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3 years ago

How much force is required to move a sled 5 meters if a person uses 60 j of work? 300N 12N 65N 30N

1 answer:

4 0

The answer is b 12N because
We know that<span>
W = F × d × c o s(θ)</span>
assuming theta=0 we then solve and have<span>
F=<span>W/d</span></span>
substitute known values to get:<span><span>
F=<span><span>60J/</span><span>5m</span></span>=12N</span></span>

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Jupiter has the strongest gravity of any planet in our solar system. Which quantity will be a maximum there?

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3 years ago

Physics question, help please!

vova2212 [387]

The change in potential energy when the block falls to ground is -480J.

The maximum change in kinetic energy of the ball is 480 J.

The initial kinetic energy of the ball is 0 J.

The final kinetic energy of the ball is 0.148J.

The initial potential energy of the ball is 0.187 J.

The final potential energy of the ball is 0 J.

The work done by the air resistance is 0.039 J.

<h3>Change in potential energy when the block falls to ground</h3>

ΔP.E = -mgh

ΔP.E = -Wh

ΔP.E = - 40 x 12

ΔP.E = -480 J

<h3>Maximum change in kinetic energy of the ball</h3>

ΔK.E = - ΔP.E

ΔK.E = - (-480 J)

ΔK.E = 480 J

<h3>Initial kinetic energy of the ball</h3>

K.Ei = 0.5mv²

where;

v is zero since it is initially at rest

K.Ei = 0.5m(0) = 0

<h3>Final kinetic energy</h3>

K.Ef = 0.5mv²

K.Ef = 0.5(0.0091)(5.7)²

K.Ef = 0.148 J

<h3>Initial potential energy of the ball</h3>

P.Ei = mghi

P.Ei = 0.0091 x 9.8 x 2.1

P.Ei = 0.187 J

<h3>Final potential energy</h3>

P.Ef = mghf

P.Ef = 0.0091 x 9.8 x 0

P.Ef = 0

<h3>Work done by the air resistance</h3>

W = ΔE

W = P.E - K.E

W = 0.187 J - 0.148 J

W = 0.039 J

Learn more about potential energy here: brainly.com/question/1242059

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1 year ago

A substance has life of 40 years. In how many years will the substance disintegrate to 1/8 of its initial size

tia_tia [17]

Answer:

5 years

Explanation:

$\frac{1}{8}(40)\\\\= 5$

8 0

2 years ago

Consider a simple tension member that carries an axial load of P=22.44N. Find the total elongation in the member due to the load

rodikova [14]

Answer:

The total elongation for the tension member is of 0.25mm

Explanation:

Assuming that material is under a linear deformation then the relation between the stress and the specific elongation is given as:

$\sigma=E*\epsilon$ (1)

Where E is the modulus of elasticity, σ the stress and ε the specific deformation. Also, the total longitudinal elongation can be expressed as:

$\delta L=L*\epsilon$ (2)

Here L is the member extension and δL the change total longitudinal elongation.

Now if the stress is found then the deformation can be calculated by solving the stress-deformation equation (1). The stress applied sigama is computed dividing the axial load P by the cross-sectional area A:

$\sigma=P/A$