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3 years ago

A rock is dropped from a bridge. What happens

1 answer:

5 0

When a rock is dropped from a bridge in this way, in terms of the magnitude of the acceleration and the speed of the rock as it falls, the best option is "<span>(1) Both acceleration and speed increase."</span>

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___________________people can see objects that are close more clearly than objects that are far away.

kicyunya [14]

Near-sighted is the answer

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3 years ago

Lucy boards a crowded bus as it sits in the station. There are no available seats, so Lucy stands in the center aisle next to a

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4 years ago

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3 years ago

Read 2 more answers

1. A person of mass 55 kg is in a bumper car that has a mass of 75 kg. They are together at a velocity of 8 m/s.

ExtremeBDS [4]

The momentum of the bumper car after the collision is 1,040 kgm/s.

<h3>Momentum of bumper</h3>

The change in momentum of the bumper is calculated as follows;

P = v(m1 + m2)

P = 8(55 + 75)

P = 1,040 kgm/s

The momentum of the bumper car after the collision is 1,040 kgm/s.

The direction is still the same.

Learn more about momentum here: brainly.com/question/7538238

#SPJ1

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2 years ago

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 $v_{b2}^2=213.16-v_{a2}^2$ :

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