How do I do this? Help please!

You move a 25 N object 5.0 meters. How much energy did you transformed?

lilavasa [31]

You move a 25 N object 5.0 meters. How much energy did you transformed?

25n x 5m = 125 J

6 0

3 years ago

A small asteroid with a mass of 1500 kg moves near the earth. At a particular instant the asteroid’s velocity is ⟨3.5 × 104, −1.

zalisa [80]

Answer:

$P_{f}$ =(5.7 x $10^{7$ i - 2.24 x $10^{7$ j) kgm/s

Explanation:

Due to earths gravity, force on asteroid is given by:

$F= \frac{Gm_{1}m_{2} }{r^{2} }$ r^

Plugging in the values, we have

F= [(6.67x $10^{-11}$ )(1500)(5.97 x $10^{24}$ )(8x $10^{6}$ i + 9x $10^{6$ j)] / ((8x $10^{6}$ )² + (9x $10^{6$ )² $)^{1.5}$

F= 2736 i^ + 3078 j^

In order find the final momentum of the Asteroid, apply impulse momentum theorem

$P_{f}$ = $P_{i$ + FΔt

$P_{f}$ = 1500(3.5 x $10^{4$ i - 1.8x $10^{4$ j) + (2736i + 3078j)(1.5x $10^{3$ )

$P_{f}$ =(5.7 x $10^{7$ i- 2.24 x $10^{7$ j)kgm/s

4 0

3 years ago

What is force of gravity

Eduardwww [97]

Answer:

On Earth all bodies have a weight, or downward force of gravity, proportional to their mass, which Earth's mass exerts on them. Gravity is measured by the acceleration that it gives to freely falling objects. At Earth's surface the acceleration of gravity is about 9.8 metres (32 feet) per second per second.

5 0

3 years ago

Two forces, F₁ and F₂, act at a point. F₁ has a magnitude of 8.00 N and is directed at an angle of 61.0° above the negative x ax

kirill115 [55]

1) -7.14 N

2) +2.70 N

3) 7.63 N

Explanation:

1)

In order to find the x-component of the resultant force, we have to resolve each force along the x-axis.

The first force is 8.00 N and is directed at an angle of 61.0° above the negative x axis in the second quadrant: this means that the angle with respect to the positive x axis is

$180^{\circ}-61^{\circ}$

so its x-component is

$F_{1x}=(8.00)(cos (180^{\circ}-61^{\circ}))=-3.88 N$

F₂ has a magnitude of 5.40 N and is directed at an angle of 52.8° below the negative x axis in the third quadrant: so, its angle with respect to the positive x-axis is

$180^{\circ}+52.8^{\circ}$

Therefore its x-component is

$F_{2x}=(5.40)(cos (180^{\circ}+52.8^{\circ}))=-3.26 N$

So, the x-component of the resultant force is

$F_x=F_{1x}+F_{2x}=-3.88+(-3.26)=-7.14 N$

2)

In order to find the y-component of the resultant force, we have to resolve each force along the y-axis.

The first force is 8.00 N and is directed at an angle of 61.0° above the negative x axis in the second quadrant: as we said previously, the angle with respect to the positive x axis is

$180^{\circ}-61^{\circ}$

so its y-component is

$F_{1y}=(8.00)(sin (180^{\circ}-61^{\circ}))=7.00 N$

F₂ has a magnitude of 5.40 N and is directed at an angle of 52.8° below the negative x axis in the third quadrant: as we said previously, its angle with respect to the positive x-axis is

$180^{\circ}+52.8^{\circ}$

Therefore its y-component is

$F_{2y}=(5.40)(sin (180^{\circ}+52.8^{\circ}))=-4.30 N$