Answer:
Explanation:
Time is not part of the Work equation. That's the only conclusion that you can come to.
Work = Force * distance.
Not enough information is given to go any further. You don't have enough information to calculate the distance.
We don't know if the box can be moved or not. It says heavy. 50 N is really not very much.
I would guess that you are intended to answer that the box didn't move, but it's really hard to tell.
A wire long and with mass is positioned horizontally near the earth's surface and perpendicular to a horizontal magnetic field of magnitude . What current I must flow through the wire in order that the wire accelerate neither upwards nor downwards
The question is incomplete. The complete question is :
A wire 0.6 m long and with mass m = 11 g is positioned horizontally near the earth's surface and perpendicular to a horizontal magnetic field of magnitude B = 0.4 T. What current I must flow through the wire in order that the wire accelerate neither upwards nor downwards? The magnetic field is directed into the page.
Solution :
Given :
Length of the wire, L = 0.6 m
Mass of the wire length, m = 11 g
= [tex]11 \times 10^{-3}[/tex] kg
Magnetic field , B = 0.4 T
Know we know that :
ILB = mg
or [tex]$I=\frac{mg}{BL}$[/tex]
[tex]$I= \frac{(11 \times 10^{-3})(9.81)}{(0.4)(0.6)}$[/tex]
[tex]I=0.44963\ A[/tex]
[tex]I = 449.63 \ mA[/tex]
A uniform disk turns at 3.6 rev/s around a frictionless spindle. A non rotating rod, of the same mass as the disk and length equal to the disk's diameter, is dropped onto the freely spinning disk . They then both turn around the spindle with their centers superposed.
What is the angular frequency in rev/s of the combination?
please express answer in proper significant figures and rounding.
Answer:
ω₁ = 2.2 rev/s
Explanation:
Conservation of angular momentum
moment of inertia uniform disk is ½mR²
moment of inertia uniform rod about an end mL²/3
We can think of our rod as two rods of mass m/2 and length R
L = ½mR²ω₀
L = (½mR² + 2(m/2)R²/3)ω₁
ω₁ = ω₀(½mR² / (½mR² + mR²/3))
ω₁ = ω₀(½ / (½ + 1/3))
ω₁ = 0.6ω₀
ω₁ = 2.16
Develop a hypothesis regarding one factor you think might affect the period of a pendulum or an oscillating mass on a spring. Potential factors include the mass, the spring constant, and the length of the pendulum's string. Write down your hypothesis. 2. Design a controlled experiment to test your hypothesis. Take extreme care to keep all factors constant except the variable you are testing.
Answer:
A hypothesis for the period of a pendulum is:
"The period of the pendulum varies with its length"
Explanation:
A hypothesis for the period of a pendulum is:
"The period of the pendulum varies with its length"
To test this hypothesis we can carry out a measurement of a simple pendulum keeping the angle fixed, in general the angle used is about 5º since when placing this value in radiand and the sine of this angle they differ little <5%. therefore measured the time of some oscillations, for example about 10 oscillations, changing the length of the pendulum to test the hypothesis.
If the hypothesis and the model used is correct, the relationship to be tested is
T² =(4π² /g) L
by making a graph of the period squared against the length if obtaining, os a line, the hypothesis is tested.
What is the volume of a metal block 3cm long by 2cm wide by 4cm high? What would be the volume of a block twice as long, wide, and high?
Answer:
Volume of a metal block = 24 cm^3
Volume of a block twice as long, wide and high = 192 cm^3
Explanation:
Volume of a block is measured in l*w*h and in the first block, the sides are 3, 2 and 4 and 3*2*4 = 24
Second block, just double each of the lengths to get 6*4*8 = 192
how much heat is produced in one hour by an electric iron which draws 2.5ampere when connected to a 100V supply
Explanation:
I=2.5 Ampere ; V=100V ;t = 1 hour=60secs
We know Heat = VIt
H=100×2.5×60=15,000J
Choose one. 5 points
Use the equation from week 3:
frequency =
wavespeed
wavelength
and the wavelength you found in #3 to calculate the frequency of this photon (remember the speed of
light is 3E8 m/s);
7.6E14 Hz
6.0E14 Hz
4,6E14 Hz
The frequency is 4,6E14 Hz.
What is the frequency?
Frequency is the fee at which modern changes direction in step with 2nd. it's far measured in hertz (Hz), a worldwide unit of degree wherein 1 hertz is identical to 1 cycle in line with 2d. Hertz (Hz) = One hertz is the same as 1 cycle in step with the second. Cycle = One entire wave of alternating present-day voltage.
