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mixas84 [53]
3 years ago
7

Suppose a 0.04-kg car traveling at 2.00 m/s can barely break an egg. What is the min

Physics
1 answer:
Paha777 [63]3 years ago
4 0

Answer: The correct answer is option (A).

Explanation:

Momentum of the car with 0.04 kg mass , which travelling with velocity of 2.00 m/s

P_1=mass\times velocity=m_1\times v_1=0.04 kg\times 2.00 m/s=8.00 kg m/s

Then the maximum speed of the another car in order to not to break the eggs will be same as first car:

P_1=P_2

0.08 kg m/s=m_2\times v_2=0.08 kg\times v_2

v_2=1 m/s

Speed slightly more than 1 m/s will increase the momentum of second car and the eggs will break. So, from the given options the minimum speed need by the second car will be 1.42m/s.

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Juli2301 [7.4K]
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8 0
3 years ago
No two electrons in an atom can have the same set of four quantum numbers' is
miv72 [106K]

Answer:

D. Pauli's exclusion principle

Explanation:

<em>A. Newton's laws</em> are related to the motion, they state that "Every object in a state of uniform motion will remain in that state of motion unless an external force acts on it", "  Force equals mass times acceleration." and "  For every action there is an equal and opposite reaction"  

<em>B. Bohr's law </em>depicts an atom as a small, positively charged nucleus surrounded by electrons. These electrons travel in circular orbits around the nucleus.  

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I hope you find this information useful and interesting! Good luck!

6 0
3 years ago
A distant planet with a mass of (7.2000x10^26) has a moon with a mass of (5.0000x10^23). The distance between the planet and the
BARSIC [14]

Answer:

Explanation:

This is a simple gravitational force problem using the equation:

F_g=\frac{Gm_1m_2}{r^2} where F is the gravitational force, G is the universal gravitational constant, the m's are the masses of the2 objects, and r is the distance between the centers of the masses. I am going to state G to 3 sig fig's so that is the number of sig fig's we will have in our answer. If we are solving for the gravitational force, we can fill in everything else where it goes. Keep in mind that I am NOT rounding until the very end, even when I show some simplification before the final answer.

Filling in:

F_g=\frac{(6.67*19^{-11})(7.2000*10^{26})(5.0000*10^{23})}{(6.10*10^{11})^2} I'm going to do the math on the top and then on the bottom and divide at the end.

F_g=\frac{2.4012*10^{40}}{3.721*10^{23}} and now when I divide I will express my answer to the correct number of sig dig's:

Fg= 6.45 × 10¹⁶ N

8 0
3 years ago
An imaginary line perpendicular to a reflecting surface is called _________.
n200080 [17]
<span>An imaginary line perpendicular to a reflecting surface is called "a normal" (principle line)

So, Your Answer would be Option B

Hope this helps!</span>
5 0
3 years ago
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a runner covers the last straigjt stretch of a race in 4 s. during that time, he speeds up from 5m/s to 9m/s.
PtichkaEL [24]

During that final period of time,
his acceleration is
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8 0
3 years ago
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