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sleet_krkn [62]
3 years ago
8

The Indianapolis Motor Speedway has four banked curves, each of which forms a quarter of a circle. Suppose a race car speeds alo

ng one of these curves with a constant tangential speed of 75.0 m/s. Neglecting the effects due to the banking of the curve, the centripetal acceleration on the car is 22.0 m/s2. What is the radius of the curve?
Physics
1 answer:
boyakko [2]3 years ago
4 0

Answer:

r =  255.68 m

Explanation:

When a body moves in a circular path, an acceleration, due to constant change in its direction, is developed, known as centripetal acceleration. The centripetal acceleration acts towards the center of the circular path. The formula to calculate the centripetal acceleration is given as follows:

ac = v²/r

where,

ac = centripetal acceleration = 22 m/s²

v = tangential speed = 75 m/s

r = radius of curve = ?

Therefore,

22 m/s² = (75 m/s)²/r

r = (75 m/s)²/(22 m/s²)

<u>r =  255.68 m</u>

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A man can throw a ball a maximum horizontal distance of 56.1523 m. The acceleration of gravity is 9.8 m/s 2 . How far can he thr
IceJOKER [234]

Answer:

hmax= 28.2 m

Explanation:

Horizontal movement:

  • For the horizontal movement, as once released, nothing exerts any influence on the ball in the horizontal direction, it will keep moving at a constant speed, so, the equation for the horizontal displacement is as follows:

        x = v₀*t = 56.1523 m

Vertical movement:

  • For the vertical movement, once thrown upward, the ball is under the sole influence of gravity, g, which causes the ball to accelerate downward, and which magnitude is 9.8 m/s².
  • At any time, applying the definition of acceleration, we can find the value of the velocity, as follows:

        vf = vo-g*t

  • When the ball reaches to the maximum height, the ball will come momentarily at rest, so vf =0.
  • At this point, we can find the time at which  the ball reached to its highest point, as follows:

        t = \frac{vo}{g}

  • In the same way, we can find the maximum height reached by the ball, just replacing this value of time in the equation for the displacement, in the vertical direction, as follows:

        h = vo*t -\frac{1}{2} *g*t^{2}

        hmax = vo*(\frac{vo}{g}) -\frac{1}{2} *g*(\frac{vo}{g} )^{2} =\frac{vo^{2}}{2*g}

  • Now, returning to the horizontal movement, as the time must be the same for both movements, we can replace the value for time we have just found, in the equation for the horizontal displacement:

       x = v0x * t = v0x* \frac{voy}{g}

  • But as we know that vox = voy, we can rewrite the equation above as follows:

       x = v0* t = v0* \frac{vo}{g} =\frac{vo^{2} }{g} = 56.1523 m

  • We can solve for v₀, as follows:

        vo = \sqrt{56.1523 m *9.8 m/s2} =23.5 m/s

  • Now, we can replace this value in the expression for hmax, as follows:
  • hmax = \frac{vo^{2}}{2*g} =\frac{(23.5m/s)^{2}}{2*9.8 m/s2} =  28.2 m
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4 years ago
*brainlest* An object accelerates from rest with a constant acceleration of 2m/s^2. How far will it have moved after 9s? Equatio
boyakko [2]

Answer: You throw an object upwards from the ground with a velocity of 20m/s and it is subject to a downward

acceleration of 9.8 m/s2. How high does it go?

Given Formula Set up Solution

vi=20 m/s

vf=0 m/s

a=-9.8 m/s2

x=?

vf

2 = vi

2 + 2a x (0 m/s)2= (20 m/s)2 + 2 (-

9.8 m/s2) x

x=20.4 meters

Explanation: i known/ i hoped that helped.

6 0
3 years ago
Two iron blocks are the same size. One block has a higher temperature than the other. Which describes the thermal energy of thes
a_sh-v [17]
<span>Higher temperature means greater kinetic energy, which means the warmer block has greater thermal energy.

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3 years ago
Determine the thrust produced if 1.5 x 10^3 kg of gas exits the combustion chamber each second, with a speed of 4.00 x 10^3 m/s.
ozzi

Answer:

The thrust is 6\times 10^6\ N

Explanation:

Given that,

Mass of gas, m=1.5\times 10^3\ kg

The rate at which the gas is expelling, \dfrac{dv}{dt}=4\times 10^{3}\ m/s

We need to find the thrust produced by the gas.

We know that force is equal to the rate of change of momentum. So,

F=\dfrac{p}{t}

Also, p = mv

F=\dfrac{mv}{t}

So,

F=1.5\times 10^3\times 4\times 10^3\\\\F=6\times 10^6\ N

So, the thrust is 6\times 10^6\ N

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