Answer:
The earth's gravitational force on the sun is equal to the sun's gravitational force on the earth
Explanation:
Newton's third law (law of action-reaction) states that:
"When an object A exerts a force (called action) on an object B, then object B exerts an equal and opposite force (called reaction) on object A"
In other words, when two objects exert a force on each other, then the magnitude of the two forces is the same (while the directions are opposite).
In this problem, we can call the Sun as "object A" and the Earth as "object B". According to Newton's third law, therefore, we can say that the gravitational force that the Earth exerts on the Sun is equal (in magnitude, and opposite in direction) to the gravitational force that the Sun exerts on the Earth.
Answer:
The boat won't be able to move if the oars were out and there was no thruster. If there was a flow of the water then yes there would be a moving boat.
The answer is photocoagulation.
The use of a laser beam to seal leaky blood vessels and to prevent the growth of new ones in diabetic retinopathy is called laser <u>photocoagulation.</u>
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What is photocoagulation?
A minimally invasive method used to treat numerous retinal illnesses is photocoagulation of the retina, also known as retinal laser photocoagulation. The retina may expand due to aberrant leaky blood vessels developing across it in a number of disorders. Laser photocoagulation uses thermal energy above 65 °C to burn the retinal tissue by creating thermal burns. This can prevent the retina from being damaged by the bleeding blood vessels. In addition to causing fibrosis, laser photocoagulation can also seal retinal tears. Laser photocoagulation is typically unable to recover already lost vision in cases of retinal disease, but it can slow the progression of the condition, lower the chance of further vision loss, and preserve residual vision. The likelihood of problems following the operation is quite minimal.
To learn more about photocoagulation click on the link below:
brainly.com/question/16016898
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Since the total distance for all three of the planets is 196.2 million miles, you would add planet i and planet iii, then subtract that number from 196.2
planet i and planet iii equaled 54.9 million miles, that subtracted from 196.2 equals 141.3
So your answer is 141.3 million miles
Answer:
P = 981 N
Explanation:
Given
Angle of the incline θ = 10°
Angle of the towing force φ =20°
Weight of the crate W = 3433.5 N
The coefficient of static friction µ = 0.5
Solution
Forces Acting along the ramp
Forces acting perpendicular to the ramp
Substituting the value of N we get
![\mu[Wcos10^{0}-Psin30^{o}] =Pcos30^{o}+Wsin10^{0}\\\\\mu Wcos10^{0}- \mu Psin30^{o} =Pcos30^{o}+Wsin10^{0}\\\\\mu Wcos10^{0}- Wsin10^{0}=Pcos30^{o}+\mu Psin30^{o} \\\\P=W\frac{\mu cos10^{0}- sin10^{0}}{cos30^{o}+\mu sin30^{o}} \\\\P=3433.5\frac{0.5 \times cos10^{0}- sin10^{0}}{cos30^{o}+0.5 \times sin30^{o}}\\\\P=980.66\\\\ P = 981 N](https://tex.z-dn.net/?f=%5Cmu%5BWcos10%5E%7B0%7D-Psin30%5E%7Bo%7D%5D%20%3DPcos30%5E%7Bo%7D%2BWsin10%5E%7B0%7D%5C%5C%5C%5C%5Cmu%20Wcos10%5E%7B0%7D-%20%5Cmu%20Psin30%5E%7Bo%7D%20%3DPcos30%5E%7Bo%7D%2BWsin10%5E%7B0%7D%5C%5C%5C%5C%5Cmu%20Wcos10%5E%7B0%7D-%20Wsin10%5E%7B0%7D%3DPcos30%5E%7Bo%7D%2B%5Cmu%20Psin30%5E%7Bo%7D%20%5C%5C%5C%5CP%3DW%5Cfrac%7B%5Cmu%20cos10%5E%7B0%7D-%20sin10%5E%7B0%7D%7D%7Bcos30%5E%7Bo%7D%2B%5Cmu%20sin30%5E%7Bo%7D%7D%20%5C%5C%5C%5CP%3D3433.5%5Cfrac%7B0.5%20%5Ctimes%20cos10%5E%7B0%7D-%20sin10%5E%7B0%7D%7D%7Bcos30%5E%7Bo%7D%2B0.5%20%5Ctimes%20%20sin30%5E%7Bo%7D%7D%5C%5C%5C%5CP%3D980.66%5C%5C%5C%5C%20P%20%3D%20981%20N)
