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3 years ago

A body moving with a velocity of 20 m/s begins to accelerate at 3 m/s2. How far does the body move in 5 seconds?

2 answers:

5 0

Initial velocity = 20m/s
Accelerations = 3m/s^2
Time taken = 5s

Using second equation of motion
ut+(1/2)at^2

Where, s is the distance covered by the body
u is initial velocity
a is acceleration and t is time taken by that body

= (20)(5)+(1/2)(3)(5)^2
=100 + 37.5
= 137.5

So, the total distance covered by the body in next 5s would be 137.5m

Arlecino [84]3 years ago

5 0

The Correct answer to this question for Penn Foster Students is: 137.5 m

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In Newtons first law of motion how is an object at rest affected

Korvikt [17]

Answer:

An object at rest remains at rest, or if in motion, remains in motion at a constant velocity unless acted on by a net external force.

Explanation:

Meaning that if an object is just sitting somewhere unless something else causes it to move it's not going anywhere.

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3 years ago

If the velocity of blood flow in the aorta is normally about 0.32 m/s, what beat frequency would you expect if 4.40-MHz ultrasou

dusya [7]

Answer:

The beat frequency is 0.0019 MHz.

Explanation:

Given that,

Velocity = 0.32 m/s

Frequency = 4.40 MHz

Speed of wave = 1540 m/s

We need to calculate the frequency

Case (I),

Observer is moving away from the source

Using Doppler's effect

$f'=\dfrac{v-v'}{v}f$

Where, v' = speed of observer

Put the value into the formula

$f'=\dfrac{1540-0.32}{1540}\times4.40$

$f'=4.399\ MHz$

Case (II),

Cell is as the source of sound of frequency f' and it moving away from the observer.

Using formula of frequency

$f''=\dfrac{v-v_{s}}{v+v_{s}}\times f$

$f''=\dfrac{1540-0.32}{1540+0.32}\times4.399$

$f''=4.3971\ MHz$

We need to calculate the beat frequency

$\Delta f= f'-f''$

$\Delta f=4.399-4.3971=0.0019\ MHz$

Hence, The beat frequency is 0.0019 MHz.

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3 years ago

A rock is stuck deep in the dirt. When you pull on the rock to remove it, what type of force are you exerting?

aleksandrvk [35]

It is Tension as the other 3 answer choices would not make sense. Compression would mean you are pressing the rock on both sides or in this case, pushing it into the dirt. It can't be nuclear force as you are pulling out a rock. Air resistance would not make sense either as there is no air involved in the scenario at all.

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3 years ago

Read 2 more answers

A 500 N weight sits on the small piston of a hydraulic machine. The small piston has area 2.0 cm2. If the large piston has area

Ghella [55]

Answer:

d. 10000N

Explanation:

When a force ( $F_1$ ) is exerted on the smaller area piston ( $A_1$ ), the pressure that originates therein is transmitted to the larger area piston( $A_2$ ). According to Pascal's principle the pressure on the smallest piston ( $P_1=\frac{F_1}{A_1}$ ) will be equal to the pressure on the largest piston ( $P_2=\frac{F_2}{A_2}$ ):

$\frac{F_1}{A_1}=\frac{F_2}{A_2}\\F_2=F_1\frac{A_2}{A_1}\\F_2=500N\frac{40cm}{2cm}\\F_2=10000N$

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increased with an increased current flow

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