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vitfil [10]
2 years ago
7

How does size of an object affect its gravity

Physics
1 answer:
irakobra [83]2 years ago
6 0

Answer:

the more particles packed together the faster it falls

Explanation:

the mass + the 1 constant g-force = the speed without adding air resistance

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An object travels 7.5 m/s toward the west . Under the influence of a constant net force of 5.2 kN, it comes to rest in 3.2 s. Wh
Softa [21]

Answer:

m = 2218.67 kg

Explanation:

It is given that,

Initial velocity, u = 7.5 m/s

Final speed of an object, v = 0 (at rest)

Force, F = 5.2 kN

Time, t = 3.2 s

We need to find the mass of the object. Force acting on an object is given by :

F = ma

m is mass, a is acceleration

F=\dfrac{m(v-u)}{t}\\\\m=\dfrac{Ft}{v-u}\\\\m=\dfrac{5.2\times 10^3\times 3.2}{0-7.5}\\\\m=2218.67\ kg

So, the mass of the object is 2218.67 kg

5 0
3 years ago
An object is dropped from a height of 75.0 m above ground level. (a) determine the distance traveled during the first second.
barxatty [35]

The relevant formula we can use in this case would be:

h = v0 t + 0.5 g t^2

where,

h = height or distance travelled

v0 = initial velocity = 0 since it was dropped

t = time = 1 seconds

g = 9.8 m/s^2

 

So calculating for height h:

h = 0 + 0.5 * 9.8 m/s^2 * (1 s)^2

<span>h = 4.9 meters</span>

6 0
2 years ago
You stand17.5 m from a wall holding a baseball. You throw the baseball at the wall at an angle of 20.5∘ from the ground with an
Lelechka [254]

Answer: 8.8 m

Explanation:

The movement of the baseball to the wall is a example of parabolic motion. While the baseball aproach the wall it is affected by gravity.

In this case, because the Initial Velocity of the ball is a vecotr, it can be defined using its two directionals compounds. One, on the Y-axis and another on the X-axis. This can be related, on how a hypotenuse is the product of two legs of the triangle. Because of this, each one of this, can be know using the following equation:

Vx = Vo * cos(∅)

Vy = Vo * sin(∅)

Where Vo is the initial velocity 27.5 m/s, and ∅ is the angle which is 20.5°. So we calculate:

Vx = 27.5m/s * cos(20.5)

Vx = 27.5m/s * 0.936

Vx = 25.74m/s

Vy = 27.5m/s  * sin(20.5)

Vy = 27.5m/s * sin(20.5)

Vy = 9.62m/s

Now the movement is divided on two parts, one under the effect of gravity and another one with a constant velocity.

To know how tall does the baseball hit the wall, we need to know first how much time it takes the ball to reach the wall on the X-axis. The wall is 17.5m away, we velocity on Vx that is constant we can calculate as it follow:

Time (T) = Distance (D) / Velocity (V)

Where Distance is 17.5m and our Velocity is the Vx calculated before.

T = 17.5 m / 25.74m/s

T = 0.68s

This is the time it takes the ball to reach the wall.

Know with the time, we can calculate the how tall it got on that time with the following equation:

x = (Vo*t) + (\frac{1}{2} *a*t^{2} )

Where Vo is the Y-Compound of the Initial Velocity.

a is the aceleration, in this case the Gravity. Which, will be negative because is oposing the movement. Gravity is equal to 9.81 m/s^{2}

And t is the time it takes the ball to get to the wall.

x = (Vy*T) + (\frac{1}{2} *g*T^{2} )

x = ((9.62m/s)*0.68s) + (\frac{1}{2} *(9.81m/s^{2})*(0.68s)^{2} )

x = 8.8 m

This is the height that the baseball touch the wall.

6 0
3 years ago
A horizontal 649 N merry-go-round of radius 1.05 m is started from rest by a constant horizontal force of 61.3 N applied tangent
nexus9112 [7]

Answer:

The kinetic energy of the merry-go-round is 632.82 J

Explanation:

Given;

weight of the merry-go-round, W = 649 N

radius of the merry-go-round, r = 1.05 m

applied horizontal force, F =  61.3 N

acceleration due to gravity, g = 9.8 m/s²

mass of  merry-go-round, m = W/g

                                               = 649/9.8  = 66.225 kg

moment of inertia of merry-go-round, I = ¹/₂mr²

                                                                 = ¹/₂ x 66.225 x (1.05)²

                                                                 = 36.507 kg.m²

Angular acceleration of the merry-go-round, α

τ = Iα = Fr

α = Fr / I

Where;

α is angular acceleration

α = (61.3 x 1.05) / 36.507

α = 1.763 rad/s²

Angular velocity of the merry-go-round, ω

ω = αt

ω = 1.763 x 3.34

ω = 5.888 rad/s

Finally, the kinetic energy of the merry-go-round, K.E

K.E = ¹/₂Iω²

K.E = ¹/₂ x 36.507 x (5.888)²

K.E = 632.82 J

4 0
2 years ago
Forces that are equal but oppistes are_____________ forces
Dima020 [189]
Balanced forces is the answer.
4 0
2 years ago
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