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Alik [6]
3 years ago
13

How much force is needed to accelerate a 1,800kg car at rate of 1.5 m/s2?

Physics
1 answer:
topjm [15]3 years ago
7 0

<u>Given data:</u>

acceleration (a) = 1.5 m/s² ,

           mass (m) = 1800 Kg ,

        Determine F = ?

         <em>From Newtons II law</em>

                          F = m.a   N

                              = 1800× 1.5

                             = 2700 N

<em>2700 N force needed to accelerate the car</em>

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Answer:

Hey there

Where trying to say that:

Newton's first law gives the concept of force and momentum?

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Newton's first law tells us that objects in motion will remain in motion and objects at rest will remain at rest.

Newton's second law gives us the concept of force and momentum.

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A proton travels with a speed of 5.02×10⁶ m/s in a direction that makes an angle of 60.0° with the direction of a magnetic ficld
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The magnitude of the magnetic force on the proton is 1.25× 10—¹³ N.

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Angel of between the velocity and the magnetic force = 60 °

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The magnitude of the magnetic force on the proton is,

F = q(v \times B)

F = qvB \: sin \:  θ

F = 1.6 \times 10 ^{ - 19}  \times 5.02 \times 10 ^{6}  \times 0.180 \times  \: sin \: 60°

= 1.25 \times 10 ^{ - 13}  \: N

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A student performs a reaction twice. In the second trial, she raises the temperature by 20¨¬C and notices that the reaction take
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Suppose the ski patrol lowers a rescue sled carrying an injured skier, with a combined mass of 97.5 kg, down a 60.0-degree slope
Kitty [74]

a. 1337.3 J work, in joules, is done by friction as the sled moved 28 m along the hill.

b.21,835 J work, in joules, is done by the rope on the sled this distance.

c. 23,170 J   the work, in joules done by the gravitational force on the sled d. The net work done on the sled, in joules is 43,670 J.

       

<h3>What is friction work?</h3>

The work done by friction is the force of friction times the distance traveled times the cosine of the angle between the friction force and displacement

a. How much work is done by friction as the sled moves 28m along the hill?

ans. We use the formula:

friction work = -µ.mg.dcosθ

  = -0.100 * 97.5 kg * 9.8 m/s² * 28 m * cos 60

= -1337.3 J

-1337.3 J work, in joules, is done by friction as the sled moved 28 m along the hill.

b. How much work is done by the rope on the sled in this distance?

We use the formula:

Rope work = -m.g.d(sinθ - µcosθ)

rope work = - 97.5 kg * 9.8 m/s² * 28 m (sin 60 – 0.100 * cos 60)

                     = 26,754 (0.816)

                     = 21,835 J

21,835 J work, in joules, is done by the rope on the sled this distance.

c.  What is the work done by the gravitational force on the sled?

By using  the formula:

Gravity work = mgdsinθ

                    = 97.5 kg * 9.8 m/s² * 28 m * sin 60

                    = 23,170 J

23,170 J   the work, in joules done by the gravitational force on the sled .

       

D. What is the total work done?

By adding all the values

work done =  -1337.3 + 21,835 + 23,170

                 = 43,670 J

The net work done on the sled, in joules is 43,670 J.

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