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melomori [17]
3 years ago
7

A force of 5.25 newtons acts on an object of unknown mass at a distance of 6.9 × 108 meters from the center of Earth. To increas

e the force to 2.5 times its original value, how far should the object be from the center of Earth?
Physics
2 answers:
alexira [117]3 years ago
7 0

Answer:

4.36 x 10^8 m

Explanation:

Let the mass of earth is M and unknown mass is m.

According to the Newton's law of gravitation, force between two objects is given by

F = G\frac{M m}{d^{2}}

Here, F = 5.25 N, d = 6.9 x 10^8 m

5.25 = G\frac{M m}{(6.9\times 10^{8})^{2}}       .... (1)

Now, F' = 2.5 F and d be the distance

2.5\times 5.25 = G\frac{M m}{(d)^{2}}            ..... (2)

Divide equation (1) by equation by (2)

\frac{1}{2.5} = \left ( \frac{d}{6.9\times 10^{8}} \right )^{2}

d = 4.36 x 10^8 m

NemiM [27]3 years ago
5 0
The force between two objects is calculated through the equation, 
                        F = Gm₁m₂/d²
where m₁ and m₂ are the masses of the objects. In this case, an unknown mass and Earth. d is the distance between them and G is universal gravitation constant. 

In the second case, if the force is to become 2.5 times the original and all the variables are constant except d then,
                       2.5F = Gm₁m₂ / (D²)
                                D = 0.623d

Subsituting the known value of d,
                                D = 0.623(6.9 x 10^8) = <em>4.298 x 10^8 m</em>
                           
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Three masses are located in the x-y plane as follows: a mass of 6 kg is located at (0 m, 0 m), a mass of 4 kg is located at (3 m
Natali [406]
<h2>Answer:</h2>

D. (1m, 0.5m)

<h2>Explanation:</h2>

The center of mass (or center of gravity) of a system of particles is the point where the weight acts when the individual particles are replaced by a single particle of equivalent mass. For the three masses, the coordinates of the center of mass C(x, y) is given by;

x = (m₁x₁ + m₂x₂ + m₃x₃) / M       ----------------(i)

y = (m₁y₁ + m₂y₂ + m₃y₃) / M       ----------------(ii)

Where;

M = sum of the masses

m₁ and x₁ = mass and position of first mass in the x direction.

m₂ and x₂ = mass and position of second mass in the x direction.

m₃ and x₃ = mass and position of third mass in the x direction.

y₁ , y₂ and y₃ = positions of the first, second and third masses respectively in the y direction.

From the question;

m₁ = 6kg

m₂ = 4kg

m₃ = 2kg

x₁ = 0m

x₂ = 3m

x₃ = 0m

y₁ = 0m

y₂ = 0m

y₃ = 3m

M = m₁ + m₂ + m₃ = 6 + 4 + 2 = 12kg

Substitute these values into equations (i) and (ii) as follows;

x = ((6x0) + (4x3) + (2x0)) / 12

x = 12 / 12

x = 1 m  

y = (6x0) + (4x0) + (2x3)) / 12

y = 6 / 12

y = 0.5m

Therefore, the center of mass of the system is at (1m, 0.5m)

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3 years ago
A skier leaves the horizontal end of a ramp with a velocity of 25.0 m/s and lands 70.0 m from the base of the ramp. How high is
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Answer:

<em>The end of the ramp is 38.416 m high</em>

Explanation:

<u>Horizontal Motion </u>

When an object is thrown horizontally with an initial speed v and from a height h, it follows a curved path ruled by gravity.

The maximum horizontal distance traveled by the object can be calculated as follows:

\displaystyle d=v\cdot\sqrt{\frac  {2h}{g}}

If the maximum horizontal distance is known, we can solve the above equation for h:

\displaystyle h=\frac  {d^2g}{2v^2}

The skier initiates the horizontal motion at v=25 m/s and lands at a distance d=70 m from the base of the ramp. The height is now calculated:

\displaystyle h=\frac  {70^2\cdot 9.8}{2\cdot 25^2}

\displaystyle h=\frac  {4900\cdot 9.8}{2\cdot 625}

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The end of the ramp is 38.416 m high

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alexgriva [62]
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v(2.0s)=22.5-9.8\cdot2.0 =2.9 m/s
v(4.0s)=22.5-9.8\cdot 4.0s=-16.7 m/s
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2) The projectile reaches its maximum height when the speed is equal to zero:
v(t)=0
So we have
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And solving this we find
t=2.30 s

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h(2.30 s)=2+22.5\cdot 2.30 -4.9 \cdot (2.30s)^2 = 27.83 m

4) The time at which the projectile hits the ground is the time at which the height is zero: h(t)=0. So, this translates into
2+22.5t -4.9 t^2 = 0
This is a second-order equation, and if we solve it we get two solutions: the first solution is negative, so we can ignore it since it's physically meaningless; the second solution is
t=4.68 s
And this is the time at which the projectile hits the ground.

5) The velocity of the projectile when it hits the ground is the velocity at time t=4.68 s:
v(4.68 s)=22.5-9.8\cdot 4.68 =-23.36  m/s
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A

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