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jenyasd209 [6]
3 years ago
14

An object initially traveling at a velocity of 52 m/s experiences an acceleration of 9.8 m/s^2 how much time will it take to com

e to rest
Physics
1 answer:
Goshia [24]3 years ago
8 0

Answer:

5.3 seconds

Explanation:

Vx = Vx0 + Ax .T

A= de-acceleration

Vx =0   final at rest velocity

Vx0 = initial velocity

T= time to bring the object to a stop

therefore

0 = 52 m/s - (9.8 m/s^2) . T

-52 m/s/ -9.8 m/sec^2 =T

5.3 seconds= T

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6 0
4 years ago
Read 2 more answers
When a parachute opens, the air exerts a large drag force on it. This upward force is initially greater than the weight of the s
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Explanation:

Assuming there is only force in the y-component, the total net force F_{y} acting on the parachute and the sky diver is:

F_{y}=F_{D}-W   (1)

Where:

F_{D}=1032N is the drag force acting upwards

W=936N is the weight of the sky diver acting downwards, hence with negative sign

Then:

F_{y}=1032N-936N=96N   (2) This is the total net force excerted on the system parachute-sky diver, and the fact it is positive means is upwards

Now, according Newton's 2nd Law of Motion the force is directly proportional to the mass m and to the acceleration a of a body:

F_{y}=m.a (3)

Where m=95.5kg is the mass of the diver.

Substituting the known values and finding a:

a=\frac{F_{y}}{m} (4)

a=\frac{96N}{95.5kg} (5)

Finally:

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3 years ago
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