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Vikentia [17]
3 years ago
8

An ice skater accelerates backwards for 5.0 seconds to a final speed of 12 m/s. If the acceleration backwards was at a rate of 1

.5 m/s^2 what was the skaters initial speed?
Physics
1 answer:
Lostsunrise [7]3 years ago
4 0

Answer:

Vo = 4.5 [m/s]

Explanation:

In order to solve this problem, we must use the following equation of kinematics.

v_{f}=v_{o}+a*t

where:

Vf = final velocity = 12 [m/s]

Vo = initial velocity [m/s]

a = acceleration = 1.5 [m/s²]

t = time = 5 [s]

Now replacing:

12=v_{o}+1.5*5\\v_{o}=12- (7.5)\\v_{o}= 4.5[m/s]

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Answer:

Part 1)

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Part 2)

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Part 3)

\tau_3 = 1.4 N m

Part 4)

Since torque on right side is more so here it will turn and slip over it

Explanation:

As we know that the block A is placed at distance

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So torque due to weight of A is given as

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the block B is placed at distance

d = 30 cm from the hinge at 70 cm mark

So torque due to weight of B is given as

\tau_2 = 14 \times (0.30) = 4.2 N m

Now torque due to weight of the scale is given as

\tau_3 = 7(0.20)

\tau_3 = 1.4 N m

now torque on left side of scale is given as

\tau_{left} = \tau_1 + \tau_3

\tau_{left} = 2.5 + 1.4 = 3.9 N m

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\tau_{right} = \tau_2 = 4.2 Nm

Since torque on right side is more so here it will turn and slip over it

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A plate drops onto a smooth floor and shatters into three pieces of equal mass.Two of the pieces go off with equal speeds v at r
Firlakuza [10]

Answer:

Speed of the this part is given as

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Also the direction of the velocity of the third part of plate is moving along 135 degree with respect to one part of the moving plate

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