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Tresset [83]
3 years ago
12

Convection currents produce the heat in the Earth’s interior.

Physics
2 answers:
DaniilM [7]3 years ago
8 0

Answer:

False

Explanation:

Convection currents do not produce heat. In fact, convection current are a method of transfer of heat, not of production. Convection occurs when there is a fluid which is heated from bottom, from an external source of heat (such as a pot of boiling water over a flame): the bottom part of the fluid becomes warmer, and so less dense than the colder part, therefore it starts moving up, and it is replaced by the colder parts of the fluid, which go down. Later, these colder parts become warmer, so they start going up, being replaced by new colder parts, etc... in a cycle. This is known as convection current, but it requires an external source of heat, it does not produce heat by itself.

Nookie1986 [14]3 years ago
3 0

Answer: false

Explanation:   Convection currents do not produce heat.

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A positively charged particle of mass 7.2 x 10-8 kg is traveling due east with a speed of 88 m/s and enters a 0.6-T uniform magn
Marianna [84]

Answer:

q = 8.57 10⁻⁵ mC

Explanation:

For this exercise let's use Newton's second law

         F = ma

where force is magnetic force

        F = q v x B

the bold are vectors, if we write the module of this expression we have

         F = qv B sin θ

as the particle moves perpendicular to the field, the angle is θ= 90º

        F = q vB

the acceleration of the particle is centripetal

        a = v² / r

we substitute

        qvB = m v² / r

         qBr = m v

          q =\frac{m\  v}{B\  r}

The exercise indicates the time it takes in the route that is carried out with constant speed, therefore we can use

          v = d / t

the distance is ¼ of the circle,

          d = \frac{1}{4} \  2\pi  r

           d =\frac{\pi }{2r}

we substitute

           v =  \frac{\pi  r}{2t}

           r = \frac{2 \ t  \ v}{\pi }

           

let's calculate

           r =\frac{2 \ 2.2  \ 10^{-3} \ 88}{\pi } 2 2.2 10-3 88 /πpi

           r = 123.25 m

         

let's substitute the values

           q = \frac{ 7.2 \ 10^{-8} \ 88}{ 0.6 \ 123.25}7.2 10-8 88 / 0.6 123.25

            q = 8.57 10⁻⁸ C

Let's reduce to mC

           q = 8.57 10⁻⁸ C (10³ mC / 1C)

           q = 8.57 10⁻⁵ mC

4 0
3 years ago
Suppose a small planet is discovered that is 16 times as far from the Sun as the Earth's distance is from the Sun. Use Kepler's
mamaluj [8]

Answer:

23376 days

Explanation:

The problem can be solved using Kepler's third law of planetary motion which states that the square of the period T of a planet round the sun is directly proportional to the cube of its mean distance R from the sun.

T^2\alpha R^3\\T^2=kR^3.......................(1)

where k is a constant.

From equation (1) we can deduce that the ratio of the square of the period of a planet to the cube of its mean distance from the sun is a constant.

\frac{T^2}{R^3}=k.......................(2)

Let the orbital period of the earth be T_e and its mean distance of from the sun be R_e.

Also let the orbital period of the planet be T_p and its mean distance from the sun be R_p.

Equation (2) therefore implies the following;

\frac{T_e^2}{R_e^3}=\frac{T_p^2}{R_p^3}....................(3)

We make the period of the planet T_p the subject of formula as follows;

T_p^2=\frac{T_e^2R_p^3}{R_e^3}\\T_p=\sqrt{\frac{T_e^2R_p^3}{R_e^3}\\}................(4)

But recall that from the problem stated, the mean distance of the planet from the sun is 16 times that of the earth, so therefore

R_p=16R_e...............(5)

Substituting equation (5) into (4), we obtain the following;

T_p=\sqrt{\frac{T_e^2(16R_e)^3}{(R_e^3}\\}\\T_p=\sqrt{\frac{T_e^24096R_e^3}{R_e^3}\\}

R_e^3 cancels out and we are left with the following;

T_p=\sqrt{4096T_e^2}\\T_p=64T_e..............(6)

Recall that the orbital period of the earth is about 365.25 days, hence;

T_p=64*365.25\\T_p=23376days

4 0
3 years ago
A wave has a frequency of 35 Hz and a wavelength of 15 meters,<br> what is the speed of the wave?
Mila [183]

Answer:

f lamda = c

Explanation:

525 m/s is the speed

5 0
2 years ago
If a dog walks north for 10 meters and then east for 10 meters, what is the direction of its displacement?
Katyanochek1 [597]

The direction of its displacement wil be

c.northeast

In fact, the dog walks north for 10 meters and east for another 10 meters. The path of the dog can be represented with two vectors, A pointing north (of magnitude 10 meters) and B pointing east (of magnitude 10 meters). The direction of the resultant vector (due to east) will be given by

tan \theta =\frac{A}{B}=\frac{10}{10}=1

\theta=tan^{-1} (1)=45^{\circ}

and the direction will be north-east.

5 0
3 years ago
Work occurs when
andrew-mc [135]
The answer is the FIRST OPTION 
Work occurs when a force is applied to an object and the object moves in the direction of the force applied <span />
6 0
3 years ago
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