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lianna [129]
4 years ago
8

Can someone help me plzzzz

Physics
1 answer:
Anit [1.1K]4 years ago
6 0

Lets use an ambulance as an example:

Lets say we are stationary on a sidewalk and a ambulance is coming towards us, at first we hear the sound at a lower pitch as it is further away, but as the ambulance travels towards us the frequency, which denotes how many waves pass a given point per unit time (we are the point), will get higher, meaning an increase in pitch as the waves become "denser" or more compressed. As the ambulance travels away you will hear the sound at a lower pitch because the frequency of sound waves passing you per unit time is less and it will continue to lessen until the ambulance can no longer be heard.

Essentially--> higher frequency means higher pitch

The waves become more compressed as a it moves with velocity v towards a certain point.

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An astronaut with a mass of 91 kg is 0.30 m above the moons surface. The astronauts potential energy is 46 J. Calculate the free
Blababa [14]

Answer:

the free-fall acceleration on the moon is 1.68 m/s^2

Explanation:

recall the formula for the gravitational potential energy (under acceleration of gravity "g"):

PE = m * g * h

replacing with our values for the problem:

46 J = 91 * g * 0.3

solve for the "g" on the Moon:

g = 46 / (91 * 0.3)

g = 1.68  m/s^2

3 0
3 years ago
can you help me create a sketch of two different objects with one that has a greater density than the other?
sergey [27]

I don't know how good you are at sketching ... I'm terrible. 
But you can put the point across in a dramatic way if you
can sketch a bowling ball and a basketball ... you'll need
to clearly identify them with the markings you sketch on
each ball. 

They're the same shape and nearly the same size, but
there's a huge difference in their densities.

6 0
3 years ago
A 150 kg line backer sacks the 120 kg quarterback. With what force is the quarterback sacked if the line backer has an accelerat
Gekata [30.6K]

Answer:

The force required to move the quarterback with linebacker is <u>1215 N</u>

Explanation:

\text { Mass of linebacker } \mathrm{m}_{2}=150 \mathrm{kg}

\text { Mass of quarterback } \mathrm{m}_{2}=120 \mathrm{kg}

\text { Moved at an acceleration }(a)=4.5 \mathrm{m} / \mathrm{s}^{2}

Using Newton's second law, it is established that  F = Ma

Where F is net force acting on the system, a is the acceleration and M is mass of the two object \left(m_{1}+m_{2}\right)

Now consider both \mathrm{m}_{1} \text { and } \mathrm{m}_{2}as a system, so net force acting on the system is \text { Force }=\left(m_{1}+m_{2}\right) a

Substitute the given values in the above formula,

\text { Force }=(150+120) \mathrm{kg} \times 4.5 \mathrm{m} / \mathrm{s}^{2}

\text { Force }=270 \mathrm{kg} \times 4.5 \mathrm{m} / \mathrm{s}^{2}

Force = 1215 N

<u>1215 N </u>is the force required to move the quarterback with linebacker.

5 0
3 years ago
Your friend's 10.8 g graduation tassel hangs on a string from his rearview mirror. When he accelerates from a stoplight, the tas
Nitella [24]

The tension in the string holding the tassel and the vertical will the tension in the string

  • T = 0.1953 N
  • Ф = 34.4 °

<h3>What is the tension in the string holding the tassel. ?</h3>

Generally, the equation for Tension is  mathematically given as

TCos\theta = mg

Therefore

TCos6.58^{o} = 19.8*10^{-3}*9.8

T = 0.1953 N

b).

Where

T* sin \theta = ma

0.1953*Sin6.58 \textdegree  = 19.8*10^{-3}*a

a = 1.13 m/s^2

In conclusion

T* sinФ = ma

2msinФ = ma

2sinФ = a

sin\theta = \frac{a}{2}

\theta = sin^{-1}\frac{a}{2} \\\\\theta= sin^{-1}\frac{1.13}{2}

Ф = 34.4 °

In conclusion, The tension in the string holding the tassel and the vertical will the tension in the string

T = 0.1953 N

Ф = 34.4 °

Read more about tension

brainly.com/question/15880959

#SPJ1

4 0
2 years ago
Someone help me please i need to finish this
Ilya [14]

Answer:

D. A zinc Zn contains 30 protons inside the nucleus and 30 electron outside the nucleus

8 0
3 years ago
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