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Novay_Z [31]
3 years ago
5

Suppose that a sled is accelerating at a rate of 2 m/s2. If the net force is tripled and the mass is halved, then what is the ne

w acceleration of the sled?
(show how you solved it)
Physics
1 answer:
yanalaym [24]3 years ago
3 0

Start with Newton's 2nd law of motion:

                             Force = (mass) x (acceleration) .

Since we're going to be talking about acceleration,
lets divide each side by (mass):

                             Acceleration = (force) / (mass) .

OK.  We start out with a certain acceleration 'A₀'.  It's the result of
a certain force 'F₀' and a certain mass 'M₀'.  (I used the little subscripts " ₀ "
to show that these are the originals, before any changes.)

             Original acceleration = (Original force) / (original mass)

                              A₀             =          F₀            /        M₀ .

Now you want to triple the force and cut the mass in half: 

                     New acceleration  A₁ = (3 F₀) / (1/2 M₀) .

Divide each side by 3:           A₁ / 3  =  F₀ / (1/2 M₀) .

Multiply each side by  1/2 :   (1/2 A₁) / 3  =  F₀/M₀

                                                   A₁ / 6  =  F₀/M₀

Take a look at the right side of that equation . . .  F₀/M₀ .
That's just the original acceleration  A₀ .
So now, after the change, we have                 A₁ / 6  =  A₀ .

You asked "What is the new acceleration ?"
OK.  Multiply each side by  6  :                        A₁ = 6 A₀ .

Whatever the original acceleration was, the
new acceleration is 6 times as much.

If it was originally  2 m/s², then after the change, it becomes  (6 x 2)  =  12 m/s² .

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Answer:\theta =9.96^{\circ} North of west

Explanation:

Given

Plane wishes to fly in west

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Plane absolute speed=195 km/h

To fly towards west velocity in Y direction should be zero

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\theta =9.96^{\circ}

so Plane should head towards 9.96^{\circ} North of west in order to fly in west.

So plane

actual velocity is

v=-195cos9.96\hat{i}+195sin9.96\hat{j}

5 0
3 years ago
12. A concave lens has a focal length of 10 cm. An object 2.5 cm high is placed 30 cm from the lens. Determine the position and
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Answer:

I think 9.5

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At the surface of Jupiter's moon Io, the acceleration due to gravity is 1.81 m/s2 . A watermelon has a weight of 54.0 N at the s
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A soccer player takes a corner kick, lofting a stationary ball 33.0° above the horizon at 15.0 m/s. If the soccer ball has a mas
Alexxandr [17]

Explanation:

It is given that,

Mass of the soccer ball, m = 0.425 kg

Speed of the ball, u = 15 m/s

Angle with horizontal, \theta=33^{\circ}

Time for which the player's foot is in contact with it, \Delta t = 5.1\times 10^{-2}\ s

Part A,

The x component of the soccer ball's change in momentum is given by :

\Delta p_x=mv\ cos\theta

\Delta p_x=0.425\times 15\ cos(33)

p_x=5.34\ kg-m/s

The y component of the soccer ball's change in momentum is given by :

\Delta p_y=mv\ sin\theta

\Delta p_y=0.425\times 15\ sin(33)

p_y=3.47\ kg-m/s

Hence, this is the required solution.

3 0
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