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Mariana [72]
3 years ago
8

Which statement about elements and atoms is true?

Physics
1 answer:
Vika [28.1K]3 years ago
8 0
Atoms are the smallest unit of an element
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The circuits, P and Q, show two different ammeter-voltmeter methods of measuring resistance. Suppose the ammeter has a resistanc
qaws [65]

Answer:

Uncorrected values for

For circuit P

R = 2.4 ohm

For circuit Q

R = 2.4 ohm

Corrected values

for circuit P

R = 12 OHM

For circuit Q

R = 2.3 ohm

Explanation:

Given data:

Ammeter resistance 0.10 ohms

Resister resistance 3.0 ohms

Voltmeter read 6 volts

ammeter reads 2.5 amp

UNCORRECTED VALUES FOR

1) circuit P

we know that IR =V

R = \frac{6}{2.5} - 2.4 ohm

2) circuit Q

R = 2.4 ohm as no potential drop across ammeter

CORRECTED VALUES FOR

1) circuit p

IR = V

\frac{3R}{R+3} \times 2.5 = 6

R= 12 ohm

2) circuit Q

I\times (R+0.1) =V

R+0.1 =\frac{6}{2.5}

R = 2.3 ohm

5 0
3 years ago
.Why does it take double the applied force to move a mass double the size?
Likurg_2 [28]

Answer:

NVIDIA GeForce RTX 3080 10GB GDDR6X PCI Express 4.0 Graphics Card

Explanation:

6 0
2 years ago
A solid conducting sphere of radius 2 cm has a charge of 8microCoulomb. A conducting spherical shell of inner radius 4 cm andout
nika2105 [10]

Answer:

C) 7.35*10⁶ N/C radially outward

Explanation:

  • If we apply the Gauss'law, to a spherical gaussian surface with radius r=7 cm, due to the symmetry, the electric field must be normal to the surface, and equal at all points along it.
  • So, we can write the following equation:

       E*A = \frac{Q_{enc} }{\epsilon_{0}} (1)

  • As the electric field must be zero inside the conducting spherical shell, this means that the charge enclosed by a spherical gaussian surface of a radius between 4 and 5 cm, must be zero too.
  • So, the +8 μC charge of  the solid conducting sphere of radius 2cm, must be compensated by an equal and opposite charge on the inner surface of the conducting shell of total charge -4 μC.
  • So, on the outer surface of the shell there must be a charge that be the difference between them:

        Q_{enc} = - 4e-6 C - (-8e-6 C) = + 4 e-6 C

  • Replacing in (1) A = 4*π*ε₀, and Qenc = +4 μC, we can find the value of E, as follows:

      E = \frac{1}{4*\pi*\epsilon_{0} } *\frac{Q_{enc} }{r^{2} } = \frac{9e9 N*m2/C2*4e-6C}{(0.07m)^{2} } = 7.35e6 N/C

  • As the charge that produces this electric field is positive, and the electric field has the same direction as the one taken by a positive test charge under the influence of this field, the direction of the field is radially outward, away from the positive charge.
6 0
3 years ago
Need help with this plz
Lisa [10]

Answer:

1 - amplification

3- actinide

5 - radioactive decay ( im not really sure on this one )

7- alternating current

Explanation:

7 0
3 years ago
The first American to orbit the earth was:
alexdok [17]
John Glenn is the first American to orbit the Earth.

<span>♡♡Hope I helped!!! :)♡♡
</span>
3 0
3 years ago
Read 2 more answers
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