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ki77a [65]
3 years ago
5

if the forces acting upon an object are balanced, then an object must a) be moving. b) be accelerating. c) be beginning to slow

down. d) be moving at a constant velocity.
Physics
1 answer:
Dennis_Churaev [7]3 years ago
4 0

The correct answer to the question is : D) Be moving at a constant velocity.

EXPLANATION:

As per Newton's first laws of motion, every body continues to be  at state of rest or of uniform motion in a straight line unless and until it is compelled by some external unbalanced forces acting on it.

Hence, it is the unbalanced force which changes the state of rest or motion of a body. Balanced force is responsible for keeping the body to be either in static equilibrium or in dynamic equilibrium.

As per the options given in the question, the last one is true for an object under balanced forces.


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Suppose the initial position of an object is zero, the starting velocity is 3 m/s and the final velocity was 10 m/s. The object
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Explanation:

We have,

The initial position of an object is zero.

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8 0
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A pot on the stove contains 200 g of water at 20°C. An unknown mass of ice that is originally at −10°C is placed in an identical
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Answer:

a) The mass of the ice is smaller than the mass of the water

b) The ice reaches first 80°C ,

Explanation:

Since the heat Q that should be provided to ice

Q = sensible heat to equilibrium temperature (as ice) + latent heat + sensible heat until final temperature ( as water)

m ice * c ice * ( T equil -T initial  ) + m ice* L + m ice* c water * ( T final - T equil)

and the heat Q that should be provided to water is

Q= m water * c water * ( T final - T equil )

since the rate of heat addition q = constant and the time t taken to reach the final temperature is the same , then the heat absorbed Q=q*t is the same for both, therefore

m water * c water *  ( T final - T equil ) = m ice* [c ice *( T equil -T initial  ) + L + c water * ( T final - T equil)]

m water/ m ice =  [c ice * ( T equil -T initial  )  + L + c water * ( T final - T equil)]/ [ c water * ( T final - T equil)]

m water/ m ice = [c ice * ( T equil -T initial  )  + L ]/[c water * ( T final - T equil) ] + 1

since  [c ice * ( T equil -T initial  )  + L ]/[c water * ( T final - T equil) ] >0 , then

m water/ m ice > 1

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b) Since the heat Q that should be provided to the ice, starting from 55°C mass would be

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and for the water mass

Q water = m water * c water * ( T final2 - T final1 )

dividing both equations

Q water / Q ice = m water / m ice >1

thus

Q water > Q ice

since the heat addition rate is constant

Q water = q* t water and Q ice=q* t ice

therefore

q* t water > q* t ice

t water >  t ice

so the time that takes to reach 80°C is higher for water , thus the ice mass reaches it first.

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3 years ago
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