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PolarNik [594]
3 years ago
9

Two objects having masses m1 and m2 are connected to each other as shown in the figure and are released from rest. There is no f

riction on the table surface or in the pulley. The masses of the pulley and the string connecting the objects are completely negligible. What must be true about the tension t in the string just after the objects are released?

Physics
2 answers:
GrogVix [38]3 years ago
5 0

Answer:

T<m2hh

Explanation:

Two objects having masses m1 and m2 are connected to each other as shown in the figure and are released from rest. There is no friction on the table surface or in the pulley. The masses of the pulley and the string connecting the objects are completely negligible. What must be true about the tension t in the string just after the objects are released?

Two masses  m1 and m2  is released and there is tension T in the string

a is the acceleration of the system

Therefore

For m1

T-m1g=m1a

T=m1(g+a)

m2

this is so because te weight is going downward wile tension in te rope is facing upward

m2g-T=m2a

T=m2(g-a)

from the equation ,

T<m2

tankabanditka [31]3 years ago
4 0

Answer:m_1g

Explanation:

Given

Two masses m_1  and m_2  is released and there is tension T in the string

Suppose a is the acceleration of the system

Therefore from Diagram

For m_1

T-m_1g=m_1a

T=m_1(g+a)------1

for m_2 body

m_2g-T=m_2a

T=m_2(g-a)-------2

From above two Equation it is said that Tension is greater than m_1g and less than m_2g

m_1g

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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
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Answer:

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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

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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)

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F = 1628.2 N = 1.63 kN

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(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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