Answer:
a)[tex]V=1.067\: m/s[/tex]
b)[tex]v=434.65\: m/s [/tex]
Explanation:
a)
Using the conservation of energy between the moment when the bullet hit the block and the maximum compression of the spring.
[tex]\frac{1}{2}MV^{2}=\frac{1}{2}k\Delta x^{2}[/tex]
Where:
M is the bullet-block mass (0.00535 kg + 2.174 kg = 2.17935 kg)V is the speed of the systemk is the spring constant (6.17*10² N/m)Δx is the compression of the spring (0.0634 m)Then, let's find the initial speed of the bullet-block system.
[tex]V^{2}=\frac{k\Delta x^{2}}{M}[/tex]
[tex]V=\sqrt{\frac{6.17*10^{2}*0.0634^{2}}{2.17935}}[/tex]
[tex]V=1.067\: m/s[/tex]
b)
Using the conservation of momentum we can find the velocity of the bullet.
[tex]mv=MV[/tex]
[tex]v=\frac{MV}{m}[/tex]
[tex]v=\frac{2.17935*1.067}{0.00535}[/tex]
[tex]v=434.65\: m/s [/tex]
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A 40-turn coil has a diameter of 11 cm. The coil is placed in a spatially uniform magnetic field of magnitude 0.40 T so that the face of the coil and the magnetic field are perpendicular. Find the magnitude of the emf induced in the coil (in V) if the magnetic field is reduced to zero uniformly in the following times.
(a) 0.30 s V
(b) 3.0 s V
(c) 65 s V
Answer:
(a) emf = 0.507 V
(b) emf = 0.0507 V
(c) emf = 0.00234 V
Explanation:
Given;
number of turns of the coil, N = 40 turns
diameter of the coil, d = 11 cm
radius of the coil, r = 5.5 cm = 0.055 m
magnitude of the magnetic field, B = 0.4 T
The magnitude of the induced emf is calculated as;
[tex]emf = - N\frac{d\phi}{dt} \\\\where;\\\\\phi \ is \ magnetic \ flux= BA \\\\A \ is the \ area \ of \ the \ coil = \pi r^2 = \pi (0.055)^2 = 0.0095 \ m^2\\\\emf = - N \frac{dB.A}{dt} = -NA\frac{dB}{dt} \\\\emf = -NA\frac{(B_2 - B_1)}{t} \\\\emf = NA \frac{(B_1 - B_2)}{t} \\\\the \ final \ magnetic \ field \ is \ reduced \ to \ zero;\ B_2 = 0\\\\emf = \frac{NAB_1}{t}[/tex]
(a) when the time, t = 0.3 s
[tex]emf = \frac{NAB_1}{t} = \frac{40\times 0.0095\times 0.4}{0.3} = 0.507 \ V[/tex]
(b) when the time, t = 3.0 s
[tex]emf = \frac{NAB_1}{t} = \frac{40\times 0.0095\times 0.4}{3} = 0.0507 \ V[/tex]
(c) when the time, t = 65 s
[tex]emf = \frac{NAB_1}{t} = \frac{40\times 0.0095\times 0.4}{65} = 0.00234 \ V[/tex]
S.I unit for distance =______
(A) m (B)cm
(c) km (d) mm
Answer:
opinion a
Explanation:
the si units of distance is metre (m)
Answer:
A
Explanation:
Open the sash half way up, take the beaker containing the dry ice / water out of the hood, and slowly move it from right in front of the hood all the way down to the floor. At what point do the fumes stop getting sucked up by the fume hood?
Answer:
The fumes stop getting sucked up by the fume hood once the beaker is pulled out of the hood.
a vector starts at the point (0.0) and ends at (2,-7) what is the magnitude of the displacement
Answer:
|x| = √53
Explanation:
We are told that the vector starts at the point (0.0) and ends at (2,-7) .
Thus, magnitude of displacement is;
|x| = √(((-7) - 0)² + (2 - 0)²)
|x| = √(49 + 4)
|x| = √53
A friend lends you the eyepiece of his microscope to use on your own microscope. He claims that since his eyepiece has the same diameter as yours but twice the focal length, the resolving power of your microscope will be doubled. Is his claim valid? Explain.
Answer:
The resolving power remains same.
Explanation:
The resolving power of the lens is directly proportional to the diameter of the lens not on the focal length.
