Ask question

Ask question

All categories

4 years ago

9

if chlorine has a mass of 35.5 atomic mass units and a sodium atom has a mass of 23.0 atomic mass units, what is the mass of one

NaCI unit?

1 answer:

Hitman42 [59]4 years ago

4 0

35.5+23.0=58.5 ......................

You might be interested in

Define accleration due to gravity

kaheart [24]

Therefore as you move around the U.S. the acceleration due to gravity (g) varies from about 9.79 to 9.81 meters per second squared. The Earth’s average is 9.80 m/s2 which is generally reported as the acceleration of gravity on Earth.

6 0

3 years ago

Read 2 more answers

A 5.58 kg object with a speed of 35 m/s strikes a steel plate at an angle of 45.0 degrees with the normal to the plate, and rebo

MAVERICK [17]

Answer:

magnitude of vector is 276.19 kg m/s

Explanation:

The initial momentum is vector of magnitude

$5.58 \times 35 = 195.3 (kg m/s)$ And driven in a coherent manner with initial vector.

same magnitude is momentum after the impact, but it is oriented perpendicularly to initial momentum vector.

So, you have 2 momentum vector of specified magnitude perpendicular to one another.

The contrast between such two vectors is a right angle triangle hypotenuse of 195.3 sides

magnitude of vector is $\sqrt{ 195.3^2 + 195.3^2} = 276.19$

8 0

3 years ago

3. Nearsighted people can undergo a medical procedure called keratotomy to produce normal (or at least more-normal) vision. In t

Mademuasel [1]

Answer:

P= -1.83D

Explanation:

Given,

u=-6 m

v=-50 cm = - 0.5 m

the power of the lens

P = 1/f = 1/v = 1/u

P = 1/ -0.50 + 1/6

P = -1.83D

7 0

3 years ago

Four identical masses of 2.5 kg each are located at the corners of a square with 1.0-m sides. What is the net force on any one o

Ghella [55]

Answer:

F=8.0*10^{-10}N

Explanation:

See the attached file for the masses distributions

The force between two masses at distance r is expressed as

$F=\frac{Gm_{1}m_{2} }{r^{2} }\\ G=Gravitional constant \\$

since the masses are of the same value, the above formula can be reduce to

$F=\frac{Gm^{2}}{r^{2} }\\$

using vector notation,Let use consider the force on the lower left corner of the mass due to the upper left side of the mass is

$F_{12} =\frac{Gm^{2}}{r^{2} }j\\$

The force on the lower left corner of the mass due to the lower right side of the mass is

$F_{14} =\frac{Gm^{2}}{r^{2} }i\\$

The force on the lower left corner of the mass due to the upper right side of the mass is

$F_{13} =\frac{Gm^{2}}{d^{2} }cos\alpha i +\frac{Gm^{2}}{d^{2} }sin\alpha j\\$

The net force can be express as

$F=\frac{Gm^{2}}{r^{2} }j +\frac{Gm^{2}}{r^{2} }i +\frac{Gm^{2}}{d^{2} }cos\alpha i +\frac{Gm^{2}}{d^{2} }sin\alpha j\\\\F=Gm^{2}[\frac{1}{r^{2}}+ \frac{1}{d^{2}cos\alpha }]i + Gm^{2}[\frac{1}{r^{2}}+ \frac{1}{d^{2}sin\alpha }]j\\\alpha=45^{0}, G=6.67*10^{-11}Nmkg^{-2}$

if we insert values we arrive at

$F=6.67*10^{-11}*2.5^{2}[\frac{1}{1^{2}}+ \frac{1}{\sqrt{2}^{2}cos45 }]i + 6.67*10^{-11}*2.5^{2}[\frac{1}{1^{2}}+ \frac{1}{\sqrt{2}^{2}sin45}]j\\F=5.643*10^{-10}i+5.643*10^{-10}j$

if we solve for the magnitude, we arrive at

$F=5.643*10^{-10}i+5.643*10^{-10}j \\F=\sqrt{(5.643*10^{-10})^{2} +(5.643*10^{-10})}^{2} \\F=8.0*10^{-10}$

Hence the net force on one of the masses is

$F=8.0*10^{-10}N$

8 0

3 years ago

A loudspeaker having a diaphragm that vibrates at 910 Hz is traveling at 75.0 m/s directly toward a pair of holes in a very larg

Alina [70]

To solve this problem we need to apply doppler equation,

Our values are:

$f_0 = 910Hz\\c = 75m\\v=344m/s\\\theta= 12.9\°\\$

Doppler equation is given by,

$f= \frac{v}{v-c} f_0$

Substituting,

$f= \frac{344}{344-75}*910$

$f = 1163Hz$

$\lambda = \frac{344}{1163}$

Wavelenght is equal to,

$\lambda = \frac{c}{f}$

$\lambda = 0.2956$

We can now find the distance by,

$d= \frac{1}{2}\frac{\lambda}{sin\theta}$

$d=\frac{1}{2}\frac{0.2956}{sin(12.9)}$

$d= 0.66m$

b) We can find at what angles would sound first cancel through the same equation of wavelenght,

$\lambda = \lambda{c}{f} = \frac{344}{1163}$

$d* sin\theta = \frac{1}{2}*\lambda$

$\theta = arcsin( 0.5*(344/1163)/0.66)=12.94\°$

5 0

3 years ago