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Vlad [161]
3 years ago
10

Describe a situation where Newton's third law of motion can be observed

Physics
1 answer:
yarga [219]3 years ago
3 0
An example would be a swimmer pushes on the wall with her feet, which causes the wall to push back on her feet due to Newton's third law
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Haileys comet is a comet that passes close to earth every 75 years. based on this passage, which gravitational attraction has th
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gravity lol

Explanation:

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An object of mass m is traveling in a circle with centripetal force f c. If the velocity of the object is v, what is the radius
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Answer:

r = mv^2/ Fc

Explanation:

Got it right

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3 years ago
it is required that an object weighing 8N be moved up ward at an acceleration of 2m/s what is the upward force needed ​
sesenic [268]

Answer:

F = 1.6 N

Explanation:

Newton's Second Law of Motion states that whenever an unbalanced force is applied on a body. It produces an acceleration in the body, in its own direction. The magnitude of the force is given by the following formula:

F = ma

where,

F = Force Required to Move the Object Upward = ?

m = mass of object

a = upward acceleration = 2 m/s²

but, we know that weight of object is given as:

Weight = mg

8 N = m(9.8 m/s²)

m = 0.8 kg

using values in the equation:

F = (0.8 kg)(2 m/s²)

<u>F = 1.6 N</u>

4 0
3 years ago
Calculate the speed of a plane which travels a distance of 2100 miles in 3 hours
Zolol [24]

it would be 700 mile each hour if he goes 2100 miles in 3 hours

5 0
3 years ago
You set a tuning fork into vibration at a frequency of 683 Hz and then drop it off the roof of the Physics building where the ac
mel-nik [20]

Answer:

The distance traveled by the tuning fork is 9.37 m

Explanation:

Given;

source frequency, f_s = 683 Hz

observed frequency, f_o = 657 Hz

The speed at which the tuning fork fell is calculated by applying Doppler effect formula;

f_o = f_s [\frac{v}{v + v_s} ]

where;

v is speed of sound in air = 343 m/s

v_s is the speed of the falling tuning fork

657 = 683[\frac{343}{343 + v_s} ]\\\\\frac{657}{683} = \frac{343}{343 + v_s}\\\\0.962 = \frac{343}{343 + v_s}\\\\0.962(343 + v_s) = 343\\\\343 + v_s = \frac{343}{0.962} \\\\343 + v_s = 356.55\\\\v_s = 356.55 - 343\\\\v_s = 13.55 \ m/s

The distance traveled by the tuning fork is calculated by applying kinematic equation as follows;

v_s^2 = v_o^2 + 2gh

where;

v_o is the initial speed of the tuning fork = 0

g is acceleration due to gravity = 9.80 m/s²

v_s^2 = 0 + 2gh\\\\h = \frac{v_s^2}{2g} \\\\h = \frac{13.55^2 }{2\times 9.8} \\\\h = 9.37 \ m

Therefore, the distance traveled by the tuning fork is 9.37 m

5 0
3 years ago
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