All categories

1 year ago

Need some help with these two physics problems!

Physics

1 answer:

Juliette [100K]1 year ago

5 0

The force that keeps the puck moving is 0.25 N while the velocity of the puck is 3.7 m/s.

<h3>What is the centripetal force?</h3>

We know that the centripetal force is the force that acts on a body that is moving along a circular path. In this case, we are told that the puck is moving along a circular path hence it is acted upon by the centripetal force that acts on it.

The centripetal force in this case would be supplied by the weight of the object that is moving in the circular path. Thus we can write in our equation that;

Centripetal force = Weight of object = mg

m = mass of the object

g = acceleration due to gravity

Then;

W = 0.026 Kg * 9.8 m/s^2

W = 0.25 N

To obtain the velocity of the object;

FT = mv^2/r

v = √ FT r/m

v = √0.25 * 1.4/0.026

v = 3.7 m/s

Learn more about centripetal force:brainly.com/question/11324711

#SPJ1

You might be interested in

Directions: Read each question carefully and answer

Lady bird [3.3K]

I have all the answers here so take this

8 0

2 years ago

A sound source A and a reflecting surface B move directly toward each other. Relative to the air, the speed of source A is 28.7

aleksandrvk [35]

(a) 1440.5 Hz

The general formula for the Doppler effect is

$f'=(\frac{v+v_r}{v+v_s})f$

where

f is the original frequency

f is the apparent frequency

$v$ is the velocity of the wave

$v_r$ is the velocity of the receiver (positive if the receiver is moving towards the source, negative otherwise)

$v_s$ is the velocity of the source (positive if the source is moving away from the receiver, negative otherwise)

Here we have

f = 1110 Hz

v = 334 m/s

In the reflector frame (= on surface B), we have also

$v_s = v_A = -28.7 m/s$ (surface A is the source, which is moving towards the receiver)

$v_r = +62.2 m/s$ (surface B is the receiver, which is moving towards the source)

So, the frequency observed in the reflector frame is

$f'=(\frac{334 m/s+62.2 m/s}{334 m/s-28.7 m/s})1110 Hz=1440.5 Hz$

(b) 0.232 m

The wavelength of a wave is given by

$\lambda=\frac{v}{f}$

where

v is the speed of the wave

f is the frequency

In the reflector frame,

f = 1440.5 Hz

So the wavelength is

$\lambda=\frac{334 m/s}{1440.5 Hz}=0.232 m$

(c) 1481.2 Hz

Again, we can use the same formula

$f'=(\frac{v+v_r}{v+v_s})f$

In the source frame (= on surface A), we have

$v_s = v_B = -62.2 m/s$ (surface B is now the source, since it reflects the wave, and it is moving towards the receiver)

$v_r = +28.7 m/s$ (surface A is now the receiver, which is moving towards the source)

So, the frequency observed in the source frame is

$f'=(\frac{334 m/s+28.7 m/s}{334 m/s-62.2 m/s})1110 Hz=1481.2 Hz$

(d) 0.225 m

The wavelength of the wave is given by

$\lambda=\frac{v}{f}$

where in this case we have

v = 334 m/s

f = 1481.2 Hz is the apparent in the source frame

So the wavelength is

$\lambda=\frac{334 m/s}{1481.2 Hz}=0.225 m$

8 0

3 years ago

Which of the following is a sample of igneous rock?

Blizzard [7]

Answer:

I would say it's B. But just in case here is some information if I'm wrong.

Explanation:

Igneous rocks are very dense and hard. They may have a glassy apprearance. Metamorphic rocks may also have a glassy appearance. You can distinguish these from igneous rocks based on the fact that metamorphic rocks tend to be brittle, lightweight, and an opaque black color.

Hope this helps!

Please mark me Brainliest! :)

5 0

3 years ago

Three moles of an ideal gas with a molar heat capacity at constant volume of 4.9 cal/(mol∙K) and a molar heat capacity at consta

Sergeu [11.5K]

Answer:

The heat flows into the gas during this two-step process is 120 cal.

Explanation:

Given that,

Number of moles = 3

Heat capacity at constant volume = 4.9 cal/mol.K