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2 years ago

How much work is done when a very large force is applied to an object but does not move the object?

Physics

2 answers:

yarga [219]2 years ago

7 0

Answer:

A. None

(took the test)

Ierofanga [76]2 years ago

4 0

Work = (force) x (distance)

If the distance is zero (the object doesn't move), then it
doesn't matter how great the force is. The work is still zero.

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How many feet are in a mile

klemol [59]

Hello There!

There are 5,280 feet in 1 mile.

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2 years ago

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What force does a trampoline have to apply to a 45.0-kg gymnast to accelerate her straight up at 7.50 m/s^2? (a) 104N (b) 338 N

Brilliant_brown [7]

Answer:

b) 338 N

Explanation: let m be the mass of the gymnast and a be the acceleration of the gymnast.

the force required to accelerate the gymnast is given by:

F = m×a

= (45.0)×(7.50)

= 337.5 N

Therefore, the force a trampoline has to apply is 138 N.

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3 years ago

An ambulance driver traveling at 31.0 m/s (69.3 mph) honks his horn as he sees a motorist ahead on the highway traveling in the

IrinaVladis [17]

Answer:

fo = 378.52Hz

Explanation:

Using Doppler effect formula:

$f'=\frac{C-Vb}{C-Va}*fo$

where

f' = 392 Hz

C = 340m/s

Vb = 20m/s

Va = 31m/s

Replacing these values and solving for fo:

fo = 378.52Hz

4 0

3 years ago

PLEASE HELP DUE IN MINS. Thanks

rjkz [21]

Answer:

"C" I think....

Explanation:

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4 0

2 years ago

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The electric field between the plates of the cathode ray tube of an older television set can be as high as 2.5x104 N/C. Determin

pentagon [3]

Answer:

(a) $F = -4.01 * 10^{-15} N$

(b) $a = 4.40 * 10^{15} m/s^2$

Explanation:

Parameter given:

Electric field, E = $2.5 * 10^4 N/C$

(a) Electric force is given (in terms of electric field) as a product of electric charge and electric field.

Mathematically:

$F = qE$

Electric charge, q, of an electron = $- 1.602 * 10^{-19} C$

$F = -1.602 * 10^{-19} * 2.5 * 10^4\\\\\\F = -4.01 * 10^{-15} N$

(b) This electrostatic force causes the electron to accelerate with an equivalent force:

F = -ma

where m = mass of an electron

a = acceleration of electron

(Note: the force is negative cos the direction of the force is opposite the direction of the electron)

Therefore:

$-ma = -4.01 * 10^{-15} N\\\\\\a = \frac{-4.01 * 10^{-15}}{-m}$

Mass, m, of an electron = $9.11 * 10^{-31} kg$

=> $a = \frac{-4.01 * 10^{-15}}{-9.11 * 10^{-31}}\\\\\\a = 4.40 * 10^{15} m/s^2$

The acceleration of the electron is $4.40 * 10^{15} m/s^2$

5 0

2 years ago