All categories

3 years ago

Curves on some test tracks and race courses, such as the Daytona International Speedway in Florida, are very steeply banked.

Physics

1 answer:

docker41 [41]3 years ago

8 0

Answer:

Speed, v = 45.84 m/s

Explanation:

Given that,

Radius of the curve, r = 100 m

Banking of the curve, $\theta=65^{\circ}$

On the banking of curve, the speed of the vehicle is given by :

$v=\sqrt{rg\ tan\theta}$

$v=\sqrt{100\times 9.8\times \ tan(65)}$

v = 45.84 m/s

So, the speed at which a 100 m radius curve banked is 45.84 m/s. Hence, this is the required solution.

You might be interested in

1 question, brainliest for whoever gets it right.

Kaylis [27]

Answer:

im pretty sure the answer is c please mark me brainliest

6 0

2 years ago

-Plz Help Meh-<br> Calculate the acceleration of a car if the force is 450N and the mass is 1300kg.

Allisa [31]

Answer:

Explanation:

F=Ma

450=1300×a

Ans=0.346m/s²

4 0

2 years ago

HELP ASAP! GIVING BRAINLIEST!

Blababa [14]

Answer:

1. Emma standing on top of mountain

Since she is at the rest position and at some height from the ground so here energy is due to gravitational potential energy

So we have

gravitational potential energy

$U = mgH$

2. Emma jumping down from mountain top

Due to free fall Emma will start moving with some speed in downwards direction so here we have

$KE = \frac{1}{2}mv^2$

motion energy

3. tension in rope at Emma’s lowest position

Due to stretch in the rope here position come to the lowest end and speed comes to zero so whole energy is converted into elastic potential energy

$U = \frac{1}{2}kx^2$

elastic potential energy

4. Emma bouncing back

Due to bouncing back she will again have its kinetic energy with some speed upwards

$KE = \frac{1}{2}mv^2$

motion energy

8 0

2 years ago

Read 2 more answers

Calculate the minimum average power output necessary for a person to run up a 12.0 m long hillside, which is inclined at 25.0° a

Viktor [21]

Answer:

Power, P = 924.15 watts

Explanation:

Given that,

Length of the ramp, l = 12 m

Mass of the person, m = 55.8 kg

Angle between the inclined plane and the horizontal, $\theta=25^{\circ}$

Time, t = 3 s

Let h is the height of the hill from the horizontal,

$h=l\ sin\theta$

$h=12\times \ sin(25)$

h = 5.07 m

Let P is the power output necessary for a person to run up long hill side as :

$P=\dfrac{E}{t}$

$P=\dfrac{mgh}{t}$

$P=\dfrac{55.8\times 9.8\times 5.07}{3}$

P = 924.15 watts

So, the minimum average power output necessary for a person to run up is 924.15 watts. Hence, this is the required solution.

3 0

2 years ago

A person with mass mp = 74 kg stands on a spinning platform disk with a radius of R = 2.31 m and mass md = 183 kg. The disk is i

timurjin [86]

Answer:

1) 883 kgm2

2) 532 kgm2

3) 2.99 rad/s

4) 944 J

5) 6.87 m/s2

6) 1.8 rad/s

Explanation:

1)Suppose the spinning platform disk is solid with a uniform distributed mass. Then its moments of inertia is:

$I_d = m_dR^2/2 = 183*2.31^2/2 = 488 kgm^2$

If we treat the person as a point mass, then the total moment of inertia of the system about the center of the disk when the person stands on the rim of the disk:

$I_{rim} = I_d + m_pR^2 = 488 + 74*2.31^2 = 883 kgm^2$

2) Similarly, he total moment of inertia of the system about the center of the disk when the person stands at the final location 2/3 of the way toward the center of the disk (1/3 of the radius from the center):

$I_{R/3} = I_d + m_p(R/3)^2 = 488 + 74*(2.31/3)^2 = 532 kgm^2$

3) Since there's no external force, we can apply the law of momentum conservation to calculate the angular velocity at R/3 from the center:

$I_{rim}\omega_{rim} = I_{R/3}\omega_{R/3}$

$\omega_{R/3} = \frac{I_{rim}\omega_{rim}}{I_{R/3}}$

$\omega_{R/3} = \frac{883*1.8}{532} = 2.99 rad/s$

4)Kinetic energy before:

$E_{rim} = I_{rim}\omega_{rim}^2/2 = 883*1.8^2/2 = 1430 J$

Kinetic energy after:

$E_{R/3} = I_{R/3}\omega_{R/3}^2/2 = 532*2.99^2/2 = 2374 J$

So the change in kinetic energy is: 2374 - 1430 = 944 J

5) $a_c = \omega_{R/3}^2(R/3) = 2.99^2*(2.31/3) = 6.87 m/s^2$

6) If the person now walks back to the rim of the disk, then his final angular speed would be back to the original, which is 1.8 rad/s due to conservation of angular momentum.

3 0

3 years ago