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Serhud [2]
3 years ago
15

What is electric current?

Physics
1 answer:
DerKrebs [107]3 years ago
4 0

Answer:

the answer to the question is the number of electric charges that pass a given point

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When a guitar string plays the note "a," the string vibrates at 440 hz ?
Rama09 [41]
Yes, that's correct. The note "A" (which is used to tune the other strings of the guitar) corresponds to a frequency of 440 Hz.
8 0
3 years ago
Do you think seismographs predict earthquakes or measure earthquakes, explain your answer?
Yuki888 [10]

Answer:

No. Neither the USGS nor any other scientists have ever predicted a major earthquake. We do not know how, and we do not expect to know how any time in the foreseeable future.

5 0
3 years ago
Read 2 more answers
Where are electrons located in an atom?
Scilla [17]

Answer:

B. Outside the nucleus.

Explanation:

Electrons orbit the nucleus of the atom.

7 0
2 years ago
A 5.5kg radio is pushed across the table. If the acceleration is 5m/s to the right, find the net force exerted on the radio.
Sholpan [36]

Answer:

Net force exerted on the radio is 27.5 Newton.

Given:

Mass = 5.5 kg

Acceleration = 5 \frac{m}{s^{2} }

To find:

Force exerted on the radio = ?

Formula used:

F = ma

Where F = net force

m = mass

a = acceleration

Solution:

According to Newton's second law of motion,

F = ma

Where F = net force

m = mass

a = acceleration

F = 5.5 × 5

F = 27.5 Newton

Hence, Net force exerted on the radio is 27.5 Newton.


4 0
4 years ago
A baseball of mass m = 0.31 kg is spun vertically on a massless string of length L = 0.51m. The string can only support a tensio
natulia [17]

Given data:

* The mass of the baseball is 0.31 kg.

* The length of the string is 0.51 m.

* The maximum tension in the string is 7.5 N.

Solution:

The centripetal force acting on the ball at the top of the loop is,

\begin{gathered} T+mg=\frac{mv^2}{L}_{} \\ v^2=\frac{L(T+mg)}{m} \\ v=\sqrt[]{\frac{L(T+mg)}{m}} \end{gathered}

For the maximum velocity of the ball at the top of the vertical circular motion,

v_{\max }=\sqrt[]{\frac{L(T_{\max }+mg)}{m}}

where g is the acceleration due to gravity,

Substituting the known values,

\begin{gathered} v_{\max }=\sqrt[]{\frac{0.51(7.5_{}+0.31\times9.8)}{0.31}} \\ v_{\max }=\sqrt[]{\frac{0.51(10.538)}{0.31}} \\ v_{\max }=\sqrt[]{17.34} \\ v_{\max }=4.16\text{ m/s} \end{gathered}

Thus, the maximum speed of the ball at the top of the vertical circular motion is 4.16 meters per second.

8 0
1 year ago
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