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Wewaii [24]
3 years ago
10

how does the electric force between two charged particles change if the distance between them is increased by a factor of 3? a.

it is reduced by a factor of 3 b. it is increased by a factor of 3 c. it is increased by a factor of 9 d. it is decreased by a factor of 9
Physics
2 answers:
Anna007 [38]3 years ago
6 0
D. it is decreased by a factor of 9.

Force = kqq ÷ r^2

The distance is squared so if it is increased or decreased by any factor, then it must be squared too. Because the distance is on the bottom of the equation, you divide the force by the increasing or decreasing factor.
seropon [69]3 years ago
5 0

Answer: C. It increased by a factor of 9

Explanation:

A P E X

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A certain car's drive-train produces a force of 5300 N as it accelerates from 0
lakkis [162]

The power is 7.1\cdot 10^4 W

Explanation:

First of all, we need to find the acceleration of the car, which is given by

a=\frac{v-u}{t}

where

v = 60 mph = 26.8 m/s is the final velocity

u = 0 is the initial velocity

t = 10.0 s is the time

Substituting,

a=\frac{26.8-0}{10.0}=2.68 m/s^2

Now we can find the mass of the car by using Newton's second law:

F=ma

where

F = 5300 N is the force applied

m is the mass

a=2.68 m/s^2 is the acceleration

Solving for m,

m=\frac{F}{a}=\frac{5300}{2.68}=1978 kg

Now we can use the work-energy theorem, which states that the work done is equal to the change in kinetic energy of the car, to find the work:

W=K_f - K_i = \frac{1}{2}mv^2-\frac{1}{2}mu^2

And substituting,

W=\frac{1}{2}(1978)(26.8)^2-0=7.10\cdot 10^5 J

Finally, we can find the power output of the car:

P=\frac{W}{t}

where

W is the work

t = 10.0 s is the time elapsed

Substituting,

P=\frac{7.1\cdot 10^5}{10.0}=7.1\cdot 10^4 W

Learn more about power:

brainly.com/question/7956557

#LearnwithBrainly

4 0
4 years ago
Which type of radiation detection device uses a phosphor-coated surface to detect radiation?
labwork [276]
<span>scintillation counter

</span>
8 0
4 years ago
Read 2 more answers
An attacker at the base of a castle wall 3.95 m high throws a rock straight up with speed 5.00 m/s from a height of 1.60 m above
s344n2d4d5 [400]

Answer:

a) No

b) the rock must have a minimum initial speed of 6.79m/s for it to reach the top of the building.

Explanation:

Given:

Height of the wall = 3.95m

Initial height = 1.60m

Initial speed = 5.00m/s

distance between the initial height and wall top = 3.95 - 1.60 = 2.35m

Using the formula;

v^2 = u^2 + 2as ....1

Where v = final velocity, u = initial velocity, a = acceleration, s = distance travelled

From equation 1

s = (v^2 - u^2)/2a ...2

Since the rock t moving up,

the acceleration = -g = -9.8m/s2

s = maximum height travelled

v = 0 (at maximum height velocity is zero)

Substituting into equation 2

s = (0 - 5^2)/(2×-9.8) = 1.28m

Therefore, the maximum height is 1.28 from his initial height Which is less than the 2.35m of the wall from his initial height. So the rock will not reach the top of the wall

b) Using equation 1:

u^2 = v^2 - 2as

v = 0

a = -9.8m/s

s = 2.35m. (distance between the initial height and wall top)

u^2 = 0 - 2(-9.8 × 2.35)

u^2 = 46.06

u = √46.06

u = 6.79m/s

Therefore, the rock must have a minimum initial speed of 6.79m/s

8 0
3 years ago
what is the position of the earth in which direction is the axis tilted when it is summer in the northern hemisphere and the sou
Alinara [238K]
Hello!

When the Northern Hemisphere is tilted toward the sun, latitudes between the equator and 90°N (the North Pole) are experiencing summer.
3 0
3 years ago
The solar noon solar radiation values at the top of the atmosphere above maun are 1366 wm−2 on february 2 and 1074 wm−2 on augus
PilotLPTM [1.2K]

You haven't stated any numbers showing that the intensity of solar radiation at the surface is lower than at the top of the atmosphere. Your data only show that the value at the top of the atmosphere is different on different dates.

From our vast experience, however, we do know that the solar intensity at the surface IS lower than it is at the top of the atmosphere, simply because the atmosphere absorbs some solar radiation ... different amounts of it at different wavelengths.

That's the main reason, for example, why the sky is red at sunrise and sunset and blue the rest of the day, and why the temperature of the air is so much higher than 3° absolute, and why we aren't broiled by X-rays all day. Also the reason why it's worth the tremendous cost and makes such a difference to build astronomical observatories on mountain tops and in low-Earth-orbit, instead of in convenient deep valleys.

8 0
3 years ago
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