angular displacement Δθ of the tub during a spin of 98.3 s, expressed both in radians and in revolutions.

Answers

Answer:

[tex]\Delta \theta = 3116.11\,rad[/tex] and [tex]\Delta \theta = 495.944\,rev[/tex]

Explanation:

The tub rotates at constant speed and the kinematic formula to describe the change in angular displacement ([tex]\Delta \theta[/tex]), measured in radians, is:

[tex]\Delta \theta = \omega \cdot \Delta t[/tex]

Where:

[tex]\omega[/tex] - Steady angular speed, measured in radians per second.

[tex]\Delta t[/tex] - Time, measured in seconds.

If [tex]\omega = 31.7\,\frac{rad}{s}[/tex] and [tex]\Delta t = 98.3\,s[/tex], then:

[tex]\Delta \theta = \left(31.7\,\frac{rad}{s} \right)\cdot (98.3\,s)[/tex]

[tex]\Delta \theta = 3116.11\,rad[/tex]

The change in angular displacement, measured in revolutions, is given by the following expression:

[tex]\Delta \theta = (3116.11\,rad)\cdot \left(\frac{1}{2\pi} \frac{rev}{rad} \right)[/tex]

[tex]\Delta \theta = 495.944\,rev[/tex]

A person is being pulled by gravity with a force of 500 N. What is the force with which the person pulls Earth?
1,000 N
O100 N
500 N
0 250 N

Answers

Answer:

The correct answer is 500 N

Explanation:

This is an exercise in Newton's third law or law of action and reaction

The Earth exerts a force on the person, which we call a weight of 500 N directed downwards, we can call this action and the person exerts a force on the Earth of equal magnitude 500N and in the opposite direction, that is directed upwards.

Which force we call action does not matter, the analysis and conclusions are the same

The correct answer is 500N

g Assume you are a farsighted person who has a near point distance of 40 (cm). If you use a converging contact lens with focal length of 10 (cm). What is nearest distance you can vision with you contacts now?

Answers

Answer:

 object distance  p = 13.33 cm

Explanation:

For this problem of finding the image of an object we must use the constructor equation

         1 / f = 1 / p + 1 / q

where f is the focal length, p and q are the distances to the object and the image, respectively.

In this case they indicate the focal length f = 10 cm, since the person has hyperopia, the image must be formed q = 40 cm, let's find where the object is (p)

        1 / p = 1 / f - 1 / q

        1 / p = 1/10 - 1/40

        1 / p = 0.075

        p = 13.33 cm

In the lab, you shoot an electron towards the south. As it moves through a magnetic field, you observe the electron curving upward toward the roof of the lab. You deduce that the magnetic field must be pointing:_______.
a. to the west.
b. upward.
c. to the north.
d. to the east.
e. downward.

Answers

Answer:

a. to the west.

Explanation:

An electron in a magnetic field always experience a force that tends to change its direction of motion through the magnetic field. According to Lorentz left hand rule (which is the opposite of Lorentz right hand rule for a positive charge), the left hand is used to represent the motion of an electron in a magnetic field. Hold out the left hand with the fingers held out parallel to the palm, and the thumb held at right angle to the other fingers. If the thumb represents the motion of the electron though the field, and the other fingers represent the direction of the field, then the palm will push in the direction of the force on the particle.

In this case, if we point the thumb (which shows the direction we shot the electron) to the south (towards your body), with the palm (shows the direction of the force) facing up to the roof, then the fingers (the direction of the field) will point west.

Matter's resistance to a change in motion is called _____ and is directly proportional to the mass of an object

Answers

Answer:

Matter's resistance to a change in motion is called INERTIA and is directly proportional to the mass of an object.

