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3 years ago

8

How are oceans connected to all bodies of water

1 answer:

inn [45]3 years ago

7 0

Oceans are collected to all bodies of water because

- All Rivers, streams, or swamps end up flowing out into an ocean somehow
- The oceans are the largest bodies of water, so most of the water on earth comes from them, or has been in the oceans.

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EXTREME HELP!!!!

mrs_skeptik [129]

the anwser is c is always the best anwser

8 0

3 years ago

Cheetahs, the fastest of the great cats, can reach 50.0 miles/hour in 2.22 s starting from rest. Assuming that they have constan

elena55 [62]

Answer:

$10.07m/s^2$

Explanation:

Information we have:

velocities:

initial velocity: $v_{i}=0mph$ (starts from rest)

final velocity: $v_{f}=50mph$

time: $t=2.22s$

Since we need the answer in $m/s^2$ , we nees to convert the speed to meters per second:

$v_{f}=\frac{50miles}{1hour}(\frac{1hour}{3,600s} )(\frac{1609.34meters}{1mile} ) \\\\v_{f}=\frac{50*1609.34}{3600} m/s\\\\v_{f}=22.35m/s$

We find the acceleration with the following formula:

$a=\frac{v_{f}-v_{i}}{t}$

substituting the known values:

$a=\frac{22.35m/s-0m/s}{2.22s}\\ \\a=10.07m/s^2$

the acceleration is 10.07 $m/s^2$

8 0

2 years ago

Read 2 more answers

A black, totally absorbing piece of cardboard of area a = 2.1 cm2 intercepts light with an intensity of 8.8 w/m2 from a camera s

PolarNik [594]

Answer:

The value of radiation pressure is $2.933 \times 10^{-8}$ Pa

Explanation:

Given:

Intensity $I = 8.8$ $\frac{W}{m^{2} }$

Area of piece $A = 2.1 \times 10^{-4}$ $m^{2}$

From the formula of radiation pressure in terms of intensity,

$P = \frac{I}{c}$

Where $P =$ radiation pressure, $c =$ speed of light

We know value of speed of light,

$c = 3 \times 10^{8}$ $\frac{m}{s}$

Put all values in above equation,

$P = \frac{8.8}{3 \times 10^{8} }$

$P = 2.933 \times 10^{-8}$ Pa

Therefore, the value of radiation pressure is $2.933 \times 10^{-8}$ Pa

8 0

3 years ago

Suppose a particle moves along a straight line with velocity v(t)=t2e−2tv(t)=t2e−2t meters per second after t seconds. How many

dimulka [17.4K]

Explanation:

It is given that,

Velocity of the particle moving in straight line is :

$v(t)=t^2e^{-2t}\ m/s$

We need to find the distance (x) traveled by the particle during the first t seconds. It is given by :

$x=\int\limits {v.dt}$

$x=\int\limits {t^2e^{-2t}dt}$

Using by parts integration, we get the value of x as :

$x=\dfrac{-(2t^2+2t+1)e^{-2t}}{4}\ meters$

Hence, this is the required solution.

6 0

3 years ago

1*2+2+1+5 by how much?

kozerog [31]

Answer:

10

Explanation:

1*2+2+1+5=10

1x2+2+1+5=10

6 0

3 years ago