As the diameter is same but the focal length is doubled so the resolving power remains same.
In a game of pool, the cue ball moves at a speed of 2 m/s toward the eight ball. When the cue ball hits the eight ball, the cue ball bounces off with a speed of 0.8 m/s at an angle of 20', as shown in the diagram below. Both balls have a mass of 0.6 kg.
a) what is the momentum of the system before the collision
b) what is the momentum after the collision
c) what angle dose the right ball travel after the collision
d) what is the magnitude of the eight balls velocity after the collision
Answer:
a) p₀ = 1.2 kg m / s, b) p_f = 1.2 kg m / s, c) θ = 12.36, d) v_{2f} = 1.278 m/s
Explanation:
For this exercise we define a system formed by the two balls, which are isolated and the forces during the collision are internal, therefore the moment is conserved
a) the initial impulse is
p₀ = m v₁₀ + 0
p₀ = 0.6 2
p₀ = 1.2 kg m / s
b) as the system is isolated, the moment is conserved so
p_f = 1.2 kg m / s
we define a reference system where the x-axis coincides with the initial movement of the cue ball
we write the final moment for each axis
X axis
p₀ₓ = 1.2 kg m / s
p_{fx} = m v1f cos 20 + m v2f cos θ
p₀ = p_f
1.2 = 0.6 (-0.8) cos 20+ 0.6 v_{2f} cos θ
1.2482 = v_{2f} cos θ
Y axis
p_{oy} = 0
p_{fy} = m v_{1f} sin 20 + m v_{2f} cos θ
0 = 0.6 (-0.8) sin 20 + 0.6 v_{2f} sin θ
0.2736 = v_{2f} sin θ
we write our system of equations
0.2736 = v_{2f} sin θ
1.2482 = v_{2f} cos θ
divide to solve
0.219 = tan θ
θ = tan⁻¹ 0.21919
θ = 12.36
let's look for speed
0.2736 = v_{2f} sin θ
v_{2f} = 0.2736 / sin 12.36
v_{2f} = 1.278 m / s
A space ship has four thrusters positioned on the top and bottom, and left and right as shown below. The thrusters can be operated independently or together to help the ship navigate in all directions.
Initially, the Space Probe is floating towards the East, as shown below, with a velocity, v. The pilot then turns on thruster #2.
Select one:
a.
Space ship will have a velocity to the West and will be speeding up.
b.
Space ship will have a velocity to the East and will be speeding up.
c.
Space ship will have a velocity to the East and will be slowing down.
d.
Space ship will have a velocity to the West and will be slowing down.
e.
Ship experiences no change in motion.
Answer:
The correct answer is - c. Spaceship will have a velocity to the East and will be slowing down.
Explanation:
In this case, if turned on thruster #2 then it will exert force on the west side as thruster 2 is on the east side and it can be understood by Newton's third law that says each action has the same but opposite reaction.
As the spaceship engine applies force on the east side then according to the law the exhauster gas applies on towards west direction. It will try to decrease the velocity of the spaceship however, the direction of floating still be east side initally.
A block of mass 2 kg is launched by compressing a spring of force constant 1200 N/m. The block slides on a frictionless surface, up a 1 meter tall ramp, then it enters a region of rough surface. It comes to a stop after traveling 3 meters over the rough surface. The coefficient of kinetic friction between the block and the rough surface is 0.40.
Required:
a. How many forces end up doing work on the block from release to stop?
b. What is the total non-conservative work done on the block?
c. What is the change in the spring potential energy of the block?
Answer:
zzyibgsdwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwe
Explanation:
Q)what are convex mirrors?
Answer:
A curved mirror is a mirror with a curved reflecting surface. The surface may be either convex or concave. Most curved mirrors have surfaces that are shaped like part of a sphere, but other shapes are sometimes used in optical devices.
A convex mirror (or lens) is one constructed so that it is thicker in the middle than it is at the edge.
An astronaut on the moon drops a rock from rest. The rock falls 0.8m in one second of falling time. If the dropped rock fell for a total of two seconds of time instead of 1 second, then the distance traveled would be:____.
A) The same.
B) Doubled.
C) Tripled.
D) Quadruple.
E) None of the above.
Answer:
D) Quadruple.
