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Snowcat [4.5K]
3 years ago
12

Planet X has a moon similar to Earth’s moon. Which path would this moon’s orbit take?

Physics
1 answer:
victus00 [196]3 years ago
3 0
What are the choices?
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When the displacement of a mass on a spring is 1/2a the half of the amplitude, what fraction of the mechanical energy is kinetic
USPshnik [31]
Total energy is a spring:
E = \frac{1}{2} kx^2 +  \frac{1}{2} mv^2 =  \frac{1}{2} ka^2

At x = 0.5a:
\frac{1}{2} k \frac{a}{2} ^2 +  \frac{1}{2} mv^2 =  \frac{1}{2} ka^2 \\  \frac{1}{2} mv^2 =  \frac{1}{2} ka^2 -  \frac{1}{8} ka^2 =  \frac{3}{8} ka^2

The ration:
\frac{ \frac{3}{8}ka^2 }{ \frac{1}{2} ka^2} =  \frac{3}{4}
5 0
3 years ago
Read 2 more answers
How are systems different from industries? Use an example to support your answer.
jeka94

Answer:

Explanation:

An industrial system consists of inputs, processes and outputs. The inputs are the raw materials, labor and costs of land,transport, power and other infrastructure. The processes include a wide range of activities that convert the raw material into finished products.

7 0
3 years ago
Read 2 more answers
A point charge q = +4.50 nC moves through a potential difference ΔV = Vf − Vi = +27.0 V. What is the change in the electric pote
olganol [36]

Change in electric potential energy: 121.5 nJ

Explanation:

For a charged particle moving in an electric field, the change in electric potential energy of the particle is given by

\Delta U = q \Delta V

where:

q is the charge of the particle

\Delta V is the potential difference between the initial and final position of the particle

For the point charge in this problem, we have:

q=+4.50 nC is the charge

\Delta V=+27.0 V is the potential difference

Therefore, the change in electric potential energy is

\Delta U=(+4.50)(+27.0)=121.5 nJ

Learn more about electric fields:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

6 0
3 years ago
In a stunt, three people jump off a platform and fall 8.5 m onto a large air bag. A fourth person at the other end of the air ba
docker41 [41]

Answer:

They Died Period NO MORE

8 0
3 years ago
A rectangular block floats in pure water with 0.400 in. above the surface and 1.60 in. below the surface. When placed in an aque
labwork [276]

Answer:

specific gravity = 0.8

specific gravity of  solution  = 2

Explanation:

given data

rectangular block above water  = 0.400 in

rectangular block below water = 1.60 in

material floats below water = 0.800 in

solution

first we get here specific gravity of block  that is

specific gravity = block vol below ÷ total block vol × specific gravity  water   ..............1

put here value we get

specific gravity =  \frac{1.60}{1.60+0.400}  × 1

specific gravity = 0.8

and now we get here specific gravity of  solution  that is express as

specific gravity of  solution  = total block vol ÷ block vol below × specific gravity  block   ........................2

put here value we get

specific gravity of  solution  = \frac{1.60+0.400}{0.800} × 0.8

specific gravity of  solution  = 2

6 0
3 years ago
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