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AleksAgata [21]
2 years ago
13

The precision of a balance is ±0.02 g. The accepted value for a measurement is 26.86 g. Which measurement is in the accepted ran

ge?
26.02 g
26.82 g
26.87 g
26.90 g
Physics
2 answers:
blagie [28]2 years ago
8 0
  • \sf{ 26.87~g}

Explanation:

The precision of a balance is ±0.02 g.

The accepted value for a measurement is 26.86 g.

  • The accepted range is:
  • =[26.86-0.02, 26.86+0.02]
  • = [26.84, 26.88]  

So,26.87 g is correct

HACTEHA [7]2 years ago
7 0

Answer:

Ben türküm sizi anlamıyom kb

hepiniz malsınız amk

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spin [16.1K]

Answer:

0.11 kg

Explanation:

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Ft = momentum 5.22kg m/s

M = mass

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Therefore

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Divide both sides by 48.3

5.22/48.3 = M x 48.3/48.3

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3 years ago
A spherical, conducting shell of inner radius r1= 10 cm and outer radius r2 = 15 cm carries a total charge Q = 15 μC . What is t
lutik1710 [3]

a) E = 0

b) 3.38\cdot 10^6 N/C

Explanation:

a)

We can solve this problem using Gauss theorem: the electric flux through a Gaussian surface of radius r must be equal to the charge contained by the sphere divided by the vacuum permittivity:

\int EdS=\frac{q}{\epsilon_0}

where

E is the electric field

q is the charge contained by the Gaussian surface

\epsilon_0 is the vacuum permittivity

Here we want to find the electric field at a distance of

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Here we are between the inner radius and the outer radius of the shell:

r_1 = 10 cm\\r_2 = 15 cm

However, we notice that the shell is conducting: this means that the charge inside the conductor will distribute over its outer surface.

This means that a Gaussian surface of radius r = 12 cm, which is smaller than the outer radius of the shell, will contain zero net charge:

q = 0

Therefore, the magnitude of the electric field is also zero:

E = 0

b)

Here we want to find the magnitude of the electric field at a distance of

r = 20 cm = 0.20 m

from the centre of the shell.

Outside the outer surface of the shell, the electric field is equivalent to that produced by a single-point charge of same magnitude Q concentrated at the centre of the shell.

Therefore, it is given by:

E=\frac{Q}{4\pi \epsilon_0 r^2}

where in this problem:

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Substituting, we find:

E=\frac{15\cdot 10^{-6}}{4\pi (8.85\cdot 10^{-12})(0.20)^2}=3.38\cdot 10^6 N/C

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frez [133]

Answer:

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Explanation:

Please look at the solution in the attached Word file

Download docx
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