Frequency describes the number of waves that pass a hard and fast place in a given quantity of time. So if the time it takes for a wave to skip is half of 2d, the frequency is 2 per 2nd. If it takes 1/one hundred of an hour, the frequency is a hundred in step with hour.
Learn more about frequency here:-https://brainly.com/question/254161
#SPJ2
During World War II, mass spectrometers were used to separate the radioactive uranium isotope U-235 from its far more common isotope, U-238. Estimate the radius of the circle traced out by a singly ionized lead atom moving at the same speed.
Answer:
21.55 m
Explanation:
If the mass of an object is 10 kg and the
velocity is -4 m/s, what is the momentum?
A. 4 kgm/s
B. -40 kgm/s
C.-4 kgm/s
D. 40 kgm/s
Answer:
B. -40 kgm/s is the answer
12. A concave lens has a focal length of 10 cm. An object 2.5 cm high is placed 30 cm from the lens. Determine the position and size of the image. (3)
Answer:
I think 9.5
Explanation:
............
An electron is released from rest at a distance of 9.00 cm from a fixed proton. How fast will the electron be moving when it is 3.00 cm from the proton
Answer:
the speed of the electron at the given position is 106.2 m/s
Explanation:
Given;
initial position of the electron, r = 9 cm = 0.09 m
final position of the electron, r₂ = 3 cm = 0.03 m
let the speed of the electron at the given position = v
The initial potential energy of the electron is calculated as;
[tex]U_i = Fr = \frac{kq^2}{r^2} \times r = \frac{kq^2}{r} \\\\U_i = \frac{(9\times 10^9)(1.602\times 10^{-19})^2}{0.09} \\\\U_i = 2.566 \times 10^{-27} \ J[/tex]
When the electron is 3 cm from the proton, the final potential energy of the electron is calculated as;
[tex]U_f = \frac{kq^2}{r_2} \\\\U_f = [\frac{(9\times 10^9)\times (1.602 \times 10^{-19})^2}{0.03} ]\\\\U_f = 7.669 \times 10^{-27} \ J \\\\\Delta U = U_f -U_i\\\\\Delta U = (7.699\times 10^{-27} \ J ) - (2.566 \times 10^{-27} \ J)\\\\\Delta U = 5.133 \times 10^{-27} \ J[/tex]
Apply the principle of conservation of energy;
ΔK.E = ΔU
[tex]K.E_f -K.E_i = \Delta U\\\\initial \ velocity \ of \ the \ electron = 0\\\\K.E_f - 0 = \Delta U\\\\K.E_f = \Delta U\\\\\frac{1}{2} mv^2 = \Delta U\\\\where;\\\\m \ is \ the \ mass \ of\ the \ electron = 9.1 1 \times 10^{-31} \ kg\\\\v^2 = \frac{ 2 \Delta U}{m} \\\\v = \sqrt{\frac{ 2 \Delta U}{m}} \\\\v = \sqrt{\frac{ 2 (5.133\times 10^{-27})}{9.11\times 10^{-31}}}\\\\v = \sqrt{11268.935} \\\\v = 106.2 \ m/s[/tex]
Therefore, the speed of the electron at the given position is 106.2 m/s
A block of mass 0.260 kg is placed on top of a light, vertical spring of force constant 5 200 N/m and pushed downward so that the spring is compressed by 0.090 m. After the block is released from rest, it travels upward and then leaves the spring. To what maximum height above the point of release does it rise
After being released, the restoring force exerted by the spring performs
1/2 (5200 N/m) (0.090 m)² = 12.06 J
of work on the block. At the same time, the block's weight performs
- (0.260 kg) g (0.090 m) ≈ -0.229 J
of work. Then the total work done on the block is about
W ≈ 11.83 J
The block accelerates to a speed v such that, by the work-energy theorem,
W = ∆K ==> 11.83 J = 1/2 (0.260 kg) v ² ==> v ≈ 9.54 m/s
Past the equilibrium point, the spring no longer exerts a force on the block, and the only force acting on it is due to its weight, hence it has a downward acceleration of magnitude g. At its highest point, the block has zero velocity, so that
0² - v ² = -2gy
where y is the maximum height. Solving for y gives
y = v ²/(2g) ≈ 4.64 m
A 700N marine in basic training climbs a 10m vertical rope at constant speed in 8sec. what is power put
Answer:
875 Watts
Explanation:
P = W/t = mgh/t = 700(10)/8 = 875 Watts
You are to connect resistors R1 andR2, with R1 >R2, to a battery, first individually, then inseries, and then in parallel. Rank those arrangements according tothe amount of current through the battery, greatest first. (Useonly the symbols > or =, for exampleseries>R1=R2>parallel.)
