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3 years ago

How do I get number of atoms?

Physics

1 answer:

Zina [86]3 years ago

4 0

To calculate the number of atoms in a sample, divide its weight in grams by the amu atomic mass from the periodic table, then multiply the result by Avogadro's number: 6.02 x 10^23.

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A hot-air balloon is descending at a rate of 2.3 m>s when a pas- senger drops a camera. If the camera is 41 m above the groun

zheka24 [161]

Answer:

a) t = 2.64s

b) Vf = -28.7m/s

Explanation:

If the balloon is descending, the velocity is -2.3m/s. So the equation to describe the postion of the falling camera is:

$Y = Vo*t - \frac{g*t^{2}}{2}$

$-41 = -2.3*t - \frac{10*t^{2}}{2}$ Solving for t, we get:

t1 = -3.1s and t2 = 2.64s We discard the negative time and use the positive one.

The velocity of the camera will be:

Vf = Vo - g*t = -2.3 - 10*2.64 = -28.7m/s

6 0

3 years ago

What is the momentum of an 18-kg object moving at 0.5 m/s?

Otrada [13]

sorry Idk the answer ..???

6 0

3 years ago

A 3-kg ball is thrown with a speed of 8 m/s at an unknown angle above the horizontal. The ball attains a maximum height of 2.8 m

Alina [70]

Answer:

13.5 J

Explanation:

mass of ball, m = 3 kg

maximum height, h = 2.8 m

initial speed, u = 8 m/ s

Angle of projection, θ

use the formula of maximum height

$H = \frac{u^{2}Sin^{2}\theta }{2g}$

$2.8 = \frac{8^{2}Sin^{2}\theta }{2\times 9.8}$

Sin θ = 0.926

θ = 67.8°

The velocity at maximum height is u Cosθ = 8 Cos 67.8 = 3 m/s

So, kinetic energy at maximum height

$K=\frac{1}{2}mv^{2}$

K = 0.5 x 3 x 3 x 3

K = 13.5 J

8 0

4 years ago

(TCO-8) A band-pass filter has fC1 = 5 kHz and fC2 = 88 kHz. Calculate Bandwidth (BW) and center frequency (fo) for this filter.

Temka [501]

Bandwidth is the difference between the upper and lower frequencies in a continuous band of frequencies.

Mathematically can be expressed as,

$BW = F_{c2}-F_{c1}$

The upper frequency is 88Hz and the lower frequency is 6, therefore the Bandwidth would be,

$BW = (88-5)kHz$

$BW = 83kHz$

The center frequency is given on the basis of the square root of the multiplication of the two reference frequencies, then,

$f_0 = \sqrt{f_{c1}*f_{c2}}$

$f_0 = \sqrt{88*5}$

$f_0 = 20.97kHz$

7 0

3 years ago

A spaceship negotiates a circular turn of radius 2925 km at a speed of 29960 km/h. (a) What is the magnitude of the angular spee

emmainna [20.7K]

a) 0.0028 rad/s

b) $23.68 m/s^2$

c) $0 m/s^2$

Explanation:

When an object is in circular motion, the angular speed of the object is the rate of change of its angular position. In formula, it is given by

$\omega = \frac{\theta}{t}$

where

$\theta$ is the angular displacement

t is the time interval

The angular speed of an object in circular motion can also be written as

$\omega = \frac{v}{r}$ (1)

where

v is the linear speed of the object

r is the radius of the orbit

For the spaceship in this problem we have:

$v=29,960 km/h$ is the linear speed, converted into m/s,

$v=8322 m/s$

$r=2925 km = 2.925\cdot 10^6 m$ is the radius of the orbit

Subsituting into eq(1), we find the angular speed of the spaceship:

$\omega=\frac{8322}{2.925\cdot 10^6}=0.0028 rad/s$

When an object is in circular motion, its direction is constantly changing, therefore the object is accelerating; in particular, there is a component of the acceleration acting towards the centre of the orbit: this is called centripetal acceleration, or radial acceleration.

The magnitude of the radial acceleration is given by

$a_r=\omega^2 r$

where

$\omega$ is the angular speed

$r$ is the radius of the orbit

For the spaceship in the problem, we have

$\omega=0.0028 rad/s$ is the angular speed

$r=2925 km = 2.925\cdot 10^6 m$ is the radius of the orbit

Substittuing into the equation above, we find the radial acceleration:

$a_r=(0.0028)^2(2.925\cdot 10^6)=23.68 m/s^2$

When an object is in circular motion, it can also have a component of the acceleration in the direction tangential to its motion: this component is called tangential acceleration.

The tangential acceleration is given by

$a_t=\frac{\Delta v}{\Delta t}$