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insens350 [35]
3 years ago
5

What is the name of a variable that you change in an experiment

Physics
2 answers:
a_sh-v [17]3 years ago
7 0

Answer:

d) Independent variable

Explanation:

In an experiment we are generally finding which variable changes with respect to another variable.

The variable which is changing irrespective of the other variable is called the independent variable i.e., generally considered as the input.

The variable which is changing with respect of the independent variable is called the dependent variable i.e., generally considered as the output.

stealth61 [152]3 years ago
3 0
Independent variable
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Please help on this one
mixas84 [53]
Ep=mgh
h= Ep/mg
h=57÷(3.3×9.8)
h= 57÷32.34
h= 1.8m
So; the answer is B. 1.8m
6 0
3 years ago
The capacitor can withstand a peak voltage of 590 volts. If the voltage source operates at the resonance frequency, what maximum
kirill115 [55]

Answer:

The maximum voltage is 41.92 V.

Explanation:

Given that,

Peak voltage = 590 volts

Suppose in an L-R-C series circuit, the resistance is 400 ohms, the inductance is 0.380 Henry, and the capacitance is 1.20×10^{-2}\ \mu F.

We need to calculate the resonance frequency

Using formula of frequency

f=\dfrac{1}{2\pi\sqrt{LC}}

Put the value into the formula

f=\dfrac{1}{2\pi\sqrt{0.380\times1.20\times10^{-8}}}

f=2356.88\ Hz

We need to calculate the maximum current

Using formula of current

I=\dfrac{V_{c}}{X_{c}}

I=2\pi\times f\times C\times V_{c}

I=2\pi\times2356.88\times1.20\times10^{-8}\times590

I=0.1048\ A

Impedance of the circuit is

z=\sqrt{R^2+(X_{L}^2-X_{C}^2)}

At resonance frequency X_{L}=X_{C}

Z=R

We need to calculate the maximum voltage

Using ohm's law

V=I\times R

V=0.1048\times400

V=41.92\ V

Hence, The maximum voltage is 41.92 V.

4 0
2 years ago
Select the arrangement of electromagnetic radiation which starts with the lowest wavelength and increases greatest wavelength. M
Scorpion4ik [409]

Answer:

gamma rays, ultraviolet, infrared, radio

Explanation:

Because on the Electromagnetic spectrum wavelength increases from the gamma end to radio end and frequency decrease in that order

8 0
2 years ago
Read 2 more answers
A space ship to the moon covered the 216,000 km in 72 hours. What was it’s average velocity
Dahasolnce [82]

Answer:3000km/h

Explanation:

speed=distance/time

216,000km/72hrs

=3000km/h

8 0
2 years ago
Read 2 more answers
A bicycle rider has a speed of 19.0 m/s at a height of 55.0 m above sea level when he begins coasting down hill. The mass of the
lukranit [14]

Answer:

The mechanical energy of the rider at any height will be 6.34 × 10⁴ J.

Explanation:

Hi there!

The mechanical energy of the rider is calculated as the sum of the gravitational potential energy plus the kinetic energy. Since there are no dissipative forces (like friction), the mechanical energy of the rider at a height of 55.0 m above the sea level will be the same at a height of 25.0 m (or at any height), because the loss in potential energy will be compensated by a gain in kinetic energy, according to the law of conservation of energy.

Then, calculating the potential and kinetic energy at 55.0 m and 19 m/s, we can obtain the mechanical energy that will be constant:

Mechanical energy = PE + KE

Where:

PE = potential energy.

KE = kinetic energy.

The potential energy is calculated as follows:

PE = m · g · h

Where:

m = mass of the object.

g = acceleration due to gravity.

h = height.

Then, the potential energy of the rider will be:

PE = 88.0 kg · 9.81 m/s² · 55.0 m = 4.75 × 10⁴ J

The kinetic energy is calculated as follows:

KE = 1/2 · m · v²

Where "m" is the mass of the object and "v" its velocity. Then:

KE = 1/2 · 88.0 kg · (19.0 m/s)²

KE = 1.59 × 10⁴ J

The mechanical energy of the rider will be:

Mechanical energy = PE + KE = 4.75 × 10⁴ J + 1.59 × 10⁴ J = 6.34 × 10⁴ J

This mechanical energy is constant because when the rider coast down the hill, its potential energy is being converted into kinetic energy, so that the sum of potential energy plus kinetic energy remains constant.

5 0
3 years ago
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