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ASHA 777 [7]
3 years ago
12

a bag of cement is pulled along a horizontal plane by a constant force of 15n calculate the work done in moving the bag through

a distance of 20m​
Physics
1 answer:
morpeh [17]3 years ago
4 0

Answer:

300J

Explanation:

Given parameters:

Constant force = 15N

Distance  = 20m

Unknown:

Work done   = ?

Solution:

Work done is the force applied to move a body through a particular distance.

 Work done  = Force x distance

 So;

  Work done  = 15 x 20  = 300J

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Why don't we use more renewable energy sources​
sattari [20]

Answer:

Renewable energy sources 1.Cost more then nonrenewable resources and 2.Are so far not as reliable as nonrenewable energy.

Explanation:

4 0
3 years ago
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A uniform continuous line charge with net positive charge Q and length L lies on the x-axis from −L2 to +L2. This problem asks a
irakobra [83]

Answer:

(C) zero (there is no net horizontal component of the E-field)

Explanation:

If we subdivide the bar into small pieces, each piece (dx) contains a charge (dq), the electric field of each piece is equivalent to the field of a punctual electric charge, and has a direction as shown in the attached figure. For each piece (dx) in the negative axis there is another symmetric piece (dx) in the positive axis, and as we see in the figure for symmetry the sum of their electric fields gives a resultant in the Y axis (because its components in X are cancelled by symmetry).

Then the resultant of the electric field will be only in Y.

(C) zero (there is no net horizontal component of the E-field)

5 0
3 years ago
In the diagram, 91, 92, and q3 are in a straight line. Each of these particles has a charge of -2.35 x 10-6 C. Particles q₁ and
Strike441 [17]

The net force on particle q₃ is  6.2128125 N.

<h3>What is electrostatic force?</h3>

The electrostatic force F between two charged objects placed distance apart is directly proportional to the product of the magnitude of charges and inversely proportional to the square of the distance between them.

F = kq₁q₂/d²

where k = 9 x 10⁹ N.m²/C²

Given is the diagram in which each of the particles has a charge of -2.35 x 10⁻⁶ C. Particles q₁ and q₂ are separated by 0.100 m and particles q₂ and q₃ are separated by 0.100 m.

Force acting on q₃ due to q₁

F₃₁  = 9 x 10⁹ x (-2.35 x 10⁻⁶)²/(0.1)²

F₃₁ = 4.97025 N(in right direction)

Force acting on q₃ due to q₂

F₃₂ = 9 x 10⁹ x (-2.35 x 10⁻⁶)²/(0.1+0.1)²

F₃₂ = 1.2425 N (in right direction)

Net force acting on particle q₃ is

F₃ = F₃₁ +F₃₂

F₃ = 4.97025 N + 1.2425 N

F₃ = 6.2128125 N

Thus the net force on a charged particle is  6.2128125 N to the right.

Learn more about electrostatic force.

brainly.com/question/9774180

#SPJ1

3 0
2 years ago
The CERN particle accelerator is circular with a circumference of 7.0 km.
Contact [7]

Answer:

a_c=2.0196\times 10^{13}\ m/s^2

F=3.37273\times 10^{-14}\ N

Explanation:

m = Mass of proton = 1.67\times 10^{-27}\ kg

v = Speed of proton = 0.5c = 0.5\times 3\times 10^8=1.5\times 10^8\ m/s

Circumference of the colider is 7 km

P=2\pi r\\\Rightarrow r=\frac{P}{2\pi}\\\Rightarrow r=\frac{7000}{2\pi}\ m

a_c=\frac{v^2}{r}\\\Rightarrow a_c=\frac{\left(1.5\times 10^8\right)^2}{\frac{7000}{2\pi}}\\\Rightarrow a_c=2.0196\times 10^{13}\ m/s^2

Centripetal acceleration is 2.0196\times 10^{13}\ m/s^2

F_c=ma_c\\\Rightarrow F_c=1.67\times 10^{-27}\times 2.0196\times 10^{13}\\\Rightarrow F=3.37273\times 10^{-14}\ N

Force on protons is 3.37273\times 10^{-14}\ N

8 0
4 years ago
Which skill will help you be able to move around other players in a soccer game?
Pani-rosa [81]
I️ would say agility. Although, speed could also be an answer.
6 0
3 years ago
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