The statement is <u>false</u> because the sky <u>can change </u>colors during sunsets, sun rises, etc. The sky is not always blue.
R = 2.06 mm = 2.06 x 10^(-3) m
Q = 1.6 x 10^(-19) C
v = 2.5 x 10^(-5) m/s
I = 8 A = 8 C/s
A = r² π = ( 2.06 x 10^(-3) ) ² x 3.14 = 13.325 x 10^(-6 ) m² =
= 1.3325 x 10^(-5) m²
I = n Q v A
n = I / (Q v A)
n = 8 C/s / ( 1.6 x 10 ^(-19) * 5.4 x 10^(-5) * 1.3325 x 10^(-5) ) =
= 0.694 x 10^(29) m^(-3)
n = 6.94 x 10^(28) m^(-3)
Answer:
A 60 kg person standing on a platform at the surface of Saturn and they jumped, they would have to push with a force greater than 540 N
Explanation:
The gravitational attraction between an object on the surface of a planet and the planet is given by the weight of the object
Therefore the force needed to be applied for an object to lift off the surface of a planet = The weight of the object
The weight of the object on the surface of a planet = m × g
Where;
m = The mass of the object
g = The strength of gravity on the planet's surface in N/kg
The given parameters are;
The mass of the person standing on a platform at the surface of Saturn, m = 60 kg
The strength of gravity on the surface of Saturn = 9 N/kg
Therefore, we have;
The weight of the person = The force greater than which the person would have to push on the surface of Saturn so as to Jump = The weight of the person on the surface of Saturn = 60 kg × 9 N/kg = 540 N
Therefore, for a 60 kg person standing on a platform at the surface of Saturn and they jumped, they would have to push with a force greater than 540 N.
<u>We are given:</u>
constant speed of the car (u) = 36.12 m/s
time in question (t) = 12 seconds
<u>Solving for the Distance and Displacement:</u>
from the second equation of motion:
s = ut + 1/2 at^2
since we have 0 acceleration:
s = ut
<em>replacing the variables</em>
s = 36.12 * 12
s = 433.44 m
Since the car is travelling in a straight line towards the same direction, it's Distance will be equal to its Displacement
Hence, both the Displacement and <u>Distance covered by the car is </u>
<u>433.44 m</u>
but since Displacement also has a direction vector along with it,
the <u>Displacement will be 433.44 m due west</u>
