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3 years ago

12

Wave A has a high frequency and a short wavelength. Wave B has a lower frequency and a longer wavelength. Which wave carries gre

ater energy?

1 answer:

dusya [7]3 years ago

4 0

Answer:

wave A

Explanation:

shortest wavelength carry the most energy

more energy in a wave, the higher its frequency. The lower the frequency is, the less energy in the wave.

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How many nutrons are in an atom of copper

xeze [42]

Answer:

35 neutrons are in an atom of copper

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3 years ago

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A bicycle is heading west. It goes 500m in 5 minutes. What is its velocity?

icang [17]

Speed = 500m/5min = 100 m/min. Direction = west. Velocity = 100 m/min west.

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3 years ago

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A kayaker needs to paddle north across a 100-m-wide harbor. The tide is going out, creating a tidal current that flows to the ea

Savatey [412]

Answer:

$41.81^{\circ}$

Explanation:

The tidal current flows to the east at 2.0 m/s and the speed of the kayaker is 3.0 m/s.

Let Vector $\overrightarrow{OA}$ is the tidal current velocity as shown in the diagram.

In order to travel straight across the harbor, the vector addition of both the velocities (i.e the resultant velocity, $\vec {R}$ must be in the north direction.

Let $\overrightarrow{AB}$ is the speed of the kayaker having angle \theta measured north of east as shown in the figure.

For the resultant velocity in the north direction, the tail of the vector $\overrightarrow {OA}$ and head of the vector $\overrightarrow{AB}$ must lie on the north-south line.

Now, for this condition, from the triangle OAB

$|\overrightarrow{AB}|\sin \theta=|\overrightarrow{OA}|$

$\Rightarrow \sin\theta=\frac{|\overrightarrow{OA}|}{|\overrightarrow{AB}|}=\frac 2 3$

$\Rightarrow \theta=\sin^{-1}\frac23$

$\Rightarrow \theta=41.81^{\circ}$

Hence, the kayaker must paddle in the direction of $41.81^{\circ}$ in the north of east direction.

3 0

4 years ago

Please help on this one?

storchak [24]

C) Physical Model I think
BUT because it’s in a computer it may be a computer model

4 0

4 years ago

A magnetic field perpendicular to the plane of a wire loop with an area of 0.40 m^2 decreases by 0.2 T in 10^-3 sec. What is the

SVETLANKA909090 [29]

Answer:

Induced EMF in the loop is 80 volts.

Explanation:

Given that,

Magnetic field, $\Delta B=0.2\ T$

Time, $\Delta t=10^{-3}\ s$

Area of the loop, $A=0.4\ m^2$

We need to find the magnitude of the average value of the emf induced in the loop. Due to change in magnetic field, an emf is induced in the loop which is given by :

$\epsilon_o=\dfrac{\Delta\phi}{dt}$

$\epsilon=A\dfrac{\Delta B}{\Delta t}$

$\epsilon=0.4\times \dfrac{0.2}{10^{-3}}$

$\epsilon=80\ V$

So, the value of induced emf is 80 V. Hence, this is the required solution.

5 0

3 years ago