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2 years ago

WHAT IS THE POTENIAL ENERGY OF A 3 KILOGRAM BALL THAT IS ON THE GROUND

Physics

1 answer:

Ratling [72]2 years ago

5 0

Answer:

147.15 Joules.

Explanation:

A 3 kilogram mass at a height of 5 meters, while acted on by Earth's gravity would have 147.15 Joules of potential energy, PE = 3kg * 9.81 m/s2 * 5m = 147.15 J.

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Look at the velocity versus time graph below. What is the magnitude of the displacement of the object after it travels for three

Elza [17]

Answer:

C. 12m

Explanation:

$veocity = \frac{displacement}{time}$

from the graph v = 4m/s and t = 3 s

d = vt = 4 × 3 = 12 m

5 0

3 years ago

The generator at a power plant produces AC at 20,000 V. A transformer steps this up to 355,000 V for transmission over power lin

Masja [62]

Answer:

Number of coil in the output is 39938

Explanation:

We have given a step up transformer

Input voltage of transformer, that is primary voltage $v_p=20000volt$

Output voltage, that is secondary voltage $v_s=355000volt$

Number of turns in primary $N_p=2250$

For transformer we know that $\frac{V_p}{V_s}=\frac{N_p}{N_s}$

$\frac{20000}{355000}=\frac{2250}{N_s}$

$N_s=39937.5$

As the number of turns can not be in fraction so number of turns in the output coil is 39938

7 0

3 years ago

At an instant when a particle of mass 80 g has a velocity of 25 m/s in the positive y direction, a 75-g particle has a velocity

dalvyx [7]

Answer:

16 m/s

Explanation:

Given that

y momentum = 0.080 *25 = 2

x momentum = 0.075*20 = 1.5

total momentum = √(4 + 2.25)

Total momentum = √6.25

Total momentum = 2.5

total mass = mass of x and y momentum = 0.075 + 0.080 = 0.155

speed of mass center = total momentum / total mass = 2.5/0.155 = 16.

And thus, the speed of the center of mass of this two-particle system at this instant is 16 m/s

7 0

3 years ago

A m = 94.2 kg object is released from rest at a distance h = 1.15134 R above the Earth’s surface. The acceleration of gravity is

OleMash [197]

Answer:

v= 4055.08m/s

Explanation:

This is a problem that must be addressed through the laws of classical mechanics that concern Potential Gravitational Energy.

We know for definition that,

$U = \frac{GMm}{r}$

We must find the highest point and the lowest point to identify the change in energy, so

Point a)

The problem tells us that an object is dropped at a distance of h = 1.15134R over the earth.

That is to say that the energy of that object is equal to,

$U_1=-\frac{(6.6738 * 10^{-11})(5.98 * 10^{24})(94.2)}{(1.15134)(6.38*10^6)}$

$U_2= - 5.1180*10^9J$

Point B )

We now use the average radius distance from the earth.

$U_2=-\frac{(6.6738 * 10^{-11})(5.98 * 10^{24})(94.2)}{(6.38*10^6)}$

$U_2= -5.8925*10^9J$

Then,

$\Delta U = U_2 - U_1 = -5.1180*10^9J - ( -5.8925*10^9J)$

$\Delta U = 774.5*10^6$

By the law of conservation of energy we know that,

$\Delta U = \frac{1}{2}mv^2$

clearing v,

$v= \sqrt{2 \Delta U/m}$

$v= \sqrt{2*774.5*10^6 /94.2}$

$v= 4055.08m/s$

Therefore the speed of the object when it strikes the Earth’s surface is 4055.08m/s

8 0

3 years ago

Calculate the resistance 50 feet of awg no. 12 solid copper wire round to nearest 0. 01 ohms.

emmasim [6.3K]

The resistance of the conductor is 0.07940 Ohms.

<h3>What is the relation between resistance and area of wire?</h3>

The wire's resistance is inversely related to its cross-sectional area; as the area drops, so does the resistance.

and it is formulated as:

$R=p\frac{l}{s}$

where,

<em>p </em>- resistivity of the conductor (0.0214-ohm mm²/m)

R - resistance

l- length of conductor (50 feet) (15.24 m)

s - the area of the wire

Thus the resistance can be calculated as

R = 0.07940 Ohms.

Learn more about resistance here:

brainly.com/question/11431009

#SPJ4

6 0

1 year ago