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2 years ago

How do streams change as they flow from mountains down to plains? Plz help URGENT I need it today

Physics

1 answer:

lisov135 [29]2 years ago

5 0

They start as one stream, then get curvier, then empy into a delta. A delta is a ton of streams of water. It is one, then empties out into a ton of streams.
~Deceptiøn

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How many calories are needed to raise the temperature of 2g of water 3 C?

olga nikolaevna [1]

Answer:

The number of calories needed is 6c.

Explanation:

The amount of energy $Q$ needed to raise the temperature $\Delta T$ of water of mass $m$ is

$Q = mC\Delta T$

where $C = 1cal/g\cdot C$ is the specific heat capacity of water.

Putting in numbers into equation (1), we get:

$Q = (2g)(1)(3^oC )\\$

$\boxed{Q = 6\:cal }$

which is the number of calories needed.

8 0

3 years ago

The maximum distance from the Earth to the Sun (at aphelion) is 1.521 1011 m, and the distance of closest approach (at perihelio

LUCKY_DIMON [66]

Answer:

29274.93096 m/s

$2.73966\times 10^{33}\ J$

$-5.39323\times 10^{33}\ J$

$2.56249\times 10^{33}\ J$

$-5.21594\times 10^{33}$

Explanation:

$r_p$ = Distance at perihelion = $1.471\times 10^{11}\ m$

$r_a$ = Distance at aphelion = $1.521\times 10^{11}\ m$

$v_p$ = Velocity at perihelion = $3.027\times 10^{4}\ m/s$

$v_a$ = Velocity at aphelion

m = Mass of the Earth = 5.98 × 10²⁴ kg

M = Mass of Sun = $1.9889\times 10^{30}\ kg$

Here, the angular momentum is conserved

$L_p=L_a\\\Rightarrow r_pv_p=r_av_a\\\Rightarrow v_a=\frac{r_pv_p}{r_a}\\\Rightarrow v_a=\frac{1.471\times 10^{11}\times 3.027\times 10^{4}}{1.521\times 10^{11}}\\\Rightarrow v_a=29274.93096\ m/s$

Earth's orbital speed at aphelion is 29274.93096 m/s

Kinetic energy is given by

$K=\frac{1}{2}mv_p^2\\\Rightarrow K=\frac{1}{2}\times 5.98\times 10^{24}(3.027\times 10^{4})^2\\\Rightarrow K=2.73966\times 10^{33}\ J$

Kinetic energy at perihelion is $2.73966\times 10^{33}\ J$

Potential energy is given by

$P=-\frac{GMm}{r_p}\\\Rightarrow P=-\frac{6.67\times 10^{-11}\times 1.989\times 10^{30}\times 5.98\times 10^{24}}{1.471\times 10^{11}}\\\Rightarrow P=-5.39323\times 10^{33}$

Potential energy at perihelion is $-5.39323\times 10^{33}\ J$

$K=\frac{1}{2}mv_a^2\\\Rightarrow K=\frac{1}{2}\times 5.98\times 10^{24}(29274.93096)^2\\\Rightarrow K=2.56249\times 10^{33}\ J$

Kinetic energy at aphelion is $2.56249\times 10^{33}\ J$

Potential energy is given by

$P=-\frac{GMm}{r_a}\\\Rightarrow P=-\frac{6.67\times 10^{-11}\times 1.989\times 10^{30}\times 5.98\times 10^{24}}{1.521\times 10^{11}}\\\Rightarrow P=-5.21594\times 10^{33}$

Potential energy at aphelion is $-5.21594\times 10^{33}\ J$

6 0

2 years ago

WHAT are the highest frequency and the lowest frequency pets of the EM spectrum?

Rus_ich [418]

Highest frequency EM waves: cosmic rays and gama rays

Lowest frequency EM waves:
Radio and Tv waves

8 0

2 years ago

When you trace the outline of your palm how do you find its area

Tcecarenko [31]

Answer:

you count the squares or messure it

Explanation:

you can raw equal squares about 1 cm wide if possible all equal and count the squares eg theres 10 squares (small hand) so that would be 10cm squared

7 0

2 years ago

Read 2 more answers

I need HELP Please !!

aliya0001 [1]

You have to figure it out

5 0

2 years ago