Explanation:
We will use the second equation of motion to solve this problem:
[tex]s = v_it + \frac{1}{2}gt^2[/tex]
where,
s = distance travelled by the rock
vi = initial speed of rock = 0 m/s
t = time taken
g = acceleration due to gravity on the surface of the moon
Therefore,
[tex]s = (0\ m/s)t+\frac{1}{2}gt^2\\\\s =\frac{1}{2}gt^2[/tex]----------- equation (1)
Now, we double the time:
[tex]s' = \frac{1}{2}g(2t)^2\\\\s' = 4(\frac{1}{2}gt^2)[/tex]
using equation (1):
s' = 4s
Hence, the correct option is:
D) Quadruple.
how can you relazie a perfect balck body in pratice
Help please!!!!!!!!! I will mark brainliest!!!
Answer:
solving for: velocity
equation: velocity = distance / time
substitution: velocity = 1425 km / 12.5 hrs
answer: 114 km/hr
Cuando el pistón tiene un volumen de 2x10^-4 m^3, el gas en el pistón está a una presión de 150 kPa. El área del pistón es 0.00133 m^2. Calcular la fuerza que el gas ejerce sobre el embolo del pistón.
Answer:
F = 1.128 10⁸ Pa
Explanation:
Pressure is defined by
P = F / A
If the gas is ideal for equal force eds on all the walls, so on the piston area we have
F = P A
We reduce the pressure to the SI system
P = 150 kpa (1000 Pa / 1kPa = 150 103 Pa
we calculate
F = 150 10³ / 0.00133
F = 1.128 10⁸ Pa
A object of mass 3.00 kg is subject to a force Fx that varies with position as in the figure below. A coordinate plane has a horizontal axis labeled x (m) and a vertical axis labeled Fx (N). There are three line segments. The first segment runs from the origin to (4,3). The second segment runs from (4,3) to (11,3). The third segment runs from (11,3) to (17,0). (a) Find the work done by the force on the object as it moves from x = 0 to x = 4.00 m. J (b) Find the work done by the force on the object as it moves from x = 4.00 m to x = 11.0 m. J (c) Find the work done by the force on the object as it moves from x = 11.0 m to x = 17.0 m. J (d) If the object has a speed of 0.450 m/s at x = 0, find its speed at x = 4.00 m and its speed at x = 17.0 m.
Answer:
Explanation:
An impulse results in a change of momentum.
The impulse is the product of a force and a distance. This will be represented by the area under the curve
a) W = ½(4.00)(3.00) = 6.00 J
b) W = (11.0 - 4.00)(3.00) = 21.0 J
c) W = ½(17.0 - 11.0)(3.00) = 9.00 J
d) ASSUMING the speed at x = 0 is in the direction of applied force
½(3.00)(v₄²) = ½(3.00)(0.450²) + 6.00
v₄ = 2.05 m/s
½(3.00)(v₁₇²) = ½(3.00)(0.450²) + 6.00 + 21.0 + 9.00
v₁₇ = 4.92 m/s
If the initial speed is NOT in the direction of applied force, the final speed will be slightly less in both cases.
a point object is 10 cm away from a plane mirror while the eye of an observer(pupil diameter is 5.0 mm) is 28 cm a way assuming both eye and the point to be on the same line perpendicular to the surface find the area of the mirror used in observing the reflection of the point
Answer:
1.37 mm²
Explanation:
From the image attached below:
Let's take a look at the two rays r and r' hitting the same mirror from two different positions.
Let x be the distance between these rays.
[tex]d_o =[/tex] distance between object as well as the mirror
[tex]d_{eye}[/tex] = distance between mirror as well as the eye
Thus, the formula for determining the distance between these rays can be expressed as:
[tex]x = 2d_o tan \theta[/tex]
where; the distance between the eye of the observer and the image is:
[tex]s = d_o + d_{eye}[/tex]
Then, the tangent of the angle θ is:
[tex]tan \theta = \dfrac{R}{d_o+d_{eye}}[/tex]
replacing [tex]tan \theta = \dfrac{R}{d_o+d_{eye}}[/tex] into [tex]x = 2d_o tan \theta[/tex], we have:
[tex]x = 2d_o \Big( \dfrac{R}{d_o+d_{eye}}\Big)[/tex]
[tex]x = 2(10) \Big( \dfrac{0.25}{10+28}\Big)[/tex]
[tex]x = 20\Big( \dfrac{0.25}{38}\Big) cm[/tex]
x = (0.13157 × 10) mm
x = 1.32 mm
Finally, the area A = π r²
[tex]A = \pi(\frac{x}{2})^2[/tex]
[tex]A = \pi(\frac{1.32}{2})^2[/tex]
A = 1.37 mm²
Explain how blood circulation takes place in humans?