Explanation:

an electron starts from rest from a fixed point charge with q. what total potential difference accelerates the electron to being very far away from Q

Answers

Answer:

V = (k*Q)/R

Explanation:

Total potential difference accelerates the electron to being very far away from Q is;

V = (k*Q)/R

Where,

V is the Potential Difference in Joules per Coulomb

k is the constant

Q is the charge in Coulomb

R is Electron distance in cm or m

Example

An electron starts from rest 66.1 cm from a fixed point charge with Q = -0.120 μC. What total potential difference accelerates the electron from being very far away from Q

For k = 9.0*10^9 N*m^2/C^2

V = (k*Q)/R

V = (9.0*10^9 * -0.120*10^-6)/0.661

V = -1633.9 Volt.

The answer will change to positive because V = (k*Q)/R is negative at the outset and Zero far away.

The electron (with a negative charge) has a positive energy in the beginning and that gets converted into a positive kinetic energy "far away".

An unstable particle at rest spontaneously breaks into two fragments of unequal mass. The mass of the first fragment is 3.00 10-28 kg, and that of the other is 1.86 10-27 kg. If the lighter fragment has a speed of 0.844c after the breakup, what is the speed of the heavier fragment

Answers

Answer: Speed = [tex]3.10^{-31}[/tex] m/s

Explanation: Like in classical physics, when external net force is zero, relativistic momentum is conserved, i.e.:

[tex]p_{f} = p_{i}[/tex]

Relativistic momentum is calculated as:

p = [tex]\frac{mu}{\sqrt{1-\frac{u^{2}}{c^{2}} } }[/tex]

where:

m is rest mass

u is velocity relative to an observer

c is light speed, which is constant (c=[tex]3.10^{8}[/tex]m/s)

Initial momentum is zero, then:

[tex]p_{f}[/tex] = 0

[tex]p_{1}-p_{2}[/tex] = 0

[tex]p_{1} = p_{2}[/tex]

To find speed of the heavier fragment:

[tex]\frac{mu_{1}}{\sqrt{1-\frac{u^{2}_{1}}{c^{2}} } }=\frac{mu_{2}}{\sqrt{1-\frac{u^{2}_{2}}{c^{2}} } }[/tex]

[tex]\frac{1.86.10^{-27}u_{1}}{\sqrt{1-\frac{u^{2}_{1}}{(3.10^{8})^{2}} } }=\frac{3.10^{-28}.0.844.3.10^{8}}{\sqrt{1-\frac{(0.844c)^{2}}{c^{2}} } }[/tex]

[tex]\frac{1.86.10^{-27}u_{1}}{\sqrt{1-\frac{u^{2}_{1}}{(3.10^{8})^{2}} } }=1.42.10^{-19}[/tex]

[tex]1.86.10^{-27}u_{1} = 1.42.10^{-19}.{\sqrt{1-\frac{u^{2}_{1}}{(3.10^{8})^{2}} } }[/tex]

[tex](1.86.10^{-27}u_{1})^{2} = (1.42.10^{-19}.{\sqrt{1-\frac{u^{2}_{1}}{(3.10^{8})^{2}} } })^{2}[/tex]

[tex]3.46.10^{-54}.u_{1}^{2} = 2.02.10^{-38}.(1-\frac{u_{1}^{2}}{9.10^{16}} )[/tex]

[tex]3.46.10^{-54}.u_{1}^{2} = 2.02.10^{-38} -[2.02.10^{-38}(\frac{u_{1}^{2}}{9.10^{16}} )][/tex]

[tex]3.46.10^{-54}.u_{1}^{2} = 2.02.10^{-38} -2.24.10^{-23}.u^{2}_{1}[/tex]

[tex]3.46.10^{-54}.u_{1}^{2}+2.24.10^{-23}.u^{2}_{1} = 2.02.10^{-38}[/tex]

[tex]2.24.10^{-23}.u^{2}_{1} = 2.02.10^{-38}[/tex]

[tex]u^{2}_{1} = \frac{2.02.10^{-38}}{2.24.10^{-23}}[/tex]

[tex]u_{1} = \sqrt{9.02.10^{-62}}[/tex]

[tex]u_{1} = 3.10^{-31}[/tex]

The speed of the heavier fragment is [tex]u_{1} = 3.10^{-31}[/tex]m/s.

