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Flauer [41]
3 years ago
8

How does the force the Earth exerts on you compare with the force you exert on it?

Physics
1 answer:
Pie3 years ago
6 0

Answer:

C.Earth and you exert equal and opposite forces on each other.

Explanation:

Hello, I can help you with this

The force that the earth exerts on any object is due to the force of gravity and is known as weight, F = mg, this force is directed towards the center of the earth, in addition, according to the laws of physics for any action must have a reaction, in this case it is a force that goes out  from the ground and it is directed upwards, this force is called normal and it has the same magnitude of your weight but the opposite direction.

Have a great day.

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Calculate the percentage increase in speed of the cyclist when the power output changes from 200 W to 300 W.
Rom4ik [11]

Answer:

<em>50%</em>

Explanation:

Given

Initial power = 200W

Final power = 300W

Increment = 300 - 200 = 100W

percentage increase = increment/initial power * 100

percentage increase = 100/200 * 100%

percentage increase = 0.5 * 100

percentage increase = 50%

<em>Hence the percentage increase in speed is 50%</em>

6 0
3 years ago
Celine has raised the temperature of water to 100°C in order to cook pasta for her lasagna. What physical property is Celine dem
LuckyWell [14K]

Answer:

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Explanation:

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5 0
3 years ago
The isobars in the conventional series that will be needed to complete the pressure analysis between the lowest and highest valu
Tresset [83]

The isobars in the conventional series that will be needed to complete the pressure analysis between the lowest and highest values on this map are: 1008, 1012, 1016, 1020.

 

To add, an isobar is <span>a line on a map connecting points having the same atmospheric pressure at a given time or on average over a given period.</span>

7 0
3 years ago
You launch a cannonball at an angle of 35° and an initial velocity of 36 m/s (assume y = y₁=
velikii [3]

Answer:

Approximately 4.2\; {\rm s} (assuming that the projectile was launched at angle of 35^{\circ} above the horizon.)

Explanation:

Initial vertical component of velocity:

\begin{aligned}v_{y} &= v\, \sin(35^{\circ}) \\ &= (36\; {\rm m\cdot s^{-1}})\, (\sin(35^{\circ})) \\ &\approx 20.6\; {\rm m\cdot s^{-1}}\end{aligned}.

The question assumed that there is no drag on this projectile. Additionally, the altitude of this projectile just before landing y_{1} is the same as the altitude y_{0} at which this projectile was launched: y_{0} = y_{1}.

Hence, the initial vertical velocity of this projectile would be the exact opposite of the vertical velocity of this projectile right before landing. Since the initial vertical velocity is 20.6\; {\rm m\cdot s^{-1}} (upwards,) the vertical velocity right before landing would be (-20.6\; {\rm m\cdot s^{-1}}) (downwards.) The change in vertical velocity is:

\begin{aligned}\Delta v_{y} &= (-20.6\; {\rm m\cdot s^{-1}}) - (20.6\; {\rm m\cdot s^{-1}}) \\ &= -41.2\; {\rm m\cdot s^{-1}}\end{aligned}.

Since there is no drag on this projectile, the vertical acceleration of this projectile would be g. In other words, a = g = -9.81\; {\rm m\cdot s^{-2}}.

Hence, the time it takes to achieve a (vertical) velocity change of \Delta v_{y} would be:

\begin{aligned} t &= \frac{\Delta v_{y}}{a_{y}} \\ &= \frac{-41.2\; {\rm m\cdot s^{-1}}}{-9.81\; {\rm m\cdot s^{-2}}} \\ &\approx 4.2\; {\rm s} \end{aligned}.

Hence, this projectile would be in the air for approximately 4.2\; {\rm s}.

8 0
1 year ago
Read 2 more answers
What is analogy a powerful rhetorical device?
wolverine [178]
The other person who answered this is wrong btw
5 0
3 years ago
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