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3 years ago

Figure 1 shows a wave movement during one second. What is the frequency of the wave

Physics

2 answers:

AURORKA [14]3 years ago

8 0

2 hertz, as explained previously- I'm afraid m/s is a completely different measurement

Andreyy893 years ago

4 0

This is 2 hertz. You can mark out 2 full wavelengths in the second of time.

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Observations of galaxies and clusters of galaxies indicate that about ____ percent of the matter in the universe is dark matter.

Setler79 [48]

All the stars, spheres and galaxies that can be perceived nowadays make up just 5 percent of the universe. The former 95 percent is prepared of stuff stargazers can't see, notice or even understand. These secretive substances are called dark energy and dark matter. Experts determine their presence grounded on their gravitational influence on what little bits of the universe can be perceived.

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2 years ago

How can we balance the use of fossil fuels with other forms of energy?

KiRa [710]

We can use renewable sources

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3 years ago

Calculate the force of gravity on a 1–kilogram box located at a point 1.3 × 107 meters from the center of Earth if the force on

Sati [7]

Since you already gave us the weight of the 2.5-kg box,
we don't even need to know what the distance is, just
as long as it doesn't change.

Look at the formula for the gravitational force:

F = G m₁ m₂ / R² .

If 'G', 'm₁' (mass of the Earth), and 'R' (distance from the Earth's center)
don't change, then the Force is proportional to m₂ ... mass of the box,
and you can write a simple proportion:

(6.1 N) / (2.5 kg) = (F) / (1 kg)

Cross-multiply: (6.1 N) (1 kg) = (F) (2.5 kg)

Divide each side by (2.5 kg): F = (6.1N) x (1 kg) / (2.5 kg) = 2.44 N .

5 0

3 years ago

The pressure exerted by a gas is 2.0 atm while it has a volume of 350 mL. What would be the volume of this sample of gas at stan

34kurt

Answer:

700 mL or 0.0007 m³

Explanation:

P₁ = Initial pressure = 2 atm

V₁ = Initial volume = 350 mL

P₂ = Final pressure = 1 atm

V₂ = Final volume

Here the temperature remains constant. So, Boyle's law can be applied here.

P₁V₁ = P₂V₂

$\frac{P_1V_1}{P_2}=V_2\\\Rightarrow V_2=\frac{2\times 350}{1}\\\Rightarrow V_2=700\ mL$

So, volume of this sample of gas at standard atmospheric pressure would be 700 mL or 0.0007 m³

7 0

3 years ago

Two charged particles are located on the x axis. The first is a charge 1Q at x 5 2a. The second is an unknown charge located at

sergejj [24]

Answer:

Q_2 = +/- 295.75*Q

Explanation:

Given:

- The charge of the first particle Q_1 = +Q

- The second charge = Q_2

- The position of first charge x_1 = 2a

- The position of the second charge x_2 = 13a

- The net Electric Field produced at origin is E_net = 2kQ / a^2

Find:

Explain how many values are possible for the unknown charge and find the possible values.

Solution:

- The Electric Field due to a charge is given by:

E = k*Q / r^2

Where, k: Coulomb's Constant

Q: The charge of particle

r: The distance from source

- The Electric Field due to charge 1:

E_1 = k*Q_1 / r^2

E_1 = k*Q / (2*a)^2

E_1 = k*Q / 4*a^2

- The Electric Field due to charge 2:

E_2 = k*Q_2 / r^2

E_2 = k*Q_2 / (13*a)^2

E_2 = +/- k*Q_2 / 169*a^2

- The two possible values of charge Q_2 can either be + or -. The Net Electric Field can be given as:

E_net = E_1 + E_2

2kQ / a^2 = k*Q_1 / 4*a^2 +/- k*Q_2 / 169*a^2

- The two equations are as follows:

1: 2kQ / a^2 = k*Q / 4*a^2 + k*Q_2 / 169*a^2

2Q = Q / 4 + Q_2 / 169

Q_2 = 295.75*Q

2: 2kQ / a^2 = k*Q / 4*a^2 - k*Q_2 / 169*a^2

2Q = Q / 4 - Q_2 / 169

Q_2 = -295.75*Q

- The two possible values corresponds to positive and negative charge Q_2.

7 0

2 years ago