A 10kg block with an initial velocity of 10 m/s slides 1o m across a horizontal surface and comes to rest. it takes the block 2 seconds to stop. The stopping force acting on the block is about

Answers

Answer:

-50N

Explanation:

F=ma=m(Vf-Vi)/t

m=10kgVf=0m/sVi=10m/st=2s

F=(10)(-10)/(2)=-50N

So the force acting on the block is -50N, where the negative sign simply tells us that the force is opposite to the direction of movement.

a 2.0 kg block slides on the horizontal, frictionless surface until it counters a spring force constant with

Answers

Complete question:

a 2.0 kg block slides on the horizontal, frictionless surface until it counters a spring with force constant of  955 N/m. The block comes to rest after compressing the spring a distance of 4.6 cm. Find the initial speed (in m/s) of the block.

Answer:

The initial speed of the block is 1.422 m/s

Explanation:

Given;

mass of the block, m = 2.0 kg

force constant of the spring, K = 955 N/m

compression of the spring, x = 4.6 cm = 0.046 m

Apply Hook's law to determine applied force on the spring;

F = Kx

F = (955 N/m)(0.046 m)

F = 43.93 N

Apply Newton's 2nd law to determine the magnitude of deceleration of the block when it encounters the spring;

F = ma

a = F / m

a = 43.93 / 2

a = 21.965 m/s²

Apply kinematic equation to determine the initial speed of the block;

v² = u² + 2ax

where;

v is the final speed of the block = 0

u is the initial speed of the block

x is the distance traveled by the block = compression of the spring

a is the block deceleration = -21.965 m/s²

0 = u² + 2(-21.965 )(0.046)

0 = u²  - 2.021

u² =  2.021

u = √2.021

u = 1.422 m/s

Therefore, the initial speed of the block is 1.422 m/s

A ball is thrown from the ground so that it’s initial vertical and horizontal components of velocity are 40m/s and 20m/s respectively. Find the ball’s total time of flight and distance it traverses before hitting the ground.

Answers

Answer:

8 seconds

160 meters

Explanation:

Given in the y direction:

Δy = 0 m

v₀ = 40 m/s

a = -10 m/s²

Find: t

Δy = v₀ t + ½ at²

0 m = (40 m/s) t + ½ (-10 m/s²) t²

0 = 40t − 5t²

0 = 5t (8 − t)

t = 0 or 8

Given in the x direction:

v₀ = 20 m/s

a = 0 m/s²

t = 8 s

Find: Δx

Δx = v₀ t + ½ at²

Δx = (20 m/s) (8 s) + ½ (0 m/s²) (8 s)²

Δx = 160 m

Ellen says that whenever the acceleration is directly proportional to the displacement of an object from its equilibrium position, the motion of the object is simple harmonic motion. Mary says this is true only if the acceleration is opposite in direction to the displacement. Which one, if either, is correct

Answers

Answer:

Both Ellen and Mary are correct.

Explanation:

Both are correct, it's just different ways of saying the same thing.

When the acceleration is always opposite in direction to the displacement, then, the acceleration is directly proportional to the displacement of an object from its equilibrium position

The kinetic energy of an object with a mass of 6.8 kg and a velocity of 5.0 m/s is J. (Report the answer to two significant figures.)

Answers

Answer:

[tex] \boxed{\sf Kinetic \ energy \ (KE) = 85 \ J} [/tex]

Given:

Mass (m) = 6.8 kg

Speed (v) = 5.0 m/s

To Find:

Kinetic energy (KE)

Explanation:

Formula:

[tex] \boxed{ \bold{\sf KE = \frac{1}{2} m {v}^{2} }}[/tex]

Substituting values of m & v in the equation:

[tex] \sf \implies KE = \frac{1}{2} \times 6.8 \times {5}^{2} [/tex]

[tex] \sf \implies KE = \frac{1}{ \cancel{2}} \times \cancel{2} \times 3.4 \times 25 [/tex]

[tex] \sf \implies KE =3.4 \times 25 [/tex]

[tex] \sf \implies KE = 85 \: J[/tex]

The kinetic energy of the object reported to two significant figures is: 85 Joules.

