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MAVERICK [17]
3 years ago
10

What is studying or working i only one field (this is for science 7th grade)

Physics
1 answer:
Rama09 [41]3 years ago
3 0

Answer:

Studying is reviewing based on what you learned .Working is doing what you got to do while learning.

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A solid cylinder of mass 10 kg is pivoted about a frictionless axis thought the center O. A rope wrapped around the outer radius
taurus [48]
Torque acting dowward = 6  x 0.5 = 3 Nm

Torque acting to the right = 5 x  1 = 5 Nm

5 - 3 = 2 Nm

inertia = 1/2 mr^2

0.5 x 10  x 1^2 = 5 kg-m^2
2/5 = alpha  = 0.4 rad /s^2

Hope this helps
5 0
3 years ago
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It is recommended that children have no more than two sets of dental X-rays per year. Although X-rays have many benefits, they c
guapka [62]

Although X-rays have many benefits, they can be dangerous to human health because they are short wavelength, high frequency, with photons sufficiently energetic to damage DNA molecules.

4 0
3 years ago
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The fundamental frequency of a standing wave on a 1.0-m-long string is 440 Hz. What would be the wave speed of a pulse moving al
Nana76 [90]

Answer: v = 880m/s

Explanation: The length of a string is related to the wavelength of sound passing through the string at the fundamental frequency is given as

L = λ/2 where L = length of string and λ = wavelength.

But L = 1m

1 = λ/2

λ = 2m.

But the frequency at fundamental is 440Hz and

V = fλ

Hence

v = 440 * 2

v = 880m/s

4 0
3 years ago
How many significant digits are in the measurement 4,032,010 m/s?
choli [55]
Since there is no decimal point in the number given above, the counting for the number of the significant figures will start from the left. Then, the first zero from the left is insignificant. Therefore, in this number there are 6 significant figures. 
5 0
3 years ago
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Assume that you have a rectangular tank with its top at ground level. The length and width of the top are 14 feet and 7 feet, re
ch4aika [34]

To solve this problem we will use the work theorem, for which we have that the Force applied on the object multiplied by the distance traveled by it, is equivalent to the total work. From the measurements obtained we have that the width and the top are 14ft and 7ft respectively. In turn, the bottom of the tank is 15ft. Although the weight of the liquid is not given we will assume this value of 62 lb / ft ^ 3 (Whose variable will remain modifiable until the end of the equations subsequently presented to facilitate the change of this, in case be different). Now the general expression for the integral of work would be given as

W = \gamma A * \int_0^15 dy

Basically under this expression we are making it difficult for the weight of the liquid multiplied by the area (Top and widht) under the integral of the liquid path to be equivalent to the total work done, then replacing

W = (62)(14*7)\int^{15}_0 (15-y)dy

W = (14*7*62)\big [15y-\frac{y^2}{2}\big ]^{15}_0

W = (14*7*62)[15(15)-\frac{(15)^2}{2}]

W = 683550ft-lbs

Therefore the total work in the system is 683550ft-lbs

6 0
3 years ago
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