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3 years ago

What is the momentum of a bus that weighs 13,500,000 grams traveling at 65 mph?

Physics

1 answer:

Goryan [66]3 years ago

5 0

The momentum of the bus is $3.93\cdot 10^5 kg m/s$

Explanation:

The momentum of an object is given by

$p=mv$

where

m is the mass of the object

v is its velocity

For the bus in this problem, we have

$m=13,500,000 g = 13,500 kg$ is its mass

$v=65 mph \cdot \frac{1609 m/mi}{3600 s/h}=29.1 m/s$ is the velocity, converted into SI units

Substituting into the equation, we find the momentum of the bus:

$p=(13,500)(29.1)=3.93\cdot 10^5 kg m/s$

Learn more about momentum:

brainly.com/question/7973509

brainly.com/question/6573742

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brainly.com/question/9484203

#LearnwithBrainly

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Answer:

When a ray of light passes through a glass slab of a certain thickness, the ray gets displaced or shifted from the original path. This is called lateral shift/displacement.

Explanation:

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Describe what happens to the energy of a wave if the frequency decreases and the frequency increased

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3 years ago

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How long would it take for a car to travel a distance of 220 kilometers if it is traveling at a velocity of 55 km/hr South?

vagabundo [1.1K]

220/55=4 hours in total

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3 years ago

A 5.0-kg block of wood is placed on a 2.0-kg aluminum frying pan. How much heat is required to raise the temperature of both the

Shalnov [3]

Heat required to raise the temperature of a given system is

$Q = ms\Delta T$

here we know that

m = mass

s = specific heat capacity

$\Delta T$ = change in temperature

now as we know that

mass of wood = 5 kg

mass of aluminium pan = 2 kg

change in temperature = 45 - 20 = 25 degree C

specific heat capacity of wood = 1700 J/kg C

specific heat capacity of aluminium = 900 J/kg C

now here we will find the total heat to raise the temperature of both

$Q = m_1s_1\Delta T_1 + m_2s_2\Delta T_2$

$Q = 5 * 1700 * 25 + 2 * 900 * 25$

$Q = 212500 + 45000$

$Q = 257500 J$

So heat required to raise the temperature of the system is 257500 J

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3 years ago

Challenge Question (This is supposed to be hard, think critically and take it one step at a time):

aivan3 [116]

Answer:

You must travel at an average speed of 67.06 m/s to be on time to the lecture.

Explanation:

From the question, the total distance from your house to the science lecture is 75 miles.

Also, you get halfway there before you stop for a gas, that is, you have covered half of 75 miles, which 37.5 miles and you also have to cover 37.5 miles to get to the science lecture.

After filling up, you only have 15 minutes before the lecture starts,

To determine how fast you must drive to be on time to the lecture,

we will determine the average speed you need to travel.

From

Average speed = Distance / Time

Distance = 37.5 miles (Convert to meters)

(NOTE: 1 mile = 1609.344 meters)

Hence, 37.5 miles = 37.5 × 1609.344 miles = 60350.4 meters

∴ Distance = 60350.4 meters

Time = 15 minutes (Convert to seconds)

(NOTE: 1 minute = 60 seconds)

Hence, 15 minutes = 15 × 60 seconds = 900 seconds

Now, from

Average speed = Distance / Time

Average speed = 60350.4 m / 900 s

Average speed = 67.06 m/s

Hence, you must travel at an average speed of 67.06 m/s to be on time to the lecture.

8 0

4 years ago