Given the following data:

Mass = 6.8 kg Velocity = 5.0 m/s.

To find the kinetic energy of the object:

Kinetic energy refers to an energy that is possessed by a physical object or body due to its motion.

Mathematically, kinetic energy is calculated by using the formula;

[tex]K.E = \frac{1}{2} MV^2[/tex]

Where:

K.E is the kinetic energy. M is the mass of an object. V is the velocity of an object.

Substituting the parameters into the formula, we have;

[tex]K.E = \frac{1}{2}[/tex] × [tex]6.8[/tex] × [tex]5^2[/tex]

[tex]K.E = 3.4[/tex] × [tex]25[/tex]

Kinetic energy = 85 Joules.

Therefore, the kinetic energy of the object is 85 Joules.

Read more: https://brainly.com/question/23153766

You make a telephone call from New York to a friend in London. Estimate how long it will take the electrical signal generated by your voice to reach London, assuming the signal is (a) carried on a telephone cable under the Atlantic Ocean, and (b) sent via satellite 36,000 km above the ocean. Would this cause a noticeable delay in either case

Answers

Answer:

a)  t = 2 10⁻² s ,  t = 2.4 10-1 s

Explanation:

In this exercise they indicate the delay in two signals

a) A signal travels on an electrical cable, between New York and London.

The wave formed in this wire for the signal. This wave travels at the speed of light, c = 3 108 m / s, so the delay is very small

                t = d / c

               t = 6000 10³/3 10⁸

               t = 2 10⁻² s

b) The signal points to a satellite in geostationary orbit

                   distance traveled = √ (2 36000 10³)² + 6000²

                  distance = 72 10⁶ m

the waiting time is

                t = d / c

                t = 72 10⁶/3 10⁸

               t = 2.4 10-1 s

    we can see that the signal sent by the satellites has more delay because its distance is much greater

In an electromagnetic wave in free space, the ratio of the magnitudes of electric and magnetic field vectors E and B is equal:_____.

Answers

Answer:

In an electromagnetic wave in free space, the ratio of the magnitudes of electric and magnetic field vectors E and B is equal:  speed of light(c)

Explanation:

Generally the ratio of the E(electric field ) and  the B(magnetic field ) is  equal to the speed of the electromagnetic wave i.e the speed of  light (c) the value is

    [tex]c = 3.0 *10^{8} \ m/s[/tex]

Test Bank, Question 18.83 Inside a room at a uniform comfortable temperature, metallic objects generally feel cooler to the touch than wooden objects do. This is because: a given mass of wood contains more heat than the same mass of metal the human body, being organic, resembles wood more closely than it resembles metal metal conducts heat better than wood heat tends to flow from metal to wood

Answers

Answer:

metal conducts heat better than wood.

Explanation:

Metals are generally good conductors of heat, and they usually conduct heat at a relatively rapid rate. Inside the room with a uniform temperature, a metal when touched will rapidly conduct the heat from your hand, leaving your hand with a cooler feeling. Wood on the other hand is a poor heat conductor, so the heat is not conducted from your hand fast enough to cool it up to the point that your hand feels cool.

Water has a specific heat capacity of 1.00 cal/g °C, and copper has a specific heat capacity of 0.092 cal/g °C. If both are heated to 100 °C, which takes longer to cool?

Answers

Answer:

The water takes longer, because it is the better insulator here.

Explanation:

Conductors and insulators work similarly in "reverse".

If something is a good heat conductor, then it's good at both absorbing heat energy and giving it away. Insulators are good at resisting temperature changes, but also take longer to cool down once they are heated up.

So because copper is the better conductor here, it will cool faster than the water at the same temperature.

A tank whose bottom is a mirror is filled with water to a depth of 19.6 cm. A small fish floats motionless a distance of 6.40 cm under the surface of the water.
A) What is the apparent depth of the fish when viewed at normal incidence?
B) What is the apparent depth of the image of the fish when viewed at normal incidence?

