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hichkok12 [17]
3 years ago
11

A uniform disk, a uniform hoop, and a uniform solid sphere are released at the same time at the top of an inclined ramp. They al

l roll without slipping. In what order do they reach the bottom of the ramp

Physics
1 answer:
pav-90 [236]3 years ago
8 0

Answer:

sphere, disk, hoop

Explanation:

See attached file

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Body composition refers to the amount of fat and muscle that the body has. ... Muscular strength is the amount of force a muscle or group of muscles can exert. true. Muscular endurance is fueled by anaerobic energy.
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A car (mass of 830 kg) is sitting on a car lift in a shop (neglect the mass of the lift itself). While the car is being lifted u
OLEGan [10]
There are two force acting on an object that is being lifted. (1) the weight of the car, (2) the upward force. The difference of these force should be equal to the product of the mass and the acceleration. (This is the content of Newton's 2nd Law of Motion). If we let the lifting force be F,
                       F - (830)(9.8) = (830)(3.8)
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3 years ago
When the electron in a hydrogen atom moves from n = 6 to n = 1, light with a wavelength of ________ nm is emitted?
Oduvanchick [21]
First we find the energy level with the following formula, where a is the energy level, n1 is the final energy level, n2 is the starting energy level and r is Rydberg's constant in Joules
a = r \times ( \frac{1}{n1}  -  \frac{1}{n2} )
We insert the values

a = 2.18 \times  {10}^{ - 18}  \times ( \frac{1}{ {1}^{2} } -  \frac{1}{ {6}^{2} }  )
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The wavelength is found with this formula, where h is Planck's constant and c is the speed of light
wavelength =  \frac{h \times c}{a}
Finally we insert the values
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3 0
3 years ago
A 0.750kg block is attached to a spring with spring constant 13.5N/m . While the block is sitting at rest, a student hits it wit
vazorg [7]

Answer:

A)A=0.075 m

B)v= 0.21 m/s

Explanation:

Given that

m = 0.75 kg

K= 13.5 N

The natural frequency of the block given as

\omega =\sqrt{\dfrac{K}{m}}

The maximum speed v given as

v=\omega A

A=Amplitude

v=\sqrt{\dfrac{K}{m}}\times A

0.32=\sqrt{\dfrac{13.5}{0.75}}\times A

A=0.075 m

A= 0.75 cm

The speed at distance x

v=\omega \sqrt{A^2-x^2}

v=\sqrt{\dfrac{K}{m}}\times \sqrt{A^2-x^2}

v=\sqrt{\dfrac{13.5}{0.75}}\times \sqrt{0.075^2-(0.075\times 0.75)^2}

v= 0.21 m/s

5 0
3 years ago
245289457u2895785/+55891235723895718957=
Bingel [31]

<em>why Is there just a random "u" in the middle of the equation </em><em>awa</em><em> </em>

<em> </em><em>t</em><em>h</em><em>e</em><em>question is impossible to do btw </em>

<em>Imao the u in the middle doesn't makes sense </em>

if<em> you edit the question I'll help you w it)</em>

have a good day bro cya

6 0
2 years ago
Read 2 more answers
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