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4 years ago

Explain when a falling object is in free fall.

Physics

2 answers:

lapo4ka [179]4 years ago

5 0

A falling object is in free fall if the only force acting on it is gravity.

Ksenya-84 [330]4 years ago

3 0

There is no gravity

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If it is 9 am in San Francisco, is it also 9 am in New York? Why or why not?

Sedaia [141]

Answer:

It wouldn't be 9am in New York because San Francisco is 3 hours behind of New York

Explanation:

7 0

3 years ago

Which of the following best describes seamounts and islands of the deep ocean basins?

Rina8888 [55]

Answer: The correct option is (d)

lava flows built up from the ocean floor by multiple, summit and flank eruptions

Explanation:

Piles of baseltic lava flows built up from the ocean floor by multiple summit and flank eruptions describes seamounts and islands of the deep ocean basins.

4 0

4 years ago

A 0.17kg ball rolls at 0.75m/s to the right on a frictionless surface and collides with a 0.17kg ball rolling to the left at 0.6

mario62 [17]

Both momentum and kinetic energy are conserved in elastic collisions (assuming that this collision is perfectly elastic, meaning no net loss in kinetic energy)

To find the final velocity of the second ball you have to use the conversation of momentum:

*i is initial and f is final*

Δpi = Δpf

So the mass and velocity of each of the balls before and after the collision must be equal so

Let one ball be ball 1 and the other be ball 2

m₁ = 0.17kg

v₁i = 0.75 m/s

m₂ = 0.17kg

v₂i = 0.65 m/s

v₂f = 0.5

m₁v₁i + m₂v₂i = m₁v₁f + m₂v₂f

Since the mass of the balls are the same we can factor it out and get rid of the numbers below it so....

m(v₁i + v₂i) = m(v₁f + v₂f)

The masses now cancel because we factored them out on both sides so if we divide mass over to another side the value will cancel out so....

v₁i + v₂i = v₁f + v₂f

Now we want the final velocity of the second ball so we need v₂f

so...

(v₁i + v₂i) - v₁f = v₂f

Plug in the numbers now:

(0.75 + 0.65) - 0.5 = v₂f

v₂f = 0.9 m/s

8 0

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

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