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geniusboy [140]
3 years ago
8

Finally, you are ready to answer the main question. Cheetahs, the fastest of the great cats, can reach 50.0 miles/hourmiles/hour

in 2.22 ss starting from rest. Assuming that they have constant acceleration throughout that time, find their acceleration in meters per second squared. Ente
Physics
1 answer:
slavikrds [6]3 years ago
5 0

Answer:

The acceleration of the cheetahs is 10.1 m/s²

Explanation:

Hi there!

The equation of velocity of an object moving along a straight line with constant acceleration is the following:

v = v0 + a · t

Where:

v = velocity of the object at time t.

v0 = initial velocity.

a = acceleration.

t = time

We know that at t = 2.22 s, v = 50.0 mi/h. The initial velocity, v0, is zero.

Let's convert mi/h into m/s:

50.0 mi/h · (1609.3 m / 1 mi) · (1 h / 3600 s) = 22.4 m/s

Then, using the equation:

v = v0 + a · t

22.4 m/s = 0 m/s + a · 2.22 s

Solving for a:

22.4 m/s / 2.22 s = a

a = 10.1 m/s²

The acceleration of the cheetahs is 10.1 m/s²

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• 2. A 70 kg man on a 100 kg boat throws a ball. The boat moves backwards 5 meters in 10 seconds. What is
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A hawk is flying horizontally at 18.0 m/s in a straight line, 230 m above the ground. A mouse it has been carrying struggles fre
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Answer:

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b) θ = 80.9°

c) t = 6.8 s

Explanation:

In the situation of the problem, you can assume that the trajectory of the hawk and the trajectory of the mouse form a rectangle triangle.

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h=\sqrt{(36m)^2+(227m)^2}=229.83m    (1)

Next, it is necessary to calculate the falling down time of the mouse, this can be done by using the following formula:

y=y_o+v_ot+\frac{1}{2}gt^2    (2)

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g: gravitational acceleration = -9.8m/s^2

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v_d=\frac{229.83m}{4.8s}=47.88\frac{m}{s}

The diving speed of the Hawk is 47.88m/s

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Then angle between the horizontal and the trajectory of the Hawk when it is descending is 80.9°

(c) The mouse is falling down during 6.8 s

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