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3 years ago

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A small model car with mass m travels at a constant speed on the inside of a track that is a vertical circle radius 5.00 m. If t

he normal force exerted by the track on the car when it is at the bottom of the track (point A) is equal to 2.50 mg, how much time does it take the car to complete one revolution around the track?

1 answer:

dolphi86 [110]3 years ago

6 0

One point to another

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The bigclaw snapping shrimp shown in (Figure 1) is aptly named--it has one big claw that snaps shut with remarkable speed. The p

leva [86]

1) $1.86\cdot 10^6 rad/s^2$

2) 2418 rad/s

3) $27000 m/s^2$

4) 36.3 m/s

Explanation:

1)

The angular acceleration of an object in rotation is the rate of change of angular velocity.

It can be calculated using the following suvat equation for angular motion:

$\theta=\omega_i t +\frac{1}{2}\alpha t^2$

where:

$\theta$ is the angular displacement

$\omega_i$ is the initial angular velocity

t is the time

$\alpha$ is the angular acceleration

In this problem we have:

$\theta=90^{\circ} = \frac{\pi}{2}rad$ is the angular displacement

t = 1.3 ms = 0.0013 s is the time elapsed

$\omega_i = 0$ is the initial angular velocity

Solving for $\alpha$ , we find:

$\alpha = \frac{2(\theta-\omega_i t)}{t^2}=\frac{2(\pi/2)-0}{0.0013}=1.86\cdot 10^6 rad/s^2$

2)

For an object in accelerated rotational motion, the final angular speed can be found by using another suvat equation:

$\omega_f = \omega_i + \alpha t$

where

$\omega_i$ is the initial angular velocity

t is the time

$\alpha$ is the angular acceleration

In this problem we have:

t = 1.3 ms = 0.0013 s is the time elapsed

$\omega_i = 0$ is the initial angular velocity

$\alpha = 1.86\cdot 10^6 rad/s$ is the angular acceleration

Therefore, the final angular speed is:

$\omega_f = 0 + (1.86\cdot 10^6)(0.0013)=2418 rad/s$

3)

The tangential acceleration is related to the angular acceleration by the following formula:

$a_t = \alpha r$

where

$a_t$ is the tangential acceleration

$\alpha$ is the angular acceleration

r is the distance of the point from the centre of rotation

Here we want to find the tangential acceleration of the tip of the claw, so:

$\alpha = 1.86\cdot 10^6 rad/s$ is the angular acceleration

r = 1.5 cm = 0.015 m is the distance of the tip of the claw from the axis of rotation

Substituting,

$a_t=(1.86\cdot 10^6)(0.015)=27900 m/s^2$

4)

Since the tip of the claw is moving by uniformly accelerated motion, we can find its final speed using the suvat equation:

$v=u+at$

where

u is the initial linear speed

a is the tangential acceleration

t is the time elapsed

Here we have:

$a=27900 m/s^2$ (tangential acceleration)

u = 0 m/s (it starts from rest)

t = 1.3 ms = 0.0013 s is the time elapsed

Substituting,

$v=0+(27900)(0.0013)=36.3 m/s$

5 0

3 years ago

As shown in the picture above, a sled of mass 67 kilograms is pulled

Scorpion4ik [409]

Explanation:

F net of sled = Tension force by rope - Kinetic friction between ground.

F normal of sled = mg = (67kg)(9.81kg/m^2) = 657.27N.

Kinetic friction = 0.18 (I cannot see the value) * Normal force of sled = 0.18 * 657.27N = 118.31N

So F net of sled = 800N - 118.31N = 681.69N.

(I cannot see what the question is asking for, please check on your own!)

3 0

3 years ago

A 1500 kg car traveling at 15.0 m/s to the south collides with a 4500 kg truck that is at rest at a stopligt. The car comes to a

Burka [1]

Answer:

5 m/s, moving to the South.

Explanation:

Parameters given:

Mass of car, m = 1500 kg

Initial velocity of car, u = 15 m/s

Mass of truck, M = 4500 kg

Initial velocity of truck, v = 0 m/s (Truck is at rest)

Final velocity of car, U = 0 m/s (Car comes to a stop)

Final velocity of truck = V

Because the collision is elastic, we can apply the principle of conservation of momentum, we have that:

Total initial momentum = Total final momentum

m*u + M*U = m*v + M*V

(1500 * 15) + (4500 * 0) = (1500 * 0) + (4500 * V)

22500 + 0 = 0 + 00V

=> V = 22500/4500

V = 5 m/s

The velocity carries a positive sign, hence, it's moving in the same direction as the car was moving initially.

That is, it's moving to the South.

8 0

3 years ago

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A sprinter set a high school record in track and field, running 200.0 m in 20.6 s . what is the average speed of the sprinter in

Paraphin [41]

Answer : The average speed of the sprinter is, 34.95 Km/hr

Solution :

Average velocity : It is defined as the distance traveled by the time taken.

Formula used for average velocity :

$v_{av}=\frac{d}{t}$

where,

$v_{av}$ = average velocity

d = distance traveled = 200 m

t = time taken = 20.6 s

Now put all the given values in the above formula, we get the average velocity of the sprinter.

$v_{av}=\frac{200m}{20.6s}\times \frac{3600}{1000}=34.95Km/hr$

conversion :

(1 Km = 1000m)

(1 hr = 3600 s)

Therefore, the average speed of the sprinter is, 34.95 Km/hr

8 0

3 years ago

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أسقط عامل حجرة من سطح بناية سقوطأ حرة. أوجد ما يلي :

Ilia_Sergeevich [38]

Answer:

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Explanation:

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3 years ago