How do I find the specific heat?

What type of glass absorbs light and does not allow light to pass through it?

sesenic [268]

opaque glass does not allow light to pass through it.

8 0

2 years ago

An Atwood's machine consists of two different masses, both hanging vertically and connected by an ideal string which passes over

Tasya [4]

Answer:

V₁ = √ (gy / 3)

Explanation:

For this exercise we will use the concepts of mechanical energy, for which we define energy n the initial point and the point of average height and / 2

Starting point

Em₀ = U₁ + U₂

Em₀ = m₁ g y₁ + m₂ g y₂

Let's place the reference system at the point where the mass m1 is

y₁ = 0

y₂ = y

Em₀ = m₂ g y = 2 m₁ g y

End point, at height yf = y / 2

$E_{mf}$ = K₁ + U₁ + K₂ + U₂

$E_{mf}$ = ½ m₁ v₁² + ½ m₂ v₂² + m₁ g $y_{f}$ + m₂ g $y_{f}$

Since the masses are joined by a rope, they must have the same speed

$E_{mf}$ = ½ (m₁ + m₂) v₁² + (m₁ + m₂) g $y_{f}$

$E_{mf}$ = ½ (m₁ + 2m₁) v₁² + (m₁ + 2m₁) g $y_{f}$

How energy is conserved

Em₀ = $E_{mf}$

2 m₁ g y = ½ (m₁ + 2m₁) v₁² + (m₁ + 2m₁) g $y_{f}$

2 m₁ g y = ½ (3m₁) v₁² + (3m₁) g y / 2

3/2 v₁² = 2 g y -3/2 g y

3/2 v₁² = ½ g y

V₁ = √ (gy / 3)

5 0

2 years ago

Examples of fixed pulley?

Neko [114]

Answer:

One example of a fixed pulley is a Flag Pole

Explanation:

A good example of a fixed pulley is a flag pole: When you pull down on the rope, the direction of force is redirected by the pulley, and you raise the flag. Other examples of movable pulleys include construction cranes, modern elevators, and some types of weight lifting machines at the gym.

4 0

2 years ago

At 20∘C20 ∘ C, the hole in an aluminum ring is 2.500 cm in diameter. You need to slip this ring over a steel shaft that has a ro

Feliz [49]

Answer:

238.75⁰C .

Explanation:

coefficient of linear thermal expansion of aluminum and steel is 23 x 10⁻⁶ K⁻¹ and 12 x 10⁻⁶ K⁻¹ respectively .

Rise in temperature be Δ t .

Formula for linear expansion due to heat is as follows

l = l₀ ( 1 + α x Δt )

l is expanded length , l₀ is initial length , α is coefficient of linear expansion and Δt is increase in temperature .

For aluminum

l = 2.5 ( 1 + 23 x 10⁻⁶ Δt )

For steel

l = 2.506 ( 1 + 12 x 10⁻⁶ Δt )

Given ,

2.5 ( 1 + 23 x 10⁻⁶ Δt ) = 2.506 ( 1 + 12 x 10⁻⁶ Δt )

1 + 23 x 10⁻⁶ Δt = 1.0024 ( 1 + 12 x 10⁻⁶ Δt )

1 + 23 x 10⁻⁶ Δt = 1.0024 + 12.0288 x 10⁻⁶ Δt

10.9712 x 10⁻⁶ Δt = .0024

Δt = 218.75

Initial temperature = 20⁰C

final temperature = 218.75 + 20 = 238.75⁰C .

8 0

2 years ago

A charged paint is spread in a very thin uniform layer over the surface of a plastic sphere of diameter 18.0 cm , giving it a ch

vladimir1956 [14]

A) The electric field inside the paint layer is zero

B) The electric field just outside the paint layer is $3.2\cdot 10^7 N/C$ (radially inward)

C) The electric field at 6.00 cm from the surface is $1.2\cdot 10^7 N/C$ (radially inward)

Explanation:

A)

We can solve the problem by applying Gauss Law, which states that the electric flux through a Gaussian surface must be equal to the charge contained in the surface divided by the vacuum permittivity:

$\int EdS = \frac{q}{\epsilon_0}$

where

E is the magnitude of the electric field

dS is the element of the surface

q is the charge contained within the surface

$\epsilon_0$ is the vacuum permittivity

By taking a sphere centered in the origin,

$\int E dS = E \cdot 4\pi r^2$

where $4\pi r^2$ is the surface of the Gaussian sphere of radius r.

In this problem, we want to find the electric field just inside the paint layer, so we take a value of r smaller than

$R=9.0 cm = 0.09 m$ (radius of the plastic sphere is half of the diameter)

Since the charge is all distributed over the plastic sphere, the charge contained within the Gaussian sphere is zero:

$q=0$

And therefore,

$E4\pi r^2 = 0\\\rightarrow E = 0$

So, the electric field inside the plastic sphere is zero.

B)

Here we apply again Gauss Law:

$E\cdot 4 \pi r^2 = \frac{q}{\epsilon_0}$

In this case, we want to calculate the electric field just outside the paint layer: this means that we take r as the radius of the plastic sphere, so

$r=R=0.18 m$

The charge contained within the Gaussian sphere is therefore

$q=-29.0 \mu C = -29.0\cdot 10^{-6}C$

Therefore, the electric field is

$E=\frac{q}{4\pi \epsilon_0 R^2}=\frac{-29.0\cdot 10^{-6}}{4\pi (8.85\cdot 10^{-12})(0.09)^2}=-3.2\cdot 10^7 N/C$

And the negative sign indicates that the direction of the field is radially inward (because the charge that generates the field is negative). However, the text of the question says "Enter positive value if the field is directed radially inward and negative value if the field is directed radially outward", so the answer to this part is

$E=3.2\cdot 10^7 N/C$

C)

For this part again, we apply Gauss Law:

$E\cdot 4 \pi r^2 = \frac{q}{\epsilon_0}$

In this case, we want to calculate the field at a point 6.00 cm outside the surface of the paint layer; this means that the radius of the Gaussian sphere must be

r = 9 cm + 6 cm = 15 cm = 0.15 m

While the charge contained within the sphere is again

$q=-29.0 \mu C = -29.0\cdot 10^{-6}C$

Therefore, the electric field in this case is

$E=\frac{q}{4\pi \epsilon_0 R^2}=\frac{-29.0\cdot 10^{-6}}{4\pi (8.85\cdot 10^{-12})(0.15)^2}=-1.2\cdot 10^7 N/C$