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astraxan [27]
3 years ago
10

A 780g, 50-cm-long metal rod is free to rotate about a frictionless axle at one end. While at rest, the rod is given a short but

sharp 1000N hammer blow at the CENTER of the rod, aimed in a direction that causes the rod to rotate on the axle. The blow lasts a mere 2.5ms.
What is the rod's angular velocity immediately after the blow? The book gives the answer 8.0rad/s
Physics
1 answer:
Semmy [17]3 years ago
5 0

Answer:

9.6 rad/s

Explanation:

L = length of the metal rod = 50 cm = 0.50 m

M = Mass of the long metal rod = 780 g = 0.780 kg

Moment of inertia of the rod about one end is given as

I = \frac{ML^{2}}{3} = \frac{(0.780)(0.50)^{2}}{3} = 0.065 kgm^{2}

F = force applied by the hammer blow = 1000 N

Torque produced due to the hammer blow is given as

\tau = \frac{FL}{2}

\tau = \frac{(1000)(0.50)}{2}

\tau = 250 Nm

t = time of blow = 2.5 ms = 0.0025 s

w = Angular velocity after the blow

Using Impulse-change in angular momentum, we have

I w = \tau t\\(0.065) w = (250) (0.0025)\\w = 9.6 rads^{-1}

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Answer:

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3 years ago
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Which of the following processes could NOT be used to separate dissolved particles from the liquid in a solution? *
DIA [1.3K]

Answer:

Vaporation

Explanation:

In the vaporization or boiling, the passage of particles from the liquid state to the gaseous state occurs completely

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Consider a wet banked roadway, where there is a coefficient of static friction of 0.30 and a coefficient of kinetic friction of
gulaghasi [49]

Answer:

a) v = 20.9 m/s

b) v = 8.46 m/s

Explanation:

Given:-

- The coefficient of static friction is us = 0.30

- The coefficient of static friction is uk= 0.25

- The radius of the curve R = 50m

- The bank Angle β = 25

Find:-

a) If the banking angle is β=25∘, what is the maximum speed the automobile can have before sliding up the banking?  

b) What is the minimum speed the automobile can have before sliding down the banking?

Solution:-

- We will investigate the sliding-up case first. Develop a FBD as given in (attachment).

- Use Newton's second law of motion vertical to slope of bank where the car is in equilibrium:

                       Sum ( F_n ) = 0

                       N*cos(β) - m*g - Ff*sin(β) = 0

Where,            Frictional Force Ff = us*N

                      N (cos(β) - us*sin(β)) = mg   ... Eq 1

- Use Newton's second law of motion horizontal to slope of bank where the car is accelerating:

                       Sum ( F_h ) = m*a

                       Ff*cos(β) + Nsin(β) = m*v^2 / R

                       N (us*cos (β) + sin (β) ) = m*v^2 / R  .... Eq 2

- Divide the two equations:

                    v^2 / gR = [ us*cos (β) + sin (β) ] / [ cos (β) - us*sin (β) ]

                    v^2 = [ 0.25*cos (25) + sin (25) ]*9.81*50 / [ cos (25) - 0.25*sin (25) ]

                    v = 20.9 m/s

- For the slip down case. We have, friction force Ff reversed hence us = -us. Then the v can be given as:

                    v^2 / gR = [ -us*cos (β) + sin (β) ] / [ cos (β) + us*sin (β) ]

                    v^2 = [ -0.25*cos (25) + sin (25) ]*9.81*50 / [ cos (25) + 0.25*sin (25) ]

                    v = 8.46 m/s

6 0
3 years ago
I really need help
Alina [70]

Answer:

Find the velocity of a projectile at the highest point, if it is projected with a speed 15ms-1 in the direction 45∘ above horizontla. (take g = 10ms-2) Answer : 15√2ms-1 along horizontal

Explanation:

6 0
3 years ago
A 4812 kg boulder sits on top of a 203 m cliff. What is its PE?
Bond [772]

Answer:

<h2>9572992.8 J</h2>

Explanation:

The potential energy of a body can be found by using the formula

PE = mgh

where

m is the mass

h is the height

g is the acceleration due to gravity which is 9.8 m/s²

From the question we have

PE = 4812 × 9.8 × 203

We have the final answer as

<h3>9572992.8 J</h3>

Hope this helps you

7 0
3 years ago
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