PHYSICS COLLEGE

A piston above a liquid in a closed container has an area of 0.75m^2, and the piston carries a load of 200kg. What will be the external pressure on the upper surface of the liquid?

Answers

Answer 1

Answer:

Answer:

2613.3 pa

Explanation:

p=F/A

p=ma/A

p=200×9.8/0.75

p=2613.3

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Ultraviolet light is typically divided into three categories. UV-A, with wavelengths between 400 nm and 320 nm, has been linked with malignant melanomas. UV-B radiation, which is the primary cause of sunburn and other skin cancers, has wavelengths between 320 nm and 280 nm. Finally, the region known as UV-C extends to wavelengths of 100 nm. (a) Find the range of frequencies for UV-B radiation. (b) In which of these three categories does radiation with a frequency of 7.9 * 1014 Hz belong

Answers

Answer:

a) The UV-B has frequencies between $9.375x10^(14)Hz$ and $1.071x10^(15)Hz$

b) The radiation with a frequency of $7.9x10^(14)Hz$ belong to the UV-A category.

Explanation:

(a) Find the range of frequencies for UV-B radiation.

Ultraviolet light belongs to the electromagnetic spectrum, which distributes radiation along it in order of different frequencies or wavelengths.

Higher frequencies:

Gamma ray

X ray

Ultraviolet rays

Visible region

Lower frequencies:

Infrared

Microwave

Radio waves

That radiation is formed by electromagnetic waves, which are transverse waves formed by an electric field and a magnetic field perpendicular to it. Any of those radiations will have a speed of $3x10^{8]m/s$ in vacuum.

The velocity of a wave can be determined by means of the following equation:

$c = \nu \cdot \lambda$ (1)

Where c is the speed of light, $\nu$ is the frequency and $\lambda$ is the wavelength.

Then, from equation 1 the frequency can be isolated.

$\nu = (c)/(\lambda)$ (2)

Before using equation 2 to determine the range of UV-B it is necessary to express $\lambda$ in units of meters in order to match with the units from c.

$\lambda = 320nm . (1m)/(1x10^(9)nm)$ ⇒ $3.2x10^(-7)m$

$\lambda = 280nm . (1m)/(1x10^(9)nm)$ ⇒ $2.8x10^(-7)m$

$\nu = (3x10^(8)m/s)/(3.2x10^(-7)m)$

$\nu = 9.375x10^(14)s^(-1)$

$\nu = 9.375x10^(14)Hz$

$\nu = (3x10^(8)m/s)/(2.8x10^(-7)m)$

$\nu = 1.071x10^(15)Hz$

Hence, the UV-B has frequencies between $9.375x10^(14)Hz$ and $1.071x10^(15)Hz$

(b) In which of these three categories does radiation with a frequency of $7.9x10^(14)Hz$ belong.

The same approach followed in part A will be used to answer part B.

Case for UV-A:

$\lambda = 400nm . (1m)/(1x10^(9)nm)$ ⇒ $4x10^(-7)m$

$\nu = (3x10^(8)m/s)/(4x10^(-7)m)$

$\nu = 7.5x10^(14)s^(-1)$

$\nu = 7.5x10^(14)Hz$

Hence, the UV-A has frequencies between $7.5x10^(14)Hz$ and $9.375x10^(14)Hz$ .

Therefore, the radiation with a frequency of $7.9x10^(14)Hz$ belongs to UV-A category.

A small space probe of mass 170 kg is launched from a spacecraft near Mars. It travels toward the surface of Mars, where it will eventually land. At a time 22.9 seconds after it is launched, the probe is at location <5600, 7200, 0> m, and at this same instant its momentum is <51000, -7000, 0> kg·m/s. At this instant, the net force on the probe due to the gravitational pull of Mars plus the air resistance acting on the probe is <-4000, -780, 0> N.Assuming that the net force on the probe is approximately constant during this time interval, what is the change of the momentum of the probe in the time interval from 22.6 seconds after the probe is launched to 22.9 seconds after the launch?

Answers

Answer:

The change in momentum is $\Delta p = <-1200 , -234 , 0> kg \cdot m/s$

Explanation:

From the question we are told that

The mass of the probe is $m = 170 kg$

The location of the prob at time t = 22.9 s is $A = <5600, 7200,0>$

The momentum at time t = 22.9 s is $p = < 51000, -7000, 0> kg m/s$

The net force on the probe is $F = <-4000 , -780 , 0> N$

Generally the change in momentum is mathematically represented as

$\Delta p = F * \Delta t$

The initial time is 22.6 s

The final time is 22.9 s

Substituting values

$\Delta p = <-4000 , -780 , 0> * (22.9 - 22.6)$

$\Delta p = <-4000 , -780 , 0> * (0.3)$

$\Delta p = <-1200 , -234 , 0> kg \cdot m/s$

How do you define 'heat' and 'temperature'

Answers

Temperature -a measure of the average kinetic energy of the individual particles in an object

Heat is the movement of thermal energy from a substance of higher temperature to a substance of lower temperature

Answer:

Heat and temperature are related but very different.

Explanation:

Heat: The total energy of molecular motion in a substance

Temperature: A measure of the average energy of molecular motion in a substance

For further help:

Examples

Heat Temperature

-Heat is a form of energy that can -The degree of hotness and

transfer from hot body to cold body coldness of the body

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-Heat flows from hot body to cold -It rises when heated and falls down

body when an object is cooled down

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-Total kinetic energy and potential -Temp. is the average kinetic

energy obtained by molecules in energy of molecules in a

an object substance

A 1300-turn coil of wire that is 2.10 cm in diameter is in a magnetic field that drops from 0.130 T to 0 T in 12.0 ms. The axis of the coil is parallel to the field.Question: What is the emf of the coil? (in V)

Answers

Answer:

4.875 V

Explanation:

N = 1300

diameter = 2.10 cm

radius = half of diameter = 1.05 cm

B1 = 0.130 T

B2 = 0 T

t = 12 ms

According to the law of electromagnetic induction,

$e = - N(d\phi )/(dt)$

Where, Ф be the magnetic flux linked with the coil

$e = - NA (dB )/(dt)$

$e = -1300*3.14*{1.05* 1.05* 10^(-4)*(0-0.130)/(12*10^(-3))=$

e = 4.875 V

Experts in model airplanes develop a supersonic plane to scale, it moves horizontally in the air while it is conducting a flight test. The development team defines that the space that the airplane travels as a function of time is given by the function: e (t) = 9t 2 - 6t + 3 Determine what acceleration the scale airplane has (Second derivative).

Answers

Explanation:

e(t) = 9t² − 6t + 3

The velocity is the first derivative:

e'(t) = 18t − 6

The acceleration is the second derivative:

e"(t) = 18

A 62.0 kg skier is moving at 6.90 m/s on a frictionless, horizontal, snow-covered plateau when she encounters a rough patch 4.50 m long. The coefficient of kinetic friction between this patch and her skis is 0.300. After crossing the rough patch and returning to friction-free snow, she skis down an icy, frictionless hill 2.50 m high.How fast is the skier moving when she gets to the bottom ofthe hill?