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Lady_Fox [76]
3 years ago
14

A race car drives around a turn on a flat race track at constant speed. Identify the force that allows the race car to travel ar

ound the turn?
A)
Air resistance applies the force needed.


B)
No force is required to go around the turn.


C)
The weight of the car applies the force needed.


D)
The friction between the tires and the track applies the force needed
Physics
1 answer:
lidiya [134]3 years ago
7 0
D) The friction between the tires and the track applies the force needed.
Without the friction, the car wouldn't handle the curve and would fly off the track.
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the reaction force in this situation would be B

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The action is the wings pushing down whilst the reaction is the air pushing up which allow the bird to fly .

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An artillery shell is fired with an initial velocity of 300 m/s at 52.0° above the horizontal. To clear an avalanche, it explode
sasho [114]

The x- and y-coordinates are 9142.57 m and -304.425 m

<u>Explanation:</u>

As the motion of the shell is in a plane (two dimensional space) and the acceleration is that due to gravity which is vertically downward, we resolve initial velocity of the shell v_{0} in horizontal and vertical directions. If the initial velocity of the shell is making angle with the horizontal, the horizontal component of initial velocity will be

                v_{x}=v_{0} \times \cos \theta

As the acceleration of the shell is vertical having no horizontal component, the shell may be considered to move horizontally with constant velocity of v_{x} and hence the horizontal distance covered (or the x coordinate of the shell with point of projection as origin) is given by

           v_{x}=v_{o} \times \cos \theta=300 \times \cos \left(52^{\circ}\right)=184.69 \mathrm{m} / \mathrm{s}

           v_{y}=v_{o} \times \sin \theta==300 \times \sin \left(52^{\circ}\right)=236.4 \mathrm{m} / \mathrm{s}

For motion with constant acceleration, we know

            s=s_{0}+v_{0} t+\left(\frac{(1)}{2}\right) a t^{2}

Along the horizontal, x-axis, we might write this as

            x=x_{0}+v_{x 0} t+\left(\frac{1}{2}\right) a_{x} t^{2}

Measuring distances relative to the firing point means

               x_{0}=0

we know that,

              a_{x}=0

or,

             v_{x}=v_{x 0}=\text { constant }

By applying the values, we get,

           x=0+(184.69 \times 49.5)+\left(\left(\frac{1}{2}\right) \times 0 \times(49.5)^{2}\right)=9142.57 \mathrm{m}

The acceleration of gravity is vertically downward and is g=-9.8 \mathrm{m} / \mathrm{s}^{2} , hence the vertical distance covered (or y coordinate of the shell) is given by the second equation of motion

           y=y_{0}+v_{y 0} t+\left(\frac{1}{2}\right) a_{y} t^{2}

we know, y_{0}=0 and a_{y}=-9.8 \mathrm{m} / \mathrm{s}^{2}, so,

          y=0+(236.4 \times 49.5)+\left(\left(\frac{1}{2}\right) \times(-9.8) \times(49.5)^{2}\right)

                 y = 11701.8 - 4.9(2450.25)= 11701.8 - 12006.225 = - 304.425 m

7 0
3 years ago
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