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emmainna [20.7K]
3 years ago
11

A soccer ball is kicked horizontally off a bridge with a height of 36 m. The ball travels 25 m horizontally before it hits the p

avement below. What was the soccer ball’s speed when it was first kicked
Physics
1 answer:
kodGreya [7K]3 years ago
4 0

Answer:

9.23 m/s

Explanation:

Let, the initial velocity be ux.

The horizontal velocity remans the same. So, time required, t = 25/ux

For vertical component, we know,

h = uy*t + (1/2)*g*t^2    [ g is positive because ball is falling downward ]

Putting in the values, we get,

36 = 0*(25/ux) + 1/2 * 9.81 * (25/u)^2

36 = 3065.625/u^2

u^2 = 85.15625

u = 9.23

[ If there's a problem with the solution, please inform me ]

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Rank these temperatures from hottest to coldest: 32° F,32° C, and 32 K 320 F> 32° C>32 K 32°C 32° F 32 K 32° F 32 K 32° c
Leviafan [203]

Answer:

32 C > 32 F > 32 K

Explanation:

32 F, 32 C, 32 K

Let T1 = 32 F

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T3 = 32 K

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(F - 32) / 9 = C / 100

so, (32 - 32) / 9 = C / 100

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So, T1 = 32 F = 0 C

The relation between c and K is given by

C = K - 273 = 32 - 273 = - 241

So, T3 = 32 K = - 241 C

So, T 1 = 0 C, T2 = 32 c, T3 = - 241 C

Thus, T2 > T1 > T3

32C > 32 F > 32 K

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3 years ago
3. In the wave models, you imagined adding energy into the wave by
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7 0
3 years ago
A man works on a striaght road from his home to market 2.5km away dwith a speed of 5km/h.Finding the market closed, he instantly
umka21 [38]

Answer:

5.625km/h

Explanation:

We are given that

Distance between home and market, d=2.5 km

Speed, v1=5km/h

Speed, v2=7.5 km/h

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Time,t=\frac{distance}{speed}

Using the formula

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t_2=\frac{2.5}{7.5}=\frac{1}{3}hour=\frac{60}{3}=20min

Distance traveled by man in 10 min with speed 7.5 km/h=\frac{2.5}{20}\times 10=2.5/2km

Therefore,

Total distance covered=2.5+2.5/2=3.75 km

Time=40 min=40/60=2/3 hour

Average speed=\frac{total\;distance}{total\;time}

Average speed=\frac{3.75}{2/3}=5.625km/h

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