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Vadim26 [7]
3 years ago
9

Earth's gravitational pull just got threetimes stronger! what happens to your weight/

Physics
1 answer:
OLga [1]3 years ago
8 0
Hold on and let's discuss this realistically.

Because of gravity, there are two forces between the Earth and me. One draws me toward the Earth. The strength of that force is what I call my "weight". The other force draws the Earth toward me, and has the same strength.

The strength of these forces depends on the masses of the Earth and me. If the strength just tripled, that means that at least one of us just picked up a lot more mass. If the Earth suddenly became three times as massive, then the weight of everything and everybody on it would suddenly triple, and I'm pretty sure it would be the end of all of us before too long.

If it was only MY mass that suddenly tripled, that would mean that I had gone tearing through my house and the neighbour's house, eating everything in sight including the 2 couches, 3 dogs, and 6 TVs. Naturally, just as you would expect, my weight changed from 207 to 621, and my skin is stretched really tight.
ooohhh
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It's a cloudy and rainy day. The air pressure is most likely _____.
kramer
The air pressure ( or atmospheric pressure ) is the force of air over a unit of area. Changes in the air pressure causes the weather changes. High pressure usually brings good weather with dry and cool air. But in a low pressure zone warm air is rising up. This vertical movements are caused by winds high in the troposphere. Water molecules stay as a gas in warmer air. After the vertical movement they condense and bring steady continuous rain. Therefore the low pressure brings cloudly and rainy weather. Answer: The air pressure is most likely low<span>. </span>
5 0
3 years ago
Question 1 (1 point)
Dmitry [639]

1). trajectory

2). person sitting in a chair

3). 490 meters

4). 65 m/s

5). False.  The projectile's displacement, velocity, and acceleration have vertical and horizontal components, but the projectile doesn't.

6). False

7). The vertical component of a projectile doesn't change due to gravity, but the vertical components of its displacement, velocity, and acceleration do.

The vertical components do NOT equal the horizontal components.

8). Decreasing if you include the effects of air resistance.  Constant if you don't.  Gravity has no effect on horizontal velocity.

9). We can't see the simulation.  But if the projectile doesn't have jets on it, then as it travels upward, its vertical velocity must decrease, because gravity is trying to not let it get away.

10). We can't see the simulation.  But if the projectile is traveling downward, we would call that "falling", and its vertical velocity must increase, because gravity is pulling it downward.

6 0
3 years ago
Read 2 more answers
An arrow is shot at an angle of 35° and a velocity of 50 m/s. How long does it take to return to its original starting height?
Ostrovityanka [42]

Answer:

4.02 s

Explanation:

From the question given above, the following data were obtained:

Angle of projection (θ) = 35°

Initial velocity (u) = 50 m/s

Acceleration due to gravity (g) = 10 m/s²

Time of flight (T) =?

The time of flight of the arrow can be obtained as follow:

T = 2uSineθ / g

T = 2 × 35 × Sine 35 / 10

T = 70 × 0.5736 / 10

T = 7 × 0.5736

T = 4.02 s

Therefore, the time taken for the arrow to return is 4.02 s

4 0
3 years ago
in 4 seconds, a cheetah can run 112 meters. How fast does this cheetah have to be running? A: 448 m/s || B: 28 m/s || C: 56 m/s
Vladimir [108]

Answer:

B: 28m/s

Explanation:

Use the speed formula

Speed =distance / time

8 0
3 years ago
Read 2 more answers
A 2.0 kg otter starts from rest at the top of a muddy incline 85 cm long and slides down to the bottom in 0.50 s. What net force
Julli [10]

Answer:

The net force acting on the otter along the incline is 13.96 N.

Explanation:

It is given that,

Mass of the otter, m = 2 kg

Distance covered by otter, d = 85 cm = 0.85 m

It takes 0.5 seconds.

We need to find the net force acts on the otter along the incline. If a is the acceleration of the otter. It can be calculated using second equation of motion as :

d=ut+\dfrac{1}{2}at^2

Here, u = 0 (at rest)

d=\dfrac{1}{2}at^2

a=\dfrac{2d}{t^2}

a=\dfrac{2\times 0.85}{0.5^2}

a=6.8\ m/s^2

The net force acting on the otter along the incline is given by :

F = ma

F=2\ kg\times 6.8\ m/s^2

F = 13.6 N

So, the net force acting on the otter along the incline is 13.96 N. Hence, this is the required solution.

5 0
3 years ago
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