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koban [17]
2 years ago
12

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

Physics
1 answer:
VARVARA [1.3K]2 years ago
4 0

Answer:

yuhhh

Explanation:

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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
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