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2 years ago

What is the frequency of a wave having a period Equal to 18 seconds

Physics

1 answer:

madreJ [45]2 years ago

5 0

The frequency is. 1/18. per second.

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Which of the following is an example of a plasma

k0ka [10]

<span>The sun, lightning, the ionosphere, fluorescent light bulb, stars, fire, tv, and solar winds. I don't know your other options but these are my examples.</span>

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3 years ago

A force F is used to raise a 4-kg mass M from the ground to a height of 5 m. What is the work done by the force F? (Note: sin 60

miskamm [114]

Answer:

Answer:196 Joules

Explanation:

Hello

Note: I think the text in parentheses corresponds to another exercise, or this is incomplete, I will solve it with the first part of the problem

the work is the product of a force applied to a body and the displacement of the body in the direction of this force

assuming that the force goes in the same direction of the displacement, that is upwards

W=F*D (work, force,displacement)

the force necessary to move the object will be

$F=mg(mass *gravity)\\F=4kgm*9.8\frac{m}{s^{2} }\\ F=39.2 Newtons\\replace\\\\W=39.2 N*5m\\W=196\ Joules$

Answer:196 Joules

I hope it helps

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2 years ago

Which two optical devices use convex lenses to focus and enlarge images?

Andreas93 [3]

It is a reflecting telescope and a compound microscope. I know this for sure

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2 years ago

An ore sample weighs 17.50 N in air. When the sample

zysi [14]

Answer with Explanation:

We are given that

Weight of an ore sample=17.5 N

Tension in the cord=11.2 N

We have to find the total volume and the density of the sample.

We know that

Tension, T= $W-F_b$

$F_b$ =buoyancy force

T=Tension force

W=Weight

By using the formula

$11.2=17.5-F_b$

$F_b=17.5-11.2=6.3$ N

$F_b=V_{object}\times \rho_{water}\cdot g$

Where

$V_{object}$ =Volume of object

$\rho_{water}=1000 kgm^{-3}$ =Density of water

$g=9.8 ms^{-2}$ =Acceleration due to gravity

Substitute the values then we get

$6.3=9.8\times 1000\times V_{object}$

$V_{object}=\frac{6.3}{9.8\times 1000}=6.43\times 10^{-4} m^3$

Volume of sample= $6.43\times 10^{-4} m^3$

Density of sample, $\rho_{object}=\frac{Mass}{volume_{object}}$

Where mass of ore sample=1.79 kg

Substitute the values then, we get

$\rho_{object}=\frac{1.79}{6.43\times 10^{-4}}=2.78\times 10^3 kg/m^3$

Density of the sample= $2.78\times 10^{3} kgm^{-3}$

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3 years ago

What is the field outside the capacitor plates in a parallel capacitor?

FrozenT [24]

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2 years ago

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