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Nadusha1986 [10]

3 years ago

15

A bowling ball is traveling at 3 m/s. It starts to roll up a ramp. How high above the ground will the ball be when it stops roll

ing? Neglect friction and assume the ramp is plenty long enough to do this. (For this problem your initial kinetic energy can be set equal to your final potential energy. 1/2mv2 = mgh).

2 answers:

Zepler [3.9K]3 years ago

8 0

Answer:

FREE FREE FREE FREEE FREEEEEEEEEEE FREEEEEEEEEEEEEEEE FREEEEEEEEEEEEEEEEEE

Explanation:

FREE FREE FREE FREEE FREEEEEEEEEEE FREEEEEEEEEEEEEEEE FREEEEEEEEEEEEEEEEEE

vivado [14]3 years ago

5 0

Answer:

h= 0.45 m

Explanation:

PE= 1/2 mv^2 KE= mgh

v= 3m/s

vf= 0 m/s

h=?

PE= 1/2(1kg)(3m/s)^2

PE= 4.5 J

4.5 J/ 1kg(9.8 m/s^2)

h=0.45 m

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mina [271]

A) 50 cm

B) 10000 cm/s

Explanation

Step 1

A)

If you know the distance between nodes and antinodes then use this equation:

$\begin{gathered} \frac{\lambda}{2}=D \\ \text{where}\lambda\text{ is the wavelength} \\ D\text{ is the distance betw}een\text{ nodes} \end{gathered}$

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now, replace to find the wavelength

$\begin{gathered} \frac{\lambda}{2}=25 \\ \text{Multiply both sides by 2} \\ \frac{\lambda}{2}\cdot2=25\cdot2 \\ \lambda=50\text{ Cm} \end{gathered}$

so, the wavelength is

A) 50 cm

Step 2

The speed of a wave can be found using the equation

$v=\lambda f$

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then,let

$\begin{gathered} \lambda=50\text{ cm} \\ f=200\text{ Hz} \end{gathered}$

replace and evaluate

$\begin{gathered} v=\lambda f \\ v=50\text{ cm }\cdot200\text{ HZ} \\ v=10000\text{ }\frac{\text{cm}}{s} \end{gathered}$

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B) 10000 cm/s

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A seagull flies at a velocity of 9.00 m/s straight into the wind. (a) if it takes the bird 20.0 min to travel 6.00 km relative t

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Here we will the speed of seagull which is v = 9 m/s

this is the speed of seagull when there is no effect of wind on it

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if effect of wind is in opposite direction then it travels 6 km in 20 min

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Part b)

now if bird travels in the same direction of wind then we will have

$v_{net}= v_{bird} + v_{wind}$

$v_{net} = 9 + 4 = 13 m/s$

now we can find the time to go back

$time = \frac{distance}{speed}$

$time = \frac{6000}{13}$

$time = 7.7 minutes$

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SashulF [63]

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