Answer:
If this is a multiple choice question, then the answer is D.
Explanation:
Compounds are composed of atoms, which are composed of subatomic particles and consist of matter. Since they are composed of atoms rather than vice-versa, compounds cannot be found inside atoms and are not the most basic form of matter (ruling out A and C). A pure substance contains atoms, which are each composed of subatomic particles. Therefore, a pure substance must have atoms if it contains subatomic particles (ruling out B). The only answer left is D.
Answer:

Explanation:
For answer this we will use the law of the conservation of the angular momentum.

so:

where
is the moment of inertia of the merry-go-round,
is the initial angular velocity of the merry-go-round,
is the moment of inertia of the merry-go-round and the child together and
is the final angular velocity.
First, we will find the moment of inertia of the merry-go-round using:
I = 
I = 
I = 359.375 kg*m^2
Where
is the mass and R is the radio of the merry-go-round
Second, we will change the initial angular velocity to rad/s as:
W = 0.520*2
rad/s
W = 3.2672 rad/s
Third, we will find the moment of inertia of both after the collision:



Finally we replace all the data:

Solving for
:

Answer:
b) True. the force of air drag on him is equal to his weight.
Explanation:
Let us propose the solution of the problem in order to analyze the given statements.
The problem must be solved with Newton's second law.
When he jumps off the plane
fr - w = ma
Where the friction force has some form of type.
fr = G v + H v²
Let's replace
(G v + H v²) - mg = m dv / dt
We can see that the friction force increases as the speed increases
At the equilibrium point
fr - w = 0
fr = mg
(G v + H v2) = mg
For low speeds the quadratic depended is not important, so we can reduce the equation to
G v = mg
v = mg / G
This is the terminal speed.
Now let's analyze the claims
a) False is g between the friction force constant
b) True.
c) False. It is equal to the weight
d) False. In the terminal speed the acceleration is zero
e) False. The friction force is equal to the weight
Answer:
1250 J
Explanation:
Work is said to be done when a force causes an object to move over a distance. The amount of work done (W) is calculated by multiplying the force by the distance traveled.
That is;
W = F × d
Where;
W = work done (J or N/m)
F = force (N)
d = distance (m)
Based on the information provided in this question, F = 5000N, d = 0.25m
Hence;
W = F × d
W = 5000 × 0.25
W = 1250J
Therefore, 1250Joules of work is done by the jack.
Answer:
A. The electric field points to the left because the force on a negative charge is opposite to the direction of the field.
Explanation:
The electric force exerted on a charge by an electric field is given by:
where
F is the force
q is the charge
E is the electric field
We see that if the charge is negative, q contains a negative sign, so the force F and the electric field E will have opposite signs (which means they have opposite directions). This is due to the fact that the direction of the lines of an electric field shows the direction of the electric force experienced by a positive charge in that electric field: therefore, a negative charge will experience a force into opposite direction.