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3 years ago

which of the following refers to the motion of an object falling straight down with only the force of gravity acting on it

Physics

1 answer:

MrRissso [65]3 years ago

6 0

Answer:

Since you haven't provided any choices, then the answer is "Free Fall Motion."

Explanation:

In order to learn more about the answer, let's discuss what free fall motion is.

Free Fall- In Physics, this refers to any body motion that is acted upon solely by "gravity." The acceleration in free fall is always downward and there's the absence of other forces. Take note that the acceleration should be the same and is independent of the object's mass. This acceleration is called "acceleration due to gravity."

Gravity- This refers to the force that pulls any object towards the center of the earth.

Examples of Objects in Free Fall Motion

1. A ball dropped at the top of a building.

2. Dropping a coin from a table.

The ball and the coin are both in free fall motion because they are being pulled by gravity towards the earth. Their acceleration is also constant and there are no other forces acting upon them.

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If we double the mass of the sun, what would happen to the gravitational force between the sun and earth?

Sphinxa [80]

It would increase
because the force is directly proportional to the Value of masses given

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4 years ago

Issac and Blaise decide to race. They both start at the same position at the same time. Issac runs at 2m/s but decides to take a

FromTheMoon [43]

Let the Blaise runs for time "t" to complete the race

so the total distance he moved is given by

$d_1 = 1* t$

Now Issac runs for time t = "t - 2*60"

because it took rest for 2 minutes

$d_2 = 2*(t - 120)$

now it is given that Blaise wins by 10 m distance

$d_1 - d_2 = 10$

$1* t - 2*(t - 120) = 10$

$t - 2t + 240 = 10$

$t = 230 s$

now the distance moved by Blaise is given by

$d_1 = 1*230 = 230 m$

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3 years ago

When passing from a low index of refraction medium to a high index of refraction medium which way does the refracted ray bend: t

Vaselesa [24]

Answer:

toward the normal

Explanation:

Light travels at different speed in different mediums.

Refractive index is equal to velocity of the light 'c' in empty space divided by the velocity 'v' in the substance.

Or ,

n = c/v.

Light travels at a slower speed in water as compared to air because there are more number of interfering molecules in the path of the light in case of water as compared to liquid.

When a light travels from lower denser medium say water to higher denser medium say water, it bends towards the perpendicular (normal) as its speed reduces in that medium.

6 0

4 years ago

A stone is thrown vertically upward with a speed of 15.5 m/s from the edge of a cliff 75.0 m high .

rjkz [21]

a) 2.64 s

We can solve this part of the problem by using the following SUVAT equation:

$s=ut+\frac{1}{2}at^2$

where

s is the displacement of the stone

u is the initial velocity

t is the time

a is the acceleration

We must be careful to the signs of s, u and a. Taking upward as positive direction, we have:

- s (displacement) negative, since it is downward: so s = -75.0 m

- u (initial velocity) positive, since it is upward: +15.5 m/s

- a (acceleration) negative, since it is downward: so a= g = -9.8 m/s^2 (acceleration of gravity)

Substituting into the equation,

$-75.0 = 15.5 t -4.9t^2\\4.9t^2-15.5t-75.0 = 0$

Solving the equation, we have two solutions: t = -5.80 s and t = 2.84 s. Since the negative solution has no physical meaning, the stone reaches the bottom of the cliff 2.64 s later.

b) 10.4 m/s

The speed of the stone when it reaches the bottom of the cliff can be calculated by using the equation:

$v=u+at$

where again, we must be careful to the signs of the various quantities:

- u (initial velocity) positive, since it is upward: +15.5 m/s

- a (acceleration) negative, since it is downward: so a = g = -9.8 m/s^2

Substituting t = 2.64 s, we find the final velocity of the stone:

$v = 15.5 +(-9.8)(2.64)=-10.4 m/s$

where the negative sign means that the velocity is downward: so the speed is 10.4 m/s.

c) 4.11 s

In this case, we can use again the equation:

$s=ut+\frac{1}{2}at^2$

where

s is the displacement of the package

u is the initial velocity

t is the time

a is the acceleration

We have:

s = -105 m (vertical displacement of the package, downward so negative)

u = +5.40 m/s (initial velocity of the package, which is the same as the helicopter, upward so positive)

a = g = -9.8 m/s^2

Substituting into the equation,

$-105 = 5.40 t -4.9t^2\\4.9t^2 -5.40 t-105=0$

Which gives two solutions: t = -5.21 s and t = 4.11 s. Again, we discard the first solution since it is negative, so the package reaches the ground after

t = 4.11 seconds.

5 0

3 years ago

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Natasha_Volkova [10]

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