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alukav5142 [94]

3 years ago

9

A 5kg object is moving downward at a speed of 12m/s. If it is currently 2.6m above the ground, what is its potential energy? Use

g=10m/s^2

1 answer:

motikmotik3 years ago

4 0

Don’t trust the people that give you these file things to click on it’s a scam!

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An open system starts with 52 J of mechanical energy. The energy changes

spayn [35]

Answer:

I think the answer is D,54 joules

5 0

2 years ago

Evaluate u+xy where U=3 X=4 Y=7

Reptile [31]

Answer:

31

Explanation:

Given:

U=3

X=4

Y=7

u + xy

Substitute the given values to the equation:

3 + (4)(7)

3 + 28

31

6 0

2 years ago

If you were thinking about a washing machine as a system which of the following represents the inputs?

Lostsunrise [7]

Answer:

The answer is a, the dirty cloths, water and detergent.

Explanation:

The answer is the above selected because the inputs basically represent the data that are passed through the system to generate the output.

In this case, the inputs are the aforementioned in the answer while the possible output would literally be the clean cloths.

4 0

3 years ago

What force is needed to accelerate an object 5 m/s2 if the object has a mass of 10 kg?

Helga [31]

We know, F = m * a
F = 10 * 5
F = 50 N

In short, Your Answer would be 50 Newtons

Hope this helps!

5 0

3 years ago

In a thunderstorm, electric charge builds up on the water droplets or ice crystals in a cloud. Thus, the charge can be considere

muminat

Answer:

$2.1\cdot 10^{21}$ electrons

Explanation:

The magnitude of the electric field outside an electrically charged sphere is given by the equation

$E=\frac{kQ}{r^2}$

where

k is the Coulomb's constant

Q is the charge stored on the sphere

r is the distance (from the centre of the sphere) at which the field is calculated

In this problem, the cloud is assumed to be a charged sphere, so we have:

$E_b=3.00\cdot 10^6 N/C$ is the maximum electric field strength tolerated by the air before breakdown occurs

$r=1.00 km = 1000 m$ is the radius of the sphere

Re-arranging the equation for Q, we find the maximum charge that can be stored on the cloud:

$Q=\frac{Er^2}{k}=\frac{(3.00\cdot 10^6)(1000)^2}{9\cdot 10^9}=333.3 C$

Assuming that the cloud is negatively charged, then

$Q=-333.3 C$

And since the charge of one electron is

$e=-1.6\cdot 10^{-19}C$

The number of excess electrons on the cloud is

$N=\frac{Q}{e}=\frac{-333.3}{-1.6\cdot 10^{-19}}=2.1\cdot 10^{21}$

5 0

2 years ago