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3 years ago

What would be the acceleration in a bus moving with uniform velocity and why

Physics

1 answer:

jeka57 [31]3 years ago

3 0

Answer:

Then its acceleration is zero.

Explanation:

Velocity → Uniform → Constant

→ Change in Velocity = 0

A= $\frac{v-m}{e}$ = (v=m)

a=0

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A 0.700-kg particle has a speed of 1.90 m/s at point circled A and kinetic energy of 7.20 J at point circled B. (a) What is its

Savatey [412]

Answer:

a). $E_{kA}=1.2635 J$

b). $V_{B}=4.535\frac{m}{s}$

c). Δ $E_{t}=8.4635 J$

Explanation:

ΔE=kinetic energy

a).

$E_{kA}=\frac{1}{2}*m*v_{A} ^{2} \\ v_{A}=1.9 \frac{m}{s}\\ m=0.70kg\\E_{kA}=\frac{1}{2}*0.70kg*(1.9 \frac{m}{s})^{2} \\E_{kA}=1.2635 J$

b).

$E_{kB}=\frac{1}{2}*m*v_{B} ^{2}$

$V_{B}^{2}=\frac{E_{kB}*2}{m} \\V_{B}=\sqrt{\frac{E_{kB}*2}{m}} \\V_{B}=\sqrt{\frac{7.2J*2}{0.70kg}} \\V_{B}=4.53 \frac{m}{s}$

c).

net work= EkA+EkB

$E_{t}=E_{kA}+ E_{kB}\\E_{t}=1.2635J+7.2J\\E_{t}=8.4635J$

3 0

3 years ago

The distance from home plate to the pitchers mound is 18.5 meters. For a pitcher capable of throwing at 3.85m/s(86mi/hr), how mu

Taya2010 [7]

Given that:

Distance , s = 18.5 m

Velocity , v = 3.85 m/s

Time , t =?

Since,

Velocity = distance/time

Time= distance/velocity

time= 18.5/ 3.85

time= 4.8 s

So the time elapse between the release of the ball and the ball passing home plate is 4.8 seconds.

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3 years ago

How far does a 1.2g bullet with kinetic energy of 1.2 j go in 2 seconds ?<br><br><br>

almond37 [142]

Answer:

To calculate the energy in joules, simply enter the mass of ammunition (in grams) that you use, and the fps that you've read from your Chrono unit.

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2 years ago

A high ____will have a short wavelength.

Sphinxa [80]

Question:

A high ____will have a short wavelength

Answer:

That means that waves with a high frequency have a short wavelength, while waves with a low frequency have a longer wavelength. Light waves have very, very short wavelengths

Explanation:

Hope it help

4 0

3 years ago

Read 2 more answers

The apparent weight of a student in alift is 564N . if the mass of the student is 60.3kg, what is the acceleration of the lift ?

Yuki888 [10]

Answer:

-.457 m/s^2

Explanation:

Actual weight = 60 .3 (9.81) = 591.54 N

Accel of lift changes this to 60.3 ( 9.81 - L) where L - accel of lift

60.3 ( 9.81 - L ) = 564

solve for L = .457 m/s^2 DOWNWARD

so L = - .457 m/s^2

4 0

2 years ago

What would be the acceleration in a bus moving with uniform velocity and why​

What would be the acceleration in a bus moving with uniform velocity and why