Answer:
The current is more in the parallel combination than in the series combination.
Explanation:
two resistances, R1 and R2 are connected to a battery of voltage V.
When they are in series,
R = R1 + R2
In series combination, the current is same in both the resistors, and it is given by Ohm's law.
V = I (R1 + R2)
[tex]I = \frac{V}{R_1 + R_2}[/tex]..... (1)
When they are connected in parallel.
the voltage is same in each resistor.
The effective resistance is R.
[tex]R = \frac{R_1R_2}{R_1 + R_2}[/tex]
So, the current is
[tex]I = \frac{V(R_1+R_2)}{R_1 R_2}[/tex]..... (2)
So, the current is more is the parallel combination.
What are the systems of units? Explain each of them.
THERE ARE COMMONLY THREE SYSTEMS OF UNIT. THEY ARE:-
• CGS System- (Centimeter-Gram-Second system) A metric system of measurement that uses the centimeter, gram and second for length, mass and time.
• FPS System- (Foot–Pound–Second system).
The system of units in which length is measured in foot , mass in pound and time in second is called FPS system. It is also known as British system of units.
• MKS System- (Meter-Kilogram-Second system) A metric system of measurement that uses the meter, kilogram, gram and second for length, mass and time. The units of force and energy are the "newton" and "joule."
When you are standing on Earth, orbiting the Sun, and looking at a broken cell phone on the ground, there are gravitational pulls on the cell phone from you, the Earth, and the Sun. Rank the gravitational forces on the phone from largest to smallest. Assume the Sun is roughly 109 times further away from the phone than you are, and 1028 times more massive than you. Rank the following choices in order from largest gravitational pull on the phone to smallest. To rank items as equivalent, overlap them.
a. Pull phone from you
b. Pull on phone from earth
c. Pull on phone from sun
Answer:
The answer is "Option b, c, and a".
Explanation:
Here that the earth pulls on the phone, as it will accelerate towards Earth when we drop it.
We now understand the effects of gravity:
[tex]F \propto M\\\\F\propto \frac{1}{r^2}\\\\or\\\\F \propto \frac{M}{r^2}\\\\Sun (\frac{M}{r^2}) = \frac{10^{28}}{(10^9)^2} = 10^{10}[/tex]
The force of the sun is, therefore, [tex]10^{10}[/tex] times greater and the proper sequence, therefore, option steps are:
b. Pull-on phone from earth
c. Pull-on phone from sun
a. Pull phone from you
Two different galvanometers G1 and G2, have internal resistances r1and r2. The galvanometers G1 and G2 require the same current IC1=IC2 for a full-scale deflection of their pointers. These galvanometers G1 and G2 are used to build lab-made ammeters A1 and A2 . Both ammeters A1 and A2 have the same maximum scale reading Imax1=Imax2=Imax. To build A1 ,shunt resistor of resistance Rsh1is used and to build A2 , shunt resistor of resistance Rsh2 is used. The value of these shunt resistor resistances are such that: Rsh1=3Rsh2. What is the ratio oftheir internal resistances: r1:r2?
Answer:
there are 3 photos attached. so check
Explanation:
Cho lực F ⃗=6x^3 i ⃗-4yj ⃗ tác dụng lên vật làm vật chuyển động từ A(-2,5) đến B(4,7). Vậy công của lực là:
The work done by [tex]\vec F[/tex] along the given path C from A to B is given by the line integral,
[tex]\displaystyle \int_C \mathbf F\cdot\mathrm d\mathbf r[/tex]
I assume the path itself is a line segment, which can be parameterized by
[tex]\vec r(t) = (1-t)(-2\,\vec\imath + 5\,\vec\jmath) + t(4\,\vec\imath+7\,\vec\jmath) \\\\ \vec r(t) = (6t-2)\,\vec\imath+(2t+5)\,\vec\jmath \\\\ \vec r(t) = x(t)\,\vec\imath + y(t)\,\vec\jmath[/tex]
with 0 ≤ t ≤ 1. Then the work performed by F along C is
[tex]\displaystyle \int_0^1 \left(6x(t)^3\,\vec\imath-4y(t)\,\vec\jmath\right)\cdot\frac{\mathrm d}{\mathrm dt}\left[x(t)\,\vec\imath + y(t)\,\vec\jmath\right]\,\mathrm dt \\\\ = \int_0^1 (288(3t-1)^3-8(2t+5)) \,\mathrm dt = \boxed{312}[/tex]
If an object with constant mass is accelerating, what does Newton's second
law imply?
