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vova2212 [387]
3 years ago
9

A golfer hits her tee-shot due north towards the fairway. Her shot has an initial velocity of 60 m/s. A 15 m/s wind is blowing i

n a northwesterly direction (45 degrees west of North). Considering the initial velocity of the ball and the velocity of wind, what will be the resultant velocity (m/s) and the resultant direction of the golf ball assuming no other forces are acting on the ball
Physics
1 answer:
vovangra [49]3 years ago
7 0

Answer:

c=71.4m/s

\theta=8.54\textdegree

Explanation:

From the question we are told that

Initial velocity of 60 m/s

Wind speed V_w= 15 m/s \angle  45 \textdegree

Generally Resolving vector mathematically

  sin(45\textdegree)15=10.6\\cos(45\textdegree)15=10.6

Generally the equation Pythagoras theorem is given mathematically by

c^2=a^2+b^2

c^2=10.6^2 +(10.6+60)^2

c=\sqrt{10.6^2 +(10.6+60)^2}

Therefore Resultant velocity (m/s)

c=71.4m/s

b)Resultant direction

Generally the equation for solving Resultant direction

\theta=tan^-1(\frac{y}{x})

Therefore

\theta=tan^-1(\frac{10.6}{70.6})

\theta=8.54\textdegree

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Answer:

Put the two solid spheres on an inclined plane . Use a meter-stick to hold the spheres on the plane.  Release the two spheres at the same time and let down roll down.  Observe the two spheres as they roll down and repeat the steps.  The hollow sphere will roll last while the solid sphere will roll first. The hollow sphere has more rotational inertia than the solid sphere. This is because the mass of the hollow sphere is distributed farther from its center of gravity.

Explanation:

The description of the experiment for the two spheres is given below:

1. Put the two solid spheres on an inclined plane .

2. Use a meter-stick to hold the spheres on the plane.

3. Release the two spheres at the same time and let down roll down.

4. Observe the two spheres as they roll down and repeat the steps.

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A 1400 kg car driving at 25 m/s slams on its brakes. The coefficient of kinetic friction between the tires and the road is 0.7.
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The acceleration of the car is 6.86 m/s² and the time taken for the car to stop is 3.64 s.

The given parameters;

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The acceleration of the car is calculated as follows;

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a = 0.7 x 9.8

a = 6.86 m/s²

The time taken for the car to stop is calculated by using Newton's second law of motion;

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F = \frac{mv}{t} \\\\ma = \frac{mv}{t}\\\\a = \frac{v}{t} \\\\t = \frac{v}{a} \\\\t = \frac{25}{6.86} \\\\t = 3.64 \ s

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Explanation:

\law \: of \: conservation \: of \: momentum)

\frac{initial}{m1u1 + m2u2}  =  \frac{fi}{m 2u2}

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<em><u>PiA</u></em><em><u> </u></em><em><u>=</u></em><em><u> </u></em><em><u>m1u1</u></em>

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