Which term is defined as the change in the direction of light when it goes from one medium into a different medium?

An ideal gas is enclosed in a piston, and 1600 J of work is done on the gas. As this happens, the internal energy of the gas inc

Phantasy [73]

Answer:

- 1100 J heat flows out

Explanation:

dW = - 1600 J (as work is done on the gas)

dU = 500 J

dQ = ?

According to the first law of thermodynamics

dQ = dU + dW

dQ = 500 - 1600

dQ = - 1100 J

As heat is negative so it flows out.

3 0

3 years ago

What do you mean by physical quantity

Naddika [18.5K]

A physical quantity is something which can be measured and whose measurement is necessary because all the laws of physics are given in terms of these physical quantities. The examples of physical quantity are mass, length, time, temperature, electric current, velocity , acceleration, force, work, weight,power, energy etc.

There are some quantities which cannot be measured and hence are not classified as physical quantities , for example, love, honesty, hatred etc

Thus a physical quantity is a quantity which can be measured.

7 0

3 years ago

Read 2 more answers

An element is highly conductive, highly reactive, soft, and lustrous. The element most likely belongs to which group?(1 point)

UkoKoshka [18]

An element which is highly conductive, highly reactive, soft, and lustrous is most likely an alkali metal.

Alkali metals are in group 1 of the Periodic table which means that they have only a single valence electron.

This causes them to be soft and highly reactive because:

The single valance electron leads to weak bonds amongst the element's atoms which makes them soft
The elements want to lose the single valance electron so as to become stable so they will react with other elements to give away the electron.

Examples of alkali electrons include:

Lithium
Sodium
Potassium etc

In conclusion therefore, alkali metals are highly reactive and soft and so the element described above is most likely an alkali metal.

<em>Find out more at brainly.com/question/18722874.</em>

6 0

2 years ago

Read 2 more answers

A proton having an initial velvocity of 20.0i Mm/s enters a uniform magnetic field of magnitude 0.300 T with a direction perpend

Sonja [21]

The time interval for which the proton remains in the field is -

Δt = $$\frac{\pi R}{40}$ .

We have a proton entering a uniform magnetic field which is in a direction perpendicular to the proton's velocity.

We have to determine time interval during which the proton is in the field.

<h3>What is the magnitude of force on the charged particle moving in a uniform magnetic field?</h3>

The magnitude of force on the charged particle moving in a uniform magnetic field is given by -

F = qvB sinθ

According to the question, we have -

Entering Velocity (v) = 20 i m/s

Magnetic field intensity (B) = 0.3 T

Leaving velocity (u) = - 20 j m/s

Now -

The entering and leaving velocity vectors have 90 degrees difference between them. Therefore, only a quarter of distance of the complete circular path of radius 'R' is traced by the proton. Therefore -

d = $$\frac{2\pi r}{4}$ = $$\frac{\pi R}{2}$

Since, the radius of circular path is not given, we will assume it R.

Therefore, time for which proton remained in the field is -

t = $$\frac{\pi R}{2v} = \frac{\pi R}{40}$

Hence, the time interval for which the proton remains in the field is -

Δt = $$\frac{\pi R}{40}$

To solve more questions on Force on charged particle, visit the link below-

brainly.com/question/14597200

#SPJ4

6 0

1 year ago

(4A) The mass of Earth is 5.972 * 10^24 kg, and the radius of Earth is 6,371 km.

faltersainse [42]

Answer:

x₁ = 345100 km

Explanation:

The direction of the attraction forces between the earth and the object, and between the moon and the object, are in opposite direction and (along the straight line between the centers of earth and moon) and as gravity is always attractive, the net force will become zero when both forces are equal. According to this:

Let call "x₁" distance between center of the earth and the object, and

"x₂" the distance between center of the moon and the object, Mt mass of the earth, Ml mass of the moon, m₀ mass of the object

we can express:

F₁ ( force between earth and the object )

F₁ = K * Mt * m₀/ ( x₁)² K is a gravitational constant

F₂ (force between mn and the object)

F₂ = K * Ml * m₀ / (x₂)²

Then:

F₁ = F₂ K*Mt*m₀ / x₁² = K*Ml*m₀ /x₂²

Or simplifying the expression

Mt/ x₁² = Ml/ x₂²

We know that x₁ + x₂ = 384000 Km then

x₁ = 384000 - x₂

Mt/( 384000 - x₂)² = Ml / x₂²

Mt * x₂² = Ml *( 384000 - x₂)²

We need to solve for x₂

Mt * x₂² = Ml *[ ( 384000)² + x₂² - 768000*x₂]

By substitution:

5.972*10∧24*x₂² = 7.348*10∧22 * [ 1.47*10∧11 ] + 7.348*10∧22*x₂² -

7.348*10∧22*768000*x₂

Simplifying by 10∧22

5.972*10²*x₂² = 7.348* [ 1.47*10∧11 ] + 7.348*x₂²- 7.348*768000*x₂

Sorting out

5.972*10²*x₂²- 7.348*x₂² = 10.80*10∧11 - 56,43* 10∧5*x₂

(597,2 - 7,348 )* x₂² = 10.80*10∧11 - 56.43*10∧5*x₂

590x₂² + 56.43*10∧5*x₂ - 10.80*10∧11 = 0

Is a second degree equation

x₂ = -56.43*10∧5 ± √3184*10∧10 + 25488*10∧11 / 1160

x₂ ₁ = -56.43*10∧5 + √3184*10∧10 + 25488*10∧11 / 1160

x₂ ₁ = -56.43*10∧5 + √3184*10∧10 + 254880*10∧10 / 1160

x₂ ₁ = -56.43*10∧5 + 10∧5 [ √3184 + 254880 ] /1160

x₂ ₁ = -56.43*10∧5 + 508* 10∧5 / 1160

x₂ ₁ = 451.27*10∧5/1160

x₂ ₁ = 4512.7*10∧4 /1160

x₂ ₁ = 3.89*10∧4 km (distance between the moon and the object)

x₂ ₁ = 38900 km

x₂ = 38900 km

We dismiss the other solution because is negative and there is not a negative distance

Then the distance between the earth and the object is:

x₁ = 384000 - x₂

x₁ = 384000 - 38900

x₁ = 345100 km

5 0

3 years ago