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harkovskaia [24]
3 years ago
8

3.A golf ball is hit of it’s tee, 200m down the fairway.

Physics
1 answer:
Alinara [238K]3 years ago
4 0

<u>Answers:</u>


3. The diagrams showing the forces acting on the golf ball are in the figure attached. Let’s have a detailed look:


a) Here the ball is under 2 forces:


F1 which is called The Normal force and is perpendicular to the surface of the tee where the ball rests


-F1, related to the gravity force, to the weight of the ball, and has the same magnitude, but the opposite direction (That’s why it has a negative sign).



In this case the sum of the forces is 0



b) Here we have again forces F1 and –F1, but in this very moment the club strikes the ball and we have:


F2, the force of the strike





c) While the ball is in its flight, it is under the following forces:


F1, the force of the lift through the air


-F2, the gravity force

-F3, the force of air resistance, also called drag


F4, the tangencial force of the ball flight




4. Here are the sizes and directions of the resultant forces:


i)Two forces of the same magnitude or size are applied to this block, but in opposite directions (in the x-axis). This is expressed as:



F=-10N+10N=0 The resulting force is zero



ii) Two forces of different size and opposite directions in the y-axis are applied to this block. The sum of the forces is:



F=30N-40N=-10N This means the resulting force is 10N applied downwards



iii) In this case the only force applied to the block is -5N applied downwards



iv) Here there are four forces applied to the block.  

In the y-axis we have to forces of the same size but opposite directions:



F1=10N-10N=0 This means the applied force in the y-axis is zero



In the x-axis we have two forces of different size and opposite directions:



F2=-15N+10N=-5N This means the resulting force is applied to the –x-side



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Answer:

The distance traveled by the skier is 5.309 km

Solution:

As per the question:

Mass of the skier, m = 64 kg

Coefficient of friction between the ski and the snow, \mu_{k} = 0.50

Mass of snow, M = 5.0 kg

Now,

To calculate the distance, 's' traveled by the skier, in order to melt 5.0 kg of snow:

We know that:

f = \mu_{k}N

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\mu_{k} = coefficient of friction

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N = mg

Thus

f = \mu_{k}mg                           (1)

Also,

fs = ML_{f}                                  (2)  

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Thus from eqn (1) and (2):

s = \frac{ML_{f}}{\mu_{k}mg}

s = \frac{5\times 3.33\times 10^{5}}{0.50\times 64\times 9.8}

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5 0
3 years ago
Please just give me an answer, thanks.
Colt1911 [192]

#1

\\ \rm\dashrightarrow P=VI\implies V=\dfrac{P}{I}

\\ \rm\dashrightarrow V=\dfrac{0.4\times 10^3}{33\times 10^{-3}}

\\ \rm\dashrightarrow V=0.012\times 10^{-6}

\\ \rm\dashrightarrow V=12mV

#2

\\ \rm\dashrightarrow R=\rho \dfrac{\ell}{A}

\\ \rm\dashrightarrow \rho =\dfrac{RA}{\ell}

\\ \rm\dashrightarrow \rho=\dfrac{86.3(0.4\times 10^6)}{69}

\\ \rm\dashrightarrow \rho=0.5\times 10^{6}

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#3

Ohms law

\\ \rm\dashrightarrow R=\dfrac{V}{I}

\\ \rm\dashrightarrow R=\dfrac{7(10^3)}{6(10^{-6})}

\\ \rm\dashrightarrow R=1.167\times 10^9\Omega

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2 years ago
An airplane flies horizontally with a constant speed of 155.0 m/s at an unknown altitude. A package is released out of the airpl
vladimir1956 [14]

Answer:

 y₀ = 1020.3 m

Explanation:

This is a projectile launching exercise, in this case as the package is released its initial vertical velocity is zero.

            y = y₀ + v_{oy} t - ½ g t²

when it reaches the ground its height is zero

           0 = y₀ + 0 - ½ g t²

           y₀ = ½ g t²

           

let's calculate

         y₀ = ½ 9.8 14.43²

         y₀ = 1020.3 m

8 0
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jeka94

Answer:

1) Lightning, you see the lightning first and then hear the thunder.

2)When a person far away from you hits a ball with a bat, you can see them striking the ball first and then you will hear the sound of ball striking against the bat.

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