What measures the mount of displacement in a longitudinal esbelta?

Give two examples of direct current electricity.

ludmilkaskok [199]

Answer:

<h3>Ion beams and Electrochemical are two examples of direct current electricity.</h3>

6 0

4 years ago

The speed of light in vacuum is exactly 299,792,458 m/s. A beam of light has a wavelength of 651 nm in vacuum. This light propag

sukhopar [10]

Answer:

$v=2.58\times10^8m/s$

Explanation:

The index of refraction is equal to the speed of light c in vacuum divided by its speed v in a substance, or $n=\frac{c}{v}$ . For our case we want to use $v=\frac{c}{n}$ , which for our values is equal to:

$v=\frac{c}{n}=\frac{299792458m/s}{1.16}=258441774.138m/s$

Which we will express with 3 significant figures (since a product or quotient must contain the same number of significant figures as the measurement with the <em>least</em> number of significant figures):

$v=2.58\times10^8m/s$

4 0

3 years ago

Can some one help me ;-;

enyata [817]

Answer: The first answer for the first problem, and the 2nd answer for the second problem

Explanation: For the first one, if it is absolute zero, the molecules would not move at all.

For the second one, the temperature of the sample will increase due to the movement.

6 0

3 years ago

The earth's radius is 6.37×106m; it rotates once every 24 hours. What is the earth's angular speed? What is the speed of a point

Zarrin [17]

Answer:

a) w = 7.27 * 10^-5 rad/s

b) v1 = 463.1 m/s

c) v1 = 440.433 m/s

Explanation:

Given:-

- The radius of the earth, R = 6.37 * 10 ^6 m

- The time period for 1 revolution T = 24 hrs

Find:

What is the earth's angular speed?

What is the speed of a point on the equator?

What is the speed of a point on the earth's surface located at 1/5 of the length of the arc between the equator and the pole, measured from equator?

Solution:

- The angular speed w of the earth can be related with the Time period T of the earth revolution by:

w = 2π / T

w = 2π / 24*3600

w = 7.27 * 10^-5 rad/s

- The speed of the point on the equator v1 can be determined from the linear and rotational motion kinematic relation.

v1 = R*w

v1 = (6.37 * 10 ^6)*(7.27 * 10^-5)

v1 = 463.1 m/s

- The angle θ subtended by a point on earth's surface 1/5 th between the equator and the pole wrt equator is.

π/2 ........... s

x ............ 1/5 s

x = π/2*5 = 18°

- The radius of the earth R' at point where θ = 18° from the equator is:

R' = R*cos(18)

R' = (6.37 * 10 ^6)*cos(18)

R' = 6058230.0088 m

- The speed of the point where θ = 18° from the equator v2 can be determined from the linear and rotational motion kinematic relation.

v2 = R'*w

v2 = (6058230.0088)*(7.27 * 10^-5)

v2 = 440.433 m/s

5 0

3 years ago

The drawing shows a large cube (mass = 28.6 kg) being accelerated across a horizontal frictionless surface by a horizontal force

MrRissso [65]

Answer:

P= 454.11 N

Explanation:

Since P is the only horizontal force acting on the system, it can be defined as the product of the acceleration by the total mass of the system (both cubes).

$P= (M+m)*a\\a = \frac{P}{28.6 +4.3}\\a = \frac{P}{32.9}$

The friction force between both cubes (F) is defined as the normal force acting on the smaller cube multiplied by the coefficient of static friction. Since both cubes are subject to the same acceleration:

$F = m * a*\mu \\F= 4.3*0.710*\frac{P}{32.9}\\F=3.053*\frac{P}{32.9}$

In order for the small cube to not slide down, the friction force must equal the weight of the small cube:

$3.053*\frac{P}{32.9} = 4.3 * g\\\\P = \frac{4.3*9.8*32.9}{3.053} \\P= 454.11 N$

The smallest magnitude that P can have in order to keep the small cube from sliding downward is 454.11 N

8 0

3 years ago