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3 years ago

12

What is the relationship between speed, frequency, and wavelength?

2 answers:

jonny [76]3 years ago

6 0

I think it’s B speed equals wavelength times frequency

Nutka1998 [239]3 years ago

5 0

Answer:

B. Speed = Wavelength x Frequency.Explanation:

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Shareen finds that when she drives her motorboat upstream she can travelwith a speed of only 8 m/s, while she moves with a speed

ANEK [815]

Let vb be the velocity of the motorboat and let vs be the velocity of the stream.

We know that when she drives upstream the velocity is 8 m/s, in this scenario the velocities point in opposite directions, then we have the equations:

$v_b-v_s=8$

When she drives downstream the velocites point in the same direction then we have the equation:

$v_b+v_s=12$

hence we have the system of equations:

$\begin{gathered} v_b-v_s=8 \\ v_b+v_s=12 \end{gathered}$

Solving the first equation for the velocity of the boat we have:

$v_b=8+v_s$

Plugging this in the second equation we have:

$\begin{gathered} 8+v_s+v_s=12 \\ 2v_s=4 \\ v_s=\frac{4}{2} \\ v_s=2 \end{gathered}$

Therefore, the velocity of the stream is 2 m/s

5 0

1 year ago

You see a lightning bolt flash in the distant sky. You hear the corresponding thunder 6.8 seconds later. If the temperature is 2

Anna35 [415]

Answer:

$Distance = 2.4\ km$

Explanation:

Given

$Time (t) = 6.8s$

$Temperature = 28C$

Required

Determine the distance at which the lighting struck

First, we need to determine the speed at which the lighting struck because the peed of sound varies with temperature.

At about 28C, the speed of sound is 346m/s

So, we have the following:

$Speed = 346m/s$

$Time = 6.8s$

Distance is calculated as thus:

$Distance = Speed * Time$

$Distance = 346m/s * 6.8s$

$Distance = 2352.8$

Divide by 1000 to get distance equivalent in kilometers

$Distance = \frac{2352.8km}{1000}$

$Distance = 2.3528km$

$Distance = 2.4\ km$ ---- Approximated

7 0

3 years ago

What is the Density of the rock?

Sindrei [870]

Your answer is gone be 3

4 0

3 years ago

Two 0.40 kg soccer ball collide elastically in a head-on collision. The first ball starts at rest, and the second ball has a spe

yulyashka [42]

Explanation:

Mass of two soccer balls, $m_1=m_2=0.4\ kg$

Initial speed of first ball, $u_1=0$

Initial speed of second ball, $u_2=3.5\ m/s$

After the collision,

Final speed of the second ball, $v_2=0$

(a) The momentum remains conserved. Using the conservation of momentum to find it as :

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

$v_1$ is the final speed of the first ball

$0.4\times 0+0.4\times 3.5=0.4v_1+0.4\times 0$

$0.4\times 3.5=0.4v_1$

$v_1=3.5\ m/s$

(b) Let $E_1$ is the kinetic energy of the first ball before the collision. It is given by :

$E_1=\dfrac{1}{2}mu_1^2$

$E_1=\dfrac{1}{2}\times 0.4\times 0$

It is at rest, so, the kinetic energy of the first ball before the collision is 0.

(c) After the collision, the second ball comes to rest. So, the kinetic energy of the second ball after the collision is 0.

Hence, this is the required solution.

7 0

3 years ago

Read 2 more answers

What is a cluster?

NARA [144]

A cluster is D.) Portion of space containing many galaxies.

Hope this helps. Chao.

7 0

3 years ago