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2 years ago

A planet orbits a sun in a clockwise elliptical orbit as shown in the diagram below

Physics

1 answer:

STatiana [176]2 years ago

3 0

Answer:

Greatest gravitational energy is at "C".

The planet has to do work "against" the field to get to "C".

Also, if m v R (angular momentum) is constant then as R increases v must decrease for this term to be constant and KE = 1/2 M v^2 must decrease also to get to point C.

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Assuming you exert a constant force on the wagon, how fast is it moving after 5 seconds?

Gelneren [198K]

I would help but I’m also taking this

3 0

2 years ago

A comet is traveling through space with speed 3.01 ✕ 104 m/s when it encounters an asteroid that was at rest. The comet and the

Tcecarenko [31]

Answer: $8.493(10)^{-3} m/s$

Explanation:

According to the conservation of linear momentum principle, the initial momentum $p_{i}$ (before the collision) must be equal to the final momentum $p_{f}$ (after the collision):

$p_{i}=p_{f}$ (1)

In addition, the initial momentum is:

$p_{i}=m_{1}V_{1}+m_{2}V_{2}$ (2)

Where:

$m_{1}=1.71(10)^{14} kg$ is the mass of the comet

$m_{2}=6.06(10)^{20} kg$ is the mass of the asteroid

$V_{1}=3.01(10)^{4} m/s$ is the velocity of the comet, which is positive

$V_{2}=0 m/s$ is the velocity of the asteroid, since it is at rest

And the final momentum is:

$p_{f}=(m_{1}+m_{2})V_{f}$ (3)

Where:

$V_{f}$ is the final velocity

Then :

$m_{1}V_{1}+m_{2}V_{2}=(m_{1}+m_{2})V_{f}$ (4)

Isolating $V_{f}$ :

$V_{f}=\frac{m_{1}V_{1}}{m_{1}+m_{2}}$ (5)

$V_{f}=\frac{(1.71(10)^{14} kg)(3.01(10)^{4} m/s)}{1.71(10)^{14} kg+6.06(10)^{20} kg}$

Finally:

$V_{f}=8.493(10)^{-3} m/s$ This is the final velocity, which is also in the positive direction.

8 0

2 years ago

Based on their locations on the periodic table, which two elements would you

Nat2105 [25]

Answer:

i think it would be B. Aluminum, Al and D. Boron, B

Explanation:

since they're both in group 13 and they forms a 3+ ion

6 0

2 years ago

Read 2 more answers

4) For the situation pictured below, F1 = 20.0 N east and F2 = 30.0 N west ,

maksim [4K]

Answer:

10.0 N West

Explanation:

Imagine East is positive and West is negative. Essentially now we have 2 forces pulling opposite each other on the same plane, so we can add them up.

F1=20N

F2=-30N

F1+F2=Net Force

20+(-30)=-10

Therefore the Net Force is 10N West.

8 0

2 years ago

the kinetic energy of an object of mass in moving with a velocity of 5 MS -1 is 25j what will be its kinetic energy when its vel

MAXImum [283]

Answer:

<em>When the speed is doubled, K = 100 J, when the speed is tripled, K = 225 J</em>

Explanation:

<u>Kinetic Energy </u>

Is the type of energy an object has due to its speed. It's proportional to the square of the speed.

The equation for the kinetic energy is:

$\displaystyle K=\frac{1}{2}mv^2$

Where:

m = mass of the object

v = speed at which the object moves

The kinetic energy is expressed in Joules (J)

The object has a kinetic energy of K=25 J when moving at v=5 m/s, thus the mass can be calculated by solving for m:

$\displaystyle m=\frac{2K}{v^2}$

$\displaystyle m=\frac{2*25}{5^2}=2$

m = 2 Kg

If the speed is doubled, v=10 m/s, the new kinetic energy is:

$\displaystyle K=\frac{1}{2}2\cdot 10^2$

K = 100 J

If the speed is tripled, v=15 m/s, the new kinetic energy is:

$\displaystyle K=\frac{1}{2}2\cdot 15^2$

K = 225 J

When the speed is doubled, K = 100 J, when the speed is tripled, K = 225 J

3 0

2 years ago