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lilavasa [31]
3 years ago
10

A 0.5 kg object is whirled on the end of a string that is 1.2 m long at a speed of 7.5 m/s. Calculate the angular momentum of th

e
object.
(Round your response to the nearest tenth, if necessary, and include the base unit abbreviation with your answer.)
Physics
1 answer:
WINSTONCH [101]3 years ago
3 0

Answer:

4.5kg m^2 s−1

Explanation:

Step one:

the expression for angular momentum is given as

L = mvr

m = mass  = 0.5kg

v = velocity  = 7.5m/s

r = radius= 1.2m

Required

L = angular momentum

Step two:

substitute into the expression

L=0.5*7.5*1.2

L=4.5kg m^2 s−1

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Cart A and cart B are traveling in the same direction on a straight track. Cart A is moving at 9.25 m/s and cart B is moving at
Liula [17]

Answer:

The speed of cart B is 11.21 m/s.

Explanation:

Given that,

Speed of cart A = 9.25 m/s

Speed of cart B = 7.15 m/s

Mass of cart A = 72.0 kg

Mass of cart B = 55.0 kg

Speed of card A after collision = 6.15 m/s

We need to calculate the speed of cart B

Using conservation of momentum

m_{A}v_{A}+m_{B}v_{B}=m_{A}v_{A}+m_{B}v_{B}

Put the value into the formula

72.0\times9.25+55.0\times7.15=72.0\times6.15+55.0\times v_{B}

v_{B}=\dfrac{72.0\times9.25+55.0\times7.15-72.0\times6.15}{55.0}

v_{B}=11.21\ m/s

Hence, The speed of cart B is 11.21 m/s.

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Which of the following is an example of an acceptable time to ignore the rules and etiquette of a sport?
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A girl pushes a 1.04 kg book across a table with a horizontal applied force 10 points
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Answer:

Approximately 11.0\; \rm m \cdot s^{-1}. (Assuming that g = 9.81 \; \rm N \cdot kg^{-1}, and that the tabletop is level.)

Explanation:

Weight of the book:

W = m \cdot g = 1.04 \; \rm kg \times 9.81\; \rm N \cdot kg^{-1} \approx 10.202\; \rm N.

If the tabletop is level, the normal force on the book will be equal (in magnitude) to weight of the book. Hence, F(\text{normal force}) \approx 10.202\; \rm N.

As a side note, the F_N and W on this book are not equal- these two forces are equal in size but point in the opposite directions.

When the book is moving, the friction F(\text{kinetic friction}) on it will be equal to

  • \mu_{\rm k}, the coefficient of kinetic friction, times
  • F(\text{normal force}), the normal force that's acting on it.

That is:

\begin{aligned}& F(\text{kinetic friction}) \\ &= \mu_{\rm k}\cdot F(\text{normal force})\\ &\approx 0.35 \times 10.202\; \rm N \approx 3.5708\; \rm N\end{aligned}.

Friction acts in the opposite direction of the object's motion. The friction here should act in the opposite direction of that 15.0\; \rm N applied force. The net force on the book shall be:

\begin{aligned}& F(\text{net force})  \\ &= 15.0 \; \rm N - F(\text{kinetic friction}) \\& \approx 15.0 - 3.5708\; \rm N \approx 11.429\; \rm N\end{aligned}.

Apply Newton's Second Law to find the acceleration of this book:

\displaystyle a = \frac{F(\text{net force})}{m} \approx \frac{11.429\; \rm N}{1.04\; \rm kg} \approx 11.0\; \rm m \cdot s^{-2}.

6 0
3 years ago
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