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slava [35]
1 year ago
11

A big lump of meat of mass 5Kg is hung from a spring balance in an elevator. Find the reading of the balance of (I) the elevator

is moving with a steady speed. (iI) the elevator is moving upwards with acceleration of 0.2 ms^-2 (g=10ms^-2)
Physics
1 answer:
Mrrafil [7]1 year ago
7 0

The reading of the balance if ,

I ) If the elevator is moving with a steady speed = 50 N

II ) If the elevator is moving upwards with acceleration of 0.2 m / s² = 51 N

T = m g + m a

T = Force

m = Mass

g = Acceleration due to gravity

a = Acceleration

m = 5 kg

g = 10 m / s²

I ) If the elevator is moving with a steady speed,

At steady speed, a = 0

T = ( 5 * 10 ) + ( 5 * 0 )

T = 50 N

II ) If the elevator is moving upwards with acceleration of 0.2 m / s²,

a = 0.2 m / s²

T = ( 5 * 10 ) + ( 5 * 0.2 )

T = 50 + 1

T = 51 N

Therefore, the reading of the balance if ,

I ) If the elevator is moving with a steady speed = 50 N

II ) If the elevator is moving upwards with acceleration of 0.2 m / s² = 51 N

To know more about reading on a spring balance

brainly.com/question/14819147

#SPJ1

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What are the magnitude and direction of the acceleration of an electron at a point where the electric field has magnitude 7400 N
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Answer: a = 1.32 * 10^18m/s² due north

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The force exerted on the charge by the electric field of intensity (E) is given by

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Thus

Eq = ma

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Where a = acceleration of charge

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a = 7400 * 1.609 * 10^-16/ 9.109 * 10^-31

a = 11906.6 * 10^-16 / 9.019 * 10^-31

a = 1.19 * 10^-12 / 9.019 * 10^-31

a = 0.132 * 10^19

a = 1.32 * 10^18m/s²

As stated in the question, the direction of the electric field is due north hence, the direction of it force will also be north thus making the electron experience a force due north ( according to Newton second law of motion)

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Answer:

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<u>Conservation of Momentum </u>

The total momentum of a system of two particles is

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Where m1,m2,v1, and v2 are the respective masses and velocities of the particles at a given time. Then, the two particles collide and change their velocities to v1' and v2'. The final momentum is now

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The momentum is conserved if no external forces are acting on the system, thus

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Let's put some numbers in the problem and say

m_1=8\ kg,\ m_2=6\ kg,\ v_1=12\ m/s, v_2=4\ m/s,\ v_1'=-6\ m/s,\ v_2'=28\ m/s

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