All categories

3 years ago

The Indianapolis speedway consists of a 2.5 mile track having four turns, each 0.25 mile long and banked at 9 12'

Physics

1 answer:

blsea [12.9K]3 years ago

8 0

Answer: Your question is missing below is the question

Question : What is the no-friction needed speed (in m/s ) for these turns?

answer:

20.1 m/s

Explanation:

2.5 mile track

number of turns = 4

length of each turn = 0.25 mile

banked at 9 12'

Determine the no-friction needed speed

First step : calculate the value of R

2πR / 4 = πR / 2

note : πR / 2 = 0.25 mile

∴ R = ( 0.25 * 2 ) / π

= 0.159 mile ≈ 256 m

Finally no-friction needed speed

tan θ = v^2 / gR

∴ v^2 = gR * tan θ

v = √9.81 * 256 * tan(9.2°) = 20.1 m/s

You might be interested in

Exercise that strengthens your heart, increases your lung capacity, and helps you to burn fate and lose weight is..

olga2289 [7]

Answer : Cardio training

Explanation : Cardio exercise helps improve lung function , strengthens your heart and improve your endurance and increases lung capacity. The peoples do cardio is burn calories and lose the weight.

Cardio helps decrease your heart rate and blood pressure and improve your breathing.

8 0

3 years ago

What is the difference between the initial position and the final position of an object?

Grace [21]

The initial is where you are starting and the final postion is where the object ends up

5 0

3 years ago

Find the electric field at a point midway between two charges of 30.0×10 power -9 and 60.0×10 power -9 separated by a distance o

KATRIN_1 [288]

Answer:

The electric field at a point midway between the two charges, E = -1.8 * 10⁴ N/C

Explanation:

Let the midpoint of the two charges be considered as the origin, and charge A = 30.0 * 10⁻⁹ C be moving in the +x- axis and the charge B = 60.0 * 10⁻⁹ C be moving in the -x-axis.

Electric field, E = kQ/r² where k is a constant = 9.0 * 10⁹ N.m²/C², Q = quantity of charge, r = distance of separation

In the given question,r = 30.0 cm = 0.03 m; the midway point between A and B = 0.03/2 = 0.015 m

Electric field due to charge A

Ea = +(9.0 * 10⁹ N.m²/C² * 30.0 * 10⁻⁹ ) / ( 0.015 m)²

Ea = +1.8 * 10⁴ N/C

Electric field due to charge B

Eb = -(9.0 * 10⁹ N.m²/C² * 60.0 * 10⁻⁹ ) / ( 0.015 m)²

Eb = -3.6 * 10⁴ N/C

The resultant electric field E = Ea + Eb

E = (+1.8 * 10⁴ + -3.6 * 10⁴) N/C

E = -1.8 * 10⁴ N/C

Therefore, the electric field at a point midway between the two charges, E = -1.8 * 10⁴ N/C

7 0

3 years ago

An object of mass 20 kg is raised vertically through a distance of 8 m above ground level. If g = 10 m/s2, what is the gravitati

IgorC [24]

2400joules

Explanation:

P.E

m= 20kg h=8m g=10m

P.E= 20×8×10

=2400joules

6 0

2 years ago

A simple pendulum having a length of 1.02 m and a mass of 6.66 kg undergoes simple harmonic motion when given an initial speed o

attashe74 [19]

Answer:

Time period of pendulum is 2.02 s.

Explanation:

A simple pendulum is a device which consists of mass m hanging from the string of length L attached to the some point.When displaced and released its swings back and forth with periodic motion.

The time period of pendulum is defined as time taken by the pendulum to complete one full oscillation . it is denoted by T.

By Huygens law of period of pendulum,

T = 2π $\sqrt{\frac{L}{g} }$ eqn 1

where L is the length of pendulum,

g is acceleration due to gravity

Period of pendulum is independent of the mass of pendulum,

Substituting values in eqn 1

T = 2π $\sqrt{\frac{1.02}{9.8\\} }$

T = 2.02 s

Time period of pendulum is 2.02 s.

4 0

3 years ago