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2 years ago

6

A boy holds a toy soldier in from of a concave mirror. The focal length of the mirror is 0.45 m and the boy holds the toy soldie

r at a distance of 0.25m from the mirror. Find the image distance.

1 answer:

Elena-2011 [213]2 years ago

5 0

Daddy is coming back to me lol

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2. physical quantities which do not depend on any physical quantities for their measurements are known as

seropon [69]

Answer:

Fundamental quantities

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3 years ago

Two firefighters are trying to break through a door. One firefighter is heavy, and the other is light. If they run at the same s

frez [133]

Answer:

The Heavier Firefighter

Explanation:

Generally, more massive objects will have more intertia than less massive objects. As such it takes more force to halt a more massive object if its moving at the same speed as a smaller object. This can also be thought of in the context of Newton's second law. The more force needed to accelerate an object means the more force the object will have.

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3 years ago

How much force is needed to accelerate a 68 kilogram-skier at a rate of 1.2 m/sec^2?

DiKsa [7]

<span>Answer: Force = 81.6 N

Explanation:
According to Newton's Second law:
F = ma --- (1)

Where F = Force = ?
m = Mass = 68 kg
a = Acceleration = 1.2 m/s^2

Plug in the values in (1):
(1) => F = 68 * 1.2
F = 81.6 N (The force needed to accelerate the skier at a rate of 1.2 m/s^2)</span>

4 0

3 years ago

Read 2 more answers

The talk test can be used to measure the __________. A. knowledge of an activity B. intensity of an activity C. length of an act

marta [7]

Answer:

B.

Explanation:

According to CDC, if you’re doing moderate-intensity activity, you can talk but not sing during the activity.

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3 years ago

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When the play button is pressed, a CD accelerates uniformly from rest to 450 rev/min in 3.0 revolutions. If the CD has a radius

Marina CMI [18]

To solve this problem it is necessary to apply the kinematic equations of angular motion.

Torque from the rotational movement is defined as

$\tau = I\alpha$

where

I = Moment of inertia $\rightarrow \frac{1}{2}mr^2$ For a disk

$\alpha =$ Angular acceleration

The angular acceleration at the same time can be defined as function of angular velocity and angular displacement (Without considering time) through the expression:

$2 \alpha \theta = \omega_f^2-\omega_i^2$

Where

$\omega_{f,i} =$ Final and Initial Angular velocity

$\alpha =$ Angular acceleration

$\theta =$ Angular displacement

Our values are given as

$\omega_i = 0 rad/s$

$\omega_f = 450rev/min (\frac{1min}{60s})(\frac{2\pi rad}{1rev})$

$\omega_f = 47.12rad/s$

$\theta = 3 rev (\frac{2\pi rad}{1rev}) \rightarrow 6\pi rad$

$r = 7cm = 7*10^{-2}m$

$m = 17g = 17*10^{-3}kg$

Using the expression of angular acceleration we can find the to then find the torque, that is,

$2\alpha\theta=\omega_f^2-\omega_i^2$

$\alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}$

$\alpha = \frac{47.12^2-0^2}{2*6\pi}$

$\alpha = 58.89rad/s^2$

With the expression of the acceleration found it is now necessary to replace it on the torque equation and the respective moment of inertia for the disk, so

$\tau = I\alpha$

$\tau = (\frac{1}{2}mr^2)\alpha$

$\tau = (\frac{1}{2}(17*10^{-3})(7*10^{-2})^2)(58.89)$

$\tau = 0.00245N\cdot m \approx 2.45*10^{-3}N\cdot m$

Therefore the torque exerted on it is $2.45*10^{-3}N\cdot m$

3 0

3 years ago