<span>Torque = I * Î± I = Â˝ * m * r^2 = Â˝ * 0.2 * 0.125^2 = 0.0015625 As the disk rotates one time, it rotates an angle of 2 Ď€ radians. Total angle = 1600 * 2 Ď€ = 3200 * Ď€ One minute is 60 seconds. To determine its initial angular velocity, divide this angle by 60. Ď‰ = 53â…“ * Ď€ This is approximately 167.55 rad/s. To determine the angular acceleration, divide by 4.1 seconds. Î± = 53â…“ * Ď€ Ă· 4.1 This is approximately 4.09 rad/s^2 Torque = 0.0015625 * (53â…“ * Ď€ Ă· 4.1) This is approximately 0.0639 N * m. electron1 Â· 2 years ago</span>

8 0

4 years ago

If you apply 150 newtons (N) of force to a solid brick wall for 5 seconds, but you do not move it, howmuch work have you done?A.

Lina20 [59]

Answer:

since there is no displacement, d = 0

Explanation:

hence work = f×d

= 150×0

= 0J

8 0

3 years ago

The block on this incline weighs 100 kg and is connected by a cable and pulley to a weight of 10 kg. If the coefficient of frict

blondinia [14]

Answer:

a. 94.54 N

b. 0.356 m/s^2

Explanation:

Given:-

- The mass of the inclined block, M = 100 kg

- The mass of the vertically hanging block, m = 10 kg

- The angle of inclination, θ = 20°

- The coefficient of friction of inclined surface, u = 0.3

Find:-

a) The magnitude of tension in the cable

b) The acceleration of the system

Solution:-

- We will first draw a free body diagram for both the blocks. The vertically hanging block of mass m = 10 kg tends to move "upward" when the system is released.

- The block experiences a tension force ( T ) in the upward direction due the attached cable. The tension in the cable is combated with the weight of the vertically hanging block.

- We will employ the use of Newton's second law of motion to express the dynamics of the vertically hanging block as follows:

$T - m*g = m*a\\\\$ ... Eq 1

Where,

a: The acceleration of the system

- Similarly, we will construct a free body diagram for the inclined block of mass M = 100 kg. The Tension ( T ) pulls onto the block; however, the weight of the block is greater and tends down the slope.

- As the block moves down the slope it experiences frictional force ( F ) that acts up the slope due to the contact force ( N ) between the block and the plane.

- We will employ the static equilibrium of the inclined block in the normal direction and we have:

$N - M*g*cos ( Q )= 0\\\\N = M*g*cos ( Q )$

- The frictional force ( F ) is proportional to the contact force ( N ) as follows:

$F = u*N\\\\F = u*M*g *cos ( Q )$

- Now we will apply the Newton's second law of motion parallel to the plane as follows:

$M*g*sin(Q) - T - F = M*a\\\\M*g*sin(Q) - T -u*M*g*cos(Q) = M*a\\$ .. Eq2

- Add the two equation, Eq 1 and Eq 2:

$M*g*sin ( Q ) - u*M*g*cos ( Q ) - m*g = a* ( M + m )\\\\a = \frac{M*g*sin ( Q ) - u*M*g*cos ( Q ) - m*g}{M + m} \\\\a = \frac{100*9.81*sin ( 20 ) - 0.3*100*9.81*cos ( 20 ) - 10*9.81}{100 + 10}\\\\a = \frac{-39.12977}{110} = -0.35572 \frac{m}{s^2}$

- The inclined block moves up ( the acceleration is in the opposite direction than assumed ).

- Using equation 1, we determine the tension ( T ) in the cable as follows:

$T = m* ( a + g )\\\\T = 10*( -0.35572 + 9.81 )\\\\T = 94.54 N$

4 0

3 years ago

A driver in a 2000 kg Porsche wishes to pass to pass a slow-moving school bus on a four-lane road. What is the average power in

creativ13 [48]

Answer:

The power require to accelerate the car is, P = 299700 watts

Explanation:

Give data,

The mass of the car, m = 2000 kg

The initial velocity of the sports car, u = 30 m/s

The final velocity of the sports car, v = 60 m/s

The time period of acceleration, t = 9 s

The acceleration of the car, a = (v-u) / t

= (60 - 30) / 9

= 3.33 m/s²

The displacement of the car,

S = ut + ½ at²

= 30 x 9 + ½ x 3.33 x 9²

= 405 m

The force acting on the car, F = m x a

= 2000 x 3.33

= 6660 N

The work done by the car, W = F S

= 6660 x 405

= 2697300 J

The power of the car, P = W / t

= 2697300 / 9

= 299700 watts

Hence, the power require to accelerate the car is, P = 299700 watts

4 0

4 years ago