PLEASE HELP!!

If all else stays the same, which would cause an increase in the gravitational force on a space shuttle?

maks197457 [2]

"decreasing the distance of the space shuttle from Earth"

F = Gm(1)m(2)/R²
where R is the distance between the 2 objects, as it decreases, the force increases.

5 0

3 years ago

Read 2 more answers

Question 15)

RUDIKE [14]

Answer:

k Nishant

Explanation:

i think option A

8 0

3 years ago

A spring has a constant of 100 N/m. What Force does the spring exert on you if you stretch it a distance of 0.5 m?

arsen [322]

Answer:

F = - K x force is opposed to direction of extension

F = -100 N / m * .5 m = -50 N

3 0

3 years ago

Physics Conversion help!!

stiv31 [10]

[Assuming that you've written 3.40 kg in 'a', and not 3.90 kg]

(a) 3,400 g x 0.001 = 3.40 kg [converting grams to kilograms]

(b) 220 cm x 0.01 = 2.2 m [converting centimeters to meters]

(c) 9.42 kg x 1000 = 9420 g [converting kilograms to grams]

(d) 6.53 m x 100 = 653 cm [converting meters to centimeters]

5 0

4 years ago

Use the work—energy theorem to solve each of these problems. You can use Newton's laws to check your answers. Neglect air resist

andreyandreev [35.5K]

Answer:

a) It is moving at $43.15\frac{m}{s^{2}}$ when reaches the ground.

b) It is moving at $101.44\frac{m}{s^{2}}$ when reaches the ground.

Explanation:

Work energy theorem states that the total work on a body is equal its change in kinetic energy, this is:

$W=K_f-K_i$ (1)

with W the total work, Ki the initial kinetic energy and Kf the final kinetic energy. Kinetic energy is defined as:

$K=\frac{mv^2}{2}$ (2)

with m the mass and v the velocity.

Using (2) on (1):

$W=\frac{mv_f^2}{2}-\frac{mv_i^2}{2}$ (3)

In both cases the total work while the objects are in the air is the work gravity field does on them. Work is force times the displacement, so in our case is weight (w=mg) of the object times displacement (d):

$W=Fd=wd=mgd$ (4)

Using (4) on (3):

$mgd=\frac{mv_f^2}{2}-\frac{mv_i^2}{2}$ (5)

That's the equation we're going to use on a) and b).

a) Because the branch started form rest initial velocity (vi) is equal zero, using this and solving (5) for final velocity:

$v_f=\sqrt{\frac{2mgd}{m}}=\sqrt{2gd}=\sqrt{2*9.8*95}$

$v_f=43.15\frac{m}{s^{2}}$

b) In this case the final velocity of the boulder is instantly zero when it reaches its maximum height, another important thing to note is that in this case work is negative because weight is opposing boulder movement, so we should use -mgd:

$-mgd=-\frac{mv_i^2}{2}$

Solving for initial velocity (when the boulder left the volcano):

$v_i=\sqrt{\frac{2mgd}{m}}=\sqrt{2gd}=\sqrt{2*9.8*525}$

$v_i=101.44 \frac{m}{s^{2}}$

3 0

3 years ago