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Goryan [66]
2 years ago
15

What is the central idea for stacking rocks vertically to test the effects of forces

Physics
1 answer:
otez555 [7]2 years ago
8 0

Answer:

You can see how forces can stack on one another by comparing the normal force of the bottom most rock to the normal force of the upper most rock.

Explanation:

Say you have 3 rocks all with mass m=3kg

These 3 rocks will all have the gravitational force of -29.4N (multiply 3kg by -9.8m/s^2)

If none of the rocks are stacked, then they all have the same normal force as well (29.4N)

However, stack them, and see how their normal forces change.

m1 will continue to have a normal force of 29.4N

m2 will now have a normal force of 58.8

m3 will now have a normal force of 88.2

Though they all retain the same gravitational force, their normal force changes depending on the forces acting down on them.

Mass 1 have no forces acting downwards other than gravity.

Mass 2 has the force of mass 1 and gravity acting down on it.

Mass 3 has the force of mass 1, mass 2, and gravity acting down on it.

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The answer is B
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A 50 Ohm resistance causes a current of 5 milliamps to flow through a circuit connected to a battery. What is the power in the c
UkoKoshka [18]

Answer:0.00125 watts

Explanation:

resistance=50 ohms

Current=5 milliamps

Current=5/1000 milliamps

Current =0.005 amps

power=(current)^2 x (resistance)

Power=(0.005)^2 x 50

Power=0.005 x 0.005 x 50

Power=0.00125 watts

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3 years ago
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Find the equivalent resistance, current, and voltage across each resistor when the specified resistors are connected across a 20
timama [110]

Answer:

Explanation:

The question is incomplete. Here is the complete question.

"Find the equivalent resistance, the current supplied by the battery and the current through each resistor when the specified resistors are connected across a 20-V battery. Part (a) uses two resistors with resistance values that can be set with the animation sliders, and you can use the animation to verify your calculation. In part (b), three resistors are specified. (a) Two resistors are connected in series across a 20-V battery. Let R1 = 1 Ω and R2 = 2 Ω. Rea = (b) Add a third resistor to the circuit in series. Let R1 = 1 Ω, R2 = 2 Ω, and R3 = 3 Ω"

Using ohms law formula to solve the problem

E = IRt

E is the supply voltage

I is the total current

Rt is the total equivalent resistant.

a) Given two resistances

R1 = 1ohms and R2 = 2ohms

If the resistors are Connected in series across a 20V supply voltage,

-Equivalent resistance = R1+R2

= 1ohms + 2ohms

= 3ohms

- In a series connected circuit, same current flows through the resistors.

Using the formula E = IRt

I = E/Rt

I = 20/3

I = 6.67A

The current in both resistors is 6.67A

- Different voltage flows across a series connected circuit.

Using the formula V = IR

V is the voltage across each resistor

I is the current in each resistor

For 1ohms resistor,

V = 6.67×1

V = 6.67Volts

For 2ohms resistor

V = 6.67×2

V = 13.34Volts

b) If the resistors are three

R1 = 1ohms, R2 = 2ohms R3 = 3ohms

- Total equivalent resistance = 1+2+3

= 6ohms

- Current in each resistor I = E/Rt

I = 20/6

I = 3.33A

Since the same current flows through the resistors, the current across each of them is 3.33A

- Voltage across them is calculated as shown:

V = IR

For 1ohm resistor

V = 3.33×1

V = 3.33volts

For 2ohms resistor

V = 3.33×2

V = 6.66volts

For 3ohms resistor

V = 3.33×3

V = 9.99volts

3 0
3 years ago
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Only 35 % of the intensity of a polarized light wave passes through a polarizing filter. What is the angle between the electric
Nana76 [90]

Answer:

The angle between the electric field and the axis of the filter is 54⁰

Explanation:

Apply the equation for intensity of light through a polarizer.

I = I_oCos^2 \theta

where;

I is the intensity of the transmitted light

I₀ is the intensity of the incident light

θ is the incident angle

If only 35 % of the intensity of a polarized light wave passes through a polarizing filter, then the ratio of the intensity of the transmitted light to that of the intensity of the incident light is given by;

\frac{I}{I_o}  = Cos^2 \theta\\\\\frac{35}{100} =  Cos^2 \theta\\\\Cos^2 \theta = 0.35\\\\Cos\theta = \sqrt{0.35} \\\\Cos\theta = 0.5916\\\\\theta = Cos^{-1}(0.5916)\\\\\theta  = 54 ^0

Therefore, the angle between the electric field and the axis of the filter is 54⁰

3 0
3 years ago
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