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hammer [34]
3 years ago
11

Your electric drill rotates initially at 5.21 rad/s. You slide the speed control and cause the drill to undergo constant angular

acceleration of 0.311 rad/s2 for 4.13 s. What is the drill's angular displacement during that time interval?
Physics
1 answer:
RUDIKE [14]3 years ago
6 0

Answer:

The drill's angular displacement during that time interval is 24.17 rad.

Explanation:

Given;

initial angular velocity of the electric drill, \omega _i = 5.21 rad/s

angular acceleration of the electric drill, α = 0.311 rad/s²

time of motion of the electric drill, t = 4.13 s

The angular displacement of the electric drill at the given time interval is calculated as;

\theta = \omega _i t \ + \ \frac{1}{2}\alpha t^2\\\\\theta = (5.21 \ \times \ 4.13) \ + \ \frac{1}{2}(0.311)(4.13)^2\\\\\theta = (21.5173 ) \ + \ (2.6524)\\\\\theta =24.17 \ rad

Therefore, the drill's angular displacement during that time interval is 24.17 rad.

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Which of the outer planets would you most likely visit? Why? What would you see on your visit?
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6 0
3 years ago
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A 4.0kg block is sliding with a constant velocity of 3.0m/s on a frictionless table that is 0.5m high. If all of the block’s ene
MArishka [77]

Answer:

Velocity = 4.33[m/s]

Explanation:

The total energy or mechanical energy is the sum of the potential energy plus the kinetic energy, as it is known the velocity and the height, we can determine the total energy.

E_{M}=E_{p}  + E_{k} \\E_{p} = potential energy [J]\\E_{k} = kinetic energy [J]\\where:\\E_{p} =m*g*h\\E_{p} = 4*9.81*0.5=19.62[J]\\E_{k}=\frac{1}{2} *m*v^{2}  \\E_{k}=\frac{1}{2} *4*(3)^{2} \\E_{k}=18[J]\\Therefore\\E_{M} =18+19.62\\E_{M}=37.62[J]

All this energy will become kinetic energy and we can find the velocity.

37.62=\frac{1}{2} *m*v^{2} \\v=\sqrt{\frac{37.62*2}{4} } \\v=4.33[m/s]

8 0
3 years ago
A 5-kg ball collides inelastically head-on with a 10-kg ball, which is initially stationary. Which of the following statements i
NARA [144]

Answer:

The magnitude of the change of velocity the 5-kg ball experiences is less than that of the 10-kg ball.

Explanation:

In inelastic collision, the total momentum is always conserved after collision but the kinetic energy is reduced.

Momentum is Mass X velocity.

5 kg ball is in motion, while 10 kg ball is stationary; that is zero velocity.

The momentum of 10 kg ball before collision is zero while the momentum of 5 kg ball before collision is more than zero. Therefore, the magnitude of change in momentum will not be equal.

Next possible options are in kinetic Energy

Initial Kinetic energy = \frac{1}{2}mu^2

Final kinetic energy =\frac{1}{2}mv^2

Change in kinetic energy = Final Kinetic Energy - Initial Kinetic Energy

Change in kinetic energy of 5kg ball = \frac{1}{2}mv^2 -\frac{1}{2}mu^2 = \frac{1}{2}m(v-u)^2

Since the 5-kg ball has initial velocity (u), the magnitude of the change in velocity will be reduced.

Change in kinetic energy of 10kg ball:

the ball is initially at rest, therefore the initial velocity (u) will be zero (0)

Δ K.E = \frac{1}{2}mv^2 -\frac{1}{2}mu^2 = \frac{1}{2}m(v-u)^2 = \frac{1}{2}m(v-0)^2 = \frac{1}{2}mv^2

From the solution above, the magnitude of the change in velocity experienced by 10 kg ball is higher than 5 kg ball.

Hence, The magnitude of the change of velocity the 5-kg ball experiences is less than that of the 10-kg ball

4 0
3 years ago
How do you think heat gets from a filament to the kernel?​
PilotLPTM [1.2K]

Answer:

Explanation:

It goes thru the fluma

8 0
3 years ago
If a penny is dropped from rest from a tower takes 2 seconds to hit the ground, how far did it travel?
zaharov [31]

Answer:

B

Explanation:

t = 2s

u = 0m/s (released from rest)

a = +g = 9.8m/s²

s = H = ?

using,

s = ut + 1/2at²

H = 0(2) + 1/2(9.8)(2²)

H = 0 + 9.8(2)

s = H = 19.6m

7 0
3 years ago
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