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3 years ago

7

A child pulls a sled up a snow-covered hill. The child does 504 J of work on the sled. If the child walks 15 m

1 answer:

Juliette [100K]3 years ago

7 0

Answer: F = 33.6 N

Explanation: work = force · distance or W = F·s

Force F = W/s = 504 J/15 m

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How do you make an electromagnet

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Putting a piece of iron or steel inside the coil makes the magnet strong enough to attract objects. The strength of an electromagnet can be increased by increasing the number of loops of wire around the iron core and by increasing the current or voltage.

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4 years ago

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Explain how muscles and bones work together to help bend the arm.

fomenos

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Bones make sure the arms structure is stable while the muscle is the one actively moving it and carrying out actions.

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3 years ago

During one trial, the acceleration is 2m/s^2 to the right. What calculation will give the tensions in actin filaments during thi

solniwko [45]

Answer: hello your question is poorly written attached below is the complete question

answer :

TA = 1.6*10^-24 * 60 * 2, TB = 1.6*10^-24 * ( 60 + 30 ) * 2 -- ( option 1 )

Explanation:

a = 2m/s^2

Ta = m₁ a = 60 * 1.6 * 10^-24 * 2 ц

Tb - Ta = m₂ a

∴ Tb = m₂ a + Ta

= ( 30 * 1.6 * 10^-24 * 2 ) + ( 60 * 1.6 * 10^-24 * 2 )

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3 years ago

A positive point charge (q = +8.65 x 10-8 C) is surrounded by an equipotential surface A, which has a radius of rA = 2.97 m. A p

ella [17]

Answer:

$r_B = 1.88 m$

Explanation:

As we know that work done by electric force is given as

$W_e = -q\Delta V$

so here we know that charge is moving from

$r_A = 2.97 m$

to another position

so we will have

$W_{AB} = \frac{kq_1q_2}{r_A} - \frac{kq_1q_2}{r_B}$

$-6.95\times 10^{-9} = (9\times 10^9)(8.65\times 10^{-8})(4.56\times 10^{-11})(\frac{1}{2.97} - \frac{1}{r_B})$

$-0.196 = (\frac{1}{2.97} - \frac{1}{r_B})$

$r_B = 1.88 m$

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4 years ago

A car traveling at a speed of 13 meters per second accelerates uniformly to a speed of 25 meters per second in 5.0 seconds. 11-

Aleks04 [339]

Answer:

$a=2.4\ m/s^2$

Explanation:

Given that,

Initial speed of a car, u = 13 m/s

Final speed of a car, v = 25 m/s

Time, t = 5 s

We need to find the acceleration of the car during this 5.0 second time interval. Let a is the acceleration. It can be calculated as :

$a=\dfrac{v-u}{t}\\\\a=\dfrac{25-13}{5}\\\\=2.4\ m/s^2$

So, the acceleration of the car is $2.4\ m/s^2$ .

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3 years ago