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Nadusha1986 [10]
3 years ago
13

A stunt driver rounds a banked, circular curve. The driver rounds the curve at a high, constant speed, such that the car is just

on the verge of skidding to the outside of the curve. A front view of a car driving on a banked curve. The cross section of the banked road is constructed like a ramp. The car drives transversely to the slope of the ramp, so that the wheels of one side of the car are lower than the wheels on the other side of the car. Which forces are directly responsible for producing the car’s centripetal acceleration? Coriolis force centripetal force frictional force normal force gravitational force
Physics
1 answer:
bagirrra123 [75]3 years ago
5 0

Answer: C

Frictional force

Explanation:

The description of the question above is an example of a circular motion.

For a car travelling in a curved path, the frictional force between the tyres and the road surface will provide the centripetal force.

Since the road is banked, and the cross section of the banked road is constructed like a ramp. The car drives transversely to the slope of the ramp, so that the wheels of one side of the car are lower than the wheels on the other side of the car, for cornering the banked road, the car will not rely only on the frictional force.

Therefore, the correct answer is option C - the frictional force.

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A fairground ride spins its occupants inside a flying saucer-shaped container. If the horizontal circular path the riders follow
ale4655 [162]

Answer: The riders are subjected to 11.5 revolutions per minute

Explanation: Please see the attachments below

3 0
2 years ago
The carbon atom with 12 protons and 12 electrons has a charge of:
olga55 [171]

Answer:

Isotopes are the elements having the same atomic number i.e same number of protons which in turn is equal to same number of electrons. Number of protons and electrons are equal as the atom is electrically neutral. So, the only option is A.

I hope it helps you !

3 0
2 years ago
A proton and an alpha particle are momentarily at rest at adistance r from each other. They then begin to move apart.Find the sp
Arte-miy333 [17]

Answer:

The unknown quantities are:

E and F

The final velocity of the proton is:

√(8/3) k e^2/(m*r)

Explanation:

Hello!

We can solve this problem using conservation of energy and momentum.

Since both particles are at rest at the beginning, the initial energy and momentum are:

Ei = k (q1q2)/r

pi = 0

where k is the coulomb constant (= 8.987×10⁹ N·m²/C²)

and q1 = e and q2 = 2e

When the distance between the particles doubles, the energy and momentum are:

Ef = k (q1q2)/2r + (1/2)m1v1^2 + (1/2)m2v2^2

pf = m1v1 + m2v2

with m1 = m,   m2 = 4m,    v1=vf_p,    v2 = vf_alpha

The conservation momentum states that:

pi = pf      

Therefore:

m1v1 + m2v2 = 0

That is:

v2 = (1/4) v1

The conservation of energy states that:

Ei = Ef

Therefore:

k (q1q2)/r = k (q1q2)/2r + (1/2)m1v1^2 + (1/2)m2v2^2

Replacing

      m1 =  m, m2 = 4m, q1 = e, q2 = 2e

      and   v2 = (1/4)v1

We get:

(1/2)mv1^2 = k e^2/r + (1/2)4m(v1/4)^2 =  k e^2/r + (1/8)mv1^2

(3/8) mv1^2 = k e^2/r

v1^2 = (8/3) k e^2/(m*r)

3 0
3 years ago
A small object begins a free-fall from a height of =81.5 m at 0=0 s . After τ=2.20 s , a second small object is launched vertica
Sergeeva-Olga [200]

Answer:

33.23 m

Explanation:

At the point where both objects will meet, the vertical height will be equal.

From the equations of motion, the vertical height of the body falling at any time is given as

(y - y₀) = ut + ½gt²

y = vertical height at any time T

y₀ = initial height of the object = 81.5 m

u = initial velocity = 0 m/s (body was dropped)

g = -9.8 m/s²

(y - 81.5) = 0 - 4.9T²

y = 81.5 - 4.9T² (eqn 1)

For an object thrown up, the vertical height of the body at any time, t, is given as

(y - y₀) = ut + ½gt²

y = vertical height of the object at any time t

y₀ = initial height of the object = 0 m

u = initial velocity = 40 m/s

g = -9.8 m/s²

y = 40t - 4.9t² (eqn 2)

At the point where the two objects meet, we equate eqn 1 and eqn 2

y = y

81.5 - 4.9T² = 40t - 4.9t²

But T = (t + 2.2) (Since object 2 was dropped 2.2 s after object 1)

81.5 - 4.9(t + 2.2)² = 40t - 4.9t²

81.5 - 4.9(t² + 4.4t + 4.84) = 40t - 4.9t²

81.5 - 4.9t² - 21.56t - 23.716 = 40t - 4.9t²

81.5 - 21.56t - 23.716 - 40t = 0

57.784 = 61.56t

t = (57.784/61.56) = 0.93866 = 0.94 s

Therefore, the vertical height at t = 0.93866 s is

y = (40×0.93866) - 4.9(0.93866²) = 33.23 m

Hope this Helps!!!

5 0
3 years ago
9. Suppose you ride your Segway to the library traveling at 0.5 km/min. It takes you 25
3241004551 [841]

Answer:

12.5 km

Explanation:

0.5 x 25 = 12.5

8 0
3 years ago
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