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scZoUnD [109]
1 year ago
10

Calculate the potential energy of a body of a mass 2kg held 4 meters above the ground if g=10m/s?​

Physics
1 answer:
Drupady [299]1 year ago
7 0

Answer:

16000

Explanation:

Mass(m)=2Kg (1kg= 1ooo g then 2 kg=2000 g)

Velocity(v)= 4 meter

Acceleration due to gravity (g)=10m/s

We know that,

P.E= 1/2 mv^2

or, 1/2 × 2ooo × 4^2

or, 1/2×2000 ×16

or, 2000×8

Therefore= 16000

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The force required to stretch a Hooke’s-law spring varies from 0 N to 63.5 N as we stretch the spring by moving one end 5.31 cm
Alika [10]

Answer:

Force constant will be 1195.85 N/m

Work done will be 1.6859 J

Explanation:

We have given the force,  F = 63.5 N

Spring is stretched by 5.31 cm

So x = 0.0531 m

Force is given , F = 63.5 N

We know that force is given by F=kx

So 63.5=k\times 0.0531

k = 1195.85 N/m

Now we have to find the work done

We know that work done is given by

W=\frac{1}{2}kx^2=\frac{1}{2}\times 1195.85\times 0.0531^2=1.6859J

8 0
3 years ago
A truck was carrying a substance in a tank. The molecules of that substance were moving away from each other. The truck parked o
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Answer:

In the morning the molecules were moving away from each other with a smaller speed than when the truck was carrying the substance.

Explanation:

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2 years ago
In the following circuit, the supplied voltage is 1.5 volts and the resistor value is 2 ohms. Use Ohm's law to determine the cur
attashe74 [19]

Answer:

0.75 Amps

Explanation:

I had this question and this was right

7 0
3 years ago
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Select the choice that best completes the following sentence. Simple machines A. reduce the amount of energy needed to do work.
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It is actually are tools that make work easier C because i just had it on study island

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3 years ago
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A dentist’s drill starts from rest. After 3.20 s of constant angu-lar acceleration, it turns at a rate of 2.51 3 104 re v/m i n.
Gekata [30.6K]

Answer:

ΔTita = 4205.6 rad

Explanation:

w_{i} means initial angular velocity, which is 0 rev/min

w_{f} means final angular velocity, which is 2.513*10^{4}rev/min

t means time t= 3.20 s

one revolution is equivalent to 2πrad so the final angular velocity is:

w_{f} = (2π/60) *2.513*10^{4} rad/s

w_{f}= 2628.5 rad/s

so the angular acceleration, α will be:

α = 2628.5 rad/s / 3.20 s

a = 821.40 rad/s^{2}

so the rotational motion about a fixed axis is:

w^{2} _{f} =w^{2} _{i} + 2αΔTita    where ΔTita is the angle in radians

so now find the ΔTita the subject of the formula

ΔTita = \frac{w^{2} _{f}-w^{2} _{i}  }{2a}

ΔTita = ((2628.5)^{2} - (0 rev/min)^{2}) / 2* 821.40

ΔTita = 4205.6 rad

7 0
3 years ago
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