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8090 [49]
3 years ago
9

How much work is done lifting a 71 kg dog 2 meters

Physics
1 answer:
alexdok [17]3 years ago
7 0
By using the potential energy rule E=mgh=71(10)(2)=1420 note i took the value of gravity 10
A change in potential energy means that there is work
Work=potential energy
Work=1420J
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4 years ago
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Define1n force ? write down cgs unit of force​
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You hang a heavy ball with a mass of 94 kg from a platinum rod 3.2 m long by 2.2 mm by 3.1 mm. You measure the stretch of the ro
vlada-n [284]

Answer:

1.67\times 10^{11} N/m^2

Explanation:

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For Young's modulus, substitute the values in the formula which is as follows:

\frac{F}{A}=Y\frac{\Delta L}{L}\\\Rightarrow Y= \frac{FL}{A\Delta L}\\\Rightarrow Y =\frac{922.14 \times 3.2}{2.2\times 10^{-3}\times 3.1\times 10^{-3} \times 0.002588}=1.67\times 10^{11} N/m^2

4 0
3 years ago
3. What is the kinetic energy of a 1300 kg car moving at 26.3 m/s?
Alexxx [7]

Answer:K.E=449598.5j

Explanation:

Kinetic energy of a moving car=1/2mv^2

Where m is the mass of the car

And V is the velocity of the car

K.E=1/2 ×1300×26.3^2

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3 0
3 years ago
A race car reaches the finish line and the driver slows down to come to a stop. If it takes the car 10 seconds to stop at a cons
boyakko [2]

Answer:

53.64 m/s

Explanation:

Applying,

a = (v-u)/t............. Equation 1

Where a = acceleration of the car, v = final velocity of the car, u = initial velocity of the car, t = time.

make u the subject of the equation

u = v-at............. Equation 2

From the question,

Given: a = -12 mph/s = -5.364 m/s², t = 10 seconds, v = 0 m/s (comes to stop)

Substitute these values into equation 2

u = 0-(-5.364×10)

u = 0+53.64

u = 53.64 m/s

6 0
3 years ago
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