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vovangra [49]
3 years ago
5

A ball rolling down a hill was displaced 21.9 m while uniformly accelerating from rest. If the final velocity was 7.14 m/s, what

was the rate of acceleration?
Physics
1 answer:
belka [17]3 years ago
4 0

Answer:

a = 1.16 m/s²

Explanation:

In order to find the acceleration of the ball we will use 3rd equation of motion.

2as = Vf² - Vi²

where,

a = acceleration = ?

s = displacement = 21.9 m

Vf = Final Velocity = 7.14 m/s

Vi = Initial Velocity = 0 m/s (Since, ball starts from rest)

Therefore, using the values, we get:

2a(21.9 m) = (7.14 m/s)² - (0 m/s)²

a = (50.97 m²/s²)/(43.8 m)

<u>a = 1.16 m/s²</u>

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A blue car pulls away from a red stop-light just after it has turned green with a constant acceleration of 0.2 m/s2. A green car
jolli1 [7]

Answer:

After 15 seconds, the green car will catch up with the blue car

Explanation:

Let the time for the green car to catch up with the blue car be T

When the green car catches up to the blue car, the distances covered by each car after time T will be equal. Also, their velocities at that instant will be equal

Distance covered by blue car after time T is given by: s = ut + 0.5 at²

Where u = 0, a = 0.2 m/s², t = T

S = 0.5 × 0.2 × T² = 0.1 T²

Velocity of blue car, v = u+ at

v = 0.2T

Distance covered by green car at T is given as: S = Velocity × time

Where v = 0.2T, t = T - 7.5 (since the blue car started 7.5 seconds earlier)

S = 0.2T (T - 7.5)

S = 0.2 T² - 1.5T

Equating the distance covered by the two cars

0.2T² - 1.5T = 0.1T²

0.1T² - 1.5T = 0

T(0.1T - 1.5) = 0

T = 0 or

T = 1.5/0.1 = 15 secs

Therefore, after 15 seconds, the green car will catch up with the blue car

8 0
2 years ago
A yet-to-be-built spacecraft starts from Earth moving at constant speed to the yet-tobe-discovered planet Retah, which is 20 lig
Gre4nikov [31]

Answer:

The  time elapsed at the spacecraft’s frame is less that the time elapsed at earth's  frame

Explanation:

From the question we are told that

The distance between earth and Retah is  d = 20 \ light \ hours =  20 * 3600 *  c =  72000c \ m

Here c is the peed of light with value c =  3.0*10^8 m/s

The time taken to reach Retah from earth is  t =  25 \ hours  =  25 * 3600 =90000 \ sec

The velocity of the spacecraft is mathematically evaluated  as

     v_s =  \frac{d }{t}

substituting values

   v_s =  \frac{72000 * 3.0*10^{8} }{90000}

    v_s =  2.40*10^{8} \ m/s

The time elapsed in the spacecraft’s frame is mathematically evaluated as

      T  =  t *  \sqrt{ 1 -  \frac{v^2}{c^2} }

substituting value

       T  =  90000 *  \sqrt{ 1 -  \frac{[2.4*10^{8}]^2}{[3.0*10^{8}]^2} }

        T = 54000 \ s

=>    T  = 15 \ hours

So  The  time elapsed at the spacecraft’s frame is less that the time elapsed at earth's  frame

       

7 0
3 years ago
Why is a minimum of three seismic stations needed to find the epicenter of an earthquake?
vazorg [7]

Each station can detect how far away the epicenter was. So each station basically has a circle made of possible epicenters. When you have three, you narrow it down to one, final point.

5 0
3 years ago
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