Answers

Answer:

A. 4.82 cm

B. 24.66 cm

Explanation:

The depth of water = 19.6 cm

Distance of fish  = 6.40 cm

Index of refraction of water = 1.33

(A). Now use the below formula to compute the apparent depth.

[tex]d_{app} = \frac{n_{air}}{n_{water}} \times d_{real} \\= \frac{1}{1.33} \times 6.40 \\= 4.82 cm.[/tex]

(B). the depth of the fish in the mirror.

[tex]d_{real} = 19.6 cm + (19.6 cm – 6.40 cm) = 32.8 cm[/tex]

Now find the depth of reflection of the fish in the bottom of the tank.

[tex]d_{app} = \frac{n_{air}}{n_{water}} \times d_{real} \\d_{app} = \frac{1}{1.33} \times 32.8 = 24.66\\[/tex]

Approximately how many calories are in one gram of carbohydrates?

Answers

Answer:

4 calories

Explanation:

Carbohydrates are known as energy giving foods and they are contained in our diets no matter how small. Other components of foods are protein, fat and oil, vitamin, water etc. A calorie is a unit of energy. It tells us the amount of energy the weight of any food contains. Knowing the amount of calories of food we are taking in helps us to monitor our weight. Each component of food have their own calories depending on the weight of the food component.

For carbohydrate as a component of food, 1gram of carbohydrates contains 4 calories. With this conversion, we can therefore calculate the calories of carbohydrate we are taking in by taking the weight of the food we want to eat.

Hence, approximately 4 calories are in one gram of carbohydrates

One gram of carbohydrates contains approximately 4 calories.

how many calories are in one gram of carbohydrates?

The amount of energy in macronutrients, like carbohydrates, is measured in calories per gram. In terms of nutrition, a "calorie" is the amount of energy needed to warm up one gram of water by one degree Celsius.

Carbohydrates are one of the three main types of nutrients that our body needs, along with proteins and fats. They give the body fuel. When you eat carbohydrates, your body breaks them down into a type of sugar called glucose.  Cells in your body use this glucose for energy.

Read more about  carbohydrates here:

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An electromagnetic flowmeter is useful when it is desirable not to interrupt the system in which the fluid is flowing (e.g. for the blood in an artery during heart surgery). Such a device is illustrated. The conducting fluid moves with velocity v in a tube of diameter d perpendicular to which is a magnetic field B. A voltage V is induced between opposite sides of the tube. Given B = 0.120 T, d = 1.2 cm., and a measured voltage of 2.88 mV, determine the speed of the blood.

Answers

Answer:

2 m/s

Explanation:

The electromagnetic flow-metre work on the principle of electromagnetic induction. The induced voltage is given as

[tex]E = Blv[/tex]

where [tex]E[/tex] is the induced voltage = 2.88 mV = 2.88 x 10^-3 V

[tex]l[/tex] is the distance between the electrodes in this field which is equivalent to the diameter of the tube = 1.2 cm = 1.2 x 10^-2 m

[tex]v[/tex] is the velocity of the fluid through the field = ?

[tex]B[/tex] is the magnetic field = 0.120 T

substituting, we have

2.88 x 10^-3 = 0.120 x 1.2 x 10^-2 x [tex]v[/tex]

2.88 x 10^-3 = 1.44 x 10^-3 x [tex]v[/tex]

[tex]v[/tex] = 2.88/1.44 = 2 m/s

A cube has a mass of 100 grams and its density is determined to be 1 g/cm3. The volume of the cube must be _____. 0.1 cm3 1 cm3 10 cm3 100 cm3

Answers

Answer:   The volume of the block will be [tex]100cm^3[/tex]

Explanation:

Density is defined as the mass contained per unit volume.

