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coldgirl [10]
3 years ago
12

Black widow (m=57000 g) shovels snow, holding the shovel at a 40 degree angle above the ground. If she uses a constant force of

180. N and pushes the snow 5.00 m. How much work does she do on the snow?
Physics
1 answer:
Dima020 [189]3 years ago
7 0

Answer:

She does a work of 689.44 J in the snow.

Explanation:

A force is said to do work when it alters the state of motion of a body. The work of the force on that body will be equivalent to the energy needed to move it.

In other words, Work is a form of energy transmission between bodies. In order to carry out work, a force must be exerted on a body and it must move.

The work is equal to the product of the force times the distance and the cosine of the angle that exists between the direction of the force and the direction that the moving point or object travels:

W= F*d* cos Ф

Work W is measured in joules (J), force is measured in newtons (N), and displacement in meters (m).

In this case:

  • F= 180 N
  • d=5 m
  • Ф= 40 degrees

Replacing:

W= 180 N*5 m* cos 40

Solving:

W= 689.44 J

<u><em>She does a work of 689.44 J in the snow.</em></u>

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The problem doesn't tell us what planet this is happening on, so let's make it easy and just assume it's on Earth.  Then the 'acceleration' is Earth gravity, and that's 9.8 m/s² .

In 5 seconds:

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A wall in a house contains a single window. The window consists of a single pane of glass whose area is 0.15 m2 and whose thickn
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Answer:

88 %

Explanation:

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P = KAΔT/d

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The total heat lost by the glass window is P₂ = K₂A₂ΔT₂/d₂ where K₂ =thermal conductivity of glass window pane wall 0.96 W/m-K, A₂ = area of glass window pane = 0.15 m², ΔT₂ = temperature gradient between inside and outside of the window and d₂ = thickness of glass window pane = 7 mm = 0.007 m

The total heat lost is P = P₁ + P₂ = K₁A₁ΔT₁/d₁ + K₂A₂ΔT₂/d₂

Now, since the temperatures of both inside and outside of both window and wall are the same, ΔT₁ = ΔT₂ = ΔT

So, P = K₁A₁ΔT/d₁ + K₂A₂ΔT/d₂

Since P₂ = K₂A₂ΔT₂/d₂ = K₂A₂ΔT/d₂is the heat lost by the window, the fraction of the heat lost by the window from the total heat lost is

P₂/P = K₂A₂ΔT/d₂ ÷ (K₁A₁ΔT/d₁ + K₂A₂ΔT/d₂)

= 1/(K₁A₁ΔT/d₁÷K₂A₂ΔT/d₂ + 1)

= 1/(K₁A₁d₂÷K₂A₂d₁ + 1)

= 1/[(0.033 W/m-K × 17 m² × 0.007 m ÷ 0.96 W/m-K × 0.15 m² × 0.20 m) + 1]

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= 1/(0.1364 + 1)

= 1/1.1364

= 0.88.

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