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scoray [572]
3 years ago
11

Match each method of transferring electric charge with the correct description

Physics
2 answers:
Mazyrski [523]3 years ago
7 0

Answer:

Friction-1

Conduction-2

Induction-3

Explanation:

Just trust me, bro

babunello [35]3 years ago
4 0

Answer:

FRICTION  OPT 2

INDUCTION OPT 3

CONDUCTION OPT 1

Explanation:

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Can I also get help on this??
Xelga [282]

Answer:

25

Explanation:

8 0
2 years ago
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Raj is trying to make a diagram to show what he has learned about nuclear fusion.
KIM [24]

No, he should place the He atom and energy on the right, and the H atoms and the heat and energy on the left.

6 0
3 years ago
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Tamsen is interested in history, and read that because of its regular period, the pendulum constituted the basis of the most acc
geniusboy [140]

We will apply the concept of period in a pendulum, defined as the product between 2\pi by the square root of the length over gravity, this is mathematically

T = 2\pi \sqrt{\frac{L}{g}}

Here,

T = Period

L = Length

g = Acceleration due to gravity

For the period to be 1 second, then we must look for the necessary length for such a requirement so

1 = 2\pi \sqrt{\frac{L}{9.8}}

(\frac{1}{2\pi})^2 = \frac{L}{9.8}

L = 9.8(\frac{1}{2\pi})^2

L = 0.2482m

The meter's length would be slight less than one-fourth of its current length. Also, the number of significant digits depends only on how precisely we know g, because the time has been defined to be exactly 1s.

Therefore the correct answer is C.

3 0
3 years ago
Tubby and his twin brother Libby have a combined mass of 200 kg and are zooming along in a 100 kg amusement park bumper car at 1
harkovskaia [24]

Answer: 14.1 m/s

Explanation:

We can solve this with the Conservation of Linear Momentum principle, which states the initial momentum p_{i} (before the elastic collision) must be equal to the final momentum p_{f} (after the elastic collision):

p_{i}=p_{f} (1)

Being:

p_{i}=m_{1}V_{i} + m_{2}U_{i}

p_{f}=m_{1}V_{f} + m_{2}U_{f}

Where:

m_{1}=200 kg +100 kg=300 kg is the combined mass of Tubby and Libby with the car

V_{i}=10 m/s is the velocity of Tubby and Libby with the car before the collision

m_{2}=25 kg + 100 kg=125 kg is the combined mass of Flubby with its car

U_{i}=0 m/s is the velocity of Flubby with the car before the collision

V_{f}=4.12 m/s is the velocity of Tubby and Libby with the car after the collision

U_{f} is the velocity of Flubby with the car after the collision

So, we have the following:

m_{1}V_{i} + m_{2}U_{i}=m_{1}V_{f} + m_{2}U_{f} (2)

Finding U_{f}:

U_{f}=\frac{m_{1}(V_{i}-V_{f})}{m_{2}} (3)

U_{f}=\frac{300 kg(10 m/s-4.12 m/s)}{125 kg} (4)

Finally:

U_{f}=14.1 m/s

8 0
3 years ago
A "spherical capacitor" is constructed of two thin, concentric spherical shells of conducting material. Let a be the radius of t
Shalnov [3]

Answer:

C=\frac{ab}{k(b-a)}

Explanation:

We can assume this problem as two concentric spherical metals with opposite charges.

We have also to take into account the formulas for the electric field and the capacitance. Hence we have

C=\frac{Q}{V}\\\\E=k\frac{Q}{r^2}\\

Where k is the Coulomb's constant. Furthermore, by taking into account the expression for the potential and by integrating

dV=Edr\\\\V=\int_{R_1}^{R_2}Edr=-\int_{R_1}^{R_2}\frac{kQ}{r^2}dr\\\\V=kQ[\frac{1}{R_2}-\frac{1}{R_1}]

Hence, the capacitance is

C=\frac{1}{k[\frac{1}{R_2}-\frac{1}{R_1}]}

but R1=a and R2=b

C=\frac{ab}{k(b-a)}

HOPE THIS HELPS!!

3 0
3 years ago
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