Answer: apparent weighlessness.
Explanation:
1) Balance of forces on a person falling:
i) To answer this question we will deal with the assumption of non-drag force (abscence of air).
ii) When a person is dropped, and there is not air resistance, the only force acting on the person's body is the Earth's gravitational attraction (downward), which is the responsible for the gravitational acceleration (around 9.8 m/s²).
iii) Under that sceneraio, there is not normal force acting on the person (the normal force is the force that the floor or a chair exerts on a body to balance the gravitational force when the body is on it).
2) This is, the person does not feel a pressure upward, which is he/she does not feel the weight: freefalling is a situation of apparent weigthlessness.
3) True weightlessness is when the object is in a place where there exists not grativational acceleration: for example a point between two planes where the grativational forces are equal in magnitude but opposing in direction and so they cancel each other.
Therefore, you conclude that, assuming no air resistance, a person in this ride experiencing apparent weightlessness.
Answer: a. F doubled
b. F reduced by one-quarter i.e
1/4*(F)
c. 1/9*(F)
d. F increased by a factor of 4 i.e 4*F
e. F reduces 3/4*(F)
Explanation: Coulombs law states the force F of attraction/repulsion experience by two charges qA and qB is directly proportional to thier product and inversely proportional to the square of distance d between them. That is
F = k*(qA*qB)/d²
a. If qA is doubled therefore the force is doubled since they are directly proportional.
b. If qA and qB are half, that means thier new product would be qA/2)*qB/2 =qA*qB/4
Which means the product of charge is divided by 4 so the force would be divided by 4 too since they are directly proportional.
c. If d is tripped that is multiplied by 3. From the formula new d would be (3*d)²=9d² but force is inversely proportional to d² so instead of multiplying by 9 the force will be divided by 9
d. If d is cut into half that is divided by 2. The new d would be (d/2)²=d²/4. So d² is divided by 4 so the force would be multiplied by 4
e. If qA is tripled that is multiplied by 3. F would be multiplied by 3 also, if at the same time d is doubled (2*d)²= 4*d² . Force would be divided by 4 at same time. So we have,
3/4*F
<u>Answer:</u>
Option: D. Gravity is pulling the crash test dummy in the direction the car is moving.
<u>Explanation:
</u>
When a car accelerates from a standing start, the crash test dummy appears to be pressed backward into the seat cushion because the gravity is pulling the crash test dummy in the direction the car is moving.
Basically when the car is starting, the person inside is in static position and the car is going to move. So it is putting a force on the person to move on the same speed. But as the person is sitting static hence gravity is pulling him behind from moving. Hence, The dummy appears to be pressed backward.
Answer:
The flux through the surface of the cube is 
Solution:
As per the question:
Edge of the cube, a = 8.0 cm = 
Volume Charge density, 
Now,
To calculate the electric flux:
(1)
where
= electric flux
= permittivity of free space
Volume Charge density for the given case is given by the formula:
(2)
Volume of cube, 
Thus
Thus from eqn (2), the total charge is given by:
Now, substitute the value of 'q' in eqn (1):
Answer:
the direction that should be walked by Ricardo to go directly to Jane is 23.52 m, 24° east of south
Explanation:
given information:
Ricardo walks 27.0 m in a direction 60.0 ∘ west of north, thus
A= 27
Ax = 27 sin 60 = - 23.4
Ay = 27 cos 60 = 13.5
Jane walks 16.0 m in a direction 30.0 ∘ south of west, so
B = 16
Bx = 16 cos 30 = -13.9
By = 16 sin 30 = -8
the direction that should be walked by Ricardo to go directly to Jane
R = √A²+B² - (2ABcos60)
= √27²+16² - (2(27)(16)(cos 60))
= 23.52 m
now we can use the sines law to find the angle
tan θ = 
= By - Ay/Bx -Ax
= (-8 - 13.5)/(-13.9 - (-23.4))
θ = 90 - (-8 - 13.5)/(-13.9 - (-23.4))
= 24° east of south
