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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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What minimum speed does a 200 g puck need to make it to the top of a frictionless ramp that is 4.1 m long and inclined at 22 ∘?
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Answer:

5.5 m/ sec

Explanation:

Because the inclined surface is frictionless so we can assume that total change of energy is zero

i-e ΔE = 0

Or we can say that difference between final and initial energy is zero i-e

Ef- Ei =0

Where,

Ef= final energy at the top of the ramp= KEf+PEf

Ei= Initial energy at the bottom of the ramp=KEi+PEi

So we have

(KEf+PEf)-(KEi+PEi)=0

==>KEf-KEi+PEf-PEi=0            -------------(1)

KEf = mgh = 200×9.8×h

Where h= Sin 22 = h/d= h/4.1

or

0.375×4.1=h

or h= 1.54 m

So, PEf= 200×9.8×1.54=3018.4 j

and KEf= 1/2 mVf^{2}= 0.5×200×0=0 j

PEi= mgh = 200×9.8×0=0 j

KEi= 1/2 mVi^{2}=0.5×200×Vi^{2}=100Vi^{2} j

Put these values in eq 1, we get;

0-100 Vi^{2}+3018.4-0=0

-100 Vi^{2}=-3018.4

==> Vi^{2}= \frac{3018.4}{100} = 30.184

==>  Vi = \sqrt{30.184}  = 5.5 m.sec

7 0
3 years ago
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