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3 years ago

Where is the centripetal force directed on a banked curve?

Physics

2 answers:

Alex777 [14]3 years ago

7 0

Answer:

The net force on a car traveling around a curve is the centripetal force, Fc = m v2 / r, directed toward the center of the curve.

Explanation:LET ME KNOW IF ITS WRONG. HAVE A NICE DAY!!!!!!!!!!

san4es73 [151]3 years ago

6 0

Answer: The net force on a car traveling around a curve is the centripetal force, Fc = m v2 / r, directed toward the center of the curve. The centripetal force is directed at right angles to the motion and also along the radius towards the centre of the circular path. In order for a car to move around a circular track, something must provide a centipetal force. One source of the centripetal force is friction between the tires and the road. This means that the tires do not need to supply any friction at all to keep the car from slipping sideways. ...

Hope this helps... stay safe and have a great day!!! :D

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What is the frequency of sound waves whose wavelength is 0.25 m on a day when the air temperature is 25 degree Celsius? Group of

Dvinal [7]

Answer:1384 Hz

Explanation:

Given

wavelength $(\lambda )$ =0.25 m

Temperature T $=25^{\circ}$

at $T=25^{\circ}$ velocity of sound is 346 m/s

and we know

$velocity=frequency\times \lambda$

$v=f\times \lambda$

$346=f\times 0.25$

f=1384 Hz

5 0

3 years ago

Read 2 more answers

found the potential spring energy. not sure how to find the height of the block by knowing only the mass. found the weight of it

evablogger [386]

(h + .16) m g = 1/2 k x^2 total PE of block relative to where it stops

(h + .16) .82 * 9.8 = .5 * 120 * .16^2 PE released = PE of spring

8.04 h + 1.29 = 1.536

h = (1.536 - 1.29) / 8.04 = .031 m = 3.1 cm

7 0

3 years ago

A 30.0-kg child sits on one end of a long uniform beam having a mass of 20.0 kg, and a 40.0-kg child sits on the other end. The

qaws [65]

let the length of the beam be "L"

from the diagram

AD = length of beam = L

AC = CD = AD/2 = L/2

BC = AC - AB = (L/2) - 1.10

BD = AD - AB = L - 1.10

m = mass of beam = 20 kg

m₁ = mass of child on left end = 30 kg

m₂ = mass of child on right end = 40 kg

using equilibrium of torque about B

(m₁ g) (AB) = (mg) (BC) + (m₂ g) (BD)

30 (1.10) = (20) ((L/2) - 1.10) + (40) (L - 1.10)

L = 1.98 m

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3 years ago

Pushing a rock from a hill top is called

wariber [46]

It's called "utter disregard for the safety and welfare
of the people standing at the bottom of the hill".

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3 years ago

A 980 kg roller coaster cart is traveling along a track at 17 m/s before it rolls down a 30 m tall hill (Point A). What will be

MrRissso [65]

The kinetic energy halfway the hill is $2.86\cdot 10^5 J$

Explanation:

If there are no friction forces acting on the cart, we can apply the law of conservation of energy: the mechanical energy of the cart (which is the sum of potential energy + kinetic energy) must be conserved. So we can write:

$U_A +K_A = U_B + K_B$