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2 years ago

A car travels 10.0 m/s. What is its velocity in km/h?

Physics

1 answer:

Vitek1552 [10]2 years ago

5 0

Since 1m/s=3.6 km/h, we can conclude that 10.0m/s = 36 km/h

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A bus driver heads south with a steady speed of

Nutka1998 [239]

ANS

{ A }

Magnitude

$5400.097 \: ✓ \: m$

Direction

$25.52 \: ✓ \: ° South \: of \: West$

{ B }

Average speed in ( m/s )

$23.43 \: ✓ \: m/s$

{ C }

Magnitude

$14.06 \: ✓ \: m/s$

Direction

$25.52 \: ✓ \: ° South \: of \: West$

5 0

2 years ago

You are riding a bicycle. If you apply a forward force of 125 N, and you and

erica [24]

Answer:

1.52g

Explanation:

Given parameters:

Force = 125N

Mass combined = 82kg

Unknown:

Acceleration of the bicycle = ?

Solution:

From Newton second law of motion suggests that:

Force = mass x acceleration

Acceleration = $\frac{force}{mass}$ = $\frac{125}{82}$ = 1.52g

8 0

2 years ago

One mole of ideal gas is slowly compressed to one-third of its original volume. in this compression, the work done on the gas ha

beks73 [17]

A boiling pot of water (the water travels in a current throughout the pot), a hot air balloon (hot air rises, making the balloon rise) , and cup of a steaming, hot liquid (hot air rises, creating steam) are all situations where convection occurs.
Read more on Brainly.com - brainly.com/question/1581851#readmore

4 0

3 years ago

During an adiabatic process an object does 100 J of work and its temperature decreases by 5 K. During another process it does 25

Liono4ka [1.6K]

Answer:

The heat capacity for the second process is 15 J/K.

Explanation:

Given that,

Work = 100 J

Change temperature = 5 k

For adiabatic process,

The heat energy always same.

$dQ=0$

$dU=-dW$

We need to calculate the number of moles and specific heat

Using formula of heat

$dU=nC_{v}dT$

$nC_{v}=\dfrac{dU}{dT}$

Put the value into the formula

$nC_{v}=\dfrac{-100}{5}$

$nC_{v}=-20\ J/K$

We need to calculate the heat

Using formula of heat

$dQ=nC_{v}(dT_{1})+dW_{1}$

Put the value into the formula

$dQ=-20\times5+25$

$dQ=-75\ J$

We need to calculate the heat capacity for the second process

Using formula of heat

$dQ=nC_{v}(dT_{1})$

Put the value into the formula

$-75=nC_{v}\times(-5)$

$nC_{v}=\dfrac{-75}{-5}$

$nC_{v}=15\ J/K$

Hence, The heat capacity for the second process is 15 J/K.

5 0

3 years ago

An imaginary line perpendicular to a reflecting surface is called _________.

n200080 [17]

<span>An imaginary line perpendicular to a reflecting surface is called "a normal" (principle line)

So, Your Answer would be Option B

Hope this helps!</span>

5 0

2 years ago

Read 2 more answers