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3 years ago

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018 10.0 points

1 answer:

-BARSIC- [3]3 years ago

6 0

Answer:

jfjcgufnfhfufm TV fifnricnrhkddufnfif km fgkfkvntfmrugrhfifnh r

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O operate this generator, a human applies a force by turning the crank. This action generates an electric current, which lights

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Answer A I’m very positive about it

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3 years ago

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The cycle that moves an important non carbon element, which organisms use to build their cells through Earth's sphere's, is the

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8 0

3 years ago

A radio have a wavelength of 0.3m and travels at a speed of 300,000,000 m/s. What is the frequency of this wave?

Ilya [14]

The frequency of the wave is $1\cdot 10^9 Hz$

Explanation:

The frequency, the wavelength and the speed of a wave are related by the following equation:

$c=f \lambda$

where

c is the speed of the wave

f is the frequency

$\lambda$ is the wavelength

For the radio wave in this problem,

$\lambda = 0.3 m$

$c=300,000,000 m/s = 3\cdot 10^8 m/s$

Therefore, the frequency is:

$f=\frac{c}{\lambda}=\frac{3\cdot 10^8}{0.3}=1\cdot 10^9 Hz$

Learn more about waves here:

brainly.com/question/5354733

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#LearnwithBrainly

4 0

3 years ago

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

3 years ago

A 50 kg bobsled slides down an ice track

STatiana [176]

Hi there!

We can use the following kinematic equation:

$v_f^2 = v_i^2 + 2ad$

The initial velocity is 0 m/s, so:

$v_f^2 = 2ad$

vf = final velocity (? m/s)
a = acceleration due to gravity (g)
d = vertical height (m)

Plug in the givens and solve:

$v_f = \sqrt{2gd} = \sqrt{2(9.8)(173)} = \boxed{58.23 \frac{m}{s}}$

8 0

3 years ago