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Hatshy [7]
2 years ago
7

I'm trying to find the potential energy of a planet using formula G M(sun) x m(planet) / r. I have given G - newton's graviation

al constant at 6.67 × 10−11 m3 kg−1 s−2 , MSun is the mass of the Sun: 1.99 × 1030 kg, m planet is the mass of the planet at 5.97 x 10^24 and r is the distance from the Sun at 147.1 x 10^6 km
Physics
1 answer:
ludmilkaskok [199]2 years ago
7 0

Answer:

-5.39\times10^{33} J

Explanation:

Potential energy =

-\frac{GMm}{r}\\=-\frac{6.67\times10^{-11}\times1.99\times10^{30}\times5.97\times10^{24}}{147/1\times10^{9}}\\=-5.39\times10^{33} J

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Question 14 (2 points)
sveticcg [70]

Answer:

after 2 seconds its velocity is -20 m/s. after 3 seconds its velocity is -30 m/s. after 10 seconds its velocity is -100 m/s.

Explanation:

This is my answer.

7 0
2 years ago
A positively charged particle moves through an electric field. As part of a complicated trajectory, the particle passes through
kow [346]

Answer:

(B) The speed is larger at A than at B.

Explanation:

Point B, the final point of the trajectory, has higher electric potential than point A, the initial point of the trajectory, so the electric potential energy of the charged particle increases, which means that its kinetic energy must be decreasing, thus the speed at B must be lower than the speed at A.

8 0
3 years ago
Mike walks 200 km in 6 hours.he then walks another 100km in 4 hours .what is his average speed?
kolbaska11 [484]

Average speed is defined as the ratio of total distance covered in total given time

speed = \frac{distance}{time}

here we know that total distance that man moved is

d_1 = 200 km

d_2 = 100 km

so total distance is

d = d_1 + d_2

d = 200 km + 100 km

d = 300 km

now here total time of the motion is

t_1 = 6 hours

t_2 = 4 hours

total time will be given as

t = t_1 + t_2

t = 6 + 4 = 10 hours

now by above formula

v_{avg} = \frac{300}{10}

v_{avg} = 30 km/h

so his average speed is 30 km/h

8 0
3 years ago
Cells are considered the smallest part of an organsim that can?​
Softa [21]
That can be considered alive.
4 0
3 years ago
A wad of clay of mass m1 = 0.49 kg with an initial horizontal velocity v1 = 1.89 m/s hits and adheres to the massless rigid bar
notka56 [123]

Answer:

<h2>The angular velocity just after collision is given as</h2><h2>\omega = 0.23 rad/s</h2><h2>At the time of collision the hinge point will exert net external force on it so linear momentum is not conserved</h2>

Explanation:

As per given figure we know that there is no external torque about hinge point on the system of given mass

So here we will have

L_i = L_f

now we can say

m_1v_1\frac{L}{2} = (m_2L^2 + m_1(\frac{L}{2})^2)\omega

so we will have

0.49(1.89)(0.45) = (2.13(0.90)^2 + 0.49(0.45)^2)\omega

\omega = 0.23 rad/s

Linear momentum of the system is not conserved because at the time of collision the hinge point will exert net external force on the system of mass

So we can use angular momentum conservation about the hinge point

6 0
3 years ago
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