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3 years ago

What force causes weight? How is weight affected as an object moves farther away from the center of the earth?

2 answers:

6 0

"What force causes weight?"
--Weight is caused by gravity. The weight of an object is determined between the gravitational force of the object and the Earth.

"How is weight affected as an object moves farther away from the center of the Earth?"
--As you travel away from the Earth's surface, your mass stays the same but your weight reduces as gravitational pull decreases.

stich3 [128]3 years ago

3 0

Gravity affects weight of an object

Its weight reduces as it moves away from the center as gravity is strongest near the core and reduces as you move away

Hope this helps C:

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Suppose that the mirror described in Part A is initially at rest a distance R away from the sun. What is the critical value of a

Sati [7]

Ans;

6.25m²/Kd

Explanation: see attached file

5 0

3 years ago

During each heartbeat, about 80 g of blood is pumped into the aorta inapproximately 0.2 s. During this time, the blood is accele

taurus [48]

Answer:

Work is done by the heart on the blood during this time is 0.04 J

Explanation:

Given :

Mass of blood pumped, m = 80 g = 0.08 kg

Initial speed of the blood, u = 0 m/s

Final speed of the blood, v = 1 m/s

Initial kinetic energy of blood is determine by the relation:

$E_{1}=\frac{1}{2} m u^{2}$

Final kinetic energy of blood is determine by the relation:

$E_{2}=\frac{1}{2} m v^{2}$

Applying work-energy theorem,

Work done = Change in kinetic energy

W = E₂ - E₁

$W=\frac{1}{2} m (v^{2}-u^{2})$

Substitute the suitable values in the above equation.

$W=\frac{1}{2}\times0.08\times (1^{2}-0^{2})$

W = 0.04 J

6 0

3 years ago

What is in the crysphere

butalik [34]

I think you mean the Cryosphere?

But the answer is D- Earths Ice

This word Cryosphere comes from the greek word "kryos" which means cold

Many people think of the cryosphere as being the north and south poles but snow and ice can be found in a lot of places on the Earth

3 0

3 years ago

Which of the following would NOT affect the level at which a canoe floats in a pond

nataly862011 [7]

Sry i was knowing the answer i forgot ;(

5 0

3 years ago

For this discussion, you will work in groups to answer the questions. In a video game, airplanes move from left to right along t

Mariulka [41]

Answer:

When fired from (1,3) the rocket will hit the target at (4,0)

When fired from (2, 2.5) the rocket will hit the target at (12,0)

When fired from (2.5, 2.4) the rocket will hit the target at $(\frac{35}{2},0)$

When fired from (4,2.25) the rocket will hit the target at (40,0)

Explanation:

All of the parts of the problem are solved in the same way, so let's start with the first point (1,3).

Let's assume that the rocket's trajectory will be a straight line, so what we need to do here is to find the equation of the line tangent to the trajectory of the airplane and then find the x-intercept of such a line.

In order to find the line tangent to the graph of the trajectory of the airplane, we need to start by finding the derivative of such a function:

$y=2+\frac{1}{x}$

$y=2+x^{-1}$

$y'=-x^{-2}$

$y'=-\frac{1}{x^{2}}$

so, we can substitute the x-value of the given point into the derivative, in this case x=1, so:

$y'=-\frac{1}{x^{2}}$

$y'=-\frac{1}{(1)^{2}}$

m=y'=-1

so we can now use this slope and the point-slope form of the line to find the equation of the line tangent to the trajectory of the airplane so we get:

$y-y_{1}=m(x-x_{1})$

$y-3=-1(x-1})$

$y-3=-1x+1$

$y=-x+1+3$

$y=-x+4$

So we can now set y=0 so find the x-coordinate where the rocket hits the x-axis.

$-x+4=0$

and solve for x

x=4

so, when fired from (1,3) the rocket will hit the target at (4,0)

Now, let's calculate the coordinates where the rocket will hit the target if fired from (2, 2.5)

so, we can substitute the x-value of the given point into the derivative, in this case x=2, so:

$y'=-\frac{1}{x^{2}}$

$y'=-\frac{1}{(2)^{2}}$

$m=y'=-\frac{1}{4}$

so we can now use this slope and the point-slope form of the line to find the equation of the line tangent to the trajectory of the airplane so we get:

$y-y_{1}=m(x-x_{1})$

$y-2.5=-\frac{1}{4}(x-2})$

$y-2.5=-\frac{1}{4}x+\frac{1}{2}$

$y=-\frac{1}{4}x+\frac{1}{2}+\frac{5}{2}$

$y=-\frac{1}{4}x+3$

So we can now set y=0 so find the x-coordinate where the rocket hits the x-axis.

$-\frac{1}{4}x+3=0$

and solve for x

x=12

so, when fired from (2, 2.5) the rocket will hit the target at (12,0)

Now, let's calculate the coordinates where the rocket will hit the target if fired from (2.5, 2.4)

so, we can substitute the x-value of the given point into the derivative, in this case x=2.5, so:

$y'=-\frac{1}{x^{2}}$

$y'=-\frac{1}{(2.5)^{2}}$

$m=y'=-\frac{4}{25}$

so we can now use this slope and the point-slope form of the line to find the equation of the line tangent to the trajectory of the airplane so we get:

$y-y_{1}=m(x-x_{1})$

$y-2.4=-\frac{4}{25}(x-2.5})$

$y-2.4=-\frac{4}{25}x+\frac{2}{5}$

$y=-\frac{4}{25}x+\frac{2}{5}+2.4$

$y=-\frac{4}{25}x+\frac{14}{5}$

So we can now set y=0 so find the x-coordinate where the rocket hits the x-axis.

$-\frac{4}{25}x+\frac{14}{5}=0$

and solve for x

$x=\frac{35}{20}$

so, when fired from (2.5, 2.4) the rocket will hit the target at $(\frac{35}{2},0)$

Now, let's calculate the coordinates where the rocket will hit the target if fired from (4, 2.25)

so, we can substitute the x-value of the given point into the derivative, in this case x=4, so:

$y'=-\frac{1}{x^{2}}$

$y'=-\frac{1}{(4)^{2}}$

$m=y'=-\frac{1}{16}$

so we can now use this slope and the point-slope form of the line to find the equation of the line tangent to the trajectory of the airplane so we get:

$y-y_{1}=m(x-x_{1})$

$y-2.25=-\frac{1}{16}(x-4})$