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3 years ago

Pleas help me phyic science

Physics

1 answer:

deff fn [24]3 years ago

3 0

Answer:

Explanation:

The amount of energy present in the matter

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our lab partner wears a new pair of sneakers to lab and, rather than performing the required experiments, you decide to measure

Dafna1 [17]

Answer:

The coefficient of static friction between your partner and the floor is 0.55

Explanation:

Given:

Mass $m = 59$ Kg

Frictional force $F_{s} = 318.3$ N

From the formula of frictional force,

$F_{s} = \mu_{s} mg$

Where $\mu _{s} =$ coefficient of static friction, $g = 9.8 \frac{m}{s^{2} }$

Put the above values and find the coefficient of static friction.

$318.3 = \mu_{s} \times 59 \times 9.8$

$\mu_{s} = 0.55$

Therefore, the coefficient of static friction between your partner and the floor is 0.55

4 0

3 years ago

Which form of energy is involved in weighting fruit on a spring scale ?

alexdok [17]

The correct answer is elastic potential energy.

I hoped this helped!

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4 years ago

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What is the main function of chloroplasts?

dangina [55]

Answer:

To convert sunlight into usable sugars

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Explanation:

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4 years ago

Which materials, including polystyrenes, can be recycled? polymers thermoplastics ceramics composites

9966 [12]

The answer is <span>thermoplastics

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4 years ago

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You are participating in a NASA traineeship, working with a group planning a new landing on Mars. Your supervisor has come up wi

aivan3 [116]

Answer:

$h=17005.8 km$

Explanation:

Newton's law of universal gravitation states that the force experimented by a satellite of mass m orbiting Mars, which has mass $M=6.39\times10^{23} kg$ at a distance r will be:

$F=\frac{GMm}{r^2}$

where $G=6.67\times10^{-11}Nm^2/kg^2$ is the gravitational constant.

This force is the centripetal force the satellite experiments, so we can write:

$F=ma_{cp}=mr\omega^2=mr(\frac{2\pi}{T})^2=\frac{4\pi^2mr}{T^2}$

Putting all together:

$\frac{GMm}{r^2}=\frac{4\pi^2mr}{T^2}$

which means:

$r=\sqrt[3]{\frac{GM}{4\pi^2}T^2}$

Which for our values is:

$r=\sqrt[3]{\frac{(6.67\times10^{-11}Nm^2/kg^2)(6.39\times10^{23} kg)}{4\pi^2}(1.026\times24\times60\times60s)^2}=20395282m=20395.3km$

Since this distance is measured from the center of Mars, to have the height above the Martian surface we need to substract the radius of Mars R=3389.5 km , which leaves us with:

$h=r-R=20395.3km-3389.5 km=17005.8 km$

6 0

3 years ago