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True [87]
3 years ago
15

A gravitational field vector points toward the earth; an electric field vector points toward an electron. why do electric field

vectors point away from protons? 1. protons have much larger mass. 2. unlike electrons, protons will produce an electric field of their own. 3. protons have more net charge than electrons. 4. protons are positively charged. 5. none of these
Physics
1 answer:
polet [3.4K]3 years ago
3 0
The correct answer is
<span> 4. protons are positively charged. 

In fact, the direction of the electric field of a charged particle depends on the sign of the charge of the particle. While a negatively charged particle (such as the electron) produces an electric field that points toward the charge, a positively charged particle (like a proton) produces an electric field that points away from the charge. This can be proofed by using a positive test charge: if we put a positive test charge near a proton, this test charge will move away from the proton (because it is repelled from it, since they have same charge), and the direction of its motion gives the direction of the electric field generated by the proton, so away from it.</span>
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Gave him good advice to others
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Which idea did Ptolemy's model use to explain why the planets appeared to move backward as they moved in their orbits? The plane
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A car is strapped to a rocket (combined mass = 661 kg), and its kinetic energy is 66,120 J.
labwork [276]

Answer:

9.4 m/s

Explanation:

According to the work-energy theorem, the work done by external forces on a system is equal to the change in kinetic energy of the system.

Therefore we can write:

W=K_f -K_i

where in this case:

W = -36,733 J is the work done by the parachute (negative because it is opposite to the motion)

K_i = 66,120 J is the initial kinetic energy of the car

K_f is the final kinetic energy

Solving,

K_f = K_i + W=66,120+(-36,733)=29387 J

The final kinetic energy of the car can be written as

K_f = \frac{1}{2}mv^2

where

m = 661 kg is its mass

v is its final speed

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v=\sqrt{\frac{2K_f}{m}}=\sqrt{\frac{2(29,387)}{661}}=9.4 m/s

4 0
4 years ago
a force is applied to a box of 10.0 kg for 4.0 s. the box goes from rest to 25 m/s in that time. What is the magnitude of that f
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Given:

m(mass of the box)=10 Kg

t(time of impact)=4 sec

u(initial velocity)=0.(as the body is initially at rest).

v(final velocity)=25m/s

Now we know that

v=u+at

Where v is the final velocity

u is the initial velocity

a is the acceleration acting on the body

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Substituting these values we get

25=0+a x 4

4a=25

a=6.25m/s^2

Now we also know that

F=mxa

F=10 x6.25

F=62.5N

8 0
4 years ago
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