Answer:
v = 384km/min
Explanation:
In order to calculate the speed of the Hubble space telescope, you first calculate the distance that Hubble travels for one orbit.
You know that 37000 times the orbit of Hubble are 1,280,000,000 km. Then, for one orbit you have:
You know that one orbit is completed by Hubble on 90 min. You use the following formula to calculate the speed:
hence, the speed of the Hubble is approximately 384km/min
Answer:
The possible thickness of the soap bubble =
Explanation:
<u>Given:</u>
- Refractive index of the soap bubble,
- Wavelength of the light taken,
Let the thickness of the soap bubble be .
It is given that the soap bubble appears very bright, it means, there is a constructive interference takes place.
For the constructive interference of light through a thin film ( soap bubble), the condition of constructive interference is given as:
where is the order of constructive interference.
Since the soap bubble is appearing very bright, the order should be 0, as order interference has maximum intensity.
Thus,
It is the possible thickness of the soap bubble.
Answer:
v_max = (1/6)e^-1 a
Explanation:
You have the following equation for the instantaneous speed of a particle:
(1)
To find the expression for the maximum speed in terms of the acceleration "a", you first derivative v(t) respect to time t:
(2)
where you have use the derivative of a product.
Next, you equal the expression (2) to zero in order to calculate t:
For t = 1/6 you obtain the maximum speed.
Then, you replace that value of t in the expression (1):
hence, the maximum speed is v_max = ((1/6)e^-1)a
-- If the field were inclined to the surface, then it would have
some component parallel to the surface.
-- Then, since we're talking about a conductor, the charges
on the object would move in response to that component
of the field, until there was no longer any component of the
field trying to move them.
Answer:
0.54
Explanation:
Draw a free body diagram. There are 5 forces on the desk:
Weight force mg pulling down
Applied force 24 N pushing down
Normal force Fn pushing up
Applied force 130 N pushing right
Friction force Fnμ pushing left
Sum of the forces in the y direction:
∑F = ma
Fn − mg − 24 = 0
Fn = mg + 24
Fn = (22)(9.8) + 24
Fn = 240
Sum of the forces in the x direction:
∑F = ma
130 − Fnμ = 0
Fnμ = 130
μ = 130 / Fn
μ = 130 / 240
μ = 0.54