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Mariana [72]
3 years ago
13

a person uses 75(joules) of work to apply a force of 5(newtons) to push a crate what distance did the crate move ?

Physics
1 answer:
Ierofanga [76]3 years ago
3 0

Hi there!

We know the equation:

W = F · d, where:

F = force (N)

d = displacement (m)

W = work (J)

Rearrange to solve for distance:

W/F = d

Plug in the given values:

75/5 = 15 m

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physics a flower pot falls from a windowsill 25.0m above the sidewalk. how much time does a passerby on the sidewalk below have
Usimov [2.4K]
v_f^2 = v_0^2 + 2*a*d
Since the flower pot is dropped, it has an initial velocity of zero. Also the flower pot accelerates due to gravity.

so

v_f = \sqrt{2*g*d}=\sqrt{2*9.81*25.0}
8 0
4 years ago
A6 cm object is 8 cm from a convex lens that has a focal length of 2.7 cm. The image is 4 cm from the lens. The height of the im
Marrrta [24]

Answer:

7.8cm

Explanation:

5 0
3 years ago
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A 0.5 kg block of aluminum (caluminum=900J/kg⋅∘C) is heated to 200∘C. The block is then quickly placed in an insulated tub of co
Genrish500 [490]

Answer: When 1.0kg of aluminium block is used, the final temperature of the mixture will be T = 36.2∘C

If 1.0kg copper block is used, T of the mixture will be = 17.4∘C

If 100g (0.1kg) of ice at 0∘C is used, T will be = 64.9∘C

If 25g (0.025Kg) of ice is used, T will be= 147.1∘C

Explanation:

H = mcΘ

heat lost by block = heat gained by water

m₁c₁Θ₁ = m₂c₂Θ₂ where m₁ is mass of aluminium, m₂ is mass of water, c₁ is cAluminium, c₂ is cWater, Θ₁ is temperature change for aluminium, Θ₂ is temperature change for water.

0.5*900*(200-20) = m₁*4186*(20-0)

m₁ = 450*180/83270

<em>m₁ = 0.973kg</em>

<em>when 1.0kg of aluminium block is used, the final temperature of the mixture will be </em><em>T</em>

heat lost by block = heat gained by water

1.0*900*(200-T) = 0.973*4186*(T-0)

180000 - 900T = 4073T

4973T = 180000

T = 180000/4973 = 36.2∘C

<em>If 1.0kg copper block is used, T of the mixture will be</em>

heat lost by block = heat gained by water

1.0*387*(200-T) = 0.973*4186*(T-0)

77400 - 387T = 4073T

4460T = 77400

T = 77400/4460 = 17.4∘C

<em>If 100g (0.1kg) of ice at 0∘C is used, T will be</em>

<em>heat lost by block = heat gained by water + heat used in melting ice to form water at 0∘C</em>

heat used in melting 0.1kg of ice, H = ml, where l= 33600J/Kg

0.5*900*(200-T) = 0.1*4186*(T-0) + 0.1*33600J/Kg

90000 - 450T =  418.6T + 33600

418.6T + 450T = 90000 - 33600

868.6T = 56400

T = 56400/868.6 = 64.9∘C

If 25g (0.025Kg) of ice is used, T will be

0.5*900*(200-T) = 0.025*4186*(T-0) + 0.025*33600J/Kg

90000 - 450T =  104.65T + 8400

104.65T + 450T = 90000 - 8400

554.65T = 81600

T = 81600/554.65 = 147.1∘C

7 0
3 years ago
What are the main characteristics of S.H.M​
agasfer [191]

Answer:

•→ The motion of a particle or body in S.H.M acts towards a fixed point.

•→ Acceleration of the body under S.H.M is proportional to its displacement.

•→ This motion is periodic.

•→ Mechanical energy is conserved in S.H.M

Explanation:

S.H.M is Simple Harmonic Motion

.

4 0
3 years ago
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A car passes point “A” and then 120 meters later. It’s velocity was measured 21 m/s. If it’s acceleration was constant at 0.853
Norma-Jean [14]

Recall that

{v_f}^2-{v_i}^2=2a\Delta x

where v_i and v_f are the initial and final velocities, respecitvely; a is the acceleration; and \Delta x is the change in position.

So we have

\left(21\dfrac{\rm m}{\rm s}\right)^2-{v_i}^2=2\left(0.853\dfrac{\rm m}{\mathrm s^2}\right)(120\,\mathrm m)

\implies v_i\approx\boxed{15.4\dfrac{\rm m}{\rm s}}

(Normally, this equation has two solutions, but we omit the negative one because the car is moving in one direction.)

7 0
3 years ago
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