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Papessa [141]
3 years ago
15

Using information about natural laws, explain why some car crashes produce minor injuries and others produce catastrophic injuri

es.
Physics
1 answer:
Nataly_w [17]3 years ago
5 0
Don't they have to pay insurance company if something happens to there car.
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Just a question anybody playing minècraft here?
yan [13]

Answer:

yes/nope

Explanation:

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2 years ago
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What is the velocity of an object that has a momentum of 4,000 kg-m/s and a mass of 115 kg? Round to the nearest hundredth.
julia-pushkina [17]
<h2>Answer: 34.78 m/s</h2>

Explanation:

The momentum p is given by the following equation:

p=m.V   (1)

Where:

m is the mass of the object

V is the velocity of the object

Finding the velocity from (1):

V=\frac{p}{m}   (2)

V=\frac{4000kg.m/s}{115kg}  

<u>Finally:</u>

V=34.78m/s >>>This is the velocity of the object

4 0
3 years ago
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A girl pushes a 1.04 kg book across a table with a horizontal applied force 10 points
mr Goodwill [35]

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}.

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2 years ago
Can someone help me with physics;((
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Answer: vf= 51 m/s and d= 112 m

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