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TEA [102]
2 years ago
15

Banked Curves: A 600-kg car is going around a banked curve with a radius of 110 m at a steady speed of 24.5 m/s. What is the app

ropriate banking angle so that the car stays on its path without the assistance of friction
Physics
1 answer:
Aleks04 [339]2 years ago
8 0

The angle of baking from the calculation is obtained as 30°.

<h3>What is banking?</h3>

The term banking refers to a means of preventing vehicles from skidding off the road at curves.

We know that the banking angle is obtained from;

θ = tan-1(v^2/rg)

v = 24.5 m/s

r = 110 m

g = 9.8 m/s^2

θ = tan-1(25^2/9.8 * 110)

θ = tan-1(625 /1078)

θ = 30°

Learn more about the banking angle:brainly.com/question/26759099?r

#SPJ1

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PUBLIC SCHOOLS
padilas [110]

Answer:

D. The cart is moving at a constant speed or velocity

Explanation:

Equilibrium is a state of body in which it is either at rest or moves with uniform velocity. The sum of forces acting on such a body is always zero and the sum of all the torques acting on it is also zero.

There are two types of equilibrium as follows:

Static Equilibrium: When a body is at rest it is said to be in static equilibrium.

Dynamic Equilibrium: When a body is moving with constant velocity, then it is said to be in dynamic equilibrium.

Hence, the correct option here will be:

<u>D. The cart is moving at a constant speed or velocity</u>

4 0
3 years ago
When serving a tennis ball, a player hits the ball when its velocity is zero (at the highest point of a vertical toss). The racq
Lilit [14]

Answer:60 gm

Explanation:

Given

initial velocity of ball u=0

Force exerted by racquet F=540 N

time period of force t=5\ ms

final velocity of ball v=45\ m/s

Racquet imparts an impulse to the ball which is given by

J=F\Delta t=\Delta P

J=540\times \Delta t=m(45-0)

m=\frac{540\times 5\times 10^{-3}}{45}

m=60\ gm

8 0
3 years ago
What is the energy of the photon emitted when an electron in a mercury atom drops from energy level f to energy level b?
xeze [42]

The energy of the photon emitted when an electron in a mercury atom drops from energy level f to energy level b is 3.06 eV.

<h3>Change in energy level of the electron</h3>

When photons jump from a higher energy level to a lower level, they emit or radiate energy.

The change in energy level of the electrons is calculated as follows;

ΔE = Eb - Ef

ΔE = -2.68 eV - (-5.74 eV)

ΔE = 3.06 eV

Thus, the energy of the photon emitted when an electron in a mercury atom drops from energy level f to energy level b is 3.06 eV.

Learn more about energy level here: brainly.com/question/14287666

#SPJ1

7 0
2 years ago
According to the kinetic theory, collisions between molecules in a gas _____.a.are inelasticb.never occurc. cause a loss of tota
KonstantinChe [14]

Answer:

d. perfectly elastic

Explanation:

According to the kinetic theory for collisions of gas molecules:

1.The loss of energy is negligible or we can say that it is zero.

2.Molecules of the gas move in a random manner.

3.The collision between molecules and with the wall of the container is perfectly elastic.That is why loss in the energy is zero.

Therefore the correct answer will be d.

d. perfectly elastic

3 0
3 years ago
In an energy pyramid, the total energy available for consumption from one trophic level to the next one it is
Flura [38]

Answer: Conserved

Explanation:

5 0
3 years ago
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