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9966 [12]
3 years ago
13

What is the charge of an atom with 21 protons and 6 electrons

Physics
2 answers:
enot [183]3 years ago
6 0

Answer:

i think it's 15

Explanation:

because the charge of a proton is +1 and for an electron it's -1

Svetach [21]3 years ago
4 0

Answer:

Scandium with an ion charge of +3

Explanation:

You might be interested in
A body is oscillating up and down at the end of a spring. Let’s consider when the body is at the top of its up-and-down motion.
Klio2033 [76]

The velocity of the body is zero; option A

<h3>What is the motion of an oscillating body?</h3>

The motion of an oscillating body is known as simple harmonic motion.

Simple harmonic motion involves a periodical motion of a body whose acceleration is directed towards a fixed point.

For a body that is oscillating up and down at the end of a spring, considering when the body is at the top of its up-and-down motion, the velocity of the body at the top and down is zero since the body comes to rest at the top and down position of its motion.

In conclusion, oscillating bodies undergo simple harmonic motion.

Learn more about simple harmonic motion at: brainly.com/question/24646514

#SPJ1

3 0
1 year ago
Assume that the amplifiers on the stage operate at a combined power of 5,000 Watts, and that 80% of this is converted to sound.
rusak2 [61]

Answer:

0.04

Explanation:

Fraction of power converted to sound = 80% = 0.08

Fraction incident upon each eardrum onstage = 0.08/2 = 0.04

6 0
3 years ago
• ¿Qué rapidez tendrá un sonido de 150 Hz con una longitud de onda de 2 metros?
kobusy [5.1K]

Answer:

Speed = 300 m/s

Explanation:

Given the following data;

Frequency = 150 Hz

Wavelength = 2 meters

To find the speed of the wave;

Mathematically, the speed of a wave is given by the formula:

Speed = wavelength * frequency

Substituting into the formula, we have;

Speed = 2 * 150

Speed = 300 m/s

7 0
2 years ago
The 1.53-kg uniform slender bar rotates freely about a horizontal axis through O. The system is released from rest when it is in
OlgaM077 [116]

Answer:

The spring constant = 104.82 N/m

The angular velocity of the bar when θ = 32° is 1.70 rad/s

Explanation:

From the diagram attached below; we use the conservation of energy to determine the spring constant by using to formula:

T_1+V_1=T_2+V_2

0+0 = \frac{1}{2} k \delta^2 - \frac{mg (a+b) sin \ \theta }{2}  \\ \\ k \delta^2 = mg (a+b) sin \ \theta \\ \\ k = \frac{mg(a+b) sin \ \theta }{\delta^2}

Also;

\delta = \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2}

Thus;

k = \frac{mg(a+b) sin \ \theta }{( \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2})^2}

where;

\delta = deflection in the spring

k = spring constant

b = remaining length in the rod

m = mass of the slender bar

g = acceleration due to gravity

k = \frac{(1.53*9.8)(0.6+0.2) sin \ 64 }{( \sqrt{0.6^2 +0.6^2 +2*0.6*0.6 sin \ 64} - \sqrt{0.6^2 +0.6^2})^2}

k = 104.82\ \  N/m

Thus; the spring constant = 104.82 N/m

b

The angular velocity can be calculated by also using the conservation of energy;

T_1+V_1 = T_3 +V_3  \\ \\ 0+0 = \frac{1}{2}I_o \omega_3^2+\frac{1}{2}k \delta^2 - \frac{mg(a+b)sin \theta }{2} \\ \\ \frac{1}{2} \frac{m(a+b)^2}{3}  \omega_3^2 +  \frac{1}{2} k \delta^2 - \frac{mg(a+b)sin \ \theta }{2} =0

\frac{m(a+b)^2}{3} \omega_3^2  + k(\sqrt{h^2+a^2+2ah sin \theta } - \sqrt{h^2+a^2})^2 - mg(a+b)sin \theta = 0

\frac{1.53(0.6+0.6)^2}{3} \omega_3^2  + 104.82(\sqrt{0.6^2+0.6^2+2(0.6*0.6) sin 32 } - \sqrt{0.6^2+0.6^2})^2 - (1.53*9.81)(0.6+0.2)sin \ 32 = 0

0.7344 \omega_3^2 = 2.128

\omega _3 = \sqrt{\frac{2.128}{0.7344} }

\omega _3 =1.70 \ rad/s

Thus, the angular velocity of the bar when θ = 32° is 1.70 rad/s

7 0
3 years ago
A coil has 400 turns and self-inductance 7.50 mH. The current in the coil varies with time according to i = 1680 mA2 cos [πt/(0.
zhannawk [14.2K]

Answer:

(a) 1.58 V

(b) 0.0126 Wb

(c) 0.0493 V

Solution:

As per the question:

No. of turns in the coil, N = 400 turns

Self Inductance of the coil, L = 7.50 mH = 7.50\times 10^{- 3}\ H

Current in the coil, i = 1680cos[\frac{\pi t}{0.0250}] A

where

i_{max} = 1680\ mA = 1.680\ A

Now,

(a)  To calculate the maximum emf:

We know that maximum emf induced in the coil is given by:

e = \frac{Ldi}{dt}

e = L\frac{d}{dt}(1680)cos[\frac{\pi t}{0.0250}]

e = - 7.50\times 10^{- 3}\times \frac{\pi}{0.0250}\times \frac{d}{dt}(1680)sin[\frac{\pi t}{0.0250}]

For maximum emf, sin\theta should be maximum, i.e., 1

Now, the magnitude of the maximum emf is given by:

|e| = 7.50\times 10^{- 3}\times 1680\times 10^{- 3}\times \frac{\pi}{0.0250} = 1.58\ V

(b) To calculate the maximum average flux,we know that:

\phi_{m, avg} = L\times i_{max} = 7.50\times 10^{- 3}\times 1.680 = 0.0126\ Wb

(c) To calculate the magnitude of the induced emf at t = 0.0180 s:

e = e_{o}sin{\pi t}{0.0250}

e = 7.50\times 10^{- 3}\times sin{\pi \times 0.0180}{0.0250} = 2.96\times 10^{- 4} =0.0493\ V

7 0
3 years ago
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