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3 years ago

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A rocket is launched straight up with constant acceleration. Four seconds after liftoff, a bolt falls off the side of the rocket

. The bolt hits the ground 6.10s later. What was the rocket's acceleration?

1 answer:

Olenka [21]3 years ago

7 0

Answer:

1.3m/s²

Explanation:

Given values in the first part: acceleration a, time t₁:

1) velocity v₀ $= at_1$

2) height h₀ $=\frac{1}{2}at_1^2$

Given values in the second part: acceleration -g, time t₂:

3) height h $= -\frac{1}{2}gt_2^2+v_0t_2+h_0$

Combining equations 1,2,3 and setting h to zero:

$0=-\frac{1}{2}gt_2^2+(at_1)t_2+\frac{1}{2}at_1^2\\ 0=a(t_1t_2+\frac{1}{2}t_1^2)-\frac{1}{2}gt_2^2$

Solve for a with t₁ = 4s and t₂=2.1s:

$a=\frac{1}{2}gt_2^2(\frac{1}{t_1t_2+\frac{1}{2}t_1^2})$

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A Ferris wheel starts at rest and builds up to a final angular speed of 0.70 rad/s while rotating through an angular displacemen

PilotLPTM [1.2K]

Answer:

The average angular acceleration is 0.05 radians per square second.

Explanation:

Let suppose that Ferris wheel accelerates at constant rate, the angular acceleration as a function of change in angular position and the squared final and initial angular velocities can be clear from the following expression:

$\omega^{2} = \omega_{o}^{2} + 2 \cdot \alpha\cdot (\theta-\theta_{o})$

Where:

$\omega_{o}$ , $\omega$ - Initial and final angular velocities, measured in radians per second.

$\alpha$ - Angular acceleration, measured in radians per square second.

$\theta_{o}$ , $\theta$ - Initial and final angular position, measured in radians.

Then,

$\alpha = \frac{\omega^{2}-\omega_{o}^{2}}{2\cdot (\theta-\theta_{o})}$

Given that $\omega_{o} = 0\,\frac{rad}{s}$ , $\omega = 0.70\,\frac{rad}{s}$ and $\theta-\theta_{o} = 4.9\,rad$ , the angular acceleration is:

$\alpha = \frac{\left(0.70\,\frac{rad}{s} \right)^{2}-\left(0\,\frac{rad}{s} \right)^{2}}{2\cdot \left(4.9\,rad\right)}$

$\alpha = 0.05\,\frac{rad}{s^{2}}$

Now, the time needed to accelerate the Ferris wheel uniformly is described by this kinematic equation:

$\omega = \omega_{o} + \alpha \cdot t$

Where $t$ is the time measured in seconds.

The time is cleared and obtain after replacing every value:

$t = \frac{\omega-\omega_{o}}{\alpha}$

If $\omega_{o} = 0\,\frac{rad}{s}$ , $\omega = 0.70\,\frac{rad}{s}$ and $\alpha = 0.05\,\frac{rad}{s^{2}}$ , the required time is:

$t = \frac{0.70\,\frac{rad}{s} - 0\,\frac{rad}{s} }{0.05\,\frac{rad}{s^{2}} }$

$t = 14\,s$

Average angular acceleration is obtained by dividing the difference between final and initial angular velocities by the time found in the previous step. That is:

$\bar \alpha = \frac{\omega-\omega_{o}}{t}$

If $\omega_{o} = 0\,\frac{rad}{s}$ , $\omega = 0.70\,\frac{rad}{s}$ and $t = 14\,s$ , the average angular acceleration is:

$\bar \alpha = \frac{0.70\,\frac{rad}{s} - 0\,\frac{rad}{s} }{14\,s}$

$\bar \alpha = 0.05\,\frac{rad}{s^{2}}$

The average angular acceleration is 0.05 radians per square second.

4 0

3 years ago

The top pair of pliers failed to loosen a stubborn bolt, but the bottom pair successfully removed it.

Lesechka [4]

The top pair of pliers failed to loosen a stubborn bolt, but the bottom pair successfully removed it. Because the contact between the bolt and the pliers working surface is less.

<h3>What is mechanical advantage ?</h3>

Mechanical advantage is a measure of the ratio of output force to input force in a system, it is used to obtained efficiency of the given mechanical machine.

The efficiency to open the stubborn bolt depends upon the contact between the working surface of the pliers and the bolt.

The contact between the bolt and the top pair of pliers working surface is less. Its mechanical advantage is less.

Hence, the top pair of pliers failed to loosen a stubborn bolt, but the bottom pair successfully removed it.

To learn more about the mechanical advantage, refer to the link;

brainly.com/question/7638820

#SPJ1

3 0

2 years ago

A 7300 kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.20 m/s2 and feels no appreci

Simora [160]

Answer:

Explanation:

We shall first calculate the velocity at height h = 575 m .

acceleration a = 2.2 m /s²

v² = u² + 2 a s

u is initial velocity , v is final velocity , s is height achieved

v² = 0 + 2 x 2.2 x 575

v = 50.3 m /s

After 575 m , rocket moves under free fall so g will act on it downwards

If it travels further by height H

from the relation

v² = u² - 2 g H

v = 0 , u = 50.3 m /s

H = ?

0 = 50.3² - 2 x 9.8 H

H = 129.08 m

Total height attained by rocket

= 575 + 129.08

= 704.08 m .

4 0

2 years ago

A pendulum is raised to a height of 0.3m above its lowest point and released. What is the velocity of the pendulum at its lowest

enyata [817]

Answer:

v = 2,425 m / s

Explanation:

A simple pendulum has anergy stored at the highest point of the path and this energy is conserved throughout the movement.

highest point

Em₀ = U = m g y

lowest point

$Em_{f}$ = K = ½ m v²

Em₀ = Em_{f}

mg y = ½ m v²

v = √ 2gy

let's calculate

v = √ (2 9.8 0.3)

v = 2,425 m / s

3 0

3 years ago

1. What are the two factors affecting friction

bija089 [108]

Answer: a.) Roughness of the surfaces in contact with each other .
Higher the roughness of surfaces in contact with each other, greater is the friction between bodies. Force of friction will be less between smooth surfaces.

b.) Weight of the sliding/rolling body: greater the weight of the moving body on the surface, more is the force of friction on the body by the surface.

I hope this helps

5 0

1 year ago