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3 years ago

What is the term for a variable that a scientist adjusts during an experiment?

Physics

1 answer:

Advocard [28]3 years ago

3 0

B because science is to experiment and that’s y it’s b hope you have a nice day or night

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Can you please help me figure out the work for the missing quantities?

marysya [2.9K]

Answer: set up proportions

Explanation:

8 0

3 years ago

Determine the velocity of a car that travels 28km in 2.2 hours

OLga [1]

Answer:

<h2>velocity = 12.73 km/hr.</h2><h2 />

Explanation:

velocity = distance / time

=<u> 28 km </u>

2.2 hr

= 12.73 km/hr.

5 0

3 years ago

A person carries a plank of wood 2.00 m long with one hand pushing down on it at one end with a force f1 and the other hand hold

slega [8]

Answer:

F₁ = 4,120.2 N

F₂ = 3,924N

Explanation:

1) Balance of angular momentum around the end where F₁ is applied.

F₂ × 0.5m - F₁ × 0 = mass × g × 1m

⇒ F2 × 0.5 m= 20 kg × 9.81 m/s² × 1 m = 1,962 N×m

F₂ = 196.2 Nm / 0.5m = 3,924 N

2) Balance of forces

F₁ - F₂ = mg

F₁ = F₂ + mg = 3,924N + 20kg (9.81 m/s²) = 4,120.2 N

4 0

4 years ago

A flux density of 1.2Wb/m^2 is required in the 1 mm air gap of an electromagnet having an iron path of length 1.5 m. Calculate t

Mandarinka [93]

Answer:

The mmf required is $1.125$ × $10^{-3}$ A

Explanation:

The Magnetomotive force (mmf) is given by the formula below

$F_{M} = Hl\\$

where $F_{M}$ is the Magnetomotive force (mmf)

$H$ is the Magnetic field strength

$l$ is the magnetic length

The magnetic permeability μ is given by

μ = $B / H$

Where $B$ is the Magnetic flux density

and $H$ is the Magnetic field strength

From the question,

$B$ = 1.2Wb/m^2

μ = 1600m

From μ = $B / H$

∴ $H = B/$ μ

$H = 1.2 / 1600\\$

$H = 7.5$ × $10^{-4}$ A/m

Now, for the Magnetomotive force (mmf)

$F_{M} = Hl\\$

From the question

$l$ = 1.5 m

∴ $F_{M} = 7.5$ × $10^{-4}$ × $1.5$

$F_{M} = 1.125$ × $10^{-3} A$

Hence, The mmf required is $1.125$ × $10^{-3}$ A

3 0

3 years ago

Each of three tuning forks, A, B, and C, has a slightly different frequency. When A and B are sounded together, they produce a b

a_sh-v [17]

Answer:

The smallest possible beat frequency when B and C are sounded together is 3 Hz

Explanation:

Given;

the beat frequency of A and B = 2 Hz

the beat frequency of A and C = 5 Hz

Beat frequency is equal to the difference in frequency of the two notes that interfere to produce the beats.

$F_b_{(AB)} = 2 \ Hz\\\\F_A - F_B = 2 \ Hz -----equation(1)\\\\F_b_{(AC)} = 5 \ Hz\\\\F_A - F_C = 5 \ Hz-----equation(2)\\\\from \ equation(1), make \ F_A \ the \ subject \ of \ the \ formula\\\\F_A = 2H_z + F_B -----equation (3)\\\\Substitute \ in \ the \ value \ of \ F_A \ in \ equation(3) \ into \ equation(2)\\\\2H_z + F_B - F_C = 5H_z\\\\F_B-F_C = 5H_z - 2H_z\\\\F_B -F_C = 3Hz$

Therefore, the smallest possible beat frequency when B and C are sounded together is 3 Hz

7 0

3 years ago