Answers:
B.) 
C.) 
Explanation:
The image attached shows the way the force 
is acting on the block. Now, if we draw a free body diagram of the situation and write the equations for the Net Force in X and Y, we will have the following:
Net Force in X:
(1)
Where:
is the Normal force
is the magnitude of the force exerted on the block
is the angle
Net Force in Y:
(2)
Where:
is the Friction force (it is expresed with the 
sign because this force may be up or down, we cannot know because the block is at rest)
is the gravity force
Rewrittin (1):
(3) This is according to option B
Rewritting (2):
(3) This is according to option C
Is D
is D because the inner layers are the Core, Radiative Zone and Convection Zone.
Answer:
A police car with its siren on is driving towards you, and you perceive the pitch of the siren to increase.
Explanation:
In Physics, Doppler effect can be defined as the change in frequency of a wave with respect to an observer in motion and moving relative to the source of the wave.
Simply stated, Doppler effect is the change in wave frequency as a result of the relative motion existing between a wave source and its observer.
The term "Doppler effect" was named after an Austrian mathematician and physicist known as Christian Johann Doppler while studying the starlight in relation to the movement of stars.
<em>The phenomenon of Doppler effects is generally applicable to both sound and light. </em>
An example of the Doppler effect is a police car with its siren on is driving towards you, and you perceive the pitch of the siren to increase. This is so because when a sound object moves towards you, its sound waves frequency increases, thereby causing a higher pitch. However, if the sound object is moving away from the observer, it's sound waves frequency decreases and thus resulting in a lower pitch.
<em>Other fields were the Doppler effects are applied are; astronomy, flow management, vibration measurement, radars, satellite communications etc. </em>
Answer:
The object will move to Xfinal = 7.5m
Explanation:
By relating the final velocity of the object and its acceleration, I can obtain the time required to reach this velocity point:
Vf= a × t ⇒ t= (7.2 m/s) / (4.2( m/s^2)) = 1,7143 s
With the equation of the total space traveled and the previously determined time I can obtain the end point of the object on the x-axis:
Xfinal= X0 + /1/2) × a × (t^2) = 3.9m + (1/2) × 4.2( m/s^2) × ((1,7143 s) ^2) =
= 3.9m + 3.6m = 7.5m
