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3 years ago

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A block of ice with mass mA rests on a wedge of ice with mass mB=7.93 kg that is cut so that the surface of the wedge makes an a

ngle of 37.5 with the ground. There is no friction between the pieces of ice or between the ice and the ground. At time t=0 the blocks are released from rest. What is the acceleration of mA in the direction parallel to the ground.

1 answer:

Alika [10]3 years ago

3 0

Answer:

The answer is 7.52 m/s².

Explanation:

To determine acceleration, N = Ma×g ÷ cos Ф

N sin Ф = ma

a = g Tan Ф m/s²

g = 9.8 m/s² and Ф = 37.5° ( angle for the velocity and gravitational force)

so, a = 9.8 × tan (37.5°)

Acceleration has to be determined in order to get the final answer.

a = 7.5198 m/s²

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A 0.300 kg ball, moving with a speed of 2.5 m/s, has a head-on collision with at 0.600 kg ball initially at rest. Assuming a per

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An 20-cm-long Bicycle Crank Arm. With A Pedal At One End. Is Attached To A 25-cm-diameter Sprocket, The Toothed Disk Around Whic

malfutka [58]

To solve the problem, it is necessary to apply the concepts related to the kinematic equations of the description of angular movement.

The angular velocity can be described as

$\omega_f = \omega_0 + \alpha t$

Where,

$\omega_f =$ Final Angular Velocity

$\omega_0 =$ Initial Angular velocity

$\alpha =$ Angular acceleration

t = time

The relation between the tangential acceleration is given as,

$a = \alpha r$

where,

r = radius.

PART A ) Using our values and replacing at the previous equation we have that

$\omega_f = (94rpm)(\frac{2\pi rad}{60s})= 9.8436rad/s$

$\omega_0 = 63rpm(\frac{2\pi rad}{60s})= 6.5973rad/s$

$t = 11s$

Replacing the previous equation with our values we have,

$\omega_f = \omega_0 + \alpha t$

$9.8436 = 6.5973 + \alpha (11)$

$\alpha = \frac{9.8436- 6.5973}{11}$

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The tangential velocity then would be,

$a = \alpha r$

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$\omega_f^2=\omega_0^2+2\alpha\theta$

Replacing with our values and re-arrange to find $\theta,$

$\theta = \frac{\omega_f^2-\omega_0^2}{2\alpha}$

$\theta = \frac{9.8436^2-6.5973^2}{2*0.295}$

$\theta = 90.461rad$

That is equal in revolution to

$\theta = 90.461rad(\frac{1rev}{2\pi rad}) = 14.397rev$

The linear displacement of the system is,

$x = \theta*(2\pi*r)$

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