3 years ago

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Physics

2 answers:

djverab [1.8K]3 years ago

7 0

Answer:

Explanation:

SVETLANKA909090 [29]3 years ago

6 0

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A plane passes over point a with a velocity of 5000 m/s north. Sixty seconds later it passes over point b at a velocity of 10,00

andrew-mc [135]

Answer:

The plane’s acceleration from A to B is

$a =83.33\ m/s^2$

Explanation:

The acceleration of an object is a measure of how its speed changes over time.

In this case, we know that the plane initially has a velocity of 5000 m/s when it passes through point A, and when it passes through point B, 60 seconds later it has a velocity of 10,000 m/s.

Then the average acceleration is calculated as

$a =\frac{s_2-s_1}{t_2-t_1}$

Where

$s_2 = 10 000$ is the final speed

$s_1=5000$ is the initial velocity

$t_2 -t_1=60$ is the time it took for the object to reach the final speed

Then

$a =\frac{10000-5000}{60}$

$a =\frac{5000}{60}$

$a =83.33\ m/s^2$

4 0

3 years ago

A cheetah can maintain its maximum speed of 100 km/hr for 30.0 seconds. What minimum distance must a gazelle running 80.0 km/hr

german

Explanation:

Suppose the cheetah is initially positioned at x=0 (m) from the reference, and the gazelle is intially at poisiton x=d (m).

Then, at the worst case, that is when cheetah is running at the maximum case, the position of the gazelle relative to the reference must be larger than that of cheetah.

In equation form,

$0\text{ km}+\frac{100\text{ km}}{1 \text{ hr}}\cdot30\text{ s }\cdot\frac{1\text{ hr}}{3600\text{ s}}\le d \text{ km}+\frac{80\text{ km}}{1 \text{ hr}}\cdot30\text{ s }\cdot\frac{1\text{ hr}}{3600\text{ s}}$