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3 years ago

What occurs when a swimmer pushes through the water to swim?

2 answers:

6 0

When a swimmer pushes threw water to swim they are propelled forward because of the water resistance against the hand and feet.

ehidna [41]3 years ago

4 0

The correct answer to the question is that water will exert a force on the swimmer in forward direction, and swimmer will move along that direction.

EXPLANATION:

Before going to answer this question, first we have to understand Newton's third laws of motion.

Newton's third laws of motion states that, for every action, there is an equal and opposite reaction.

In the given question, the swimmer pushes the water in backward direction. This is the force of action. As per Newton's third law, the water must exert a reaction on the swimmer. The water applies a force on the swimmer in forward direction for which the swimmer can swim in the water.

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If the speed of an object doubles, how does that affect its kinetic energy? A. Halves B. Doubles C. Quarters D. Quadruples

kvasek [131]

Answer is :

D. Quadruples

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Which substance will cool down the slowest during the night, and why?

Studentka2010 [4]

Cool down the Slowest

B. Seawater, because it heats up slower and gives away heat slower than sand.

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3 years ago

A box of mass 26 kg is initially at rest on a flat floor. The coefficient of kinetic friction between the box and the floor is 0

Kazeer [188]

Answer:

$\Delta K = 52J$

Explanation:

The change in kinetic energy will be simply the difference between the final and initial kinetic energies: $\Delta K=K_f-K_i$

We know that the formula for the kinetic energy for an object is:

$K=\frac{mv^2}{2}$

where m is the mass of the object and v its velocity.

For our case then we have:

$\Delta K = K_f-K_i=\frac{mv_f^2}{2}-\frac{mv_i^2}{2}=\frac{m(v_f^2-v_i^2)}{2}$

Which for our values is:

$\Delta K = \frac{m(v_f^2-v_i^2)}{2} = \frac{(26Kg)((2m/s)^2-(0m/s)^2)}{2} = 52J$

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3 years ago

Use Kepler’s third law and the orbital motion of Earth to determine the mass of the Sun. The average distance between Earth and

dexar [7]

Kepler’s
third law formula: T^2=4pi^2*r^3/(GM)

We’re trying to find M, so:

M=4pi^2*r^3/(G*T^2)

M=4pi^2*(1.496
× 10^11 m)^3/((6.674× 10^-11N*m^2/kg^2)*(365.26days)^2)

M=1.48× 10^40(m^3)/((N*m^2/kg^2)*days^2))

Let’s work
with the units:

(m^3)/((N*m^2/kg^2)*days^2))=

=(m^3*kg^2)/(N*m^2*days^2)

=(m*kg^2)/(N*days^2)

=(m*kg^2)/((kg*m/s^2)*days^2)

=(kg)/(days^2/s^2)

=(kg*s^2)/(days^2)

So:

M=1.48× 10^40(kg*s^2)/(days^2)

Now we need to convert days to seconds in order to cancel
them:

1 day=24 hours=24*60minutes=24*60*60s=86400s

M=1.48× 10^40(kg*s^2)/((86400s)^2)

M=1.48× 10^40(kg*s^2)/(
86400^2*s^2)

M=1.48× 10^40kg/86400^2

M=1.98x10^30kg

The
closest answer is 1.99
× 10^30

(it may vary
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