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d1i1m1o1n [39]
3 years ago
12

Prove 1/P = 1/P1 + 1/P2

Physics
1 answer:
Sati [7]3 years ago
7 0

Answer:

Proof:

|----------(R1)-------------(R2)---------|

Let the resistances be R1 , R2 connected in series. And the potential difference be V ;

We know that :

P1 = V²/R1 ..........(1) and

P2 = V²/R2 ...........(2)

In series combination , the net resistance is the algebraic addition of individual resistance :

R = R1 + R2

=> V²/P = V²/P1 + V²/P2

=> 1/P = 1/P1 + 1/P2

Explanation:

Copied it from a different brainly question that is this question. Please try to find it on brainly instead of asking it'd make it much easier.

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4 0
3 years ago
A tennis ball connected to a string is spun around in a vertical, circular path at a uniform speed. The ball has a mass m = 0.15
Oksanka [162]

1) 5.5 N

When the ball is at the bottom of the circle, the equation of the forces is the following:

T-mg = m\frac{v^2}{R}

where

T is the tension in the string, which points upward

mg is the weight of the string, which points downward, with

m = 0.158 kg being the mass of the ball

g = 9.8 m/s^2 being the acceleration due to gravity

m \frac{v^2}{R} is the centripetal force, which points upward, with

v = 5.22 m/s being the speed of the ball

R = 1.1 m being the radius of the circular trajectory

Substituting numbers and re-arranging the formula, we find T:

T=mg+m\frac{v^2}{R}=(0.158 kg)(9.8 m/s^2)+(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=5.5 N

2) 3.9 N

When the ball is at the side of the circle, the only force acting along the centripetal direction is the tension in the string, therefore the equation of the forces becomes:

T=m\frac{v^2}{R}

And by substituting the numerical values, we find

T=(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=3.9 N

3) 2.3 N

When the ball is at the top of the circle, both the tension and the weight of the ball point downward, in the same direction of the centripetal force. Therefore, the equation of the force is

T+mg=m\frac{v^2}{R}

And substituting the numerical values and re-arranging it, we find

T=m\frac{v^2}{R}-mg=(0.158 kg)\frac{5.22 m/s)^2}{1.1 m}-(0.158 kg)(9.8 m/s^2)=2.3 N

4) 3.3 m/s

The minimum velocity for the ball to keep the circular motion occurs when the centripetal force is equal to the weight of the ball, and the tension in the string is zero; therefore:

T=0\\mg = m\frac{v^2}{R}

and re-arranging the equation, we find

v=\sqrt{gR}=\sqrt{(9.8 m/s^2)(1.1 m)}=3.3 m/s

7 0
3 years ago
Your friend is constructing a balancing display for an art project. She has one rock on the left ( ms=2.25 kgms=2.25 kg ) and th
Firdavs [7]

Answer:

Torque = 35.60 N.m (rounded off to 3 significant figures.

Explanation:

Given details:

The mass of the rock on the left, ms = 2.25 kg

The total mass of the rocks, mp = 10.1 kg

The distance from the fulcrum to the center of the pile of rocks, rp = 0.360 m

(a) The torque produced by the pile of rock, T = F*rp = m*g*rp

Torque = 9.8*0.360*10.1 = 35.6328

Torque = 35.60 N.m (rounded off to 3 significant figures).

5 0
3 years ago
Review. A light spring of force constant 3.85 N/m is compressed by 8.00 cm and held between a 0.250-kg block on the left and a 0
photoshop1234 [79]

The answer is 0.245N.

<h3>What is kinetic energy?</h3>
  • A particle or an item that is in motion has a sort of energy called kinetic energy. An item accumulates kinetic energy when work, which involves the transfer of energy, is done on it by exerting a net force.
  • Kinetic energy comes in five forms: radiant, thermal, acoustic, electrical, and mechanical.
  • The energy of a body in motion, or kinetic energy (KE), is essentially the energy of all moving objects. Along with potential energy, which is the stored energy present in objects at rest, it is one of the two primary types of energy.
  • Explain that a moving object's mass and speed are two factors that impact the amount of kinetic energy it will possess.

(b) 0.100

For the block on the left, f_{k} =u_{k} n= 0.100(2.45N)=0.245N.

∑F_{x}=ma_{x}

–0.308N+0.245N=(0.250kg)a

a=−0.252m/s^{2} if the force of static friction is not too large.

For the block on the right, f_{k} =u_{k} n=0.490N. The maximum force of static friction would be larger, so no motion would begin, and the acceleration is zero

To learn more about kinetic energy, refer to:

brainly.com/question/25959744

#SPJ4

7 0
2 years ago
The power lines are at a high potential relative to the ground, so there is an electric field between the power lines and the gr
Makovka662 [10]

Answer:

The tube should be held vertically, perpendicular to the ground.

Explanation:

As the power lines  of ground are equal, so its electrical field is perpendicular to the ground and the equipotential surface is cylindrical. Therefore, if we put the position fluorescent tube parallel to the ground so the both ends of the tube lie on the same equipotential surface and the difference is zero when its  potential.

And the ends of the tube must be on separate equipotential surfaces to optimize potential. The surface near the power line has a greater potential value and the surface farther from the line has a lower potential value, so the tube must be placed perpendicular to the floor to maximize the potential difference.

8 0
3 years ago
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