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3 years ago

Mathphys please notice me

Physics

2 answers:

eimsori [14]3 years ago

7 0

Answer:

24.2 s

29.2 m

Explanation:

First, convert each speed from km/h to m/s.

91.5 km/h = 25.42 m/s

77.5 km/h = 21.53 m/s

The speed of the cheetah relative to the gazelle is:

v = 25.42 m/s − 21.53 m/s

v = 3.89 m/s

So the time it takes for the cheetah to catch the gazelle is:

x = v t

94.2 m = (3.89 m/s) t

t = 24.2 s

If instead we want to find x so that t = 7.5 s, then:

x = v t

x = (3.89 m/s) (7.5 s)

x = 29.2 m

Marianna [84]3 years ago

3 0

Answer:

24.2 s

29.2 m

Explanation:

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The graph in the accompanying figure (Figure 1) shows the magnitude of the force exerted by a given spring as a function of the

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The elastic potential energy stored in the stretched spring is 1 J.

<h3>What is Hooke's law?</h3>

Hooke's law states that; provided the elastic limit is not exceeded, the extension of the spring is directly proportional to the force on the spring.

Given that;

Force on the spring = 350 Newton

Distance stretched = 7 centimeters or 0.07 m

Hence;

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2 years ago

A firefighting crew uses a water cannon that shoots water at 25.0 m/s at a fixed angle of 53.0° above the horizontal. The fire-f

zysi [14]

Answer:

8.8 m and 52.5 m

Explanation:

The vertical component and horizontal component of water velocity leaving the hose are

$v_v = vsin(\alpha) = 25sin(53^0) = 25*0.8 = 19.97 m/s$

$v_h = vcos(\alpha) = 25cos(53^0) = 25*0.6 = 15 m/s$

Neglect air resistance, vertically speaking, gravitational acceleration g = -9.8m/s2 is the only thing that affects water motion. We can find the time t that it takes to reach the blaze 10m above ground level

$s = v_vt + gt^2/2$

$10 = 19.97t - 9.8t^2/2$

$4.9t^2 - 19.97t + 10 = 0$

$t= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$t= \frac{19.9658877511823\pm \sqrt{(-19.9658877511823)^2 - 4*(4.9)*(10)}}{2*(4.9)}$

$t= \frac{19.9658877511823\pm14.24}{9.8}$

t = 3.49 or t = 0.58

We have 2 solutions for t, one is 0.58 when it first reach the blaze during the 1st shoot up, the other is 3.49s when it falls down

t is also the times it takes to travel across horizontally. We can use this to compute the horizontal distance between the fire-fighters and the building

$s_1 = v_ht_1 = 15*0.58 = 8.8 m$