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4 years ago

Why one part of the earth's surface is arid and dry and a nearby part is lush and wet is explained by the term?

1 answer:

8 0

The process that explains why one part of the earth's surface is arid and dry and a nearby part is lush and wet is areal differentiation. It is<span> an approach to geography that shows </span>the dependence of the distribution of physical and human phenomena and the relation to each other from the physical location. Areal integration on the other hand is the approach that studies how places interact with each other.

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A crate of 890 kg gets raised to a height of 2.1m. Calculate the crate's gravitational potential energy.

mars1129 [50]

Answer:

18,316.2

Explanation:

The formula for GPE is mgh, where

M = Mass of the object

G = Acceleration due to gravity (9.8 m/s^2 on Earth)

H = Height above ground

890 × 9.8 × 2.1 = 18,316.2

4 0

3 years ago

Read 2 more answers

A 62-kg person jumps from a window to a fire net 20.0 m directly below, which stretches the net 1.4 m. Assume that the net behav

gayaneshka [121]

Answer:

a) x = 0.098

b) x = 2.72 m

Explanation:

(a) To find the stretch of the fire net when the same person is lying in it, you can assume that the net is like a spring with constant spring k. It is necessary to find k.

When the person is falling down he acquires a kinetic energy K, this energy is equal to the elastic potential energy of the net when it is max stretched.

Then, you have:

$K=U\\\\\frac{1}{2}mv^2=\frac{1}{2}kx^2$ (1)

m: mass of the person = 62kg

k: spring constant = ?

v: velocity of the person just when he touches the fire net = ?

x: elongation of the fire net = 1.4 m

Before the calculation of the spring constant, you calculate the final velocity of the person by using the following formula:

$v^2=v_o^2+2gy$

vo: initial velocity = 0 m/s

g: gravitational acceleration = 9.8 m/s^2

y: height from the person jumps = 20.0m

$v=\sqrt{2gy}=\sqrt{2(9.8m/s^2)(20.0m)}=14\frac{m}{s}$

With this value you can find the spring constant k from the equation (1):

$mv^2=kx^2\\\\k=\frac{mv^2}{x^2}=\frac{(62kg)(14m/s)^2}{(1.4m)^2}=6200\frac{N}{m}$

When the person is lying on the fire net the weight of the person is equal to the elastic force of the fire net:

$W=F_e\\\\mg=kx$

you solve the last expression for x:

$x=\frac{mg}{k}=\frac{(62kg)(9.8m/s^2)}{6200N/m}=0.098m$

When the person is lying on the fire net the elongation of the fire net is 0.098m

b) To find how much would the net stretch, If the person jumps from 38 m, you first calculate the final velocity of the person again:

$v=\sqrt{2gy}=\sqrt{2(9.8m/s^2)(38m)}=27.29\frac{m}{s}$

Next, you calculate x from the equation (1):

$x=\sqrt{\frac{mv^2}{k}}=\sqrt{\frac{(62kg)(27.29m/s)^2}{6200N/m}}\\\\x=2.72m$

The net fire is stretched 2.72 m

5 0

3 years ago

If you push a box across the floor, the box experiences what kind of friction?

Talja [164]

I think it is kinetic friction not the best at physics

4 0

3 years ago

Read 2 more answers

How to solve for h using the formula eg=mgh?

SVEN [57.7K]

Divide each side of the equation by mg .

8 0

4 years ago

Astronomers have observed a small, massive object at the center of our Milky Way galaxy. A ring of material orbits this massive

vazorg [7]

Answer: Yes, it can be a single, ordinary star.

Explanation: To determine a mass of a star, we use the orbital speed formula, given by: v = $\sqrt{\frac{GM}{R} }$ , where

v is the speed;

G is a constant: G = 6.67* $10^{-11}$ $\frac{m^{3} }{kg.s^{2} }$

M is mass of a massive object;

R is the distance between the object orbiting and the massive object;

The formula can be rewritten as:

$M = \frac{v^{2}.R }{G}$

First, we change R from light years to km:

1km=1.057* $10^{-13}$

R= $\frac{15}{2*1.057.10^{-13} }$

Calculating mass:

M = $\frac{2^{2}*10^{4}*14.2*10^{13} }{6.67*10^{-11} }$

M = 4.25* $10^{28}$ kg

A solar mass is the standard unit of mass. It is approximately 2* $10^{30}$ Kg and can be used for comparison: A single star cannot be more than 50 solar masses.

50 solar masses = 50*2* $10^{30}$ = $10^{32}$ kg

Comparing the mass of the object with this parameter, we have

$\frac{10^{32} }{4.25.10^{28} }$ = 0.235. $10^{4}$ = 2.35. $10^{3}$

From this, we know that 50 solar masses is greater than the small, massive object found. So, this object can be a single, ordinary star.

3 0

3 years ago