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yanalaym [24]
3 years ago
15

A plastic tube which has been rubbed with animal fur and gained 3.8x10 electrons.

Physics
1 answer:
posledela3 years ago
4 0
Don’t it is a bot scam don’t do it
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A water bug is suspended on the surface of a pond by surface tension (water does not wet the legs). The bug has six leg, and eac
Crazy boy [7]

Answer:

m = 2.2 x 10⁻⁴ kg = 0.22 g

Explanation:

The surface tension of water is 0.072 N/m. So in order for the bug to avoid sinking, its weight per unit length of contact must be no more than the surface tension of water. Therefore,

Weight\ of bug\ per\ unit\ length = Surface\ Tension\ of\ Water\\\frac{mg}{L} = Surface\ Tension\ of Water\\m = \frac{(Surface\ Tension\ of\ Water)(L)}{g}

where,

m = mass of bug = ?

g = acceleration due to gravity = 9.81 m/s²

L = Contact length = (contact length of each leg)(No. of Legs) = (5 mm)(6)

L = 30 mm = 0.03 m

Therefore,

m = \frac{(0.072\ N/m)(0.03\ m)}{9.81\ m/s^{2}} \\

<u>m = 2.2 x 10⁻⁴ kg = 0.22 g</u>

8 0
3 years ago
A star's apparent magnitude is most closely related to which of the following? Select all that apply.
kramer
If you want the "most", then you can't have more than one.

A star's apparent magnitude is determined by both its intrinsic luminosity
and its distance from us.
5 0
3 years ago
Read 2 more answers
What is oscillating to form a light wave?
aniked [119]
I think the answer is B ( matter )
5 0
3 years ago
How much heat does a freezer need to remove from 1kg of water at 40°C to make ice at 0°C? (You will need to use the specific hea
Montano1993 [528]

Answer:

Explanation:

Q = m c Δt

heat withdrawn to lower temperature by 40° C

Q = 1 X 1000 X 40 = 40,000 calories.

Q = mL

heat withdrawn to freeze water at 0°C to ice

= 1 x 80000 = 80,000 calories

total heat to be withdrawn = 120,000 calories. = 120,000 x 4.2 = 504 ,000 Joules.

3 0
3 years ago
A research-level Van de Graaff generator has a 2.15 m diameter metal sphere with a charge of 5.05 mC on it. What is the potentia
Llana [10]

Answer:

42.3 MV

Explanation:

d = diameter of the metal sphere = 2.15 m

r = radius of the metal sphere

diameter of the metal sphere is given as

d = 2r

2.15 = 2 r

r = 1.075 m

Q = charge on sphere = 5.05 mC = 5.05 x 10⁻³ C

Potential near the surface is given as

V = \frac{kQ}{r}

V = \frac{(9\times 10^{9})(5.05\times 10^{-3})}{1.075}

V = 4.23 x 10⁷ volts

V = 42.3 MV

5 0
4 years ago
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