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Nat2105 [25]
3 years ago
5

PLS HELP!!! ILL GIVE YOU A BRAINLIEST! look carefully at the landform, if you were building a shelter, where would you build?

Physics
1 answer:
yulyashka [42]3 years ago
4 0

Answer:

I would build it at a top of a hill because water will go down and will decrease chances of flooding since y'know, height?

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A coin is dropped. What is its displacement after 0.30 s.
Morgarella [4.7K]
Assuming this coin is on earth and that it wasn’t dropped forcefully:
Use the formula d = 1/2at^2. Rewriting using a=g and solving for height h gets us h = 1/2(9.8)t^2.
In this case that would get that the change in height h is 0.5(9.8)(0.3^2) = 0.441 m.
3 0
3 years ago
Can someone help me on question 1?
grigory [225]
In question 1, both of your answers are correct, but I don't understand the process you went through in the 'a' part.

R = v/I . That's a correct formula.
But it doesn't help you in this form, because you need to find I
So turn it into a helpful form ... Solve it for I, so it says I=something.

R= v/I

Multiply each side by I : R I = V.

Now divide each side by R: I= V/R .
THERE'S the equation you want.

I = V / R

I = 1.5 / 10 = 0.15 Amp.

That's slightly cleaner, although I don't really understand what you were actually thinking in that part.

But again ... You answered both parts correctly, and your process in b is fine.
6 0
3 years ago
A 10.0 g bullet moving at 300m/s is fired into a 1.00 kg block at rest. The bullet emerges (the bullet does not get embedded in
chubhunter [2.5K]

Answer:

v' = 1.5 m/s

Explanation:

given,

mass of the bullet, m = 10 g

initial speed of the bullet, v = 300 m/s

final speed of the bullet after collision, v' = 300/2 = 150 m/s

Mass of the block, M = 1 Kg

initial speed of the block, u = 0 m/s

velocity of the block after collision, u' = ?

using conservation of momentum

 m v + Mu = m v' + M u'

 0.01 x 300 + 0 = 0.01 x 150 + 1 x v'

v' = 0.01 x 150

v' = 1.5 m/s

Speed of the block after collision is equal to v' = 1.5 m/s

5 0
3 years ago
A flag is waved 3.2 m above the surface of a flat pool of water. When viewed from under the water, what is the magnification of
jek_recluse [69]

Answer:

1.33

Explanation:

For an optical instrument, the magnification ratio of the apparent diameter of the image to that of the object.

Mathematically, from the given information;

Magnification= \dfrac{n_{water}}{n_{air}}

where;

n_{water} =1.33\\ \\ n_{air} = 1.00

= \dfrac{1.33}{1.00} \\ \\=  \mathbf{1.33}

5 0
3 years ago
Captain Kirk (80.0 kg) beams down to a planet that is the same size as Uranus and finds that he weighs
Archy [21]

The mass of the planet is 1.51\cdot 10^{26}kg

Explanation:

The weight of Captain Kirk on the surface of the planet is equal to the gravitational force between him and the planet, which is:

F=G\frac{Mm}{R^2}

where

G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2} is the gravitational constant

M is the mass of the planet

m is the mass of Captain Kirk

R is the radius of the planet

In this problem, we have:

m = 80.0 kg is the mass of Kirk

R=25,362 km = 2.54\cdot 10^7 m is the radius of the planet (same  as Uranus)

F = 1250 N is the magnitude of the gravitational force between Kirk and the planet

Solving for M, we find the mass of the planet:

M=\frac{FR^2}{Gm}=\frac{(1250)(2.54\cdot 10^7)^2}{(6.67\cdot 10^{-11})(80.0)}=1.51\cdot 10^{26}kg

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

8 0
3 years ago
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