A. It will continue to accelerate until it meets an opposing force.
B. The object is exerting an opposite but equal force.
C. A force must be acting on the object.
D. The object will be difficult to decelerate.
Answer:
C. A force must be acting on the object.
Explanation:
This is due to the action of its momentum direction.
[tex].[/tex]
An object moving with initial velocity 10 m/s is subjected to a uniform acceleration of 8 m/s ^² . The displacement in the next 2 s is: (a) 0m (b) 36 m (c) 16 m (d) 4 m
Joule is a SI unit of power
Measuring cylinder is used to measure the volume of a liquid
Answer:
The SI unit of power is watt
Polarized sunglasses:
a. block most sunlight because sunlight is polarized
b. are better but work the same way as non-polarized sunglasses
c. are polarized to filter out certain wavelengths of light
d. block reflected light because reflected light is partially polarized.
Polarized sunglasses creates filter of vertical openings for light. The light rays will reach the eyes of human vertically only.
The sun rays will not reach human eye directly which will create a shield for sun light burden on human eye.
Polarized sunglasses are best used for blocking and eliminating certain wavelengths of light.
Therefore the correct answer is option C. Polarizes Sunglasses are polarized and it filter out certain wavelengths of light.
Learn more at https://brainly.com/question/24372632
Two train 75 km apart approach each other on parallel tracks, each moving at
15km/h. A bird flies back and forth between the trains at 20km/h until the trains pass
each other. How far does the bird fly?
Answer:
The correct solution is "37.5 km".
Explanation:
Given:
Distance between the trains,
d = 75 km
Speed of each train,
= 15 km/h
The relative speed will be:
= [tex]15 + (-15)[/tex]
= [tex]30 \ km/h[/tex]
The speed of the bird,
V = 15 km/h
Now,
The time taken to meet will be:
[tex]t=\frac{Distance}{Relative \ speed}[/tex]
[tex]=\frac{75}{30}[/tex]
[tex]=2.5 \ h[/tex]
hence,
The distance travelled by the bird in 2.5 h will be:
⇒ [tex]D = V t[/tex]
[tex]=15\times 2.5[/tex]
[tex]=37.5 \ km[/tex]
on a horizontal axis whose unit is the meter, a linear load ranging from 0 to 1 ma a linear load distribution = 2 nC / m.
determine the modulus of the electric field created by the previous loaded bar at the point A of abscissa 2m (we have to find the relation between l, which is the distance between the elementary bar and the point A and x which sweeps the segment [0: 1]
Answer:
The correct answer is - 8.99N/C
Explanation:
[tex]dE=k\dfrac{dq}{x^2}\\ dq=\lambda{dx}\\ \lambda=2nC/m\\ dq=2dxnC\\ dE=k\dfrac{2dx}{x^2}\\ E=2k\int_1^2\dfrac{dx}{x^2}\\ E=2k(\frac{-1}{x})_1^2=k\times10^{-9}N/C\\ E=8.99\times10^9\times10^{-9}N/C\\ E=8.99N/C\\dE=k[/tex]
What is the magnitude of the force between a 25μC charge exerts on a -10μC charge 8.5cm away?
Answer:
Force,
[tex]F = \frac{kQ_{1} Q_{2} }{ {r}^{2} } \\ F = \frac{(9 \times {10}^{9}) \times (25 \times {10}^{ - 6}) \times (10 \times {10}^{ - 6} ) }{ {(0.85)}^{2} } \\ \\ F = 3.114 \: newtons[/tex]
The magnitude of the force between a 25μC charge exerts on a -10μC charge 8.5cm away would be 311.4 N.
What is Coulomb's Law?Coulomb's law can be stated as the product of the charges and the square of the distance between them determine the force of attraction or repulsion acting in a straight line between two electric charges.
