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3 years ago

A force of 85 N is used to push a box along the floor a distance of 15 m. How much work was done?

Physics

2 answers:

mario62 [17]3 years ago

7 0

A :-) work = force x distance
W = 85 x 15
W = 1275 joules

.:. The work done is 1275 joules

denpristay [2]3 years ago

4 0

Answer:

1275J

Explanation:

Given parameters:

Force on box = 85N

Distance moved = 15m

Unknown:

Work done = ?

Solution:

Work done is the amount of force applied on a body to move it through a specific distance.

Work done = Force x distance

Now insert the parameters and solve;

Work done = 85 x 15 = 1275J

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A tourist stands at the top of the Grand Canyon, holding a rock, overlooking the valley below. Find the final velocity and displ

Anestetic [448]

Answer:

a. -39.2 m/s; -78.4 m

b. -31.2 m/s; -46.4 m

c. -47.2 m/s; -110.4 m

Explanation:

We are given/can infer these variables:

t = 4.0 s
a = -9.8 m/s²
v_0 = 0 m/s

We want to find the displacement and the final velocity of the rock.

Δx = ?
v = ?

We can use this equation to find the final velocity:

v = v_0 + at

Plug in the known variables into this equation.

v = 0 + (-9.8)(4.0)
v = -9.8 * 4.0
v = -39.2 m/s

The final velocity of the rock is -39.2 m/s.

Now we can use this equation to find the displacement of the rock:

Δx = v_0 t + 1/2at²

Plug in the known variables into this equation.

Δx = 0 * 4.0 + 1/2(-9.8)(4.0)²
Δx = 1/2(-9.8)(4.0)²
Δx = -4.9 * 16
Δx = -78.4 m

The displacement of the rock is -78.4 m.

We are given/can infer these variables:

v_0 = 8.0 m/s
a = -9.8 m/s²
t = 4.0 s

We can use this equation to find the final velocity:

v = v_0 + at

Plug in the known variables into this equation.

v = 8.0 + (-9.8)(4.0)
v = 8.0 + -39.2
v = -31.2 m/s

The final velocity of the rock is -31.2 m/s.

We can use this equation to find the displacement:

Δx = v_0 t + 1/2at²

Plug in known variables:

Δx = 8.0(4.0) + 1/2(-9.8)(4.0)²
Δx = 32 - 4.9(16)
Δx = -46.4 m

The displacement of the rock is -46.4 m.

We are given/can infer these variables:

v_0 = -8.0 m/s
a = -9.8 m/s²
t = 4.0 s

We can use this equation to find the final velocity:

v = v_0 + at

Plug in the known variables into this equation.

v = -8.0 + (-9.8)(4.0)
v = -8.0 - 39.2
v = -47.2 m/s

The final velocity of the rock is -47.2 m/s.

We can use this equation to find the displacement:

Δx = v_0 t + 1/2at²

Plug in known variables:

Δx = -8.0(4.0) + 1/2(-9.8)(4.0)²
Δx = -32 - 4.9(16)
Δx = -110.4 m

The displacement of the rock is -110.4 m.

8 0

3 years ago

A ball thrown into the air has its greatest kinetic energy as it reaches the highest point in its path. please select the best a

postnew [5]

F - False.

Its greatest kinetic energy is at the point of release.

It has the least kinetic energy, zero, at its highest point in its path.

3 0

3 years ago

Read 2 more answers

What is used as evidence for sea-floor spreading?

raketka [301]

Answer:

Sea-floor spreading occurs in the oceanic ridges. In there, volcanic activity, together with the gradual movement of the bottom, form new oceanic crust. This allows a better understanding of the continental drift explained by the theory of plate tectonics.

The greatest evidence for Sea-floor spreading is the oceanic trenches, the oceanic ridges, the magma protruding to the surface and the new seafloor.

In previous theories, continents were assumed to be transported across the sea. Harry Hess, in the 1960s, proposed the idea that the seabed itself moves as it expands from a central point. The theory is now accepted, and the phenomenon is thought to be caused by convection currents in the upper layer of the mantle.

4 0

4 years ago

Read 2 more answers

The power of a lens is defined as the reciprocal of its focal length: P = 1/f. (Thus power is measured in inverse meters, called

Len [333]

Answer:

This the power of combination is given by

$P_{combination}=\frac{1}{F}=\frac{d+f_{2}-f_{1}}{f_{2}(d-f_{1})}$

Explanation:

Let 'F' be the focal length of the combination of the two lenses and the focal length's of individual lenses be $f_{1},f_{2}$

We know that focal length is the position of image when object is placed at infinity

Let us place the the object at infinity with respect to the first lens thus the position of image formed by the first lens shall be obtained using lens formula as

$\frac{1}{f_{1}}=\frac{1}{u}+\frac{1}{v}$

Applying values we get

$\frac{1}{f_{1}}=\frac{1}{\infty }+\frac{1}{v}\\\\\Rightarrow \frac{1}{f}=0+\frac{1}{v}\\\\\therefore v=+f_{1}$

Now this position of image formed by the first lens act's as object for the second lens, thus we have

$\frac{1}{f_{2}}=\frac{1}{-(d-f_{1})}+\frac{1}{v}\\\\\Rightarrow \frac{1}{f_{2}}+\frac{1}{d-f_{1}}=\frac{1}{v}\\\\\therefore \frac{1}{v}=\frac{(d-f_{1})+f_{2}}{f_{2}(d-f_{1})}\\\\=\frac{1}{v}=\frac{d+f_{2}-f_{1}}{f_{2}(d-f_{1})}\\\\\therefore v_{f}=\frac{f_{2}(d-f_{1})}{d+f_{2}-f_{1}}$

Since image of an object placed at infinity will be formed at $v_{f}=\frac{f_{2}(d-f_{1})}{d+f_{2}-f_{1}}$ thus the focal length of the combination of the 2 thin lenses will be $F=\frac{f_{2}(d-f_{1})}{d+f_{2}-f_{1}}$