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2 years ago

What is the potential energy of a 25 kg bicycle resting at the top of a hill 3 m high? (Formula: PE = mgh)

Physics

2 answers:

Lisa [10]2 years ago

8 0

Answer:

The answer would be 735J

Explanation:

PE=mgh

=(mass)(force of gravity)(height)

=(25kg)(9.8m/s^2)(3m)

=735J

tatiyna2 years ago

8 0

Explanation:

given, mass =25 kg

height=3m

and acceleration deu to gravity =9.8m/s^2

now, we have,

pe =mgh

=25×9.8×3

=735j.....ans

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Metals combine easily with nonmetals. true or false

Ad libitum [116K]

The answer for that is True.

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2 years ago

A girl on a spinning amusements park is 12m from the center of the ride and has a centripetal acceleration of 17 m/s^2. What is

Tanya [424]

Answer: 14.28 m/s

Explanation:

Assuming the girl is spinning with uniform circular motion, her centripetal acceleration $a_{c}$ is given by the following equation:

$a_{c}=\frac{V^{2}}{r}$ (1)

Where:

$a_{c}=17 m/s^{2}$ is the centripetal acceleration

$V$ is the tangential speed

$r=12 m$ is the radius of the circle

Isolating $V$ from (1):

$V=\sqrt{a_{c}r}$ (2)

$V=\sqrt{(17 m/s^{2})(12 m)}$

Finally:

$V=14.28 m/s$ This is the girl's tangential speed

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2 years ago

What is the force of gravity on an 100kg person standing on the earth

Bingel [31]

I think its 980.7
because 9.807 × 100

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2 years ago

A train travels 55 km south along a straight track in 34 minutes. What is the train's average velocity in kilometers per hour?

AURORKA [14]

Hopefully I’m not late and I apologize if I am, but the answer to your question would be 95.6 km/hr. You know you can look up your question as well to see if they already have a answer to that so you won’t waste your points.

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2 years ago

Read 2 more answers

In certain ranges of a piano keyboard, more than one string is tuned to the same note to provide extra loudness. For example, th

Georgia [21]

Explanation:

Given that,

Frequency in the string, f = 110 Hz

Tension, T = 602 N

Tension, T' = 564 N

We know that frequency in a string is given by :

$f=\dfrac{1}{2L}\sqrt{\dfrac{T}{m/L}}$ , T is the tension in the string

i.e.

$f\propto\sqrt{T}$

$\dfrac{f}{f'}=\sqrt{\dfrac{T}{T'}}$ , f' is the another frequency

${f'}=f\times \sqrt{\dfrac{T'}{T}}$

${f'}=110\times \sqrt{\dfrac{564}{602}}$

f' =106.47 Hz

We need to find the beat frequency when the hammer strikes the two strings simultaneously. The difference in frequency is called its beat frequency as :

$f_b=|f-f'|$

$f_b=|110-106.47|$

$f_b=3.53\ beats/s$

So, the beat frequency when the hammer strikes the two strings simultaneously is 3.53 beats per second.

8 0

2 years ago