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eduard
3 years ago
12

Convert the following: 9.87 x 108s = _____ h

Physics
2 answers:
Lynna [10]3 years ago
7 0

given that

t = 9.87 * 10^8 s

we need to convert this time into hours

as we know that

1 h = 60 min

and we also know

1 min = 60 s

so here we have

1 h = 60 * 60 = 3600 s

now we have

t = 9.87* 10^8 s* \frac{1h}{3600 s}

t = 2.74 * 10^5 h

so we can say

9.87* 10^8 s = 2.74 * 10^5 h

Oksanka [162]3 years ago
6 0

Answer:

times 9.87 x 108 and then convert the answer to hours so what would 0.2961 or 10^{5}

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Find the current through a person and identify the likely effect on her if she touches a 120 V AC source in the following circum
QveST [7]

Answer:

A) 0.4 mA

B) 0.03 mA

Explanation:

Given that

voltage source, V = 120 V

to solve this question, we would be using the very basic Ohms Law, that voltage is proportional to the current and the resistance passing through the circuit, if temperature is constant.

mathematically, Ohms Law, V = IR

V = Voltage

I = Current

R = Resistance

from question a, we were given 300kΩ, substituting this value of resistance in the equation, we have

120 = I * 300*10^3 Ω

making I the subject of the formula,

I = 120 / 300000

I = 0.0004 A

I = 0.4 mA

Question said, currents above 10 mA causes involuntary muscle contraction, this current is way below 10 mA, so nothing happens.

B, we have Resistance, R = 4000kΩ

Substituting like in part A, we have

120 = I * 4000*10^3 Ω

I = 120 / 4000000

I = 0.00003 A

I = 0.03 mA

This also means nothing happens, because 0.03 mA is very much lesser compared to in the 10 mA

5 0
3 years ago
what is the cost to run a single 100 w light bulb for 30 days if the electric company charges $0.18 for every KW-hr?​
xxTIMURxx [149]

Explanation:

power of the bulb = 100w=0.1kw

power consumption when the bulb is used for 30 days ( 30 *24 = 720 hours) =0.1*720=72kwh

cost to run the bulb for 30 days = 72*0.18

= $12.96

Answer is $12.96..................

8 0
3 years ago
A fixed 14.1-cm-diameter wire coil is perpendicular to a magnetic field 0.52 T pointing up. In 0.28 s , the field is changed to
ELEN [110]

Answer:

the average induced emf in the coil is 0.016 V.

Explanation:

Given;

diameter of the wire, d = 14.1 cm = 0.141 m

change in magnetic filed strength, dB = 0.52 T - 0.23 T = 0.29 T

change in time, dt = 0.28 s

The area of the wire is calculated as follows;

A = \frac{\pi d^2}{4} \\\\A = \frac{\pi \times (0.141)^2}{4} \\\\A = 0.0156 \ m ^2

The induced emf is calculated as follows;

emf = \frac{dBA}{dt} \\\\emf = \frac{0.29 \times 0.0156}{0.28} \\\\emf = 0.016 \ V

Therefore, the average induced emf in the coil is 0.016 V.

8 0
3 years ago
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