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1 year ago

A 2 kg stone moves with a speed of 1 m/s. A second 2 kg stone is moving twice as fast. Compare their kinetic energies.

Physics

2 answers:

alekssr [168]1 year ago

5 0

D
is the answer
Well it should be

patriot [66]1 year ago

3 0

Answer:

D is the answer

Explanation:

i did a test with this question on discovery education and i got it right

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A pitching machine is programmed to pitch baseballs horizontally at a speed of 87 mph . The machine is mounted on a truck and ai

Natalija [7]

Answer:

A) The speed of the pitching machine relative to the truck is 0 m/s

B) the speed of the pitched ball relative to the truck is 87mph

C) the speed of the pitching machine relative to him is 65mph

D) the speed of the pitched ball relative to him is 87mph - 65mph = 22mph

5 0

1 year ago

Two 7.0cm×7.0cm metal electrodes are spaced 1.0 mm apart and connected by wires to the terminals of a 9.0 v battery.

AleksandrR [38]

Answer:

$q=390.285x10^{-12}C$

Explanation:

Those kind of problems of electric physics is about capacitors, so the normal questions are:

What are the charge on each electrode?

Solve this you can get other information required in the problem or write down the other questions you need

$7.0cm*\frac{1m}{100cm}=70x10^{-3}m$

$A=70x10^{-3}m*70x10^{-3}m=4.9x10^{-3}m62$

Capacitance

$C=E_o*\frac{A}{d}$

$E_o=8.85x10^{-12}\frac{C^2}{N*m^2}$

$C=8.85x10^{-12}\frac{C^2}{N*m^2}*\frac{4.9x10^{-3}m^2}{1x10^{-3}m}$

$C=43.365x10^{-12}F$

$C=43.365pF$

The charge is find by the equation

$q=C*V$

$q=43.365pF*9V$

$q=390.285x10^{-12}C$

5 0

2 years ago

What is true of an object pulled inward in an electric field?

slava [35]

Answer:

option b

Explanation:

There is an object pulled inward in an electric field.

We have to find out of the four options given which is true.

a) The object has a neutral charge is false since when electric field pulls the object inward, there is a charge inside.

b) The object has a charge opposite that of the field, this option is correct since there will be an equal and opposite charge created by the object

c) The object has a negative charge will be correct only if the original charge was positive hence wrong

d) The object has a charge the same as that of the field is incorrect since this would be opposite the charge

So only option b is right

5 0

1 year ago

In a distant solar system, a giant planet has

sergeinik [125]

Answer:

mass of the planet: $5.9\,10^{26}\,kg$

Explanation:

When a moon keeps a circular orbit around a planet, it is the force of gravity the one that provides the centripetal force to keep it in its circular trajectory of radius R. So if we can write that in such cases (being the mass of the planet "M" and the mass of the moon "m"), we can form an equation by making the centripetal force on the moon equal the force of gravity (using the Newton's Universal Law of Gravity):

$m\frac{v^2}{R}=G\frac{M\,m}{R^2}$

where we used here the tangential velocity (v) of the moon around the planet. This equation can be further simplified by dividing both sides by "m" and multiplying both sides by the orbital radius R:

$m\frac{v^2}{R}=G\frac{M\,m}{R^2}\\v^2=G\frac{M}{R}$

Notice that the mass of the moon has actually disappeared from the equation, which tells us that the orbiting velocity and period do not depend on the mass of the moon, but on the mass of the actual planet.

We know the orbital radius R ( $5.32\,10^5\,km=5.32\,10^8\,m$ , the value of the Universal Gravitational constant G, and we can estimate the value of the tangential velocity of the moon since we know it period: 36.3 hrs = 388800 seconds.

We know that the moon makes a full circumference ( $2\,\pi\,R$ ) in 388800 seconds, therefore its tangential velocity is:

$v=\frac{2\,\pi\,5.32\,10^8}{388800} \frac{m}{s} \\v=8.6\,10^3\,\frac{m}{s}$

where we rounded the velocity to one decimal.

Notice that we have converted all units to the SI system, so when using the formula to solve for the mass of the planet, the answer comes directly in kg.

Now we use this value for the tangential velocity to estimate the mass of the planet in the first equation we made and simplified:

$v^2=G\frac{M}{R}\\M=\frac{v^2\,R}{G} \\M=\frac{(8.6\,10^3)^2\,5.32\,10^8}{6.67\,10^{-11}}kg\\M=5.9\,10^{26}\,kg$