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DedPeter [7]
3 years ago
12

Jaipal measures a circuit at 1.2 A and 24 . Using Ohm’s law, what can he calculate for the circuit? current (C) current (I) volt

age (V) voltage (A)
Physics
2 answers:
Elan Coil [88]3 years ago
6 0

Answer:

voltage

(V)

Explanation:

jenyasd209 [6]3 years ago
4 0

Answer: voltage (V)

Explanation:

Hi, in the answer the letter A (1.2 A) stands for Ampere, and the symbol next to the number 24 must be Ω 8 (Ohms)

So, applying Ohm's law:

Voltage (V) =current (I) x resistance (R)

V = I x R

Replacing with the values given;

V = 1.2 A x 24Ω

In conclusion, we can calculate the value of Voltage for the circuit.

Feel free to ask for more if needed or if you did not understand something.

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5 0
3 years ago
Two strings are adjusted to vibrate at exactly 202 Hz. Then the tension in one string isincreased slightly. Afterward, three bea
Semenov [28]

The concepts necessary to solve this problem are framed in the expression of string vibration frequency as well as the expression of the number of beats per second conditioned at two frequencies.

Mathematically, the frequency of the vibration of a string can be expressed as

f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}

Where,

L = Vibrating length string

T = Tension in the string

\mu = Linear mass density

At the same time we have the expression for the number of beats described as

n = |f_1-f_2|

Where

f_1 = First frequency

f_2 = Second frequency

From the previously given data we can directly observe that the frequency is directly proportional to the root of the mechanical Tension:

f \propto \sqrt{T}

If we analyze carefully we can realize that when there is an increase in the frequency ratio on the tight string it increases. Therefore, the beats will be constituted under two waves; one from the first string and the second as a residue of the tight wave, as well

n = f_2-f_1

f_2 = n+f_1

Replacing 3/sfor n and 202Hz for f_1,

f_2 = 3/s + 202Hz

f_2 = 3/s(\frac{1Hz}{1/s})+202Hz

f_2 = 206Hz

The frequency of the tightened is 205Hz

7 0
3 years ago
James gently releases a ball at the top of a slope But does not push the ball. Space the ball rolls down the slope. Which force
NARA [144]
On an incline, the force causing the ball to move downwards would be gravity. Additionally, the component of gravity causing this ball to move downwards would be mgsintheta.

Hope this helps!
7 0
3 years ago
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3 years ago
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CISTU U
marusya05 [52]

solution:

radius of steel ball(r)=5cm=0.05m

density of ball =8000kgm

terminal velocity(v)=25m/s^2

density of air( d) =1.29 kgm

now

volume of ball(V)=4/3pir^3=1.33×3.14×0.05^3=0.00052 m^3

density of ball= mass of ball/Volume of ball

or, 8000=m/0.00052

or, m=4.16 kg

weight of the ball (W)= mg=4.16×10=41.6 N

viscous force(F)=6 × pi × eta × r × v

=6×3.14×eta×0.05×25

=23.55×eta

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Fiscous force=Weight

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or, eta = 1.76

whete eta is the coefficient of viscosity.

5 0
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