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3 years ago

A car accelerates for 10 seconds. During this time, the angular

Physics

1 answer:

bagirrra123 [75]3 years ago

4 0

Answer:

the angular acceleration of the car is 1.5 rad/s²

Explanation:

Given;

initial angular velocity, $\omega_i$ = 10 rad/s

final angular velocity, $\omega_f$ = 25 rad/s

time of motion, t = 10 s

The angular acceleration of the car is calculated as follows;

$a_r = \frac{\omega_f - \omega_i }{t} \\\\a_r = \frac{25-10}{10} = 1.5 \ rad/s^2$

Therefore, the angular acceleration of the car is 1.5 rad/s²

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How would the mathematical model change if the direction that the object traveled was reversed

victus00 [196]

A line that is falling towards the x axis represents an object that is negatively accelerating, or slowing down. When the line hits the x axis, the object has stopped moving. If the graph continues below the x axis, the object has changed direction and is moving backwards at increasing velocity.

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3 years ago

Estimate the change in the equilibrium melting point of copper caused by a change in pressure of 10 kbar. The molar volume of co

mr Goodwill [35]

Answer:

The change in the equilibrium melting point is 4.162 K.

Explanation:

Given that,

Pressure = 10 kbar

Molar volume of copper $V=8.0\times10^{-6}\ m^3$

Volume of liquid $V=7.6\times10^{-6}\ m^3$

Latent heat of fusion $L= 13.05 kJ$

Melting point =1085°C

We need to calculate the change temperature

Using Clapeyron equation

$\dfrac{\Delta P}{\Delta T}=\dfrac{\Delta H}{T\Delta V}$

Put the value into the formula

$\dfrac{1000\times10^{5}}{\Delta T}=\dfrac{13050}{(1085+273)\times(8.0-7.6)\times10^{-6}}$

$\Delta T=\dfrac{1000\times10^{-5}\times(1085+273)\times(8.0-7.6)\times10^{-6}}{13050}$

$\Delta T=4.162\ K$

Hence, The change in the equilibrium melting point is 4.162 K.

5 0

3 years ago

A freight train has a mass of 1.5 X 10^7 kg. If the

Elena L [17]

Answer:

t=444.4s

Explanation:

m=1.5*10^7 kg

F=7.5*10^5 N

v=80km/h*(1h/3600s)*(1000m/1km)=22.22m/s

<u>Second Newton's Law:</u>

F=ma

a=F/m=7.5*10^5/(1.5*10^7)=0.05m/s^2

<u>Kinematics equation:</u>

vf=vo+at=at

vo: initial velocity equal zero

t=vf/a=22.22/0.05=444.4s

4 0

3 years ago

How dense the medium is in the compression part of the wave, and how

ivolga24 [154]

How dense the medium is in the compression part of the wave and how rare the medium is in the rarefaction part of the wave is a measure of the longitudinal wave's amplitude.

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3 years ago

Help me!!! I really need it

Alenkinab [10]

Answer:

The reading on the ammeter A₁ , should be 2 Amp.

Explanation:

Given circuit shows all the bulbs are connected in parallel and ammeter A(T) at the source reads 6 Amp.

So, as the bulbs are in parallel connection, the current gets divided equal to each bulb.

so, the reading on the ammeter A₁ , should be 2 Amp.

But it will show 6 Amp, which is three times of the required value(2 amp).

This is because there was a mistake while making the circuit connections.

Instead of making connections as given circuit, the connections were made different as uploaded circuit.

Instead of connecting ammeter A₁ to only one bulb, it is connected to all the bulbs.

To correct this, remove the wire connecting ammeter and the second bulb.

And connect the second bulb directly to the switch.

3 0

4 years ago