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Mice21 [21]
3 years ago
7

Which is the variable that the experimenter changes

Physics
1 answer:
SashulF [63]3 years ago
4 0
The independent variable
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What is the period of oscillation for a pendulum that completes 3.00 cycles every 2.00 s?
torisob [31]

Answer:

\huge\boxed{\sf T = 0.67\ seconds}

Explanation:

<u>Given Data:</u>

Cycles = c = 3 cycles

Time = t = 2 seconds

<u>Required:</u>

Time period = T = ?

<u>Formula:</u>

1) f = c/t

2) T = 1/f

<u>Solution:</u>

<u>Finding frequency first:</u>

f = c / t

f = 3/2

f = 1.5 Hz

<u>Now, finding period:</u>

T = 1/f

T = 1 / 1.5

T = 0.67 seconds

\rule[225]{225}{2}

6 0
3 years ago
What are the characteristics of a magnetic force
fomenos

Answer:

Magnetic field lines never intersect each other.

They originates(from north pole) and terminates (to south pole) perpendicularly from/to the surface of magnet. Also magnetic field lines continues through the magnet.

Magnetic field is denser at the pole and becomes less dense as we move away from the poles.

Explanation:

8 0
3 years ago
the natural process where the sun's energy is trapped by the atmosphere to make the earth warm enough for life is known as globa
alexira [117]
The natural process is called greenhouse effect. This is when the Sun's light enters the Earth, and converts to heat. Light passes through the Earth's atmosphere much easier compared to heat. Because of this, heat is trapped on Earth because it cannot exit the atmosphere as easily as light does.
6 0
4 years ago
10 points and brainliest lol goteem
SCORPION-xisa [38]
I think it's D cuz D is the heliocentric theory
6 0
3 years ago
Read 2 more answers
When a surface is illuminated with electromagnetic radiation of wavelength 480 nm, the maximum kinetic energy of the emitted ele
algol [13]

Answer:

Max kinetic energy for 340 nm wavelength will be 2.238\times 10^{-19}j

Explanation:

In first case wavelength of electromagnetic radiation \lambda =480nm=480\times 10^{-9}m

Plank's constant h=6.6\times 10^{-34}J-s

Maximum kinetic energy = 0.54 eV

Energy is given by E=\frac{hc}{\lambda }=\frac{6.6\times 10^{-34}\times 3\times 10^8}{480\times 10^{-9}}=4.125\times 10^{-19}J

We know that energy is given

E=K_{MAX}+\Phi, here \Phi is work function

So 4.125\times 10^{-19}=0.54\times 10^{-19}+\Phi

\Phi =3.585\times 10^{-19}J

Now wavelength of second radiation = 340 nm

So energy E=\frac{hc}{\lambda }=\frac{6.6\times 10^{-34}\times 3\times 10^8}{340\times 10^{-9}}=5.823\times 10^{-19}J

So K_{MAX}=5.823\times 10^{-19}-3.585\times 10^{-19}=2.238\times 10^{-19}j

6 0
4 years ago
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