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Maurinko [17]
2 years ago
8

HELP PLS .. The half-life of potassium-44 is 22 minutes. It decays into calcium-44. After

Physics
2 answers:
anzhelika [568]2 years ago
6 0

Answer:

B. 1/8

Explanation:

did the test !!

Ber [7]2 years ago
3 0

Answer:

1/4

Explanation:

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You and your friend are pushing a box at the same constant speed. You pushed 20 feet in 10 minutes, your friend was pushing for
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Answer:

60 N.C

Explanation:

The box will move in the direction of the push and pull with a force of 60 N. C.

7 0
3 years ago
Two scales on a nondigital voltmeter measure voltages up to 20.0 and 30.0 V, respectively. The resistance connected in series wi
Snowcat [4.5K]

Answer:

Resistance of the circuit is 820 Ω

Explanation:

Given:

Two galvanometer resistance are given along with its voltages.

Let the resistance is "R" and the values of voltages be 'V' and 'V1' along with 'G' and 'G1'.

⇒ V=20\ \Omega,\ V_1=30\ \Omega

⇒ G=1680\ \Omega,\  G_1=2930\ \Omega

Concept to be used:

Conversion of galvanometer into voltmeter.

Let G be the resistance of the galvanometer and I_g the maximum deflection in the galvanometer.

To measure maximum voltage resistance R is connected in series .

So,

⇒ V=I_g(R+G)

We have to find the value of R we know that in series circuit current are same.

For G=1680                                    For G_1=2930

⇒ I_g=\frac{V}{R+G}   equation (i)                ⇒ I_g=\frac{V_1}{R+G_1} equation (ii)

Equating both the above equations:

⇒ \frac{V}{R+G} = \frac{V_1}{R+G_1}

⇒ V(R+ G_1) = V_1 (R+G)

⇒ VR+VG_1 = V_1R+V_1G

⇒ VR-V_1R = V_1G-VG_1

⇒ R(V-V_1) = V_1G-VG_1

⇒ R =\frac{V_1G-VG_1}{(V-V_1)}

⇒ Plugging the values.

⇒ R =\frac{(30\times 1680) - (20\times 2930)}{(20-30)}

⇒ R =\frac{(50400 - 58600)}{(-10)}

⇒ R=\frac{-8200}{-10}

⇒ R=820\ \Omega

The coil resistance of the circuit is 820 Ω .

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3 years ago
The weight of a person can be represented by a vector that acts...
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Straight down, even if the person is standing on a hill
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Volume , mass, and weight can be measured with balances, scales, and ?
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