The angular velocity, ω=
2π/t; t = 24 hrs = 24 x 3600 seconds = 86400 s
ω = 7.27 x 10⁻⁵
v = ωr
= 7.27 x 10⁻⁵ x 3242.8 x 1.6 x 1000 (converting miles to meters)
= 377.2 m/s

4 0

3 years ago

A 6.0 g marble is fired vertically upward using a spring gun. The spring must be compressed 9.4 cm if the marble is to just reac

RoseWind [281]

Answer:

a) $\Delta U_{g} = 12.945\,J$ , b) $\Delta U_{k} = 12.945\,J$ , c) $k = 2930.059\,\frac{N}{m}$

Explanation:

a) The change in the gravitational potential energy of the marble-Earth system is:

$\Delta U_{g} = (0.06\,kg)\cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (22\,m)$

$\Delta U_{g} = 12.945\,J$

b) The change in the elastic potential energy of the spring is equal to the change in the gravitational potential energy, then:

$\Delta U_{k} = 12.945\,J$

c) The spring constant of the gun is:

$\Delta U_{k} = \frac{1}{2} \cdot k \cdot x^{2}$

$k = \frac{2\cdot \Delta U_{k}}{x^{2}}$

$k = \frac{2\cdot (12.945\,J)}{(0.094\,m)^{2}}$

$k = 2930.059\,\frac{N}{m}$

4 0

3 years ago

The graph represents the force applied on an 3.00kg crate while it moved 5.0m. A. How much total work is done on the crate? B. I

jeka57 [31]

a. We can calculate the amount of work by calculating the area under the graph.

first area (rectangular): 2.5 x 6 = 15

second area(trapezoid): 1/2 x (6+10) x 2.5 =20

total work done: 35 J

b. the force was first applied = 6 N

F = m.a

a = 6 : 3 = 2 m/s²

vf²=vi²+2as

vf²=6²+2.2.5

vf²=56

vf=7.5 m/s

8 0

3 years ago

what causes places near the equator to receive more direct sunlight than the poles? the poles are farther away from the sun than

Andru [333]

This is due to the tilt of the earth on its axis. Although the Sun shines on Earth, because of how the Earth is tilted, the equator is more directly hit compared to places found on the poles. The poles are hit at an angle, therefore the sunlight they receive is lesser than the places at the equator.

5 0

3 years ago