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fgiga [73]
3 years ago
7

A ball bounced by a basketball player on the floor bounces back up at her. Newton's First Law Newton's Second Law Newton's Third

Law
Physics
1 answer:
taurus [48]3 years ago
4 0

Answer:

Newton's Third Law of Motion

Explanation:

Newton's Third Law of Motion which states that, for every action there is an equal but opposite reaction.

This ultimately implies that, in every interaction, there is a pair of forces acting on the two interacting objects.

In this scenario, a ball bounced by a basketball player on the floor bounces back up at her.

According to Newton's Third Law of Motion, the statement above simply means that in every interaction, there is a pair of forces acting on the two interacting objects i.e the ball and floor. The size of the force on the ball equals the size of the force on the floor. These two forces are called action and reaction forces and are the subject of Newton's third law of motion.

Hence, the ball bounced by the basketball player on the floor would bounce back in equal magnitude.

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Hockey puck B rests on a smooth ice surface and is struck by a second puck A, which has the same mass. Puck A is initially trave
Marina86 [1]

Answer:

a.v_{b2}=6.8544 \ m/s\\\\b. v_{a2}=12.891 \ m/s\\\\c. \theta _b=62\textdegree

Explanation:

Puck A's initial speed is v_a_1=14.6m/s and move in a direction \theta_b=28.0\textdegree after the collision.

#P_1=P_2 since there's no external force on the system(P=mv).

#The collision equation can be written as;

m_av_a_1+m_bv_b_1=m_av_a_2+m_bv_b_2

The kinetic energies before and after the collision are expressed as:

K_a_1+K_{b1}=K_a_2+K_{b2}, \ K=0.5mv^2

0.5m_av_a_1+0.5m_b(0)=0.5m_av_a_2+0.5m_bv_b_2\\\\14.6^2=v_{a2}^2+v_{b2}^2\\\\v_{b2}^2=213.16-v_{a2}^2

Let +x along A's initial direction and +y along A's final direction makes the angle 62\textdegree

\dot v_{a1}=14.6i\\\\\dot v_{a2}=(v_{a2} \ cos \ 28\textdegree)i+(v_a_2\ sin \ 28\textdegree)j\\\\\dot v_{b2}=v_{b2x}i+v_{b2y}j

#Substitute in v_{b2}^2=213.16-v_{a2}^2:

\dot v_{b2}=(14.6i)-\dot v{a2}\\\\\dot v_{b2}.\dot v_{b2}=v_{b2}^2\\\\\#Right \ side\\\\(14.6i-\dot v{a2}).(14.6i-\dot v{a2})=(14.6i)^2+\dot v_{a2}^2.\dot v_{a2}^2-2(14.6i).\dot v_{a2}\\\\=213.16+v_{a2}^2-2(14.6i).(v_{a2}\ cos 28\  \textdegree i+v_{a2} \ sin \ 28\textdegree j)\\\\v_{b2}^2=213.16+v_{a2}^2-29.2\ cos \ 28\textdegree v_{a2}\ \  v_{b2}^2=213.16-v_{a2}^2\\\\213.16-v_{a2}^2=213.16+v_{a2}^2-29.2\ cos \ 28\textdegree v_{a2}\\\\2v_a2}^2=29.2\ cos\ 28 \textdegree v_{a2}\\\\

v_{a2}=12.891\ m/s

Hence, the the speed of puck A after the collision is 12.891  m/s

#. The velocity of A after the collision is;

\dot v_{a2}=12.891 \ cos \ 28 \textdegree i+12.891 \ sin \ 28\textdegree j\\\\=11.382i+6.052j

Substitute \dot v_{a2} into \dot v_{b2}=(14.6i)-\dot v{a2}:

\dot v_{b2}=14.6i-(11.382i+6.052j)\\\\=3.218i-6.052j

This is the velocity of puck B after the collision, it's speed is:

v_{b2}=\sqrt{v_{b2x}^2+v_{b2y}^2}\\\\=\sqrt{3.218^2+(-6.052)^2}\\\\v_{b2}=6.8544 \ m/s

The velocity of puck B after the collision is 6.8544 m/s

c. The direction of puck B after the collision is:

\theta _b=tan^{-1}\frac{v_{b2y}}{v_{b2x}}\\\\=tan^{-1} \frac{-6.052}{3.218}\\\\\approx 62\textdegree

Hence, the direction of B's velocity after the collision is 62°

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What are three physical properties of gases
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Explanation:

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Answer:

(3 ) friction hope this is correct

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State and explain the joules law​
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Answer:

Joule’s law of heating states that when a current ‘i ' passes through a conductor of resistance ‘r’ for time ‘t’ then the heat developed in the conductor is equal to the product of the square of the current, the resistance and time.

H = i 2 rt

The reason behind the heat produced is the collision of the free electrons with the lattice ions or atoms while accelerating in presence of the external electric field.

we know that

volt= work done/ total charge or v = w/q

from this we get w= vq

we know that current i = q / t

so q = it

putting this in prev equation

we get w = vit

from ohms law v = ir

putting it in prev eq we get

w = irit = i2rt joule as work is stored as heat soH = I2RT Joules

if a current carrying conductor having resistance (r) gets heated by to passing of current (i) for time (t) and the heat thereby produced be H

joules law states that

H propotional to i^ 2 when r and t are constant

H propotional to r when I and t are constant

H propotional to t when r and I constant

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