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Reptile [31]
3 years ago
15

Gauss’ law: a. Relates the surface charge density to the electric field.b. Relates the electric field at points on a closed surf

ace to the net charge enclosed by that surface.c. Only applies to point charges.d. Relates the electric field throughout space to the charges distributed through that space
Physics
1 answer:
Mamont248 [21]3 years ago
8 0

Answer:

b. Relates the electric field at points on a closed surface to the net charge enclosed by that surface

Explanation:

Gauss's law states that the flux of certain fields through a closed surface is proportional to the magnitude of the sources of that field within the same surface. The electric flux expresses the measure of the electric field that crosses a certain surface. Therefore, the electric field on a closed surface is proportional to the net charge enclosed by that surface.

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Acceleration is best defined as the rate of change of __________ of an object. A. position B. force C. speed D. velocity 2. Whic
Fittoniya [83]
Velocity!
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7 0
3 years ago
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States that there is an exchange of materials when two objects come into contact with each other
pishuonlain [190]

"Edmond Locard" states that there is an exchange of materials when two objects come into contact with each other.

<u>Explanation:</u>

A French criminologist who was popular as the "Sherlock Holmes of France," the pioneer in forensic science named as Dr. Edmond Locard. He articulated forensic science's fundamental principle "Each touch leaves a trace." This became known as Locard's philosophy of exchange. A Locard hypothesized that each and every time you touch another person, place or object, the result would be an exchange of materials. Burglars, for instance, will leave evidence of their existence behind and take traces with them too.

3 0
3 years ago
A dentist’s drill starts from rest. After 4.28 s of constant angular acceleration it turns at a rate of 28940 rev/min. Find the
mylen [45]

Answer:

Angular acceleration, is 708.07\ rad/s^2

Explanation:

Given that,

Initial speed of the drill, \omega_i=0

After 4.28 s of constant angular acceleration it turns at a rate of 28940 rev/min, final angular speed, \omega_f=28940\ rev/min=3030.58\ rad/s

We need to find the drill’s angular acceleration. It is given by the rate of change of angular velocity.

\alpha =\dfrac{\omega_f}{t}\\\\\alpha =\dfrac{3030.58\ rad/s}{4.28\ s}\\\\\alpha =708.07\ rad/s^2

So, the drill's angular acceleration is 708.07\ rad/s^2.

4 0
3 years ago
A locomotive is pulling 8 freight cars, each of which is loaded with the same amount of weight. The mass of each freight car (wi
nikitadnepr [17]

Answer:106560 N

Explanation:

Given

Let Tension between  2 and 3 car be T_{23} and between 3 &4 is T_{34}

T_{23}-T_{34}=ma

mass of freight car =37,000 kg

acceleration of car=0.48 m/s^2

T_{34} is accelerating all freights behind 3

therefore

T_{34}=5\times ma

T_{34}=5\times 37000\times 0.48=88,800 N

Thus

T_{23}=T_{34}+ma

T_{23}=88,800+37000\times 0.48=88,800+17,760=106560 N

4 0
3 years ago
If an arrow is shot upward on the moon with a velocity of 58 m/s, its height (in meters) after t seconds is given by H = 58t − 0
Alex73 [517]

Answer:

a) v = 54.7m/s

b) v = (58 - 1.66a) m/s

c) t = 69.9 s

d) v = -58.0 m/s

Explanation:

Given;

The height equation of the arrow;

H = 58t - 0.83t^2

(a) Find the velocity of the arrow after two seconds. m/s;

The velocity of the arrow v can be given as dH/dt, the change in height per unit time.

v = dH/dt = 58 - 2(0.83t) ......1

At t = 2 seconds

v = dH/dt = 58 - 2(0.83×2)

v = 54.7m/s

(b) Find the velocity of the arrow when t = a. m/s

Substituting t = a into equation 1

v = 58 - 2(0.83×a)

v = (58 - 1.66a) m/s

(c) When will the arrow hit the surface? (Round your answer to one decimal place.) t = s

the time when H = 0

Substituting H = 0, we have;

H = 58t - 0.83t^2 = 0

0.83t^2 = 58t

0.83t = 58

t = 58/0.83

t = 69.9 s

(d) With what velocity will the arrow hit the surface? m/s

from equation 1;

v = dH/dt = 58 - 2(0.83t)

Substituting t = 69.9s

v = 58 - 2(0.83×69.9)

v = -58.0 m/s

8 0
3 years ago
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