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tia_tia [17]
3 years ago
7

How long would it take Jesse with the acceleration of -2.50 m/s to bring his bicycle with an initial velocity of 13.5 m/s to a c

omplete stop
Physics
1 answer:
Anna [14]3 years ago
6 0

Answer:

Time taken by Jesse to bring his bicycle =5.4 Sec

Explanation:

acceleration of bicycle = -2.5  m/sec²

Initial velocity of by cycle u =13.5 m/sec,

final velocity of bicycle v = 0 (since bicycle completely stop finally)

Let the time taken to stop the bicycle completely = t sec

We now by equation of motion,

v= u + a\cdot t

putting the value of u,v & a in the equation

0=13.5 -2.5 x t ( a is taken as negative since velocity of bicycle is reducing)

t=\frac{13.5}{2.5} = 5.4 Sec

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Suppose a certain jet plane creates an intensity level of 124 dB at a distance of 5.01 m. What intensity level does it create on
Zolol [24]

Answer:71 dB

Explanation:

Given

sound Level \beta _1=124 dB

distance r_1=5.01 m

From sound Intensity

\beta =10dB\log (\frac{I_1}{I_0})

124=10dB\log (\frac{I_1}{I_0})

12.4=\log (\frac{I_1}{I_0})

I_1=(1\times 10^{-12})\times 10^{12.4}

I_1=2.51 W/m^2

we know Intensity I\propto ^\frac{1}{r^2}

I_1r_1^2=I_2r_2^2

I_2=I_1(\frac{r_1}{r_2})^2

I_2=2.51\cdot (\frac{5.01}{2.25\times 10^3})^2

I_2=1.24\times 10^{-5} W/m^2

Sound level corresponding to I_2

\beta _2=10\log (\frac{I_2}{I_0})

\beta _2=10\log (\frac{1.24\times 10^{-5}}{1\times 10^{-12}})

\beta _2=70.93\approx 71 dB

6 0
4 years ago
A 75 kg object is moving west at 5.6 m/s. What is the object's momentum? (p = mv)
Licemer1 [7]

Explanation:

p=mv

p=5.6×75

p= 420

<em>hope</em><em> it</em><em> was</em><em> helpful</em><em> to</em><em> you</em>

7 0
3 years ago
What is nuclear power
katrin2010 [14]

Answer:

<h2><em><u>D. All of the statements describe nuclear power</u></em></h2>

What is <em><u>NUCLEAR</u></em><em><u> </u></em><em><u>POWER</u></em><em><u>?</u></em><em><u> </u></em>

➡️ is the <u>use of nuclear</u> <u>reactions</u> to <u>produce electricity</u>.

➡️ it can be <u>obtained from nuclear</u> <u>fission, nuclear decay and nuclear fusion reactions. </u>

Hope it helps

3 0
2 years ago
A basketball player standing up with the hoop launches the ball straight up with an initial velocity of v_o = 3.75 m/s from 2.5
denis23 [38]

Answer:

a) The maximum height the ball will achieve above the launch point is 0.2 m.

b) The minimum velocity with which the ball must be launched is 4.43 m/s or 0.174 in/ms.

Explanation:

a)

For the height reached, we use 3rd equation of motion:

2gh = Vf² - Vo²

Here,

Vo = 3.75 m/s

Vf =  0m/s, since ball stops at the highest point

g = -9.8 m/s² (negative sign for upward motion)

h = maximum height reached by ball

therefore, eqn becomes:

2(-9.8m/s²)(h) = (0 m/s)² - (3.75 m/s²)²

<u>h = 0.2 m</u>

b)

To find out the initial speed to reach the hoop at height of 3.5 m, we again use 3rd eqn. of motion with h= 3.5 m - 2.5m = 1 m (taking launch point as reference), and Vo as unknown:

2(-9.8m/s²)(1 m) = (0 m/s)² - (Vo)²

(Vo)² = 19.6 m²/s²

Vo = √19.6 m²/s²

<u>Vo = 4.43 m/s</u>

Vo = (4.43 m/s)(1 s/1000 ms)(39.37 in/1 m)

<u>Vo = 0.174 in/ms</u>

<u />

6 0
3 years ago
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