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Lubov Fominskaja [6]
3 years ago
12

Find the work done by a force of 20 lbs acting in the direction north 50 degrees west in moving an object 4 feet due west?

Physics
1 answer:
zhuklara [117]3 years ago
7 0

Answer: Workdone = 61.28ft.lbs

Workdone in moving the object 4ft due west is 61.28ft.lbs

Explanation:

Given;

Force = 20lbs

Direction = N 50° W

Distance moved = 4ft

Workdone = Force × distance

W = fd

But f in this case is

Force due North = fcos50

Force due west = fsin50

Since we are to find the workdone due west then

W = fsin50 × d = fdsin50

W = 20lbs × 4ft × sin50

Workdone = 61.28ft.lbs

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An atom that has a negative four charge will have which of the following? Question 1 options: 4 more protons than electrons An e
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Electrons: negative charge
Protons: positive charge
Neutrons: negative charge

The atom would have to have more electrons than protons

Hope this helps :)
3 0
3 years ago
You move a 25 n object 4 meters. find the work you did
d1i1m1o1n [39]
In physics, "work<span>" is when a force applied to an object moves the object in the same direction as the force. If someone pushes against a wall, no </span>work<span> is done on the system. It is calculated as follows:

Work = Force x distance
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5 0
4 years ago
Read 2 more answers
What is the correct order of the layers' density from lowest density to highest?
Orlov [11]

Answer:

C. crust, mantle, core

Explanation:

density increases as you travel from the crust to the inner core

the crust is on top

next is the mantle

and then the core

6 0
3 years ago
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What is energy?
mestny [16]

Answer:

Energy is the ability for an object to do work.

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6 0
3 years ago
What is the magnitude of the electric force between a proton and an electron when they are at a distance of 4.09 angstrom from e
Zinaida [17]

Answer:

F=1.38*10^{-9}N

Explanation:

According to Coulomb's law, the magnitude of the electric force between two point charges is directly proportional to the product of the magnitude of both charges and inversely proportional to the square of the distance that separates them:

F=\frac{kq_1q_2}{d^2}

Here k is the Coulomb constant. In this case, we have q_1=-e, q_2=e and d=4.09*10^-10m. Replacing the values:

F=\frac{8.99*10^{9}\frac{N\cdot m^2}{C^2}(-1.6*10^{-19}C)(1.6*10^{-19}C)}{(4.09*10^{-10})^2}\\F=-1.38*10^{-9}N

The negative sign indicates that it is an attractive force. So, the magnitude of the electric force is:

F=1.38*10^{-9}N

5 0
3 years ago
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