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3 years ago

What do sound waves, light waves, and ocean waves all have in common?

2 answers:

6 0

I don't want to scare you, but I suggest that we look at the choices
and THINK about them:

A. No. Light waves don't need a medium.

B. No. Sound waves and ocean waves are not electromagnetic waves.

C. Yes! All waves move energy from one place to another.

D. No. Light waves are not mechyanical waves.

vekshin13 years ago

4 0

C. They all transfer energy

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What is the difference between a contact force and a non-contact force?

Mkey [24]

Hi, Illiana is correct :)

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3 years ago

Which statement is true about acceleration?

Ilia_Sergeevich [38]

The answer is C is think

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3 years ago

Two long, straight wires are parallel and 10 cm apart.One carries a current of 2.0 A, the other a current of 5.0A. (a) If the tw

lubasha [3.4K]

Answer:

Explanation:

Given

Distance between two wires $d=10 cm$

Current value $i_1=2 A$

$i_2=5 A$

(a)If current flows in opposite direction

When current Flows in opposite direction the two wires will repel each other

Force due to current carrying wire $F=\frac{\mu _0i_1i_2}{2\pi r}$

$F_{12}=F_{21}=\frac{\mu _0\times 2\times 5}{2\times \pi \times 0.1}$

$F_{12}=\frac{4\pi \times 10^{-7}\times 2\times 5}{2\pi \times 0.1}$

$F_{12}=2\times 10^{-5} N$ (Repulsive Force)

(b)If current is flowing in the same direction

When direction of current is same then force will be attractive in nature

$F_{12}=F_{21}=\frac{\mu _0i_1i_2}{2\pi r}$

$F_{12}=\frac{4\pi \times 10^{-7}\times 2\times 5}{2\pi \times 0.1}$

$F_{12}=2\times 10^{-5} N$ (attractive Force)

5 0

4 years ago

Sound waves are mechanical waves in which the particles in the medium vibrate in a direction parallel to the direction of energy

mel-nik [20]

<span>longitudinal waves

Hope this helps!</span>

6 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.

7 0

3 years ago