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4 years ago

6

Which of the following atoms is considered neutral? An atom with 7 protons, 7 neutrons & 8 electrons An atom with 3 protons,

3 neutrons & 7 electrons An atom with 3 protons, 4 neutrons & 3 electrons An atom with 3 protons, 3 neutrons & 4 electrons

1 answer:

Gwar [14]4 years ago

5 0

Answer:

The answer is the atom with 3 protons, 4 neutrons, and 3 electrons.

Explanation:

The reason why, is because protons are positive, neutrons are neutral, and electrons are negative. The protons cancel out the electrons, and the neutrons are just there.

You might be interested in

What phase does the moon have to be in for a solar eclipse

kakasveta [241]

Answer:

new moon

Explanation:

A solar eclipse take place at new moon phase, when the moon passes between the earth and the sun and its shadows fall on the Earth's surface which by definition a solar eclipse.

7 0

3 years ago

A fan at a rock concert is 50.0 m from the stage, and at this point the sound intensity level is 114 dB. Sound is detected when

Marianna [84]

Answer:

A) $P=13.92\ J.s^{-1}$

B) $v=3730.9912\ m.s^{-1}$

C) $v=74.44\ mm.s^{-1}$

D) mosquitoes speed in part B is very much larger than that of part C.

Explanation:

Given:

Distance form the sound source, $s=50\ m$

sound intensity level at the given location, $\beta=114\ dB$

diameter of the eardrum membrane in humans, $d=8.4 \times 10^{-3}\ m$

We have the minimum detectable intensity to the human ears, $I_0=10^{-12}\ W.m^{-2}$

(A)

<u>Now the intensity of the sound at the given location is related mathematically as:</u>

$\beta=10\ log(\frac{I}{I_0} )$ ..........................................(1)

$114=10\ log\ (\frac{I}{10^{-12}} )$

$11.4=log\ I+12\ log\ 10$

$I=0.2512\ W.m^{-2}$

<em>As we know :</em>

$I=\frac{P}{A}$

$0.2512=\frac{P}{\pi\times \frac{8.4^2}{4} }$

$P=13.92\ J.s^{-1}$ is the energy transferred to the eardrums per second.

(B)

mass of mosquito, $m=2\times 10^{-6}\ kg$

<u>Now the velocity of mosquito for the same kinetic energy:</u>

$KE=\frac{1}{2} m.v^2$

$13.92=\frac{1}{2}\times 2\times 10^{-6}\times v^2$

$v=3730.9912\ m.s^{-1}$

(C)

Given:

Sound intensity, $\beta = 20\ dB$

<u>Using eq. (1)</u>

$20=10\ log\ (\frac{I}{10^{-12}} )$

$2=log\ I+12\ log\ 10$

$I=10^{-10}\ W.m^{-2}$

Now, power:

$P=I.A$

$P=10^{-10}\times \pi\times \frac{8.4^2}{4}$

$P=5.54\times 10^{-9}\ J.s^{-1}$

Hence:

$KE=\frac{1}{2} m.v^2$

$5.54\times 10^{-9}=0.5\times 2\times 10^{-6}\times v^2$

$v=0.07444\ m.s^{-1}$

$v=74.44\ mm.s^{-1}$

(D)

mosquitoes speed in part B is very much larger than that of part C.

7 0

3 years ago

Which type of forces would be experienced by a sled that is being pulled by a rope (choose 4)

sukhopar [10]

Friction, wind resistance, gravity, tension force

8 0

3 years ago

A thin, uniform stick of length 1.9 m and mass 3.1 kg is pinned through one end and is free to rotate. The stick is initially ha

nirvana33 [79]

Answer:

The acceleration is $\alpha = 7.10 \ rad/s^2$

Explanation:

From the question we are told that

The length of the stick is $d = 1.9 \ m$

The mass of the stick is $m = 3.1 \ kg$

The angular displacement is $\theta = 23.4^o$

Generally the torque of this uniform stick after this displacement is mathematically represented as

$\tau = \frac{1}{2} * d * [m*g]* sin (90 - \theta)$

$\tau = \frac{1}{2} * 1.9 * [3.1*9.8]* sin (90 - 23.4)$

$\tau = 26.49 \ kg\cdot m^2 \cdot s^{-2}$

Generally the moment of inertia of the uniform stick is mathematically represented as

$I = \frac{1}{3} * m * d^2$

=> $I = \frac{1}{3} * 3.1 * 1.9 ^2$

=> $I = 3.73 \ kg \cdot m^2$

Generally the angular acceleration is mathematically represented as

$\alpha = \frac{\tau}{I}$

=> $\alpha = \frac{26.49}{3.73}$

=> $\alpha = 7.10 \ rad/s^2$

7 0

3 years ago

Un carro parte del reposo y acelera uniformemente a 4.0 m / s^2 durante 5.0 s. A continuación, mantiene la velocidad que alcanzó

Arada [10]

Answer:

Average speed of the car = 16.53 m.s

Explanation:

For first 5 seconds:

As the car starts from rest, so the initial speed of the car, u=0

Acceleration, a= 4 m/s^2

So, the final speed (speed at time, t=5 second)

v=u+at

v=0 + 4 x 5 = 20 m/s

Distance covered,

$s=ut+\frac 1 2 a t^2 \\\\s=0+\frac 1 2 \times 4 \times 5^2$

s=50 m

As the car maintained the constant speed of 20 m/s for the next 10 seconds.

Distance traveled in 10 seconds, d=20 x 10 =200 m

For the last 4 seconds:

The car slows down at the rate of 2.0 m / s ^ 2 from the speed of 20 m/s.

As the car slows down, so the acceleration will be negative, i.e a=-2 m/s^2

$s=ut+\frac 12 at^2 \\\\s=20\times 4 +\frac 12 (-2)4^2$

s=80-16=64 m.

Considering all three cases, the total distance covered=

=50+200+64=314 m

Total time= 5+10+4=19 s

So, the averate peed= Totat distance covered / total time taken

=314/19=16.53 m/s

6 0

3 years ago