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notsponge [240]
3 years ago
15

A race-car drives around a circular track of radius RRR. The race-car speeds around its first lap at linear speed v_iv i ​ v, st

art subscript, i, end subscript. Later, its speed increases to 4v_i4v i ​ 4, v, start subscript, i, end subscript. How does the magnitude of the car's centripetal acceleration change after the linear speed increases
Physics
1 answer:
OLEGan [10]3 years ago
8 0

Answer:

The magnitude of the centripetal acceleration increases by 16 times when the linear speed increases by 4 times.

Explanation:

The initial centripetal acceleration, a of the race-car around the circular track of radius , R with a linear speed v is a = v²/R.

When the linear speed of the race-car increases to v' = 4v, the centripetal acceleration a' becomes a' = v'²/R = (4v)²/R = 16v²/R.

So the centripetal acceleration, a' = 16v²/R.

To know how much the magnitude of the car's centripetal acceleration changes, we take the ratio a'/a = 16v²/R ÷ v²/R = 16

a'/a = 16

a' = 16a.

So the magnitude of the centripetal acceleration increases by 16 times when the linear speed increases by 4 times.

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Evaluate u+xy where U=3 X=4 Y=7
Reptile [31]

Answer:

31

Explanation:

Given:

U=3

X=4

Y=7

u + xy

Substitute the given values to the equation:

3 + (4)(7)

3 + 28

31

6 0
2 years ago
How many DAYS occur between a new moon and a first quarter moon?
N76 [4]

Answer:

About 29.5

Explanation:

4 0
2 years ago
Read 2 more answers
A car is traveling at a constant speed of 33 m/s on a highway. At the instant this car passes an entrance ramp, a second car ent
Paha777 [63]

Answer:

0.8712 m/s²

Explanation:

We are given;

Velocity of first car; v1 = 33 m/s

Distance; d = 2.5 km = 2500 m

Acceleration of first car; a1 = 0 m/s² (constant acceleration)

Velocity of second car; v2 = 0 m/s (since the second car starts from rest)

From Newton's equation of motion, we know that;

d = ut + ½at²

Thus,for first car, we have;

d = v1•t + ½(a1)t²

Plugging in the relevant values, we have;

d = 33t + 0

d = 33t

For second car, we have;

d = v2•t + ½(a2)•t²

Plugging in the relevant values, we have;

d = 0 + ½(a2)t²

d = ½(a2)t²

Since they meet at the next exit, then;

33t = ½(a2)t²

simplifying to get;

33 = ½(a2)t

Now, we also know that;

t = distance/speed = d/v1 = 2500/33

Thus;

33 = ½ × (a2) × (2500/33)

Rearranging, we have;

a2 = (33 × 33 × 2)/2500

a2 = 0.8712 m/s²

3 0
2 years ago
B) A skilled jet fighter flies a stunt plane in a vertical circle of 1200 ft
vodka [1.7K]

Answer:I need the ans to this one too

Explanation:

4 0
3 years ago
A cyclist rides 16.0 km east, then 8.0 km west, then 8.0 km east, then 32.0 km west, and finally 11.2 km east.
Anna71 [15]

Answer:

Time, t = 12 minutes

Explanation:

It is given that,

A cyclist rides 16.0 km east, then 8.0 km west, then 8.0 km east, then 32.0 km west, and finally 11.2 km east. Let west direction is negative and east direction is positive. The displacement of the cyclist is :

d=16-8+8-32+11.2=-4.8\ km

d = 4800 m

Let us assumed that the average speed of the cyclist is, v = 24 km/h = 6.66667 m/s

Let t is the time taken by the cyclist to complete the trip. The velocity of an object is given by :

v=\dfrac{d}{t}

t=\dfrac{d}{v}

t=\dfrac{4800\ m}{6.66667\ m/s}

t = 719.99 seconds

t = 720 seconds

or

t = 12 minutes

So, the time taken by the cyclist to complete the trip is 12 minutes. Yes, the time taken by the cyclist to complete the trip is reasonable. Hence, this is the required solution.                                      

7 0
3 years ago
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