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kolbaska11 [484]
2 years ago
15

A 2.40 μC charge is subject to a 3.00 mN force due to an Electric Field. What is the magnitude of the Electric Field at the loca

tion of the charge?
Physics
1 answer:
Bezzdna [24]2 years ago
4 0
<h2>Electric field at the location of the charge is 1250 N/C</h2>

Explanation:

Electric field is the ratio of force and charge.

Force, F = 3.00 mN = 3 x 10⁻³ N

Charge, q = 2.40 μC = 2.40 x 10⁻⁶ C

We have

                 E=\frac{F}{q}\\\\E=\frac{3\times 10^{-3}}{2.40\times 10^{-6}}\\\\E=1250N/C

Electric field at the location of the charge is 1250 N/C

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What is the centripetal force
Lelechka [254]

Answer:90N

Explanation:

Mass=30kg

Centripetal acceleration=3m/s^2

centripetal force=mass x centripetal acceleration

Centripetal force=30 x 3

Centripetal force =90

Centripetal force =90N

4 0
3 years ago
A rocket starting from its launch pad is subjected to a uniform acceleration of 100 meters/second2. Determine the time needed to
gizmo_the_mogwai [7]

Answer:

10s

Explanation:

Acceleration is a measure of a rate of change of velocity, or in other words, a measure of how quickly the velocity is changing.

If acceleration is constant, then the velocity is changing by a constant amount.

With an acceleration of 100 m/s^2, starting from the launching pad (and thus, an initial velocity of zero), we can calculate how long it will take to reach a final velocity of 1000m/s with the following formula:

v=at+v_o where "v" is the final velocity at some later time "t", "a" is the constant acceleration, and "v" sub-zero is the initial velocity.

v=at+v_o

(1000\text{ [m/s]})=(100 \text{ } [\text{m/s}^2] )t+(0\text{ [m/s]})

1000\text{ [m/s]}=100 \text{ } [\text{m/s}^2] *t

\dfrac{1000\text{ [m/s]}}{100 \text{ } [\text{m/s}^2]}=\dfrac{100 \text{ } [\text{m/s}^2] *t}{100 \text{ } [\text{m/s}^2]}

10\text{ [s]}=t

So, it will take 10 seconds for the rocket to reach 1000m/s when starting from the launching pad, with a constant velocity of 100m/s^2.

<u>Verification:</u>

In this situation, it is quick to verify that 10 seconds is correct by looking at what the velocities will be each second.

Recognizing that the acceleration is a=\dfrac{100 [\frac{m}{s}]}{1[s]}, the velocity increases by 100 units [m/s] every second.

At time 0[s], the velocity is 0[m/s]

At time 1[s], the velocity is 100[m/s]

At time 2[s], the velocity is 200[m/s]

At time 3[s], the velocity is 300[m/s]

At time 4[s], the velocity is 400[m/s]

At time 5[s], the velocity is 500[m/s]

At time 6[s], the velocity is 600[m/s]

At time 7[s], the velocity is 700[m/s]

At time 8[s], the velocity is 800[m/s]

At time 9[s], the velocity is 900[m/s]

At time 10[s], the velocity is 1000[m/s]

So, indeed, after 10 seconds, the velocity reaches 1000 m/s

5 0
2 years ago
In a nuclear station if the temperature of the superheated water starts to decrease what should the engineer in charge of the re
Sophie [7]

Answer:

They sh0uld g0 t0 the reactor and then, see what the issue is...

Explanation:

then, see if they can fix the problem, im sorry if its wr0ng.

8 0
2 years ago
021. The hydrostatic pressure 'P' of a liquid column depends upon the
Lana71 [14]

Answer:

A

Explanation:

hydrostatic pressure, P=hρg,

where 'h' is the ht of the liq column and ρ is the density of the liquid and 'g' is the effective acceleration, but as far as hydroSTATICS is concerned, g stands for the acceleration due to gravity

3 0
2 years ago
Question 20 of 31
lbvjy [14]

Answer:

C: Valence electrons are transferred from a metal to a nonmetal

Explanation:

In ionic bonds, the metal loses electrons to become a positively charged cation, whereas the nonmetal accepts those electrons to become a negatively charged anion.

hope this was helpful :)

5 0
3 years ago
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