Answer: Two celestial objects are in space: one with a mass of 8.22 x 109 kg and one with a mass of 1.38 x 108 kg. If they are separated by a distance of 1.43 km, then, the magnitude of the force of attraction (in newtons) between the objects will be 52.9kN
Explanation: To find the answer we need to know more about the Newton's law of gravitation.
<h3>What is Newton's law of gravitation?</h3>
- Gravitation is the force of attraction between any two bodies.
- Every body in the universe attracts every other body with a force.
- This force is directly proportional to the product of their masses and inversely proportional to the square of the distance between these two masses.
- Mathematically we can expressed it as,
![F=\frac{GMm}{r^2} \\where, G=6.67*10^-11Nm^2kg^-2](https://tex.z-dn.net/?f=F%3D%5Cfrac%7BGMm%7D%7Br%5E2%7D%20%5C%5Cwhere%2C%20G%3D6.67%2A10%5E-11Nm%5E2kg%5E-2)
<h3>How to solve the problem?</h3>
- Here, we have given with the data's,
![M=8.22*10^9kg\\m=1.38*10^8 kg\\r=1.43*10^3m](https://tex.z-dn.net/?f=M%3D8.22%2A10%5E9kg%5C%5Cm%3D1.38%2A10%5E8%20kg%5C%5Cr%3D1.43%2A10%5E3m)
- Thus, the force of attraction between these two bodies will be,
![F=6.67*10^-11*\frac{8.22*10^9*1.38*10^8}{1.43*10^3} =52.9kN](https://tex.z-dn.net/?f=F%3D6.67%2A10%5E-11%2A%5Cfrac%7B8.22%2A10%5E9%2A1.38%2A10%5E8%7D%7B1.43%2A10%5E3%7D%20%3D52.9kN)
Thus, if two celestial objects are in space: one with a mass of 8.22 x 109 kg and one with a mass of 1.38 x 108 kg and, If they are separated by a distance of 1.43 km, then, the magnitude of the force of attraction (in newtons) between the objects will be 52.9kN.
Learn more about the Newton's law of gravitation here:
brainly.com/question/28045318
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Answer:
v = 0
Explanation:
This problem can be solved by taking into account:
- The equation for the calculation of the period in a spring-masss system
( 1 )
- The equation for the velocity of a simple harmonic motion
( 2 )
where m is the mass of the block, k is the spring constant, A is the amplitude (in this case A = 14 cm) and v is the velocity of the block
Hence
![T = \sqrt{\frac{2 kg}{50 N/m}} = 0.2 s](https://tex.z-dn.net/?f=T%20%3D%20%5Csqrt%7B%5Cfrac%7B2%20kg%7D%7B50%20N%2Fm%7D%7D%20%3D%200.2%20s)
and by reeplacing it in ( 2 ):
![v = \frac{2\pi }{0.2s}(14cm)sin(\frac{2\pi }{0.2s}(0.9s)) = 140\pi sin(9\pi ) = 0](https://tex.z-dn.net/?f=v%20%3D%20%5Cfrac%7B2%5Cpi%20%7D%7B0.2s%7D%2814cm%29sin%28%5Cfrac%7B2%5Cpi%20%7D%7B0.2s%7D%280.9s%29%29%20%3D%20140%5Cpi%20%20sin%289%5Cpi%20%29%20%3D%200)
In this case for 0.9 s the velocity is zero, that is, the block is in a position with the max displacement from the equilibrium.
Answer:
1 m/s
Explanation:
Impulse = Change in momentum
Force × Time = Mass(Final velocity) - Mass(Initial Velocity)
(1.0)(1.0) = (1.0)(Final Velocity) - (1.0)(0)
Final velocity = <u>1 m/s</u>
Answer:
The influence of diameter of the blood vessel on peripheral resistance is significant because resistance is inversely proportional to the fourth power of the diameter.
Explanation:
The influence of diameter of the blood vessel on peripheral resistance is significant because the relation between the peripheral resistance and the diameter is given as, resistance is inversely proportional to the fourth power of the diameter. Thus, with small increase or decrease in the value of diameter, the peripheral resistance may vary by a significant amount.