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enot [183]
3 years ago
11

Ja , Daniel and Helen are pulling a metal ring. Ja pulls with a force of 100N in one direction and Daniel with a force of 140N i

n the opposite direction. If the ring doesn't move. What force does Helen exert if she pulls in the same direction as Ja.
Physics
1 answer:
drek231 [11]3 years ago
3 0

Answer:

40N

Explanation:

Given that Ja , Daniel and Helen are pulling a metal ring. Ja pulls with a force of 100N in one direction and Daniel with a force of 140N in the opposite direction. If the ring doesn't move. What force does Helen exert if she pulls in the same direction as Ja ?

Since the ring does not move, that means the forces are in equilibrium.

That is, sum of the forces is equal to zero.

Since Helen is also pulling in the direction of ja, let Helen force be equal to F. The equilibrium of forces among ja, Helen and Daniel can be expressed as:

F + 100 - 140 = 0

That is,

F + 100 = 140

Make F the subject of formula

F = 140 - 100

F = 40N.

Therefore, the force Helen exert if she pulls in the same direction as Ja is 40N.

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A car accelerates uniformly from rest for a time of 12 sec. The rate of acceleration is 4m/s^2. What is the velocity at the.end
Natali5045456 [20]

initial velocity = 0 = v₍i₎

final velocity = ? = v₍f₎

t = 12 sec

Acceleration = 4m/s²

First we have to find the distance d, for this we use the formula,

D = v₍i₎t + 1/2at²

D = 0(12) + ½ (4)(12)²

Distance = d = 288 m

Now to find the Vf use the formula,

V₍f₎² = v₍i₎² + 2ad

V₍f₎² = (0)2 + 2(4)(288)

V₍f₎² = 2304

V₍f₎ = 48 m/s

so the velocity at the end of 12 sec is 48 m/s

4 0
3 years ago
An ammeter is connected in series with a resistor of unknown resistance R and a DC power supply of known emf e. A student uses t
rosijanka [135]

Answer:

B:The actual power dissipated by the resistor is less than P because the ammeter had some resistance.

Explanation:

Here,power has been calculated using current I and total EMF \ε . So,P=EMF*current= ε I will represent total power dissipated in resistor and ammeter.

Now, this total power P has been dissipated in both resistor and ammeter. So, power dissipated in resistor must be less than P as some power is also dissipated in ammeter because it has non-zero resistance.

So, the answer is B:The actual power dissipated by the resistor is less than P because the ammeter had some resistance.

Note that option A,C and E are ruled out as they state power dissipated by resistor is greater than or equal to P which is false.

Also,option D is ruled out as ammeter is connected in series.

8 0
4 years ago
Your mother asks you to carry a load of wood to the fireplace and lower it to the hearth. In doing this you have used your arms
Evgesh-ka [11]

Answer: The correct answer is the weight of the wood.

Explanation:

Hope this helps

8 0
3 years ago
Read 2 more answers
A shooting star is actually the track of a meteor, typically a small chunk of debris from a comet that has entered the earth's a
s2008m [1.1K]

Answer:

A. Power generated by meteor = 892857.14 Watts

Yes. It is obvious that the large amount of power generated accounts for the glowing trail of the meteor.

B. Workdone = 981000 J

Power required = 19620 Watts

Note: The question is incomplete. A similar complete question is given below:

A shooting star is actually the track of a meteor, typically a small chunk of debris from a comet that has entered the earth's atmosphere. As the drag force slows the meteor down, its kinetic energy is converted to thermal energy, leaving a glowing trail across the sky. A typical meteor has a surprisingly small mass, but what it lacks in size it makes up for in speed. Assume that a meteor has a mass of 1.5 g and is moving at an impressive 50 km/s, both typical values. What power is generated if the meteor slows down over a typical 2.1 s? Can you see how this tiny object can make a glowing trail that can be seen hundreds of kilometers away? 61. a. How much work does an elevator motor do to lift a 1000 kg elevator a height of 100 m at a constant speed? b. How much power must the motor supply to do this in 50 s at constant speed?

Explanation:

A. Power = workdone / time taken

Workdone = Kinetic energy of the meteor

Kinetic energy = mass × velocity² / 2

Mass of meteor = 1.5 g = 0.0015 kg;

Velocity of meteor = 50 km/s = 50000 m/s

Kinetic energy = 0.0015 × (50000)² / 2 = 1875000 J

Power generated = 1875000/2.1 = 892857.14 Watts

Yes. It is obvious that the large amount of power generated accounts for the glowing trail of the meteor.

B. Work done by elevator against gravity = mass × acceleration due to gravity × height

Work done = 1000 kg × 9.81 m/s² × 100 m

Workdone = 981000 J

Power required = workdone / time

Power = 981000 J / 50 s

Power required = 19620 Watts

Therefore, the motor must supply a power of 19620 Watts in order to lift a 1000 kg to a height of 100 m at a constant speed in 50 seconds.

6 0
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Which body is in equilibrium?
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Which body is in equilibrium?

(1) a satellite orbiting Earth in a circular orbit .  No.  The forces on it are unbalanced.  There's only one force acting on it ... the force of gravity, pulling it toward the center of the Earth.  That's a centripetal force, and the satellite is experiencing centripetal acceleration.

(2) a ball falling freely toward the surface of Earth.  No.  The forces on it are unbalanced.  There's only one force acting on it ... the force of gravity, pulling it toward the center of the Earth.  The ball is accelerating toward the ground.  

<em> (3) a car moving with a constant speed along a  straight, level road.  YES.</em> We don't even need to analyze the forces, just look at the car.  It's moving in a straight line, and its speed is not changing.  The car's acceleration is zero !  That  right there tells us that the NET force ... the sum of all forces acting on the car ... is zero.  THAT's called 'equilibrium'.

(4) a projectile at the highest point in its trajectory.  No.  The forces on it are unbalanced.  There's only one force acting on it ... the force of gravity, pulling it toward the center of the Earth.  The projectile is accelerating toward the ground.

7 0
3 years ago
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