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4 years ago

9

The cheetah can maintain its maximum speed for only 7.5 s. What is the minimum distance the gazelle must be ahead of the cheetah

to have a chance of escape? (After 7.5 s the speed of cheetah is less than that of the gazelle.) Answer in units of m.

1 answer:

amm18124 years ago

6 0

Basically the cheetah is running 31.5km/h faster than the gazelle. So to determone how long it will take to cover 9mm at that speed, you have to a lot of work. If you skip all of that work, the answer is 1.60m seconds

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Bola bermassa 200 gram dilempar

ikadub [295]

Answer:

0.4

Explanation:

$\frac{1}{2} mv ^{2}$

kinetic energy formula , potential energy is not considered

0.5×0.2×2×2

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3 years ago

Points P and Q are located at (0, 2, 4) and (-3, 1,5). Calculate

Akimi4 [234]

Answer:

a) The position vector of P is $\vec P =(0, 2,4)$ .

b) The distance vector from P to Q is $\overrightarrow{PQ} = (-3,-1,1)$ .

c) The distance between P and Q is $\|\overrightarrow{PQ}\|=\sqrt{11}$ .

d) A vector parallel to PQ with magnitude of 10 is $\vec v = \left(-\frac{30\sqrt{11}}{11},-\frac{10\sqrt{11}}{11}, \frac{10\sqrt{11}}{11} \right)$ .

Explanation:

a) The position vector of a point is the vector displacement from the origin to the location of the point. That is:

$\vec P = (0,2,4)-(0,0,0)$

$\vec P = (0-0, 2-0, 4-0)$

$\vec P =(0, 2,4)$

The position vector of P is $\vec P =(0, 2,4)$ .

b) First, we calculate the position vector of point Q:

$\vec Q = (-3,1,5)-(0,0,0)$

$\vec Q = (-3-0,1-0,5-0)$

$\vec Q =(-3,1,5)$

The distance vector from P to Q is define by the following vectorial expression:

$\overrightarrow{PQ} = \vec Q - \vec P$ (1)

$\overrightarrow{PQ} = (-3,1,5)-(0,2,4)$

$\overrightarrow{PQ} =(-3-0,1-2,5-4)$

$\overrightarrow{PQ} = (-3,-1,1)$

The distance vector from P to Q is $\overrightarrow{PQ} = (-3,-1,1)$ .

c) There are two approaches to calculate the distance between P and Q:

First Method - Pythagorean Theorem:

$\|\overrightarrow{PQ}\| = \sqrt{(-3)^{2}+(-1)^{2}+1^{2}}$

$\|\overrightarrow{PQ}\|=\sqrt{11}$

Second Method - Dot Product:

$\|\overrightarrow{PQ}\| = \sqrt{\overrightarrow{PQ}\,\bullet\,\overrightarrow{PQ}}$ (2)

$\|\overrightarrow{PQ}\| = \sqrt{(-3,-1,1)\,\bullet (-3,-1,1)}$

$\|\overrightarrow{PQ}\|=\sqrt{11}$

The distance between P and Q is $\|\overrightarrow{PQ}\|=\sqrt{11}$ .

d) To determine a vector parallel to PQ with a given magnitude is determined by the following expression:

$\vec v = \frac{k}{\|\overrightarrow{PQ}\|} \cdot \overrightarrow{PQ}$ (3)

Where $k$ is the scale factor.

If we know that $\overrightarrow{PQ} = (-3,-1,1)$ , $\|\overrightarrow{PQ}\|=\sqrt{11}$ and $k = 10$ , then the vector is:

$\vec v = \frac{10}{\sqrt{11}}\cdot (-3,-1,1)$

$\vec v = \left(-\frac{30}{\sqrt{11}},-\frac{10}{\sqrt{11}},\frac{10}{\sqrt{11}}\right)$

$\vec v = \left(-\frac{30\sqrt{11}}{11},-\frac{10\sqrt{11}}{11}, \frac{10\sqrt{11}}{11} \right)$

A vector parallel to PQ with magnitude of 10 is $\vec v = \left(-\frac{30\sqrt{11}}{11},-\frac{10\sqrt{11}}{11}, \frac{10\sqrt{11}}{11} \right)$ .

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3 years ago

Which of the following is not a result of deforestation?

klemol [59]

Answer:

B

Explanation:

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3 years ago

Read 2 more answers

What value do we use to describe acceleration due to gravity

andrey2020 [161]

Answer:

9.8 m/s2

Explanation:

In the first equation above, g is referred to as the acceleration of gravity. Its value is 9.8 m/s2 on Earth. That is to say, the acceleration of gravity on the surface of the earth at sea level is 9.8 m/s2.

Got it from the internet, hope it helps though ^^

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3 years ago

Suppose you increase your walking speed from 7 m/s to 15 m/s in a period of 2 m. What is your acceleration?

likoan [24]

Acceleration = (change in speed) / (time for the change)

Change in speed = (end speed) - (start speed) = (15 m/s - 7 m/s) = 8 m/s

time for the change = 2 minutes = 120 seconds

Acceleration = (8 m/s) / (120 seconds)

Acceleration = 0.067 m/s²

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3 years ago