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OLga [1]
2 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]2 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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By electromagnetic waves.

Explanation:

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What is the temperature when a solid begins to liquefy
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1 year ago
A 10-turn conducting loop with a radius of 3.0 cm spins at 60 revolutions per second in a magnetic field of 0.50T. The maximum e
bogdanovich [222]

Answer:

Maximum emf = 5.32 V

Explanation:

Given that,

Number of turns, N = 10

Radius of loop, r = 3 cm = 0.03 m

It made 60 revolutions per second

Magnetic field, B = 0.5 T

We need to find maximum emf generated in the loop. It is based on the concept of Faraday's law. The induced emf is given by :

\epsilon=\dfrac{d(NBA\cos\theta)}{dt}\\\\\epsilon=NBA\dfrac{d(\cos\theta)}{dt}\\\\\epsilon=NBA\omega \sin\omega t\\\\\epsilon=NB\pi r^2\omega \sin\omega t

For maximum emf, \sin\omega t=1

So,

\epsilon=NB\pi r^2\omega \\\\\epsilon=NB\pi r^2\times 2\pi f\\\\\epsilon=10\times 0.5\times \pi (0.03)^2\times 2\pi \times 60\\\\\epsilon=5.32\ V

So, the maximum emf generated in the loop is 5.32 V.

3 0
2 years ago
Jeff of the Jungle swings on a 7.6-m vine that initially makes an angle of 32 ∘ with the vertical.
shtirl [24]

To solve this problem we will use the trigonometric concepts to find the distance h, which will allow us to find the speed of Jeff and that will finally be the variable that will indicate the total tension, since it is the variable of the centrifugal Force given in the vine at the lowest poing of the swing.

From the image:

cos (32) = \frac{(7.6-h)}{7.6}

h = 1.1548m

When Jeff reaches his lowest point his potential energy is converted to kinetic energy

PE = KE

mgh = \frac{1}{2} mv^2

v = \sqrt{2gh}

v = \sqrt{2(9.8)(1.1548)}

v= 4.75m/s

Tension in the string at the lowest point is sum of weight of Jeff and the his centripetal force

T = W+F_c

T = mg + \frac{mv^2}{r}

T = (83)(9.8)+\frac{(9.8)( 4.75)^2}{7.6}

T = 842.49N

Therefore the tension in the vine at the lowest point of the swing is 842.49N

3 0
1 year ago
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