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Aleks04 [339]
2 years ago
4

A pulley is in the form of a uniform solid cylinder of radius 7cm and mass 2kg. One end of a very light rope is fixed to wind th

e pulley and the other end is
free. When we pull the free end of the rope the pulley starts rotating from rest and accelerates uniformly. If the angular acceleration is 100rad/s so the
tangential acceleration of the rope is:
Chọn một:
a.
200 N
b.
49 N
C.
0.49 N
d.
7N
Physics
1 answer:
Nady [450]2 years ago
7 0
I think it’s, A. 200, I’m sorry if I am wrong, I really hope this helps!
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A mouse runs along a baseboard in your house. The mouse's position as a function of time is given by x(t)=pt2+qt, with p = 0.36
AlexFokin [52]

Answer:

0.60 m/s

Explanation:

The average velocity from t = a to t = b is:

v_avg = (x(b) − x(a)) / (b − a)

Given that x(t) = 0.36t² − 1.20t, and the time is from 1.0 to 4.0:

v_avg = (x(4.0) − x(1.0)) / (4.0 − 1.0)

v_avg = [(0.36(4.0)² − 1.20(4.0)) − (0.36(1.0)² − 1.20(1.0))] / 3.0

v_avg = [(5.76 − 4.8) − (0.36 − 1.20)] / 3.0

v_avg = [0.96 − (-0.84)] / 3.0

v_avg = 0.60

The average speed is 0.60 m/s.

5 0
2 years ago
An electric ceiling fan is rotating about a fixed axis with an initial angular velocity magnitude of 0.240 rev/s. The magnitude
bazaltina [42]

Explanation:

Given that,

Angular velocity = 0.240 rev/s

Angular acceleration = 0.917 rev/s²

Diameter = 0.720 m

(a). We need to calculate the angular velocity after time 0.203 s

Using equation of angular motion

\omega_{f}=\omega_{i}+\alpha t

Put the value in the equation

\omega_{f}=0.240+0.917\times0.203

\omega_{f}=0.426\ rev/s

The angular velocity is 0.426 rev/s.

(b). We need to calculate the tangential speed of the blade

Using formula of  tangential speed

v= r\omega

Put the value into the formula

v = \dfrac{0.720 }{2}\times0.426\times2\pi

v=0.963\ m/s

The tangential speed of the blade is 0.963 m/s.

(c). We need to calculate the magnitude at of the tangential acceleration

Using formula of tangential acceleration

a_{t}=r\alpha

Put the value into the formula

a_{t}=0.36\times0.917\times2\pi

a_{c}=2.074\ m/s^2

The tangential acceleration is 2.074 m/s².

Hence, This is required solution.

4 0
2 years ago
In this problem, you will analyze a system composed of two blocks, 1 and 2, of respective masses m1 and m2. To simplify the anal
Dafna1 [17]

Answer:

a) p = m1 v1 + m2 v2 , b) dp / dt = m1 a1 + m2 a2 , c) It is equivalent to force

dp / dt = 0

Explanation:

In this problem we have two blocks and the system is formed by the two bodies.

Part A. Initially they ask us to find the moment of the whole system

    p = m1 v1 + m2 v2

Part B.

Find the derivative

     dp / dt = m1 dv1dt + m2 dv2 / dt

     dp / dt = m1 a1 + m2 a2

Part C.

Let's analyze the dimensions

     m a = [kg] [m / s2] = [N]

It is equivalent to force

Part d

Acceleration is due to a net force applied

Part e

The acceleration of block 1 is due to the force exerted by block 2 during the moment change

Part f

Force of block 1 on block 2

True f12 = m1a1        f21 = m2a2

Part g

By the law of action and reaction are equal magnitude F12 = f21

Part H

     dp / dt = 0

Isolated system F12 = F21 and the masses are constant. The total moment is only redistributed

7 0
2 years ago
A satellite, orbiting the earth at the equator at an altitude of 400 km, has an antenna that can be modeled as a 1.76-m-long rod
ivann1987 [24]

Answer:

The inducerd emf is 1.08 V

Solution:

As per the question:

Altitude of the satellite, H = 400 km

Length of the antenna, l = 1.76 m

Magnetic field, B = 8.0\times 10^{- 5}\ T

Now,

When a conducting rod moves in a uniform magnetic field linearly with velocity, v, then the potential difference due to its motion is given by:

e = - l(vec{v}\times \vec{B})

Here, velocity v is perpendicular to the rod

Thus

e = lvB           (1)

For the orbital velocity of the satellite at an altitude, H:

v = \sqrt{\frac{Gm_{E}}{R_{E}} + H}

where

G = Gravitational constant

m_{e} = 5.972\times 10^{24}\ kg = mass of earth

R_{E} = 6371\ km = radius of earth

v = \sqrt{\frac{6.67\times 10^{- 11}\times 5.972\times 10^{24}}{6371\times 1000 + 400\times 1000} = 7670.018\ m/s

Using this value value in eqn (1):

e = 1.76\times 7670.018\times 8.0\times 10^{- 5} = 1.08\ V

5 0
2 years ago
A thin film of polystyrene is used as an antireflective coating for fabulite (known as the substrate). the index of refraction o
kvasek [131]

To solve this problem, we assume that the wavelength of the light in air is 500 nanometers.

For this case we only need the refractive index of the polystyrene. For an antireflective coating, we need a quarter of wave thickness at the wavelength in the air. Which means that the antireflective coating needs to be as thick as 1/4 of the wavelength, divided by the coating’s refractive index. This is expressed mathematically in the form:

x = λ / (4 * n)

where,

x = thickness

λ = wavelength of light

n = index of refraction of polystyrene

Substituting:

x = 500 nm / (4 * 1.49)
x = 500 nm / 5.96
x = 83.90 nm

6 0
2 years ago
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