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vfiekz [6]
3 years ago
15

Charge is given in microcoulombs. What must you multiply the charge by to use Coulomb's

Physics
1 answer:
iVinArrow [24]3 years ago
7 0

Answer:

Option A. 10¯⁶

Explanation:

To know which option is correct, we must bear in mind, the relationship between micro coulomb (μC) and coulomb (C). This is given below:

Recall:

1 μC = 10¯⁶ C

Therefore, to convert micro coulomb (μC) to coulomb (C), multiply the value given in micro coulomb (μC) by 10¯⁶.

Thus, option A gives the correct answer to the question.

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Studentka2010 [4]

Answer:

B, C

Explanation:

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3 years ago
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Which process causes the transfer of energy by air currents within the Earth's atmosphere?
ruslelena [56]

Answer:

Conduction, radiation and convection all play a role in moving heat between Earth's surface and the atmosphere. Since air is a poor conductor, most energy transfer by conduction occurs right near Earth's surface

8 0
3 years ago
Why is rust formed on iron
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When iron is exposed to moisture and oxygen it starts to rust.
5 0
3 years ago
A 32.0 kg wheel, essentially a thin hoop with radius 1.20 m, is rotating at 280 rev/min. It must be brought to a stop in 15.0 s.
elena-14-01-66 [18.8K]

Answer:

A) Must be done 19806.62 joules of work.

B) The average power is 1320.44 Watts.

Explanation:

A) First, we're going to use the work-energy theorem that states total work (W) done on an object is equal to the change in its kinetic energy (\Delta K):

W=\Delta K = K_{f}-K_{i} (1)

So, all we must do is to find the change on kinetic energy. Because we're working with rotational body, we should use the equation K=\frac{I\omega^{2}}{2} for the kinetic energy so:

\Delta K=\frac{I(\omega_{f})^{2}}{2}-\frac{I(\omega_{i})^{2}}{2} (2)

with \omega_{i} the initial angular velocity, \omega_{f} the final angular velocity (is zero because the wheel stops) and I the moment of inertia that for a thin hoop is I=MR^{2}, using those on (2)

\Delta K=0-\frac{MR^{2}(\omega_{i})^{2}}{2} (3)

By (3) on (1):

W= \frac{MR^{2}(\omega_{i})^{2}}{2} = \frac{(32.0)(1.2)^{2}(29.32)^{2}}{2}

W=19806.62\,J

B) Average power is work done divided by the time interval:

P=\frac{W}{\Delta t}=\frac{19806.62}{15.0}

P=1320.44\,W

NOTE: We use the relation 1rpm*\frac{2\pi}{60s}=\frac{rad}{s} to convert 280 rev/min(rpm) to 29.32 rad/s

4 0
4 years ago
A 43 kg girl and a 7.1 kg sled are on the frictionless ice of a frozen lake, 12 m apart but connected by a rope of negligible ma
trapecia [35]

Answer:

(a) Girl acceleration =  0.15 m/s², (b) Sled acceleration =  0.93 m/s²

Explanation:

Given:

M1=Mass of girl = 43 Kg

M2= mass of Sled = 7.1 Kg

Force F= 6.6 N

To Find (a) a1 = girl acceleration  (b) a2 = sled acceleration

Solution:

F1 = F2 = F   (Newton's 3rd Law)

F=ma (Newton's 2nd Law)

(a) F = M1a1

⇒a1 = F/M1 = 6.6 N / 43 Kg = 0.15 m/s²

(b) a2 = F / M2 = 6.6 N / 7.1 Kg =  0.93 m/s²

6 0
3 years ago
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