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garri49 [273]
3 years ago
9

If an object has a mass of 81 kg, what is its approximate weight on earth?

Physics
1 answer:
telo118 [61]3 years ago
8 0
The weight of an 81 kg object on earth is 793.8 N.
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A mass is placed at the end of a spring. It has starting velocity of V & allowed to oscillate freely. If the mass has a star
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Answer:

Equation for SHM can be written

V = w A cos w t        where w is the angular frequency and the velocity is a                                         maximum at t = 0

V1 = w1  A cos w1 t

V2 = w2 A cos w2 t

V2 / V1 = w2 / w1     since cos X t = 1 if t = zero

V2 / V1 = 2 pi f2 / (2 pi f1) = f2 / f1 = T1 / T2

If the velocity is twice as large the period will be 1/2 long

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______ 4. Justin was creating a chart comparing the potential energy between charged particles and the potential energy of magne
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Answer:

The correct option is;

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What are some resources used to generate electricity in energy stations?
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Read 2 more answers
Please help me with this question ASAP.
victus00 [196]

Answer:

i. 6.923 V

ii. The e.m.f. = 22.5 V

Explanation:

i. The given parameters are;

Length of potentiometer = 1 m

The resistance of the potentiometer = 10 Ω

The e. m. f. of the attached cell = 9 V

The current, I flowing in the circuit = e. m. f/(Total resistance)

The current, I flowing in the circuit = 9 V/(10 + 3) = 9/13 A

The potential difference, p.d. across the 1 m potentiometer wire = I × Resistance of the potentiometer wire

The p.d. across the potentiometer wire = 9/13×10 = 90/13 = 6.923 V

ii) Given that the 1 m potentiometer wire has a resistance of 10 Ω, 75 cm which is 0.75 m will have an e.m.f. given by the following relation;

\dfrac{E}{R_{balance}} = \dfrac{V}{R_{cell}}

Where:

E = e.m.f. of the balance point cell

R_{balance} = Resistance of 75 cm of potentiometer wire  = 0.75×10 = 7.5 Ω

R_{cell} = Resistance of the cell in the circuit = 3 Ω

V = e.m.f. attached cell = 9 V

\dfrac{E}{7.5} = \dfrac{9}{3}

E = 7.5*3 = 22.5 V

The e.m.f. = 22.5 V

7 0
4 years ago
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