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elixir [45]
4 years ago
7

Two charged objects attract each other with a certain force. If the charge on one of the objects is doubled with no change in th

e other or the separation, the force between them
A.) quadruples
B.) doubles
C.) halves
D.) increases, but we can't say how much without knowing the distance between them
Physics
1 answer:
Vanyuwa [196]4 years ago
4 0
It is B. secod try i know its dumb but im trying to cheer myself up
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This is junior year english
pychu [463]
D. is the correct answer.
6 0
4 years ago
Read 2 more answers
An airplane of mass 1.60 ✕ 104 kg is moving at 66.0 m/s. The pilot then increases the engine's thrust to 7.70 ✕ 104 N. The resis
Ivan

(a) No, because the mechanical energy is not conserved

Explanation:

The work-energy theorem states that the work done by the engine on the airplane is equal to the gain in kinetic energy of the plane:

W=\Delta K (1)

However, this theorem is only valid if there are no non-conservative forces acting on the plane. However, in this case there is air resistance acting on the plane: this means that the work-energy theorem is no longer valid, because the mechanical energy is not conserved.

Therefore, eq. (1) can be rewritten as

W=\Delta K + E_{lost}

which means that the work done by the engine (W) is used partially to increase the kinetic energy of the airplane (\Delta K) and part is lost because of the air resistance (E_{lost}).

(b) 77.8 m/s

First of all, we need to calculate the net force acting on the plane, which is equal to the difference between the thrust force and the air resistance:

F=7.70\cdot 10^4 N - 5.00 \cdot 10^4 N=2.70\cdot 10^4 N

Now we can calculate the acceleration of the plane, by using Newton's second law:

a=\frac{F}{m}=\frac{2.70\cdot 10^4 N}{1.60\cdot 10^4 kg}=1.69 m/s^2

where m is the mass of the plane.

Finally, we can calculate the final speed of the plane by using the equation:

v^2- u^2 = 2aS

where

v=? is the final velocity

u=66.0 m/s is the initial velocity

a=1.69 m/s^2 is the acceleration

S=5.00 \cdot 10^2 m is the distance travelled

Solving for v, we find

v=\sqrt{u^2+2aS}=\sqrt{(66.0 m/s)^2+2(1.69 m/s^2)(5.00\cdot 10^2 m)}=77.8 m/s

8 0
3 years ago
You are on the roof of a building, 46.0 m above the ground. Your friend, who is 1.80 m tall, is standing next to the building. W
ExtremeBDS [4]

Answer:

a.) 866 m/s

beacues

5 0
3 years ago
A car travels 24 m East in 3 s. what is the velocity of the car?
Bad White [126]

Well, since you noted the direction of travel is “due east”, that clearly means the question is asking about a frame of reference that included planetary rotation into the mix. At the equator, that’s about 460m/s in the eastward direction. Clearly, at the north pole, that’s zero—you’re effectively stationary. So, let’s assume you’re somewhere between the equator and the pole, so we’ll take the average of the two, and say the earth is contributing 230m/sec of eastward velocity. So, at time t=0s, your velocity is 230m/sec; at time t=30s, your velocity is 254m/sec. Thus, you plug those into the formula, and you get (254–230)/(30–0), or 0.8m/sec^2.

Now, if the question was to consider this from a larger frame of reference, we’d also have to take the rotation of the earth around the sun into consideration, which is about 107,000km/hr, or about 29,722m/sec. The problem is that we don’t know if we need to *add* that to the rotational velocity of the earth, and motion of the car, or *subtract* it; that all depends on whether the side of the earth that the car is on is facing the sun, or away from the sun. If we assume that sane people do math experiments on their cars only when the sun is shining, then we need to add the velocity in as well; so we get 29,722+230+24, or (29,976–29,952)(30–0). However, that only works if you do your vehicular calculations in the daytime. If, on the other hand, you’re a dark, brooding vigilante who only comes out after darkness falls to drive around, then we need to adjust our calculations to account for the fact that you’re now going retrograde with respect to the sun. Thus, the calculation would become 29,722–230–24, or (29468–29492)/(30–0), or -0.8m/sec^2.

HOPE THIS HELPS :)

8 0
3 years ago
Read 2 more answers
Nearly all of the energy that Earth receives from the Sun is used in photosynthesis. Please select the best answer from the choi
jekas [21]

Answer: False

Explanation:

Earth large amount of energy ( about 174 PW). 30% of this energy reflects back and rest of it is absorbed by clouds, oceans and land mass. Thus, we can say that only some amount of energy that Earth receives from the sun is used in photosynthesis. Photosynthesis is a process that uses carbon-dioxide and water in presence of sunlight and chlorophyll produces oxygen and sugar. Hence, the given statement is false.

7 0
3 years ago
Read 2 more answers
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