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Mademuasel [1]
3 years ago
10

To calculate work done on an object, _____. A. multiply the force in the direction of motion by the distance the object moved B.

multiply all forces applied to the object by the distance the object moved C. divide the force in the direction of motion by the distance the object moved D. divide the distance the object moved by the force in the direction of motion
Physics
2 answers:
o-na [289]3 years ago
6 0
I believe your answer would be B, hope it helps

SIZIF [17.4K]3 years ago
3 0
I know the answer, I have the answer in my notes but I forgot it, I'll check soon. But for now, I'd go with A.
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Which layer of the sun is responsible for producing the light shown in the picture above?
Eva8 [605]

Answer:

C. Photosphere

Explanation:

The lights shown in the figure comes from the outermost layer of the Sun. This layer is called photosphere.

This is the layer from where the light of the Sun is radiated, before travelling through space and reaching us.

The photosphere is the coldest layer of the Sun: its surface temperature is between 4500 and 6000 K. Its width is approximately 100 km.

A characteristic of the photosphere is the presence of the sunspots, which appear as darker spots, and are regions of lower temperature caused by a concentration of magnetic flux.

3 0
3 years ago
Read 2 more answers
What was the average velocity for the entire trip?
just olya [345]

Answer:

<u><em>3.721 m/s</em></u>

This is the explanation of the ans

6 0
2 years ago
A small sphere is at rest at the top of a frictionless semicylindrical surface. The sphere is given a slight nudge to the right
V125BC [204]

Answer:

vi = 4.77 ft/s

Explanation:

Given:

- The radius of the surface R = 1.45 ft

- The Angle at which the the sphere leaves

- Initial velocity vi

- Final velocity vf

Find:

Determine the sphere's initial speed.

Solution:

- Newton's second law of motion in centripetal direction is given as:

                         m*g*cos(θ) - N = m*v^2 / R

Where, m: mass of sphere

             g: Gravitational Acceleration

             θ: Angle with the vertical

             N: Normal contact force.

- The sphere leaves surface at θ = 34°. The Normal contact is N = 0. Then we have:

                         m*g*cos(θ) - 0 = m*vf^2 / R

                         g*cos(θ) = vf^2 / R    

                         vf^2 = R*g*cos(θ)

                         vf^2 = 1.45*32.2*cos(34)

                        vf^2 = 38.708 ft/s

- Using conservation of energy for initial release point and point where sphere leaves cylinder:

                          ΔK.E = ΔP.E

                          0.5*m* ( vf^2 - vi^2 ) = m*g*(R - R*cos(θ))

                          ( vf^2 - vi^2 ) = 2*g*R*( 1 - cos(θ))

                          vi^2 =  vf^2 - 2*g*R*( 1 - cos(θ))

                          vi^2 = 38.708 - 2*32.2*1.45*(1-cos(34))

                          vi^2 = 22.744

                           vi = 4.77 ft/s

4 0
3 years ago
Nurse wears a uniform in a plural form​
Softa [21]

Answer:

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Explanation:

6 0
3 years ago
Read 2 more answers
A hammer exerts 49.8 N of force on the head (r=0.00510 m) of a nail. How much pressure does it exert on the nail?
Kisachek [45]

Answer:

609547.12 Pa ≈ 6.10×10^5 Pa

Explanation:

Step 1:

Data obtained from the question. This include the following:

Force (F) = 49.8 N

Radius (r) = 0.00510 m

Pressure (P) =..?

Step 2:

Determination of the area of the head of the nail.

The head of a nail is circular in nature. Therefore, the area is given by:

Area (A) = πr²

With the above formula we can obtain the area as follow:

Radius (r) = 0.00510 m

Area (A) =?

A = πr²

A = π x (0.00510)²

A = 8.17×10^-5 m²

Therefore the area of the head of the nail is 8.17×10^-5 m²

Step 3:

Determination of the pressure exerted by the hammer.

This is illustrated below:

Force (F) = 49.8 N

Area (A) = 8.17×10^-5 m²

Pressure (P) =..?

Pressure (P) = Force (F) /Area (A)

P = F/A

P = 49.8/8.17×10^-5

P = 609547.12 N/m²

Now, we shall convert 609547.12 N/m² to Pa.

1 N/m² = 1 Pa

Therefore, 609547.12 N/m² = 609547.12 Pa.

Therefore, the pressure exerted by the hammer on the nail is 609547.12 Pa or 6.10×10^5 Pa

8 0
4 years ago
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