[tex]Density=\frac{mass}{volume}[/tex]

Given : Mass of cube = 100 grams

Density of cube = [tex]1g/cm^3[/tex]

Putting in the values we get:

[tex]Volume=\frac{mass}{density}[/tex]

[tex]Volume=\frac{100g}{1g/cm^3}=100cm^3[/tex]

 Thus volume of the block will be [tex]100cm^3[/tex]

a skier starts at the top of a hill this hill is 100 meters in the air the hill is pictured below the skier has a mass of about 50kg using the law of conversation of energy determine the Pe and Ke at the various points a he is at his maximum height and not moving at point E he has come to a complete stop ​

Answers

Answer:

a)  Em = Pe = 4.9 10⁴ J,  b)   K = 2.05 10⁴ J , c)     K = 3.92  104 J ,

e)  W_ friction = Em = 4.9 10⁴ J  

Explanation:

The skier goes down the slope if we assume that there is no friction, the mechanical energy is conserved

         Em = PE + K

where the potential energy is

         PE = m g h

the kinetic energy is

         K = ½ m v²

Let's write the mechanical energy at various points

a) Point A. It is the highest point of the entire system and as the skier is leaving his speed is zero

           Em = Pe

           Em = m g h

let's calculate

           Em = 50 9.8 100

           Em = 4.9 10⁴ J

b) Point B. This point is 60 m

          Em = Pe + K

          K = Em - Pe

          K = 4.9 10⁴ - m g h_B

          K = 4.9   10⁴ - 5 9.8 60

          K = 4.9 10⁴ - 2.85 10⁴

          K = 2.05 10⁴ J

c) point c. This point is 20 m

          Em = Pe + K

          K = Em -Pe = 4.9 10⁴ J - m g h_c

          K = 4.9 10⁴ - 50 9.8 20  = 4.9 10⁴ -  9800

          K = 3.92  104 J

d) point d. It is at a height of 60 m

           Em = Pe + K

           K = Em -Pe

           K = 4.9 10⁴ - m g h

           K = 4.9 10⁴ - 50 9.8 60 =4.9 104 - 2.94 10⁻⁴

           K = 4.897 104 J

e) point E. In this part they indicate that the body is stopped, therefore in this flat part it must be friction so that a device work is carried out that makes the understanding transform into heat by friction and the system stops

            W_ friction = Em = 4.9 10⁴ J

Which of the following regions of the electromagnetic spectrum have longer wavelengths than visible light? 1. infrared radiation 2. ultraviolet radiation 3. microwave radiation

Answers

Answer:infrared radiation

Explanation:

Infrared radiation and  microwave radiation  of the electromagnetic spectrum have longer wavelengths than visible light.

What is electromagnetic wave?

EM waves are another name for electromagnetic waves. When an electric field interacts with a magnetic field, electromagnetic waves are created. These electromagnetic waves make up electromagnetic radiations. It is also possible to say that electromagnetic waves are made up of magnetic and electric fields that are oscillating. The basic equations of electrodynamics, Maxwell's equations, have an answer in electromagnetic waves.

If we arrange   electromagnetic wave with decrease in wavelength, we get:

Radio waves > microwave >  Infrared >  Visible light > Ultraviolet > X-rays > Gamma radiation.

Hence,  Infrared  radiation and  microwave radiation  of the electromagnetic spectrum have longer wavelengths than visible light.

Learn more about electromagnetic wave here:

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A train is approaching you at very high speed as you stand next to the tracks. Just as an observer on the train passes you, you both begin to play the same recorded version of a Beethoven symphony on identical MP3 players. (a) According to you, whose MP3 player finishes the symphony first?
A. your player,
B. the observer's player,
C. both finish at the same time. (b) According to the observer on the train, whose MP3 player finishes the symphony first?
A. your player,
B. the observer's player,
C. both finish at the same time. (c) Whose MP3 player actually finishes the symphony first?
A. your player,
B. the observer's player,
C. each observer measures his symphony as finishing first,
D. each observer measures the other's symphony as finishing first.

Answers

Answer:

a) Your player

b) Observer's player

c) Each measures their own first

Explanation:

Because given problem is having relative velocity to each other. The person sitting on the train is moving with a very high speed relative to the person standing next to the track.

In this case, the clock situated in the train will be running slow with respect to the stationary frame of reference

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