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3 years ago

8

A

1 answer:

Arada [10]3 years ago

4 0

Answer:

i would think c

Explanation:

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At what distance above the surface of the earth is the acceleration due to the earth's gravity 0.855 m/s2 if the acceleration du

natulia [17]

Given: Normal pull of gravity g = 9.8 m/s²;

g = 0.855 m/s² (at a certain distance)

Universal gravitational constant G = 6.67 x 10⁻¹¹ N.m²/Kg²

Mass of the Earth Me = 5.98 x 10²⁴ Kg

Radius r = ?

g = GMe/r²

r = √GMe/g

r = √(6.67 x 10⁻¹¹ N.m²/Kg²)(5.98 x 10²⁴ Kg)/(0.855 m/s²)

r = 2.16 x 10⁷ m or

r = 21,610 Km

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3 years ago

Help please :)))))))))))

Kamila [148]

Don’t do the link it’s a scam

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3 years ago

What is the frequency of a wave travelling at 45.45 m/s with a wavelength of 2 m?

HACTEHA [7]

Answer:

22.73 Hz

Explanation:

Frequency = 45.45/2 = 22.73 Hz

4 0

3 years ago

If the mass of both weights is 150 gm, the first mass is located 25° north of east, the second mass is located 25° south of east

Pani-rosa [81]

Answer:

voltage is 1.38 V

Explanation:

given data

weights = 150 gm = 0.150 kg

sensitivity = 0.5 volts/Newton

to find out

large a voltage

solution

we know here in by symmetry so here in x axis

F3 will be

F3 = 2mgcos(20)

F3 = 2(0.150)(9.8)cos20

F3 = 2.76 N

so volatge is

voltage = 2.76×sensitivity

voltage = 2.76×0.5

voltage = 1.38

so voltage is 1.38 V

8 0

4 years ago

I NEED HELP ASAP! BRAINIEST TO THE CORRECT ANSWER. HELP ME NOW!

sergij07 [2.7K]

Answer:

<h3>a)</h3>

$\boxed{\mathfrak{Power(P)=\frac{Voltage(V)^2}{Resistance(R)} }}$

V = 12 V
P = 24 W

$\implies \mathsf{24=\frac{12^2}{R} }$

$\implies \mathsf{24R=12^2 }$

$\implies \mathsf{24R=144 }$

<u>=> R= 6 Ohms(Ω)</u>

<h3>b)</h3>

$\boxed{\mathfrak{Power(P)=\frac{Voltage(V)^2}{Resistance(R)} }}$

Power (P)= 100 W

<em>these lights operate at the usual 240 volts direct from the main electricity supply. Therefore,</em>

V = 240 V

$\implies \mathsf{100=\frac{240^2}{R} }$

<em>R and 100 can interchange places</em>

$\implies \mathsf{R=\frac{240^2}{100} }$

$\implies \mathsf{R=\frac{57600}{100} }$

<u>=> R = 576 Ω</u>

<u></u>

By Ohm's Law:

$\boxed{\mathsf{Voltage(V)=Current(I) \times Resistance(R)}}$

=> 240 = I × 576

=>

=> I = 0.417 A

<h3 /><h3>c)</h3>

I don't know it's resistance,... so sorry

<h3>d)</h3>

The brightness of the bulb in series is <em><u>less than</u></em> when they're placed individually.

For bulbs in series their resistance gets added to form the equivalent resistance of the two bulbs.

Their resistances are nothing but mere numbers and the sum of two numbers(positive of course) is greater than the numbers.

So, the effective resistance of some bulbs in series <u>is more</u> than the individual resistance.

And

<em>Brightness, i. e., Power</em>

$\boxed{\mathfrak{Power \propto \frac{1}{Resistance} }}$

If resistance increases, Power decreases.

Here, the effective resistance was for sure larger, therefore resistance was increasing, hence power decreased taking brightness along with it.

3 0

3 years ago