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OLga [1]
3 years ago
15

Car 1 goes around a level curve at a constant speed of 65 km/h . The curve is a circular arc with a radius of 95 m . Car 2 goes

around a different level curve at twice the speed of Car 1. How much larger will the radius of the curve that Car 2 travels on have to be in order for both cars to have the same centripetal acceleration
Physics
1 answer:
Arte-miy333 [17]3 years ago
4 0

Answer:

The radius of the curve that Car 2 travels on is 380 meters.

Explanation:

Speed of car 1, v_1=65\ km/h

Radius of the circular arc, r_1=95\ m

Car 2 has twice the speed of Car 1, v_2=130\ km/h

We need to find the radius of the curve that Car 2 travels on have to be in order for both cars to have the same centripetal acceleration. We know that the centripetal acceleration is given by :

a=\dfrac{v^2}{r}

According to given condition,

\dfrac{v_1^2}{r_1}=\dfrac{v_2^2}{r_2}

\dfrac{65^2}{95}=\dfrac{130^2}{r_2}

On solving we get :

r_2=380\ m

So, the radius of the curve that Car 2 travels on is 380 meters. Hence, this is the required solution.

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3 years ago
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A scooter is traveling at a constant speed v when it encounters a circular hill of radius r = 480 m. The driver and scooter toge
Grace [21]

Answer:

68.585m/sec , 779.1 N

Explanation:

To feel weightless, centripetal acceleration must equal g (9.8m/sec^2). The accelerations then cancel.

From centripetal motion.

F =( mv^2)/2

But since we are dealing with weightlessness

r = 480m

g = 9.8m/s^2

M also cancels, so forget M.

V^2 = Fr

V = √ Fr

V =√ (9.8 x 480) = 4704

= 68.585m/sec.

b) Centripetal acceleration = (v^2/2r) = (68.585^2/960) = 4704/960

= 4.9m/sec^2.

Weight (force) = (mass x acceleration) = 159kg x (g - 4.9)

159kg × ( 9.8-4.9)

159kg × 4.9

= 779.1N

6 0
3 years ago
A toolbox of mass 3.2kg is lowered by a rope from the roof to the ground. Find the acceleration of the toolbox when the force of
Lisa [10]

Answer:

The answer to your question is:

a) 2.7 m/s²

b) -3.6 m/s²

Explanation:

Data

mass of the toolbox = 3.2 kg

a = ?

F = 40 N and F = 20 N

g = 9.81 m/s²

Formula

Second law of motion = F = ma

                              a + g = F / m

                              a = F/m - g

a)                            a = 40/3.2 - 9.81

                              a = 2.69 ≈ 2.7 m/s²   positive up

b)                            a = 20/ 3.2 - 9.81

                              a = 6.25 - 9.81

                                  = - 3.56 ≈ - 3.6 m/s²  negative down

8 0
3 years ago
As an airplane leaves the runway and goes into the air, has acceleration occurred?
malfutka [58]
A, yes because the plane is using air resistance and acceleration is increasing while it goes up. Although you don’t know speed, still yes.
6 0
2 years ago
What is the potential energy of a rock that weighs 10.0 kg and is sitting on top of a hill 100 meters high?l
Firlakuza [10]

Answer:

9800 J

Explanation:

P.E = mgh

  • m = mass
  • g = gravitational acceleration
  • h = height

P.E = (10.0)(9.8)(100) = 9800 J

8 0
3 years ago
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