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Gnoma [55]
3 years ago
9

1. What are the different types of waves? Give examples of each type.

Physics
1 answer:
lyudmila [28]3 years ago
4 0

Answer:

See explanation

Explanation:

1) There are two main types of waves;

Mechanical waves and electromagnetic waves.

Mechanical waves usually require a medium for propagation, e.g sound, waves on a spiral spring, etc

Electromagnetic waves do not require a medium for propagation e.g microwaves, x-rays etc

2) The phenomenon is known as refraction. Refraction occurs at the boundary between two media because of the change in the speed of a wave as it moves from one medium to another.

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What is the relationship between DNA and mutation
Scrat [10]

Answer:

A mutation is a change in DNA, the hereditary material of life. An organism's DNA affects how it looks, how it behaves, and its physiology. So a change in an organism's DNA can cause changes in all aspects of its life. Mutations are essential to evolution; they are the raw material of genetic variation

Explanation:

5 0
3 years ago
Read 2 more answers
Consider identical spherical conducting space ships in deep space where gravitational fields from other bodies are negligible co
Illusion [34]

Answer:

 q = 8.61 10⁻¹¹ m

charge does not depend on the distance between the two ships.

it is a very small charge value so it should be easy to create in each one

Explanation:

In this exercise we have two forces in balance: the electric force and the gravitational force

          F_e -F_g = 0

          F_e = F_g

Since the gravitational force is always attractive, the electric force must be repulsive, which implies that the electric charge in the two ships must be of the same sign.

Let's write Coulomb's law and gravitational attraction

         k \frac{q_1q_2}{r^2} = G \frac{m_1m_2}{r^2}

In the exercise, indicate that the two ships are identical, therefore the masses of the ships are the same and we will place the same charge on each one.

          k q² = G m²

          q = \sqrt{ \frac{G}{k} }    m

we substitute

           q = \sqrt{ \frac{ 6.67 \ 10^{-11}}{8.99 \ 10^{9}} }   m

            q = \sqrt{0.7419 \ 10^{-20}}   m

           q = 0.861 10⁻¹⁰ m

           q = 8.61 10⁻¹¹ m

This amount of charge does not depend on the distance between the two ships.

It is also proportional to the mass of the ships with the proportionality factor found.

Suppose the ships have a mass of m = 1000 kg, let's find the cargo

            q = 8.61 10⁻¹¹ 10³

            q = 8.61 10⁻⁸ C

             

this is a very small charge value so it should be easy to create in each one

6 0
3 years ago
Guys I need the answer ASAP
kati45 [8]

Answer:

cnbzdbhvhndjcn bvhdbvjsdhvjsbdjcbhkavwhe4w7334856743534685347856784687367856732346356675ygafjdbvc

Explanation:

3 0
3 years ago
lightning strikes in the distance and six seconds later Thunder is heard how far away was the lightning strike
marshall27 [118]
The speed of sound really depends on the temperature and moisture in the air,
and the sound of the thunder doesn't necessarily travel straight from the lightning
to where you are.  So we can't say exactly. 

But if we use the nominal speed of 340 m/s, then 6 seconds means 2,040 meters,
or about 6,700 feet, or about 1.27 miles.
8 0
3 years ago
A 10-cm-long thin glass rod uniformly charged to 14.0 nC and a 10-cm-long thin plastic rod uniformly charged to - 14.0 nC are pl
Sonbull [250]

Answer:

A. At a point 1cm from the glass rod, E = 1.36 * 10⁶N/C

B. At a point 2cm from the glass rod, E = 5.17 * 10⁵N/C

C. At a point 3cm from the glass rod, E = 6.993 * 10⁵N/C

Explanation:

Parameters given:

Charge of glass rod, Q = 14nC = 14 * 10⁻⁹ C

Charge of plastic rod, q = 14nC = 14 * 10⁻⁹ C

Distance between both rods = 4.5cm = 0.045

A. Electric field strength at a point 1.0cm (0.01m) from the glass rod is the sum of electric field strength due to both rods i.e.

E = E₁ + E₂

Where

E₁ = electric field strength due to glass rod

E₂ = electric field strength due to plastic rod

E₁ = kQ/0.01²

E₂ = kq/(0.045 - 0.01)² = kq/(0.035)²

E = kQ/0.01² + kq/(0.035)² = k(Q/0001 + q/0.001225)

E = 9 * 10⁹ [(14 * 10⁻⁹ / 0.001) + (14 * 10⁻⁹)/0.001225]

E = 9 * 10⁹[(14 * 10⁻⁵) + (1.143 * 10⁻⁵)]

E = 9 * 10⁹ * 15.143* 10⁻⁵

E = 1.36 * 10⁶N/C

B. Electric field strength at a point 2.0cm (0.02m) from the glass rod is the sum of electric field strength due to both rods i.e.

E = E₁ + E₂

E₁ = kQ/0.02²

E₂ = kq/(0.045 - 0.02)² = kq/(0.025)²

E = kQ/0.02² + kq/(0.025)² = k(Q/0.0004 + q/0.000625)

E = 9 * 10⁹ [(14 * 10⁻⁹ / 0.0004) + (14 * 10⁻⁹)/0.000625]

E = 9 * 10⁹[(3.5 * 10⁻⁵) + (2.24 * 10⁻⁵)]

E = 9 * 10⁹ * 5.74 * 10⁻⁵

E = 5.17 * 10⁵N/C

C. Electric field strength at a point 3.0cm (0.03m) from the glass rod is the sum of electric field strength due to both rods i.e.

E = E₁ + E₂

E₁ = kQ/0.03²

E₂ = kq/(0.045 - 0.03)² = kq/(0.015)²

E = kQ/0.03² + kq/(0.015)² = k(Q/0009 + q/0.000225)

E = 9 * 10⁹ [(14 * 10⁻⁹ / 0.009) + (14 * 10⁻⁹)/0.000225]

E = 9 * 10⁹[( 1.55 * 10⁻⁵) + (6.22 * 10⁻⁵)]

E = 9 * 10⁹ * 7.77 * 10⁻⁵

E = 6.993 * 10⁵N/C

4 0
3 years ago
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