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nydimaria [60]
3 years ago
10

QUESTION 2 DOK 3 A Thompson's gazelle has a maximum acceleration of 4.5 m/s2 At this acceleration, how much time is required for

it to reach a speed of 40 km/h?​
Physics
1 answer:
Dominik [7]3 years ago
5 0

Answer:

2.47 s

Explanation:

Convert the final velocity to m/s.

  • 40 km/h → 11.1111 m/s

We have the acceleration of the gazelle, 4.5 m/s².

We can assume the gazelle starts at an initial velocity of 0 m/s in order to determine how much time it requires to reach a final velocity of 11.1111 m/s.

We want to find the time t.

Find the constant acceleration equation that contains all four of these variables.

  • v = v₀ + at

Substitute the known values into the equation.

  • 11.1111 = 0 + (4.5)t
  • 11.1111 = 4.5t
  • t = 2.469133333

The Thompson's gazelle requires a time of 2.47 s to reach a speed of 40 km/h (11.1111 m/s).

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n deep space, sphere A of mass 47 kg is located at the origin of an x axis and sphere B of mass 110 kg is located on the axis at
Volgvan

Answer:

a)-1.014x 10^{-7J

b)3.296 x  10^{-7J

Explanation:

For Sphere A:

mass 'Ma'= 47kg

xa= 0

For sphere B:

mass 'Mb'= 110kg

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a)the gravitational potential energy is given by

U_{i = -GMaMb/ d

U_{i= - 6.67 x 10^{-11} x 47 x 110/ 3.4 => -1.014x 10^{-7J

b) at d= 0.8m (3.4-2.6) and U_{i=-1.014x 10^{-7J

The sum of potential and kinetic energies must be conserved as the energy is conserved.

K_{i + U_{i= K_{f + U_{f

As sphere starts from rest and sphere A is fixed at its place, therefore K_{i is zero

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The final potential energy is

U_{f= - GMaMb/d

Solving for 'K_{f '

K_{f = U_{i + GMaMb/d => -1.014x 10^{-7 + 6.67 x 10^{-11} x 47 x 110/ 0.8

K_{f = 3.296 x  10^{-7J

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