there are 1000 g in every kilogram of a box has a mass of 483 g what is the mass of the box in kilogram

Carbon has six protons. Which model shows a neutral atom of carbon?

MatroZZZ [7]

Explanation:

Neutral carbon-12 (or any carbon atom) has 6 electrons with a total negative charge of 6e- orbiting a nucleus with a total positive charge of 6e+, so that the total net charge is zero. The nucleus is made up of 6 protons, each with a positive charge of e+, and 6 neutrons, each with zero charge.

4 0

2 years ago

An Aeolian harp (named after Aeolus, the Greek god of the wind) consists of several strings fixed to a frame or a sounding box.

Mashcka [7]

Answer:

0.434 m

Explanation:

We are given that

Mass per unit length= $\mu=\frac{m}{l}=2.4 g/m=2.4\times 10^{-3} kg/m$

1 kg=1000 g

Tension=350 N

Fundamental frequency=f=440 Hz

We have to find the length you should make it.

We know that

$l=\frac{1}{2f}\sqrt{\frac{T}{\mu}}$

Using the formula

$l=\frac{1}{2\times 440}\times\sqrt{\frac{350}{2.4\times 10^{-3}}}$

l=0.434 m

Hence, you should make it 0.434 m long.

5 0

3 years ago

Given vectors D (3.00 m, 315 degrees wrt x-axis) and E (4.50 m, 53.0 degrees wrt x-axis), find the resultant R= D + E. (a) Write

fomenos

Answer:

(a) $\vec{R}= 4.83\ m\ \hat{i}+1.47\ m\ \hat{j}$

(b) (5.05 m, 16.93 degrees wrt x-axis)

Explanation:

Given:

$\vec{D}$ = (3.00 m, 315 degrees wrt x-axis)
$\vec{E}$ = (4.50 m, 53.0 degrees wrt x-axis)

Let us first fond out vector D and E in their rectangular form.

$\vec{D} = (3\cos 315^\circ\ \hat{i}+3\sin 315^\circ\ \hat{j})\ m\\\Rightarrow \vec{D} = (2.12\ \hat{i}-2.12\ \hat{j})\ m\\$

Similarly,

$\vec{E} = (4.5\cos 53^\circ\ \hat{i}+4.5\sin 53^\circ\ \hat{j})\ m\\\Rightarrow \vec{D} = (2.71\ \hat{i}+3.59\ \hat{j})\ m\\\because \vec{R}=\vec{D}+\vec{E}\\\therefore \vec{R} = (2.12\ \hat{i}-2.12\ \hat{j})\ m+(2.71\ \hat{i}+3.59\ \hat{j})\ m\\\Rightarrow \vec{R} = (4.83\ \hat{i}+1.47\ \hat{j})\ m$

Part (a):

We can write the resultant vector R as below:

$\vec{R} = (4.83\ \hat{i}+1.47\ \hat{j})\ m$

Part (b):

$Magnitude\ of\ resultant = \sqrt{4.83^2+1.47^2}\ m = 5.05\ m\\\textrm{Direction in angle with the x-axis} = \theta = \tan^{-1}(\dfrac{1.47}{4.83})= 16.93^\circ$

Since both the components of the resultant lie on the positive x and y axes. So, the resultant makes an acute angle with the positive x-axis.

So, R = (5.05 m, 16.93 degrees wrt x-axis)

3 0

2 years ago

Two conducting spheres with diameters of 0.400 m and 1.00 m are separated by a distance that is large compared with the diameter

PSYCHO15rus [73]

Answer:

Part a)

$Q_1 = 2\mu C$

Part b)

$\Delta V = 4.5 \times 10^4 V$

Explanation:

As we know that conducting sphere is treated like spherical capacitor

so we can say

$\frac{C_1}{C_2} = \frac{0.4}{1}$

also we know that

$Q = CV$

so charge on two spheres will be in the ratio of their capacitance

Part a)

$Q_1 + Q_2 = 7 \mu C$

$\frac{Q_1}{Q_2}= 0.4$

$1.4Q_2 = 7\mu C$

$Q_2 = 5 \mu C$

$Q_1 = 2\mu C$

Part b)

Now potential difference between two sphere can be given as

$\Delta V = \frac{Q}{C}$

$\Delta V = \frac{2\mu C}{4\pi \epsilon_0 R_1}$

$\Delta V = \frac{2\mu C}{4\pi \epsilon_0(0.400)}$

$\Delta V = 4.5 \times 10^4 V$

8 0

3 years ago

Suppose a particle is accelerated through space by a constant 10-N force. Suddenly the particle encounters a second force of 10-

Harlamova29_29 [7]

Answer:

The particle will continue moving at constant velocity

Explanation:

When the particle encounters the second force of 10 N, the net force acting on the particle becomes zero, because the two forces are equal in magnitude but opposite in direction:

$F_{net}=10 N-10 N=0$

For Newton's second Law, the acceleration of the particle is proportional to the net force acting on it:

$F_{net}=ma$

Therefore, since $F_{net}=0$ , the acceleration is zero (a=0) and so the particle will keep a constant velocity.

3 0

3 years ago