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ra1l [238]
4 years ago
9

Two students make the following claims:

Physics
1 answer:
antiseptic1488 [7]4 years ago
6 0

Answer:

E. Student 1 is correct, because as θ is increased, h is the same.

Explanation:

Here we have the object of a certain mass falling under gravity so the force acting on the it will depend on mass of the object and the acceleration due to gravity.

Mathematically:

F=m.g

As we know that the work done is evaluated as the force applied on a body and the displacement of the body in the direction of the force.

And for work we have:

W=F.s\cos\theta

where:

s= displacement of the object

\theta= angle between the force and displacement vectors

Given that the height of the object is same in each trail of falling object under the gravity be it a free-fall or the incline plane.

  • In case of free-fall the angle between the force is and the displacement is zero.
  • In case when the body moves along the inclined plane the force applied by the gravity is same because it depends upon the mass of the object. And the net displacement in the direction of the gravitational force is the height of the object which is constant in both the cases.

So, the work done by the gravitational force is same in the two cases.

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4 years ago
Two charges are located in the xx – yy plane. If ????1=−4.25 nCq1=−4.25 nC and is located at (x=0.00 m,y=1.080 m)(x=0.00 m,y=1.0
Sati [7]

Answer:

Ex=  -17.1 N/C

Ey =  +26.9 N/C

Explanation:

We apply formula of electric field:

Ep=k*q/d²

Ep:  Electric field at point ( N/C)

q: Electric charge (C)

k: coulomb constant (N.m²/C²)

d: distance from charge q to point P (m)

In the attached graph we observe the directions of the electric field at P(0,0) due to q1 and q2

Calculation of the field at point P due to the load q₁

E₁=k*q₁/d₁² = 9*10⁹*4.25*10⁻⁹/1.080²= 32.8 N/C : Magnitude of E1

Direction of E₁ :Because the charge q₁ is negative the field enters the charge (+ y)

Calculation of the field at point P due to the load q₂

d_{2} = \sqrt{1.30^{2}+0.450^{2}  }

d₂=1.375 m

E₂=k*q₂/d₂² = 9*10⁹*3.80*10⁻⁹/ 1.375² = 18.09 N/C Magnitude of E₂

Direction of E₂ :Because the charge q₂ is positive the field leaves the charge in direction of angle β

, then,E₂ tiene componentes x-y  en P.

E₂x=-E₂cos β= -18.09*(1.3/1.375)= -17.1 N/C

E₂y=-E₂sin β= -18.09*(0.45/1.375)= -5.9 N/C

Calculation of the electric field at point P located at the origin(0,0)

Ex=E₂x= -17.1 N/C

Ey=E₁y+E₂y =32.8 N/C -5.9 N/C = 26.9 N/C

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3 years ago
Which of the following best describes the use of a renewable resource?
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C IS THE CORRECT ANSWER.

Hope this helps!

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The net charge of 3 mol of protons is
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The charge of 1 mole of proton is equivalent to 1.60217656x10^-19 coulombs. This is also called an elementary charge. Multiply that by 3 to get the net charge of three moles of protons.

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So the answer is <span>c) 4.81×10 −19 C</span>
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