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mixas84 [53]
3 years ago
11

Two masses of size m and 4m are connected by a massless thread and are strung over a frictionless pulley of radius R and moment-

of-inertia I. The two masses are held motionless at the same height and then released from rest. After the mass 4m has fallen a distance L and the mass m has risen the same distance L, what is the total kinetic energy of the entire system
Physics
1 answer:
tekilochka [14]3 years ago
6 0

Answer:

Total kinetic energy of entire system is 3 mgl

Explanation:

Given two masses: m and 4m.

Since the pulley is frictionless and the thread is massless, the energy here is linked to the two masses.

Total kinetic energy of entire system = decrease in gravitational potential energy of the system.

Therefore, we have :

ΔKE = Δp

ΔKE = 4mgl - mgl

= 3 mgl

Total kinetic energy of entire system is 3 mgl

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Leaning against a brick wall.

All the others use scientific forces of work.

-Steel jelly.

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A toy cart is pulled a distance of 6.00 m in a straight line across the floor. The force pulling the cart has a magnitude of 20.
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Answer:

W = 95.8J

Explanation:

Given

S = 6.00m

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W = 95.8J

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How dose an atom change if all of its electrons are removed
cluponka [151]

Answer:

The atom becomes a positively charged ion.

Explanation:

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  2. The protons and neutrons present in the core of the atom are called nucleus.
  3. The electrons are scattered in an ordered way around the nucleus.
  4. The protons are positively charged and the electrons are negatively charged particles. The neutrons do not possess any charges.
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  6. It is possible to remove the electrons of the lighter elements.
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  8. When all of the electrons are removed from the helium atom, it becomes a positively charged α particle.
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7 0
3 years ago
Each driver has mass 79.0 kg. Including the masses of the drivers, the total masses of the vehicles are 800 kg for the car and 4
Mademuasel [1]

Answer:

Force exerted on the car driver by the seatbelt = 8139.4 N = 8.14 kN

Force exerted on the truck driver by the seatbelt = 1628.2 N = 1.63 kN

It is evident that the driver of the smaller vehicle has it worse. The car driver is in way more danger in this perfectly inelastic head-on collision with a bigger vehicle (the truck).

Explanation:

First of, we calculate the velocity of the vehicles after collision using the law of conservation of Momentum

Momentum before collision = Momentum after collision

Since the collision of the two vehicles was described as a head-on collision, for the sake of consistent convention, we will take the direction of the velocity of the bigger vehicle (the truck) as the positive direction and the direction of the car's velocity automatically is the negative direction.

Velocity of the truck before collision = 6.80 m/s

Velocity of the car before collision = -6.80 m/s

Let the velocity of the inelastic unit of vehicles after collision be v

Momentum before collision = (4000)(6.80) + (800)(-6.80) = 27200 - 5440 = 21,760 kgm/s

Momentum after collision = (4000 + 800)(v) = (4800v) kgm/s

Momentum before collision = Momentum after collision

21760 = 4800v

v = (21760/4800)

v = 4.533 m/s (in the direction of the big vehicle (the truck)

So, we then apply Newton's second law of motion which explains that the magnitude change in momentum is equal to the magnitude of impulse.

|Impulse| = |Change in momentum|

But Impulse = (Force exerted on each driver by the seatbelt) × (collision time) = (F×t)

Change in momentum = (Momentum after collision) - (Momentum before collision)

So, for the driver of the truck

Initial velocity = 6.80 m/s (the driver moves with the velocity of the truck)

Final velocity = 4.533 m/s

Change in momentum of the truck driver = (79)(6.80) - (79)(4.533) = 179.1 kgm/s

(F×t) = 179.1

F × 0.110 = 179.1

F = (179.1/0.11)

F = 1628.2 N = 1.63 kN

So, for the driver of the car

Initial velocity = -6.80 m/s (the driver moves with the velocity of the car)

Final velocity = 4.533 m/s

Change in momentum of the car driver = (79)(-6.80) - (79)(4.533) = -895.3 kgm/s

(F×t) = |-895.3|

F × 0.110 = 895.3

F = (895.3/0.11)

F = 8139.4 N = 8.14 kN

Hope this Helps!!!

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Answer:

Arrange an annual service. Treat your boiler like your car. ...

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