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vova2212 [387]
2 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]2 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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3 years ago
A boy throws a stone straight upward with an initial
Dominik [7]

Answer:

14.8m

Explanation:

Given parameters:

Initial speed  = 17m/s

Unknown:

Maximum height  = ?

Solution:

At the maximum height, the final speed will be 0m/s;

 We use of the kinematics equation to solve this problem.

     V²   = U²   - 2gH

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g is the acceleration due to gravity

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         0²   = 17²  -  (2  x 9.8 x h )

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5 0
3 years ago
How does your power output in climbing the stairs compare to the power output of a 100-watt light bulb? if your power could have
cricket20 [7]
1) Assuming an adult person has an average mass of m=80 kg, and assuming it takes about 30 seconds to climb 5 meters of stairs, the energy used by the person is
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And since the estimate we made is very rough, we can say that the power output of the person is comparable to the power output of the light bulb of 100 W.

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6 0
3 years ago
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An object is dropped from a height of 75.0 m above ground level. (a) Determine the distance traveled during the first second. (b
lys-0071 [83]

Answer:

a)Distance traveled during the first second = 4.905 m.

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c)Distance traveled during the last second of motion before hitting the ground = 33.45 m

Explanation:

a) We have equation of motion

             S = ut + 0.5at²

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              S = 0.5gt²

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b)  We have equation of motion

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c) We have S = 0.5gt²

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   We need to find distance traveled last second

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   Distance traveled during the last second of motion before hitting the ground = 33.45 m

       

3 0
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Marta_Voda [28]
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3 years ago
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