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3 years ago

8

As Clinton walks he pushes his shoe against the track.

1 answer:

Valentin [98]3 years ago

4 0

Answer:A:The track pushes back on Clinton's shoe with the same force.

Explanation:According to Newton's third law of motion, for every action force there is an equal and opposite reaction force. In this case, the action force is Clinton's shoe pushing on the track. As this happens, there is an equal and opposite reaction force in which the track pushes back on Clinton's shoe with the same force.

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Scientists have observed that the average sunspots have increased in the last 50 years. Which of these statements best describes

olga2289 [7]

The right answer for the question that is being asked and shown above is that: "B) Earth's temperature has not changed over time." Scientists have observed that the average sunspots have increased in the last 50 years. The statement that best describes an effect of this increase in sunspots <span>Earth's temperature has not changed over time. </span>

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4 years ago

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Which of the following is the correct equation we get

UkoKoshka [18]

Answer:

$\frac{a}{b}=c$ which agrees with the third answer in your list of answer options

Explanation:

Start with:

$a=b\,*\,c$

divide both sides by b to isolate c on the right:

$\frac{a}{b} =\frac{b\,*\,c}{b} \\\frac{a}{b} =c$

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3 years ago

The quickest braking by acceleration is 8.8 m/s set on a car driven at 32m/s to rest. How long did it take the car to brake?

Olin [163]

3.6 I hope this helps

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3 years ago

A 2.1 ✕ 103-kg car starts from rest at the top of a 5.9-m-long driveway that is inclined at 19° with the horizontal. If an avera

Lina20 [59]

Answer:

3.9 m/s

Explanation:

We are given that

Mass of car,m= $2.1\times 10^3 kg$

Initial velocity,u=0

Distance,s=5.9 m

$\theta=19^{\circ}$

Average friction force,f= $4.0\times 10^3 N$

We have to find the speed of the car at the bottom of the driveway.

Net force, $F_{net}=mgsin\theta-f=2.1\times 10^3\times 9.8sin19-4.0\times 10^3$

Where $g=9.8 m/s^2$

Acceleration, $a=\frac{F_{net}}{m}=\frac{2.1\times 10^3\times 9.8sin19-4.0\times 10^3}{2.1\times 10^3}$

$v=\sqrt{2as}$

$v=\sqrt{2\times \frac{2.1\times 10^3\times 9.8sin19-4.0\times 10^3}{2.1\times 10^3}\times 5.9}$

v=3.9 m/s

7 0

3 years ago

A 2.4-kg cart is rolling along a frictionless, horizontal track towards a 1.7-kg cart that is held initially at rest. The carts

inessss [21]

Answer:

Explanation:

Mass of first cart M1=2.4kg

Velocity of first cart U1=4.1m/s

Mass of second cart M2=1.7kg

Second cart is initially at rest U2=0

After an instant, the velocity of the second cart is U2=-2.8m/s

Now after collision the two cart move together with the same velocity I.e inelastic collision

Using conservation of momentum

Momentum before collision, = momentum after collision

M1U1 + M2U2 = (M1+M2)V

2.4×4.1 + 1.7× -2.8 =(2.4+1.7)V

9.84 - 4.76 = 4.1V

5.08=4.1V

V=5.08/4.1

V=1.24m/s

The momentum of the two cart at that instant is

M1U1+M2U2

2.4×4.1 + 1.7× -2.8

9.84 - 4.76

5.08kgm/s

So the momentum at the instant the velocity is 4.1m/s for cart 1 and -2.8m/s for cart 2 is 5.08kgm/s

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3 years ago