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Alchen [17]
3 years ago
12

What is the medium the energy in the slinky wave travels through?

Physics
2 answers:
vlabodo [156]3 years ago
6 0

Answer:

The metal coil of the slinky

Explanation:

The wave travels on the actual slinky which we see deforming as the wave goes through.

Sergio [31]3 years ago
4 0

Answer:

A.) The metal coil of the slinky

Explanation:

Hello! This is the correct answer! Have a blessed day! :)

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Define the strip attached to the magnet?​
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Answer:

A stripe of magnetic information that is affixed to the back of a plastic credit or debit card.

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Everyone experiences a wide range of emotions, but when could they indicate a mental disorder?
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Answer:

Excessive paranoia, worry, or anxiety.

Long-lasting sadness or irritability.

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Explanation:

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7 0
3 years ago
A 60-W, 120-V light bulb and a 200-W, 120-V light bulb are connected in series across a 240-V line. Assume that the resistance o
gulaghasi [49]

A. 0.77 A

Using the relationship:

P=\frac{V^2}{R}

where P is the power, V is the voltage, and R the resistance, we can find the resistance of each bulb.

For the first light bulb, P = 60 W and V = 120 V, so the resistance is

R_1=\frac{V^2}{P}=\frac{(120 V)^2}{60 W}=240 \Omega

For the second light bulb, P = 200 W and V = 120 V, so the resistance is

R_1=\frac{V^2}{P}=\frac{(120 V)^2}{200 W}=72 \Omega

The two light bulbs are connected in series, so their equivalent resistance is

R=R_1 + R_2 = 240 \Omega + 72 \Omega =312 \Omega

The two light bulbs are connected to a voltage of

V  = 240 V

So we can find the current through the two bulbs by using Ohm's law:

I=\frac{V}{R}=\frac{240 V}{312 \Omega}=0.77 A

B. 142.3 W

The power dissipated in the first bulb is given by:

P_1=I^2 R_1

where

I = 0.77 A is the current

R_1 = 240 \Omega is the resistance of the bulb

Substituting numbers, we get

P_1 = (0.77 A)^2 (240 \Omega)=142.3 W

C. 42.7 W

The power dissipated in the second bulb is given by:

P_2=I^2 R_2

where

I = 0.77 A is the current

R_2 = 72 \Omega is the resistance of the bulb

Substituting numbers, we get

P_2 = (0.77 A)^2 (72 \Omega)=42.7 W

D. The 60-W bulb burns out very quickly

The power dissipated by the resistance of each light bulb is equal to:

P=\frac{E}{t}

where

E is the amount of energy dissipated

t is the time interval

From part B and C we see that the 60 W bulb dissipates more power (142.3 W) than the 200-W bulb (42.7 W). This means that the first bulb dissipates energy faster than the second bulb, so it also burns out faster.

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3 years ago
Which changes in an electric motor will make the motor stronger? select 3 options. using a stronger permanent magnet using a wea
Masja [62]

1. Using Strong Permanent. 2. increasing the current. 3. Decreasing the space between Magnets

Explanation:

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Which shows the order of mechanical advantage from least to greatest?
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