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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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the relation that relates the speed of wave, frequency of wave and wavelength is given as

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This 80 kg car is moving at 20m/sec at the top where the hills radius is 100m. What is the centrifugal force?
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A girl is standing 150m in front of a tall building, fires a shot with starting pistol. A boy standing 350m behind her, hears tw
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Thus;

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Multiply through by v to get;

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calculate the frequency and time period of sound wave of 35 m wave length propagating at a speed of 3500m/s​
zheka24 [161]

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5 0
2 years ago
Bill is farsighted and has a near point located 121 cm from his eyes. Anne is also farsighted, but her near point is 74.0 cm fro
Arada [10]

Answer:

Explanation:

The lens equation is

1 / f = 1 / di + 1 / do

Where

f is focal length

di is the image distance

do is the object distance

Both Annie and Billy use a glass whose near point is 25cm

Then, the object distance is

do = 25 - 2 = 23cm

The have the same object distance.

Let find the vocal length of bills eye

Given that,

Bill near point is 121cm and distance of the glass from the eye is 2cm

Then,

Image distance of bill is

di_B = -(121-2) = -119cm

object distance do = 23cm

Then,

1 / f_B = 1 / di_B + 1 / do

1 / f_B = -1 / 119 + 1 / 23

1 / f_B = -119^-1 + 23^-1

1 / f_B = 0.0351

Then, f_B = 28.51 cm

Also, let find Annie focal length

Given that,

Annie near point is 74 cm and distance of the glass from the eye is 2cm

Then,

Image distance of Annie is

di_A = -(74-2) = -72cm

object distance do = 23cm

Then,

1 / f_A = 1 / di_A + 1 / do

1 / f_A = -1 / 72 + 1 / 23

1 / f_A = -72^-1 + 23^-1

1 / f_A = 0.02959

Then, f_A = 33.8 cm

Distance of object from the lens when Annie uses Billy glass

Then,

1 / f_B = 1 / di_A + 1 / do

1 / 28.51 = -1 / 72 + 1 / do

28.51^-1 = -72^-1 + do^-1

do^-1 = 28.51^-1 + 72^-1

do^-1 = 0.048964

do = 20.42 cm

Then, the object location from the eye will be, the eye is 2cm from the glass. Then,

do_A = 20.42 + 2 = 22.42cm

do_A = 22.42 cm

Distance of object from the lens when Billy uses Annie glass

Then,

1 / f_A = 1 / di_B + 1 / do

1 / 33.8 = -1 / 119 + 1 / do

33.8^-1 = -119^-1 + do^-1

do^-1 = 33.8^-1 + 119^-1

do^-1 = 0.03799

do = 26.32 cm

Then, the object location from the eye will be, the eye is 2cm from the glass. Then,

do_B = 26.32 + 2 = 28.32 cm

do_B = 28.32 cm

7 0
3 years ago
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