Blood comes into the right atrium from the body, moves into the right ventricle and is pushed into the pulmonary arteries in the lungs. After picking up oxygen, the blood travels back to the heart through the pulmonary veins into the left atrium, to the left ventricle and out to the body's tissues through the aorta.
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Match the following properties to the type of wave.
Answer:
hi there
Explanation:
1 - III
2- 1
3-1
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PLz mark me as a BRAINLIST
Explanation:
1 . 3
2. 1
3. 2
I hope it is helpful to you.
I’m a photoelectric effect, which property of the incident light determines how much kinetic energy the ejected electrons have ?
A) brightness
B) frequency
C) size of the beam
D) none of the above
Answer:
b = frequency
What type of wave is a microwave?
O heat
O longitudinal
sound
transverse
Answer:
Microwave is a types of a electromagnetic radiation
Answer:
Transvers
Explanation:
Because microwave is electromagnetic waves and all electromagnetic waves are transvers.
Good evening everyone Help me i n my hw ,The wall of cinema hall are covered with sound absorbing materials. Why?Answer it ASAP.Good day
what do you mean about it
calculate the length of wire.
Answer:
L = 169.5 m
Explanation:
Using Ohm's Law:
V = IR
where,
V = Voltage = 1.5 V
I = Current = 10 mA = 0.01 A
R = Resistance = ?
Therefore,
1.5 V = (0.01 A)R
R = 150 Ω
But the resistance of a wire is given by the following formula:
[tex]R = \frac{\rho L}{A}[/tex]
where,
ρ = resistivity = 1 x 10⁻⁶ Ω.m
L = length of wire = ?
A = cross-sectional area of wire = πr² = π(0.6 mm)² = π(0.6 x 10⁻³ m)²
A = 1.13 x 10⁻⁶ m²
Therefore,
[tex]150\ \Omega = \frac{(1\ x\ 10^{-6}\ \Omega .m)L}{1.13\ x\ 10^{-6}\ m^2}\\\\L = \frac{150\ \Omega(1.13\ x\ 10^{-6}\ m^2)}{1\ x\ 10^{-6}\ \Omega .m}\\\\[/tex]
L = 169.5 m
If a jet travels 350 m/s, how far will it travel each second?
Answer:
350
Explanation:
Since it travels 350 meters per second, the jet will travel 350 meters in one second.
Understanding Heisenberg's uncertainty principle is one of the keys to understanding quantum mechanics.
a. True
b. False
Answer:
Very TRUE
Option A is legit
Which of the following scientists won a Nobel Prize for pioneering work in the
study of the evolution of stars?
A. Christian Doppler
B. Warren Washington
C. Charles Kuen Kao
-D. Subrahmanyan Chandrasekhar
Answer:
Subrahmanyan Chandrasekhar
Answer:
D. Subrahmanyan Chandrasekhar
Explanation:
Suppose a 60-turn coil lies in the plane of the page in a uniform magnetic field that is directed out of the page. The coil originally has an area of 0.325 m2. It is stretched to have no area in 0.100 s. What is the magnitude (in V) and direction (as seen from above) of the average induced emf if the uniform magnetic field has a strength of 1.60 T
Answer:
emf = 312 V
Explanation:
In this exercise the electromotive force is asked, for which we must use Faraday's law
emf = [tex]- N \frac{d \Phi }{dt}[/tex]- N dfi / dt
Ф = B. A = B A cos θ
bold type indicates vectors.
They indicate that the magnetic field is constant, the angle between the normal to the area and the magnetic field is parallel by local cosine values 1
It also indicates that the area is reduced from a₀ = 0.325 me² to a_f = 0 in a time interval of ΔT = 0.100 s, suppose that this reduction is linear
emf = -N B [tex]\frac{dA}{dT}[/tex]
emf = - N B (A_f - A₀) / Dt
we calculate
emf = - 60 1.60 (0 - 0.325) /0.100
emf = 312 V
The direction of this voltage is exiting the page
1. Why do only some people get addicted to
drugs?
