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serious [3.7K]
3 years ago
6

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
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