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Digiron [165]
3 years ago
10

A person who weighs 150 pounds on Earth would weigh ____ pounds on the moon.

Physics
1 answer:
enot [183]3 years ago
6 0

Answer:

25lb

Explanation:

You haven't changed (you are made up of the same atoms), but the force exerted on you is different. Physicists like to say that your mass hasn't changed, only your weight.

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Determine whether a moving tennis ball and a racket held by the player have the same momentum or different momentum. If differen
NikAS [45]

Answer:

key points:1.momentum and impulse

2.condition for conservation of momentum and why

3.how to solve collision problems

4.centre of mass

Explanation:

momentum is a vector

force of a tennis ball

for a top player,a tennis ball may leave the racket on the with a speed of 55m/s(about 120mi/h).if the ball has a mass of 0.060kg and is in contact with the racket for about 4ms(4×10)s.estimate the average force on the ball

5 0
3 years ago
A negative ion of charge -2e is located at the origin and a second negative ion of charge -3e is located nearby at x = 3.8 nm ,
Rus_ich [418]

Answer:

\vec{F}_{21}=-5.63\times 10^{-11}N\\\\\vec{F}_{21}=\\

Explanation:

Given that

Q_1 = -2e\, C\\\\Q_2=-3e\,C\\\\x= 3.8 \times 10^{-9}\,m\\\\y= 3.2 \times 10^{-9}\,m\\\\r=\sqrt{x^2+y^2}\\\\r= 4.96\times 10^{-9} m\\

As both charges are negative so there exist force of repulsion in direction as shown in figure.

F_{12}=\frac{kQ_1Q_2}{r^2}\\\\F_{12}= \frac{(9\times 10^9)(6)(1.602\times 10^{-19})^2}{(4.96\times 10^{-9})^2}\\\\F_{12}=5.63\times 10^{-11}N

Angle at which force F12 is acting is

\theta=tan^{-1}\frac{3.2}{3.8}\\\\\theta=tan^{-1}\frac{y}{x}\\\\\theta= 40.1^o

F_{x}=F_{12}cos\theta\\\\F_{x}=(5.63\times 10^{-11})cos(40.1)\\\\F_{x}=4.306\times 10^{-11}N\\\\F_{y}=F_{12}sin\theta\\\\F_{y}=(5.63\times 10^{-11})sin(40.1)\\\\F_{y}=3.62\times 10^{-11}N\\\\

\vec{F}_{12}=\vec{F}_{x}+\vec{F}_y\\\\\vec{F}_{12}=4.30\times 10^{-11}\,\hat{i} + 3.62\times 10^{-11}\,\hat{j}\\\\\vec{F}_{12}=

Force exerted on charge -2e is equal in magnitude to F12 but is in opposite direction

F_{21}=-5.63\times 10^{-11}N

\vec{F}_{21}=

7 0
4 years ago
It is 2058 and you are taking your grandchildren to Mars. At an elevation of 34.7 km above the surface of Mars, your spacecraft
Paul [167]

Answer: 1.23\ m/s^2

Explanation:

Given

At an elevation of y=34.7\ km, spacecraft is dropping vertically at a speed of u=293\ m/s

Final velocity of the spacecraft is v=0

using equation of motion i.e. v^2-u^2=2as

Insert the values

\Rightarrow 0-(293)^2=2\times a\times (34.7\times 10^3)\\\\\Rightarrow a=-\dfrac{293^2}{2\times 34.7\times 10^3}\\\\\Rightarrow a=-1.23\ m/s^2

Therefore, magnitude of acceleration is 1.23\ m/s^2.

8 0
3 years ago
7) The acceleration due to gravity near the surface of Mars is about one-third of the value
Doss [256]

Answer:

C) three times slow than on earth

8 0
3 years ago
How are CD's recorded?<br> Using digital technology<br> using analog technology
alex41 [277]
I think analog but I could be wrong
5 0
3 years ago
Read 2 more answers
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