What is the wave speed in this equation <br><br><br> Wave speed = frequency x wavelength

Hitting a home run in baseball is difficult, because it requires a lot of force. Most home runs are hit off of fastballs. Which

OLga [1]

The impact force makes sense because the impulse experienced by the body cause a great change in momentum of the body.

<h3 /><h3>What is impulse?</h3>

This is the force that acts on a body over a given period of time. The impulse experienced by a body is determined as the product of force and time of action.

J = Ft

The change in momentum of a body is equal to the impulse experienced by the body.

ΔP = Ft

Thus, the impact force makes sense because the impulse experienced by the body cause a great change in momentum of the body.

Learn more about impulse here: brainly.com/question/25700778

7 0

2 years ago

State three precautions taken when using a metre rule

marishachu [46]

Answer:

1. Our eye should be vertically abvove the point, where the measurement is to be taken.

2. For proper measurement, Keep the ruler exactly along the length , for the measurement

Explanation:

good luck

5 0

3 years ago

Which type of path do planets follow around the sun?

Elza [17]

A. ellipse...........

4 0

3 years ago

Read 2 more answers

Use the dichotomous key to classify the wading bird labelled A.

meriva

Obviously C. Search up "White Stork" and it's the same Image

4 0

3 years ago

Read 2 more answers

A positive point charge q1 = +5.00 × 10−4C is held at a fixed position. A small object with mass 4.00×10−3kg and charge q2 = −3.

Lelechka [254]

Answer:

Therefore the speed of q₂ is 1961.19 m/s when it is 0.200 m from from q₁.

Explanation:

Energy conservation law: In isolated system the amount of total energy remains constant.

The types of energy are

Kinetic energy.
Potential energy.

Kinetic energy $=\frac{1}{2} mv^2$

Potential energy = $\frac{Kq_1q_2}{d}$

Here, q₁= +5.00×10⁻⁴C

q₂=-3.00×10⁻⁴C

d= distance = 4.00 m

V = velocity = 800 m/s

Total energy(E) =Kinetic energy+Potential energy

$=\frac{1}{2} mv^2$ + $\frac{Kq_1q_2}{d}$

$=\frac{1}{2} \times 4.00\times 10^{-3}\times(800)^2 +\frac{9\times10^9\times 5\times10^{-4}\times(-3\times10^{-4})}{4}$

=(1280-337.5)J

=942.5 J

Total energy of a system remains constant.

Therefore,

E $=\frac{1}{2} mv^2$ + $\frac{Kq_1q_2}{d}$

$\Rightarrow 942.5 = \frac{1}{2} \times 4 \times10^{-3} \times V^2 +\frac{9\times10^{9}\times5\times 10^{-4}\times(-3\times 10^{-4})}{0.2}$

$\Rightarrow 942.5 = 2\times10^{-3}v^2 -6750$

$\Rightarrow 2 \times10^{-3}\times v^2= 942.5+6750$

$\Rightarrow v^2 = \frac{7692.5}{2\times 10^{-3}}$

$\Rightarrow v= 1961.19$ m/s

Therefore the speed of q₂ is 1961.19 m/s when it is 0.200 m from from q₁.

5 0

3 years ago

What is the wave speed in this equation Wave speed = frequency x wavelength