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marishachu [46]
3 years ago
10

Please solve this for me ​

Physics
1 answer:
uranmaximum [27]3 years ago
6 0

See if you get the answer with this formulae.

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A spinning disc rotating at 130 rev/min slows and stops 31 s later. how many revolutions did the disc make during this time?
gayaneshka [121]
F = 130 revs/min = 130/60 revs/s = 13/6 revs/s
t = 31s
wi = 2πf = 2π × 13/6 = 13π/3 rads/s
wf = 0 rads/s = wi + at
a = -wi/t = -13π/3 × 1/31 = -13π/93 rads/s²
wf² - wi² = 2a∅
-169π²/9 rads²/s² = 2 × -13π/93 rads/s² × ∅
∅ = 1209π/18 rads
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3 years ago
What would you do to change the volume of gas?
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4 years ago
How can you increase the sensitivity of a Mercury thermometer?​
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8 0
3 years ago
An object is placed 96.5 cm from a glass lens (n = 1.51) with one concave surface of radius 24 cm and one convex surface of radi
poizon [28]

Answer:

image is vertical at distance -203.62 cm

magnification is 2.110

Explanation:

given data

n = 1.51

distance u = 96.5 cm

concave radius r1 = 24 cm

convex radius r2 = 19.1 cm

to find out

final image distance and magnification

solution

we will apply here lens formula to find focal length f

1/f = n-1 ( 1/r1 - 1/r2)   .......................1

put here all value

1/f = 1.51 -1 ( -1/24 + 1/19.1)

f = 183.43

so from lens formula

1/f = 1/v + 1/u     .............................2

put here all value and find v

1/183.43 = 1/v + 1/96.5

so

v = −203.62 cm

so here image is vertical at distance -203.62 cm

and

magnification are = -v /u

magnification =  203.62 / 96.5

magnification is 2.110

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4 years ago
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4 0
3 years ago
Read 2 more answers
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