A large crate is at rest on a ramp at a loading dock

Calculate the force of Earth’s gravity on a spacecraft 2.00 Earth radii above the Earth’s surface (That would be 3.00 Earth radi

igomit [66]

Answer:

2014.44 N

Explanation:

mass of spacecraft, m = 1850 kg

distance r = 3 x R

where r be the radius of earth.

g be the acceleration due to gravity on the surface of earth and g' be the acceleration due to gravity at height

$\frac{g'}{g}=\left (\frac{R}{r} \right )^{2}$

$\frac{g'}{g}=\left (\frac{R}{3R} \right )^{2}$

g' = g / 9

g' = 9.8 / 9 = 1.089 m/s²

Force of gravity on the space craft

F = m g' = 1850 x 1.089

F = 2014.44 N

Thus, the force of gravity on the space craft at height is 2014.44 N.

3 0

2 years ago

A 0.305 kg book rests at an angle against one side of a bookshelf. The magnitude and direction of the total force exerted on the

tankabanditka [31]

Answer

given,

$F_L= 1.52\ N$

$\theta_L= 31^0$

mass of book = 0.305 Kg

so, from the diagram attached below

$F_L cos {\theta_L} + F_b sin {\theta_b} = m g$

$1.52 times cos {31^0} + F_b sin {\theta_b} = 0.305 \times 9.8$

$F_b sin {\theta_b} = 2.989 -1.303$

$F_b sin {\theta_b} = 1.686$

computing horizontal component

$F_b cos {\theta_b} = F_L sin {\theta_L}$

$cos {\theta_b} = \dfrac{F_L sin {\theta_L}}{F_b}$

$cos {\theta_b} = \dfrac{1.52 \times sin {31^0}}{1.686}$

$cos {\theta_b} = 0.464$

θ = 62.35°

5 0

2 years ago

A runner has a momentum of 720 kg m/s and is traveling at a velocity of 5 m/s. What is his mass?

antiseptic1488 [7]

I could be wrong, but I'm pretty sure it's 144kg.

8 0

2 years ago

Read 2 more answers

A truck pulled a car of 2,350 kg a distance of 25 meters. If the car accelerates from 3 m/s to 6 m/s, whats the average force ex

faust18 [17]

Answer:

1,269 N

Explanation:

4 0

2 years ago

A solenoid of length 18 cm consists of closely spaced coils of wire wrapped tightly around a wooden core. The magnetic field str

Kisachek [45]

Answer:

$B_2 = 1.71 mT$

Explanation:

As we know that the magnetic field near the center of solenoid is given as

$B = \frac{\mu_0 N i}{L}$

now we know that initially the length of the solenoid is L = 18 cm and N number of turns are wounded on it

So the magnetic field at the center of the solenoid is 2 mT

now we pulled the coils apart and the length of solenoid is increased as L = 21 cm

so we have

$\frac{B_1}{B_2} = \frac{L_2}{L_1}$

now plug in all values in it

$\frac{2.0 mT}{B_2} = \frac{21}{18}$

$B_2 = 1.71 mT$

3 0

2 years ago