Answer:
When drugs are taken in are body are brain release dopamine: which make us feel so pleasure and good, and for this some people are addicted to drugs which makes them feel good. on other hand damaging their health.
There are 5640 lines per centimeter in a grating that is used with light whose wavelegth is 455 nm. A flat observation screen is located 0.661 m from the grating. What is the minimum width that the screen must have so the centers of all the principal maxima formed on either side of the central maximum fall on the screen
The minimum width of the screen is 34 cm.
For a diffraction grating, dsinθ = mλ where d = grating spacing = 1/5640 lines per cm = 1/5640 cm per line = 1/5640 × 10⁻² m per line, θ = angle between principal maximum and the center axis of the grating, m = order of maxima = 1 (since we require the position of the principal maximum) and λ = wavelength = 455 nm = 455 × 10⁻⁹ m
So, sinθ = mλ/d
Also tanθ = L/D where θ = angle between principal maximum and the center axis of the grating, L = distance between central maximum and principal maximum and D = distance between grating and screen = 0.661 m.
For small angles sinθ ≈ tanθ
So, mλ/d = L/D
making L subject of the formula, we have
L = mλD/d
L = 1 × 455 × 10⁻⁹ m × 0.661 m ÷ 1/5640 × 10⁻² m per line
L = 1 × 455 × 10⁻⁹ m × 0.661 m × 5640 × 10² line per m
L = 1696258.2 × 10⁻⁷ m
L = 0.16963 m
L ≅ 0.17 m
So, for centers of all the principal maxima formed on either side of the central maximum fall on the screen, the minimum width of the screen is w = 2L.
So, w = 2 × 0.17 m
w = 0.34 m
w = 34 cm
So for the centers of all the principal maxima formed on either side of the central maximum fall on the screen, the minimum width of the screen is 34 cm.
Learn more about diffraction grating here:
https://brainly.com/question/15712101
Chameleons catch insects with their tongues, which they can rapidly extend to great lengths. In a typical strike, the chameleon's tongue accelerates at a remarkable 220 m/s^2 for 20 msms, then travels at constant speed for another 30 ms.
Required:
During this total time of 50 ms, 1/20 of a second, how far does the tongue reach?
Solution :
We know,
Distance,
[tex]$S=ut+\frac{1}{2}at^2$[/tex]
[tex]$S=ut+0.5(a)(t)^2$[/tex]
For the first 20 ms,
[tex]$S=0+0.5(220)(0.020)^2$[/tex]
S = 0.044 m
In the remaining 30 ms, it has constant velocity.
[tex]$v=u+at$[/tex]
[tex]$v=0+(220)(0.020)[/tex]
v = 4.4 m/s
Therefore,
[tex]$S=ut+0.5(a)(t)^2$[/tex]
[tex]$S'=4.4 \times 0.030[/tex]
S' = 0.132 m
So, the required distance is = S + S'
= 0.044 + 0.132
= 0.176 m
Therefore, the tongue can reach = 0.176 m or 17.6 cm
Answer:
The total distance is 0.176 m.
Explanation:
For t = 0 s to t = 20 ms
initial velocity, u = 0
acceleration, a = 220 m/s^2
time, t = 20 ms
Let the final speed is v.
Use first equation of motion
v = u + at
v = 0 + 220 x 0.02 = 4.4 m/s
Let the distance is s.
Use second equation of motion
[tex]s = u t + 0.5 at^2\\\\s = 0 + 0.5 \times 220 \times 0.02\times 0.02\\\\s = 0.044 m[/tex]
Now the distance is
s' = v x t
s' = 4.4 x 0.03 = 0.132 m
The total distance is
S = s + s' = 0.044 + 0.132 = 0.176 m
A 500 kg rocket sled is coasting in reverse at 10 m/s (to the left). It then turns on its rocket engines for 10.0 s, with a thrust of 1500 N (to the right). What is its final velocity? (Remember velocity has magnitude and direction)
Explanation:
F = ma
[tex]a = \frac{f}{m} [/tex]
[tex]a = \frac{1500}{500} = 3[/tex]
[tex]a = \frac{v2 - v1}{t} [/tex]
[tex]3 = \frac{v2 - 10}{10} [/tex]
v2 (final) = 40 m/s to the right direction
Each rarefraction on a longitudinal wave correspond to what point on a transverse wave?
