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3 years ago

What causes the movement of deep ocean currents?

Physics

1 answer:

kondaur [170]3 years ago

7 0

The moon formation causes the seas currents

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A 1200 kg sports car accelerates from 0 m/s to 30 m/s in 10 s. What is the average power of the engine?

DIA [1.3K]

Answer:

3600N

Explanation:

Given: m = 1200kg, Vo = 0m/s, Vf = 30m/s, Δt = 10s

ΣF = ma

we need to find 'a' first, using the definition of 'a' we get equation:

a = (Vf-Vo)/Δt

a = (30m/s)/10s

a = 3 m/s^2

now substitute into top equation

ΣF = ma

Fengine = (1200kg)(3m/s^2)

Fengine = 3600N

5 0

2 years ago

Read 2 more answers

……………………………………………………………..?

kotegsom [21]

Answer:

Explanation:

First find the electrical wattage

W = I^2 * R

R = 12 ohms

I = 2 amps

Wattage = 2^2 * 12

Wattage = 4* 12

Wattage = 48 watts.

Now you need to use the power formula

Work = Power * Time

Work = ?

Power = 48 watts

Time = 3 minutes = 3 * 60 = 180 seconds.

Work = 48 * 180

Work = 8640 J

That's C

3 0

2 years ago

At what angle are the electronic and the magnetic wave related in an electromagnetic signal?

VikaD [51]

Answer:

90degrees I'm pretty sure

7 0

2 years ago

Which statement is correct?

Sedbober [7]

Answer:

The answer is "Choice C ".

Explanation:

The relationship between the E and V can be defined as follows:

$\to E= -\Delta V$

Let,

$\to E= \frac{\delta V}{\delta x}$

When E=0

$\to \frac{\delta V}{\delta x}=0$

v is a constant value

Therefore, In the electric potential in a region is a constant value then the electric-field must be into zero that is everywhere in the given region, that's why in this question the "choice c" is correct.

7 0

2 years ago

Assume that the radius ????r of a sphere is expanding at a rate of 70 cm/min.70 cm/min. The volume of a sphere is ????=43???????

rodikova [14]

Answer:

the rate of change in volume with time is 280πr² cm³/min

Explanation:

Data provided in the question:

Radius of the sphere as 'r'

$\frac{d\textup{r}}{\textup{dt}}$ = 70 cm/min

Volume of the sphere, V = $\frac{\textup{4}}{\textup{3}}\pi r^3$

Surface area of the sphere as 4πr²

Now,

Rate of change in volume with time, $\frac{d\textup{V}}{\textup{dt}}$

= $\frac{d(\frac{\textup{4}}{\textup{3}}\pi r^3)}{dt}$

= $3\times\frac{\textup{4}}{\textup{3}}\pi r^2}\times\frac{dr}{dt}$

Substituting the value of $\frac{dr}{dt}$

= $3\times\frac{\textup{4}}{\textup{3}}\pi r^2}\times70$

= 280πr² cm³/min

Hence, the rate of change in volume with time is 280πr² cm³/min

4 0

3 years ago