Answer:
Acceleration is percieved, not constant velocity.
Explanation:
You are most aware when the vehicle is accelerating. At constant velocity you would not be aware of the motion. Only if the system is accelerated the dynamics must be solved considering a pseudo-force (of inertial origin) acting.
It's because of this that:
(A) False. The acceleration can be detected from the inside of a closed car.
(B) False. You would be aware of the motion, but not because humans can sense speed but acceleration.
(C) False. Constant velocity cannot be felt in a closed car.
(D) False. Again, you can't feel constant speed.
Answer:
f'=5.58kHz
Explanation:
This is an example of the Doppler effect, the formula is:

Where f is the actual frequency,
is the observed frequency,
is the velocity of the sound waves,
the velocity of the observer (which is negative if the observer is moving away from the source) and
the velocity of the source (which is negative if is moving towards the observer). For this problem:


<span>The weight of the spacecraft keeps changing.
</span>
<span>The mass of the spacecraft remains the same.
These are the correct answers</span>
Answer:
The correct answer is B
Explanation:
Let's calculate the electric field using Gauss's law, which states that the electric field flow is equal to the charge faced by the dielectric permittivity
Φ
= ∫ E. dA =
/ ε₀
For this case we create a Gaussian surface that is a sphere. We can see that the two of the sphere and the field lines from the spherical shell grant in the direction whereby the scalar product is reduced to the ordinary product
∫ E dA =
/ ε₀
The area of a sphere is
A = 4π r²
E 4π r² =
/ ε₀
E = (1 /4πε₀
) q / r²
Having the solution of the problem let's analyze the points:
A ) r = 3R / 4 = 0.75 R.
In this case there is no charge inside the Gaussian surface therefore the electric field is zero
E = 0
B) r = 5R / 4 = 1.25R
In this case the entire charge is inside the Gaussian surface, the field is
E = (1 /4πε₀
) Q / (1.25R)²
E = (1 /4πε₀
) Q / R2 1 / 1.56²
E₀ = (1 /4π ε₀
) Q / R²
= Eo /1.56
²
= 0.41 Eo
C) r = 2R
All charge inside is inside the Gaussian surface
=(1 /4π ε₀
) Q 1/(2R)²
= (1 /4π ε₀
) q/R² 1/4
= Eo 1/4
= 0.25 Eo
D) False the field changes with distance
The correct answer is B