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marishachu [46]
3 years ago
5

A diver dives off of a raft - what happens to the diver? the raft? how does this relate to newton's third law? action force: ___

____________ reaction force: _____________
Physics
2 answers:
svetlana [45]3 years ago
7 0

1.      The diver transfers forward and dives into the water.

2.      The raft transfers backwards in the water for the reason that of the reaction force.

3.      The action force is the diver pushing off of the raft; and the

4.      reaction force is the raft pushing back on the diver (causing the diver to go forward and into the water)

kondaur [170]3 years ago
5 0
<span>Actually newtons third law says for every action there is an equal and opposite reaction, Hence here in this case, the diver diving of a raft is the action, after which surely reaction should come in the form where the raft and the driver will rebound with same speed back, and hence here the action force is diving and reaction force is rebounding from the diving place, with same intensity.</span>
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These two pls :)))) ill mark brainliest :)
Anna71 [15]

Answer:

Bowling Ball: weight on Earth = 49 N

Textbook: Mass = 2 kg; weight on the moon = 3.2 N

Large dog: weight on Earth = 490 N; weight on the moon = 80 N

Law of Universal Gravitation: F_{G}=\frac{Gm_{1}m_{2}}{r^{2}}

F_{G} = gravitational force (Newtons/N)

<em>G</em> = gravitational constant, 6.67430 × 10¹¹ \frac{N*m^{2}}{kg^{2}}

<em>m</em>₁ and <em>m</em>₂ = masses of two objects (kilograms/kg)

<em>r</em>² = square of distance between centers of the two objects (meters/m)

Have a fantastic day!

4 0
2 years ago
A sealed tank containing seawater to a height of 10.5 mm also contains air above the water at a gauge pressure of 2.95 atmatm. W
weqwewe [10]

Answer:

The water is flowing at the rate of 28.04 m/s.

Explanation:

Given;

Height of sea water, z₁ = 10.5 m

gauge pressure, P_{gauge \ pressure} = 2.95 atm

Atmospheric pressure, P_{atm} = 101325 Pa

To determine the speed of the water, apply Bernoulli's equation;

P_1 + \rho gz_1 + \frac{1}{2}\rho v_1^2 = P_2 + \rho gz_2 + \frac{1}{2}\rho v_2^2

where;

P₁ = P_{gauge \ pressure} + P_{atm \ pressure}

P₂ = P_{atm}

v₁ = 0

z₂ = 0

Substitute in these values and the Bernoulli's equation will reduce to;

P_1 + \rho gz_1 + \frac{1}{2}\rho v_1^2 =  P_2 + \rho gz_2 + \frac{1}{2}\rho v_2^2\\\\P_1 + \rho gz_1 + \frac{1}{2}\rho (0)^2 =  P_2 + \rho g(0) + \frac{1}{2}\rho v_2^2\\\\P_1 + \rho gz_1 =  P_2 + \frac{1}{2}\rho v_2^2\\\\P_{gauge} + P_{atm} + \rho gz_1 = P_{atm} + \frac{1}{2}\rho v_2^2\\\\P_{gauge} +  \rho gz_1 =  \frac{1}{2}\rho v_2^2\\\\v_2^2 = \frac{2(P_{gauge} +  \rho gz_1)}{\rho} \\\\v_2 = \sqrt{ \frac{2(P_{gauge} +  \rho gz_1)}{\rho} }

where;

\rho is the density of seawater = 1030 kg/m³

v_2 = \sqrt{ \frac{2(2.95*101325 \ + \  1030*9.8*10.5 )}{1030} }\\\\v_2 = 28.04 \ m/s

Therefore, the water is flowing at the rate of 28.04 m/s.

7 0
3 years ago
If you put a drop of food coloring in water and watch the drop disperse, is entropy increasing or decreasing.
den301095 [7]

Answer:

Entropy is increasing. Entropy is decreasing.

Explanation:

The Entropy doesn't change.

4 0
3 years ago
The centripetal force acting on the space shuttle
tamaranim1 [39]

Answer:

(4) weight

Explanation:

The centripetal force acting on the space shuttle in orbit is given by:

F=m\frac{v^2}{r}

where

m is the mass of the shuttle

v is the tangential speed of the shuttle

r is the radius of its circular orbit

When the shuttle orbits the Earth, the centripetal force that keeps the shuttle in circular motion is given by the gravitational attraction between the shuttle and the Earth, which corresponds to the weight of the shuttle, and it is given by:

F=G\frac{Mm}{r^2}

where

G is the gravitational constant

M is the Earth's mass

And this force, therefore, corresponds to the centripetal force.

7 0
3 years ago
Read 2 more answers
The table below shows the right ascensions of two stars at a location.
noname [10]
Star 1 - 4 hours right ascension
 Star 2 - 3 hours right ascension
 Subtracting hours right ascension
 4 hours right ascension - 3 hours right ascension = 1 hours right ascension.
Thus,
 star 1 will rise 1 hour before star 2
7 0
2 years ago
Read 2 more answers
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