The math mathematical expression for the coulomb's law given as
F= k Q₁Q₂/r²
where F is the force between two charges
k is the electrostatic constant which is also known as the coulomb constant,it has a value of 9×10⁹
Q₁ and Q₂ are the electric charges
r is the distance between the charges
As given in the problem two charges a 25μC charge exerts on a -10μC charge 8.5cm away
By substituting the respective values in the above formula of Coulomb law
F =9×10⁹×(25×10⁻⁶)×(-10×10⁻⁶)/(8.5×10⁻²)²
F= -311.4 N
A negative sign represents that the force is attractive in nature
Thus, the magnitude of the force is 311.4 N.
Learn more about Coulomb's law from here
https://brainly.com/question/506926
#SPJ2
Which was a major effect of Pope Leo III crowning Charlemagne emperor of the Romans ?
Answer:
The crowning of Charlemagne by Pope Leo III was significant in a number of ways. For Charlemagne, it was necessary because it encouraged to give him higher reliability. It gave him the rank of a dictator, giving him the only ruler in Europe west of the Byzantine emperor in Constantinople.
Define wave length as applied to wave motion
Answer: Wavelength can be defined as the distance between two successive crests or troughs of a wave. It is measured in the direction of the wave.
Explanation:
Wavelength refers to the length or distance between two identical points of neighboring cycles of a wave signal traveling in space or in any physical medium. ... The wavelength of a signal is inversely proportional to its frequency, that is, the higher the frequency, the shorter the wavelength.
A cylindrical swimming pool has a radius 2m and depth 1.3m .it is completely filled with salt water of specific gravity 1.03.The atmospheric preassure is 1.013 x 10^5 Pa.
a.calculate the density of salt water.
Answer:
the density of the salt water is 1030 kg/m³
Explanation:
Given;
radius of the cylindrical pool, r = 2 m
depth of the pool, h = 1.3 m
specific gravity of the salt water, γ = 1.03
The atmospheric pressure, P₀ = 1.013 x 10⁵ Pa
Density of fresh water, [tex]\rho _w[/tex] = 1000 kg/m³
The density of the salt water is calculated as;
[tex]Specific \ gravity \ of \ salt\ water \ (\gamma _s_w) = \frac{density \ of \ salt \ water \ (\rho_{sw})}{density \ of \ fresh \ water \ (\rho_{w})} \\\\1.03 = \frac{\rho_{sw}}{1000 \ kg/m^3}\\\\\rho_{sw} = 1.03 \times 1000 \ kg/m^3\\\\\rho_{sw} = 1030 \ kg/m^3[/tex]
Therefore, the density of the salt water is 1030 kg/m³
Electromagnetic radiation with a wavelength of 525 nm appears as green light to the human eye. Calculate the frequency of this light. Be sure to include units in your answer.
Answer:
5.71×10¹⁴ Hz
Explanation:
Applying,
v = λf................. Equation 1
Where v = speed of the electromagnetic radiation, λ = wavelength of the electromagnetic radiation, f = frequency
make f the subject of the equation
f = v/λ............. Equation 2
From the question,
Given: λ = 525 nm = 5.25×10⁻⁷ m,
Constant: Speed of electromagnetic wave (v) = 3.0×10⁸ m/s
Substitute these values into equation 2
f = (3.0×10⁸)/(5.25×10⁻⁷)
f = 5.71×10¹⁴ Hz
Hence the frequency of light is 5.71×10¹⁴ Hz
Choose the appropriate explanation how such a low value is possible given Saturn's large mass - 100 times that of Earth.
a. This low value is possible because the magnetic field of Saturn is so strong.
b. This low value is possible because the magnetic field of Saturn is so weak.
c. This low value is possible because the density of Saturn is so high.
d. This low value is possible because the density of Saturn is so low.
Answer:
Explanation:
That is an amazing fact.
The minus sign is what you have to pay attention to. The earth has a mass of 100 times that of Saturn. As someone on here once noted, Saturn has such a low density that it would float in water.
The answer is D
In a photoelectric effect experiment, it is observed that violet light does not eject electrons from a particular metal. Next, red light with the same intensity is incident on the same metal. Which result is possible
Answer:
No ejection of photo electron takes place.
Explanation:
When a photon of suitable energy falls on cathode, then the photoelectrons is emitted from the cathode. This phenomenon is called photo electric effect.
The minimum energy required to just eject an electron is called work function.
The photo electric equation is
E = W + KE
where, E is the incident energy, W is the work function and KE is the kinetic energy.
W = h f
where. h is the Plank's constant and f is the threshold frequency.
Now, when the violet light is falling, no electrons is ejected. When the red light is falling, whose frequency is less than the violet light, then again no photo electron is ejected from the metal surface.
