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UkoKoshka [18]
3 years ago
13

Can anyone help with these questions please

Physics
1 answer:
Svet_ta [14]3 years ago
6 0
Here... I've finished this course already

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two identical steel balls mounted on wooden posts initially have different amounts of charge, one with -14uC and the other with
GrogVix [38]

According to the Law of Conservation of Charge, the net charge remains constant. If both things have different charges, upon contact, they would share the charge equally. In this case, the total charge is -16μC. The final charge for each ball would be -8 μC.

7 0
4 years ago
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(True or False Statement)
sertanlavr [38]
False is correct......
4 0
3 years ago
A fan blade is rotating with a constant angular acceleration of 11.5 rad/s2. At what point on the blade, as measured from the ax
vichka [17]

Answer:

Alpha = ω^2 R    where R is radius of blade

g = w^2 r      where r is distance from center

ω^2 R = 11.5 ω^2 r

R / r = 11.5 / 9.8 = 1.17

Or r = .852 R

Since the angular acceleration depends on both R and ω it seems that one can only get r as it depends on R

7 0
2 years ago
If a cup of water has holes in it than why does no water come out when when it is falling?
quester [9]

Because the acceleration of gravity is the acceleration of gravity.
It doesn't matter what the mass of a falling object is, and it  doesn't
matter whether a falling object is solid or liquid.  ALL falling objects
fall with the same acceleration, reach the same speed, and hit the
ground at the same time.

If there was no air in the way, then a feather, a school bus, and a
battleship would accelerate at the same rate, fall together and hit
the ground at the same time.

When you drop a cup full of water that has holes in it, the cup and
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8 0
3 years ago
Read 2 more answers
Suppose 8.41 moles of a monatomic ideal gas expand adiabatically, and its temperature decreases from 395 to 279 K. Determine (a)
sergeinik [125]

Answer:

a) W=12166.20876 J

b) U= -12166.20876 J

Explanation:

No. of moles, n = 8.41

Change of temperature, ΔT = T1 - T2

                                         = 395 - 279

                                         = 116 K

For monatomic gas, γ = 5/3

γ -1 = 2 /3

Solution:

(a)

Work done,W= \frac{nR}{\gamma-1}(T_1-T_2)

plugging values we get

W= \frac{8.314\times8.41}{2/3}(116)

Ans:   12166.20876 J

Work done, W = + 12166.20876 J

(b)

From first law of thermodynamics, dQ = U + W

but, dQ = 0 ( adiabatic process)

Hence, U = - W

                 = - 12166.20876 J

Ans:

Change in internal energy, U = - 12166.20876 J

8 0
3 years ago
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