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What is the speed of a wave that has a frequency of 45 Hz and a wavelength of 0.1 meters?

Yuri [45]

<span>λν=c
(wavelength x frequency = speed)

speed = 45 x 0.1
= 4.5 m/s</span>

6 0

3 years ago

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Question 10 (1 point)

torisob [31]

Answer:

It does not hit the students face because the speed of the balloon slows down as energy is lost through thermal.

Explanation:

3 0

2 years ago

Mrs. Paul and Dr. Mykannen ride a tandem bike on the beach. They ride along the beach for about 455 meters. Their final velocity

algol [13]

Answer:

Time = 80.91 seconds

Explanation:

Given the following data;

Velocity = 5.50 m/s.

Distance = 445 meters

To find the time;

Velocity can be defined as the rate of change in displacement (distance) with time. Velocity is a vector quantity and as such it has both magnitude and direction.

Mathematically, velocity is given by the equation;

$Velocity = \frac{distance}{time}$

Substituting into the formula, we have;

5.5 = 445/time

Time = 445/5.5

Time = 80.91 seconds

5 0

2 years ago

3. Jeremy has a solution with a mass of 228 g. He knows that the mass of the solvent was 224 g. How much is the mass of the solu

mariarad [96]

228 - 224 = 4

there is 4g of solute in the solution.

3 0

2 years ago

A uniform marble rolls down a symmetrical bowl, start- ing from rest at the top of the left side. The top of each side is a dist

Paha777 [63]

Answer:

Part a)

$h' = \frac{10}{14} h$

Part b)

if both sides are rough then it will reach the same height on the other side because the energy is being conserved.

Part c)

Since marble will go to same height when it is rough while when it is smooth then it will go to the height

$h' = \frac{10}{14} h$

so on smooth it will go to lower height

Explanation:

As we know by energy conservation the total energy at the bottom of the bowl is given as

$\frac{1}{2} mv^2 + \frac{1}{2}I\omega^2 = mgh$

here we know that on the left side the ball is rolling due to which it is having rotational and transnational both kinetic energy

now on the right side of the bowl there is no friction

so its rotational kinetic energy will not change and remains the same

so it will have

$\frac{1}{2}mv^2 = mgh'$

now we know that

$I = \frac{2}{5}mr^2$

$\omega = \frac{v}{r}$

so we have

$\frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mr^2)(\frac{v}{r})^2 = mgh$

$\frac{1}{2}mv^2 + \frac{1}{5}mv^2 = mgh$

$\frac{7}{10}mv^2 = mgh$

$\frac{1}{2}mv^2 = \frac{10}{14}mgh$

so the height on the smooth side is given as

$h' = \frac{10}{14} h$

Part b)

if both sides are rough then it will reach the same height on the other side because the energy is being conserved.

Part c)

Since marble will go to same height when it is rough while when it is smooth then it will go to the height

$h' = \frac{10}{14} h$

so on smooth it will go to lower height

6 0

3 years ago

Read 2 more answers

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