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tamaranim1 [39]
2 years ago
7

In your own words describe the motion of convection currents?

Physics
2 answers:
Marysya12 [62]2 years ago
5 0
Convection is the circular motion that occurs as hotter air or liquid increases when the cooler air or liquid drops down, and has faster moving molecules, rendering it less dense. Convection currents within the earth shift layers of magma, and currents are formed by convection in the ocean.
Black_prince [1.1K]2 years ago
5 0

Heat energy can transfer by convection when there is a significant difference in temperature between two parts of a fluid. When this temperature difference exists, hot fluids rise and cold fluids sink, and then currents, or movements, are created in the fluid.
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Weight is measured in: grams, kilograms, metric tons, or pounds
sergejj [24]
'Gram', 'kilogram', and 'metric ton' are units of mass, not weight.

'Pound' can be either a unit of mass or of force.  Most people
use it most of the time to mean weight.
8 0
2 years ago
If a 4 engine jet accelerates down a runway at 8.7 m/s^2, Suppose now that all 4 engins are operational on the jet from the prev
soldi70 [24.7K]

Answer:

5.22m/s^2

Explanation:

One of the first propulsion characteristics given in the example is that all engines are equal.

In this way if we have 4 engines running at the same time, it means that its capacity is 100%.

Under this premise, if 100% is found, the Jet is capable of reaching a speed of 8.7m / s ^ 2.

However, the question is, what would happen if 2.4 "Engines" now work.

To do this then we make a simple equivalence,

If 4 engines is the equivalent of 100%, when would it be 2.4 engines?

X = \frac{2.4 * 100\%} { 4 }= 60\%

In this way it would mean that the body could be driven to 60% of its total.

So

Speed_{Decreased} = 8.7 * 60\% = 5.22 \frac{m}{s^2}

3 0
2 years ago
Astronaut Rob leaves Earth in a spaceship at a speed of 0.960crelative to an observer onEarth. Rob's destination is a star syste
Slav-nsk [51]

To solve this problem we will use the mathematical definition of the light years in metric terms, from there, through the kinematic equations of motion we will find the distance traveled as a function of the speed in proportion to the elapsed time. Therefore we have to

1Ly =9.4605284*10^{15}m \rightarrow 'Ly'means  Light Year

Then

14.4Ly = 1.36231609*10^{17} m

If we have that

v= \frac{x}{t} \rightarrow t = \frac{x}{t}

Where,

v = Velocity

x = Displacement

t = Time

We have that

t = \frac{1.36231609*10^{17}}{0.96c} \rightarrow c= Speed of light

t = \frac{1.36231609*10^{17}}{0.96(3*10^8)}

t= 454105363 s (\frac{1hour}{3600s})

t= 126140 hours(\frac{1day}{24hours})

t= 5255.85 days(\frac{1 year}{365days})

t = 14.399 years

Therefore will take 14.399 years

5 0
3 years ago
In the diagram, q1= -2.60*10^-9 C and
Alekssandra [29.7K]

Answer:

The magnitude of the net electric field is:

E_{net}=90.37\: N/c

Explanation:

The electric field due to q1 is a vertical positive vector toward q1 (we will call it E1).

On the other hand, the electric field due to q2 is a horizontal positive vector toward q2(We will call it E2).

Knowing this, the <u>magnitude of the net electric</u> field will be the<u> E1 + E2. </u>

Let's find first E1 and E2.

The electric field equation is given by:

|E_{1}|=k\frac{|q_{1}|}{d_{1}^{2}}

Where:

  • k is the Coulomb constant (k = 9*10^{9} Nm²/C²)
  • q1 is the first charge
  • d1 is the distance from q1 to P

|E_{1}|=(9*10^{9})\frac{|-2.60*10^{-9}|}{0.538^{2}}

|E_{1}|=80.84\: N/C

And E2 will be:

|E_{2}|=k\frac{|q_{2}|}{d_{2}{2}}

|E_{2}|=(9*10^{9})\frac{|-8.30*10^{-9}|}{1.36^{2}}

|E_{2}|=40.39\: N/C

Finally, we need to use the  Pythagoras theorem to find the magnitude of the net electric field.

E_{net}=\sqrt{E_{1}^{2}+E_{2}^{2}}

E_{net}=\sqrt{80.84^{2}+40.39^{2}}

E_{net}=90.37\: N/c

I hope it helps you!

7 0
3 years ago
What would Hubble's constant be if we found one galaxy moving away at 30,000 km/s at a distance of 600 Mpc?
lora16 [44]

Answer:

H₀ = 1.6 x 10⁻¹⁸ s⁻¹

Explanation:

The Hubble's Constant can be found by the following formula:

v = H_o D\\\\H_o = \frac{v}{D}

where,

H₀ = Hubble's Constant = ?

v = speed of galaxy = 30000 km/s = 3 x 10⁷ m/s

D = Distacance = 600 Mpc = (6 x 10⁸ pc)(3.086 x 10¹⁶ m/1 pc)

D = 18.52 x 10²⁴ m

Therefore,

H_o = \frac{3\ x\ 10^7\ m/s}{18.52\ x\ 10^{24}\ m}

<u>H₀ = 1.6 x 10⁻¹⁸ s⁻¹</u>

3 0
2 years ago
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