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2 years ago

13

A 50 kg student climbs 3m to the top of a set of stairs. Calculate the change in the student’s gravitational potential energy fr

om the bottom to the top of the stairs.

1 answer:

erastova [34]2 years ago

7 0

Gpe = mgh
gpe = 50*10*3
gpe = 1500 J

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Use a common denominator to find-2/3 + -4/5

Juliette [100K]

Answer:

-22/15

Explanation:

the least common denominator is 15 so first you multiply -2/3 by 5 in both the numerator and denominator making it -10/15

Then you do the same to -4/5 except you multiply the numerator and denominator by 3 giving you -12/15

If you add -10/15+ -12/15 you get -22/15

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3 years ago

A wave with a frequency of 500hz is traveling at a speed of 1000m/s whats the wavelength?

Nimfa-mama [501]

The answer would be 0.4m

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1 year ago

Need help please and thank you

Scilla [17]

C
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C
C

Hope this helps you!!!

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2 years ago

** ANSWER IS A** As you move from left to right across Period 3, which statement is true of the number of

Sedbober [7]

Answer:

A. The number of valence electrons increases by 1.

Explanation:

As you move across any period on the periodic table, the number of valence electrons increases by a value of 1.

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From left to right, number of valence electrons increases.
Down a group, the valence electrons are the same.
Across a period, the number of valence electrons increases.

7 0

2 years ago

Read 2 more answers

A solid circular plate has a mass of 0.25 kg and a radius of 0.30 m. It starts rolling from rest at the top of a hill 10 m long

KATRIN_1 [288]

To solve the problem it is necessary to apply the equations related to the conservation of both <em>kinetic of rolling objects</em> and potential energy and the moment of inertia.

The net height from the point where it begins to roll with an inclination of 30 degrees would be

$h=Lsin30$

$h=10sin30$

$h=5m$

In the case of Inertia would be given by

$I = \frac{mR^2}{2}$

In general, given an object of mass m, an effective radius k can be defined for an axis through its center of mass, with such a value that its moment of inertia is

$I = mk^2$

$\frac{mR^2}{2}= mk^2$

$\frac{k^2}{R^2}=\frac{1}{2}$

Replacing in Energy conservation Equation we have that

Potential Energy = Kinetic Energy of Rolling Object

$mgh = \frac{1}{2}mv^2(1+\frac{k^2}{r^2})$

$9.8*5=\frac{1}{2}v^2(1+\frac{1}{2})$

$v^2 (1.5) = 98$

$v=8.0829m/s$

Therefore the correct answer is C.

3 0

2 years ago