Answer:
The acceleration of the satellite is 
Explanation:
The acceleration in a circular motion is defined as:
(1)
Where a is the centripetal acceleration, v the velocity and r is the radius.
The equation of the orbital velocity is defined as
(2)
Where r is the radius and T is the period
For this particular case, the radius will be the sum of the high of the satellite (
) and the Earth radius (
) :


Then, equation 2 can be used:
⇒ 


Finally equation 1 can be used:

Hence, the acceleration of the satellite is 
Explanation:
Centripetal acceleration is:
a = v² / r
a = (4.0 m/s)² / 0.60 m
a = 26.6 m/s²
Answer:
Magnitude of Vector = 79.3
Explanation:
When a vector is resolved into its rectangular components, it forms two vector components. These components are named as x-component and y-component, they are calculated by the following formulae:
x-component of vector = (Magnitude of Vector)(Cos θ)
y-component of vector = (Magnitude of Vector)(Sin θ)
where,
θ = angle of the vector with x-axis = 27°
Therefore, using the values in the equation of y-component, we get:
36 = (Magnitude of Vector)(Sin 27°)
Magnitude of Vector = 36/Sin 27°
<u>Magnitude of Vector = 79.3</u>
Answer:
-4.0 N
Explanation:
Since the force of friction is the only force acting on the box, according to Newton's second law its magnitude must be equal to the product between mass (m) and acceleration (a):
(1)
We can find the mass of the box from its weight: in fact, since the weight is W = 50.0 N, its mass will be

And we can fidn the acceleration by using the formula:

where
v = 0 is the final velocity
u = 1.75 m/s is the initial velocity
t = 2.25 s is the time the box needs to stop
Substituting, we find

(the acceleration is negative since it is opposite to the motion, so it is a deceleration)
Therefore, substituting into eq.(1) we find the force of friction:

Where the negative sign means the direction of the force is opposite to the motion of the box.
The photosynthetic wave interaction between visible light and a photosensitive part of a plant is very important t how plants use light to grow.
The frequency range and intensity levels of this light, I don't know.
Maybe the above ???