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2 years ago

Question 1

Physics

1 answer:

Alekssandra [29.7K]2 years ago

6 0

Answer:

negatively charged object is is an object that has an excess of negative electrons and its the circle with the this symbol "-" and a positively charged object is an object that has an excess of positive electrons its the circle with the this symbol "+" .

tip: just remember the circles with the + and - symbol

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Why is the moon tilted

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It’s cause by the fact that the earth is tilted 23.5 degrees

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2 years ago

What number is between 2000 and 2500

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3 years ago

The average force of a baseball is 18.9 N . It’s mass is 0.145kg fine the acceleration in m/s^2

Artyom0805 [142]

Answer:

the acceleration is 130.3m/s²

Explanation:

Given data

Force F= 18.9N

Mass of ball m= 0.145kg

Acceleration a=?

Applying the Newton's second law of motion

"The rate of change of momentum of a body is proportional to the external force".

F=ma

a= F/m

a= 18.9/0.142

a= 130.3m/s²

3 0

3 years ago

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

$\tau = I\alpha$

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