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2 years ago

What is the potential energy of a 20-kg safe sitting on a shelf 0.5 meters

Physics

1 answer:

ki77a [65]2 years ago

8 0

Explanation:

P.E=MGH

Where m is mass

Where G is Acceleration Due to Gravity

Where h is Height

So the parameters are M = 20kg

G = 9.8m/s

H = 0.5meters

P.E= 20x9.8x0.5

=98J.

So Ans. is A= 98J

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Need physics help<br> ASAP please!

Effectus [21]

Answer:

Explanation:

7 0

2 years ago

If a planet has the same mass as the earth, but has twice the radius, how does the surface gravity, g, compare to g on the surfa

shepuryov [24]

Answer:

The surface gravity g of the planet is 1/4 of the surface gravity on earth.

Explanation:

Surface gravity is given by the following formula:

$g=G\frac{m}{r^{2}}$

So the gravity of both the earth and the planet is written in terms of their own radius, so we get:

$g_{E}=G\frac{m}{r_{E}^{2}}$

$g_{P}=G\frac{m}{r_{P}^{2}}$

The problem tells us the radius of the planet is twice that of the radius on earth, so:

$r_{P}=2r_{E}$

If we substituted that into the gravity of the planet equation we would end up with the following formula:

$g_{P}=G\frac{m}{(2r_{E})^{2}}$

Which yields:

$g_{P}=G\frac{m}{4r_{E}^{2}}$

So we can now compare the two gravities:

$\frac{g_{P}}{g_{E}}=\frac{G\frac{m}{4r_{E}^{2}}}{G\frac{m}{r_{E}^{2}}}$

When simplifying the ratio we end up with:

$\frac{g_{P}}{g_{E}}=\frac{1}{4}$

So the gravity acceleration on the surface of the planet is 1/4 of that on the surface of Earth.

3 0

2 years ago

A very long uniform line of charge has charge per unit length λ1 = 4.68 μC/m and lies along the x-axis. A second long uniform li

Kitty [74]

Answer:

$E_{net} = 6.44 \times 10^5 N/C$

Explanation:

As we know that electric field due to infinite line charge distribution at some distance from it is given as

$E = \frac{2k \lambda}{r}$

now we need to find the electric field at mid point of two wires

So here we need to add the field due to two wires as they are oppositely charged

Now we will have

$E_{net} = \frac{2k\lambda_1}{r} + \frac{2k\lambda_2}{r}$

now plug in all data

$\lambda_1 = 4.68 \muC/m$

$\lambda_2 = 2.48 \mu C/m$

$r = 0.200 m$

now we have

$E_{net} = \frac{2k}{r}(4.68 + 2.48)$

$E_{net} = \frac{2(9\times 10^9)}{0.200}(7.16 \times 10^{-6})$

$E_{net} = 6.44 \times 10^5 N/C$

8 0

2 years ago

Imagine a raindrop starting from rest in a cloud 2 km in the air. If it fell with no air friction at all, it would accelerate to

LenKa [72]

Answer:

2) 433 mph

Explanation:

The final velocity of the raindrop as it reaches the ground can be found by using the equation for a uniformly accelerated motion:

$v^2 = u^2 + 2ad$

where

v is the final velocity

u = 0 is the initial velocity (the raindrop starts from rest)

a = g = 9.8 m/s^2 is the acceleration due to gravity

d = 2 km = 2000 m is the distance covered

Solving for v,

$v=\sqrt{u^2 +2gd}=\sqrt{0^2+2(9.8 m/s^2)(2000)}=198 m/s$

And keeping in mind that

1 mile = 1609 metres

1 hour = 3600 s

The speed converted into miles per hour is

$v=198 \frac{m}{s}\cdot \frac{3600 s/h}{1609 m/mi}=433 mph$

5 0

2 years ago

Which vector correctly indicates the direction of centripetal acceleration experienced by the car?

gogolik [260]

The answer is C.

I hope this helped!

5 0

2 years ago

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