All categories

2 years ago

Stars that are not very hot but give off a lot of light are

Physics

2 answers:

melomori [17]2 years ago

6 0

$\huge{\textbf{\textsf{{\color{navy}{An}}{\purple{sw}}{\pink{er}} {\color{pink}{:}}}}}$

Giants.

thanks
hope it helps.

MAVERICK [17]2 years ago

3 0

Explanation:

giants are those stars that are not so hot but give a lot of light

You might be interested in

SCALCET8 3.9.018.MI. A spotlight on the ground shines on a wall 12 m away. If a man 2 m tall walks from the spotlight toward the

Firlakuza [10]

Answer:

The length of his shadow is decreasing at a rate of 1.13 m/s

Explanation:

The ray of light hitting the ground forms a right angled triangle of height H, which is the height of the building and width, D which is the distance of the tip of the shadow from the building.

Also, the height of the man, h which is parallel to H forms a right-angled triangle of width, L which is the length of the shadow.

By similar triangles,

H/D = h/L

L = hD/H

Also, when the man is 4 m from the building, the length of his shadow is L = D - 4

So, D - 4 = hD/H

H(D - 4) = hD

H = hD/(D - 4)

Since h = 2 m and D = 12 m,

H = 2 m × 12 m/(12 m - 4 m)

H = 24 m²/8 m

H = 3 m

Since L = hD/H

and h and H are constant, differentiating L with respect to time, we have

dL/dt = d(hD/H)/dt

dL/dt = h(dD/dt)/H

Now dD/dt = velocity(speed) of man = -1.7 m/s ( negative since he is moving towards the building in the negative x - direction)

Since h = 2 m and H = 3 m,

dL/dt = h(dD/dt)/H

dL/dt = 2 m(-1.7 m/s)/3 m

dL/dt = -3.4/3 m/s

dL/dt = -1.13 m/s

So, the length of his shadow is decreasing at a rate of 1.13 m/s

5 0

3 years ago

Two metallic rods A and B of different materials have same length. The linear expansivity of A is 12×10–6 oC–1and cubical expans

uysha [10]

Answer:

The length of rod A will be <u>greater than </u>the length of rod B

Explanation:

We, know that the formula for final length in linear thermal expansion of a rod is:

L' = L(1 + ∝ΔT)

where,

L' = Final Length

L = Initial Length

∝ = Co-efficient of linear expansion

ΔT = Change in temperature

Since, the rods here have same original length and the temperature difference is same as well. Therefore, the final length will only depend upon the coefficient of linear expansion.

For Rod A:

∝₁ = 12 x 10⁻⁶ °C⁻¹

For Rod B:

∝₂ = β₂/3

where,

β₂ = Coefficient of volumetric expansion for rod B = 24 x 10⁻⁶ °C⁻¹

Therefore,

∝₂ = 24 x 10⁻⁶ °C⁻¹/3

∝₂ = 8 x 10⁻⁶ °C⁻¹

Since,

∝₁ > ∝₂

Therefore,

L₁ > L₂

So, the length of rod A will be <u>greater than </u>the length of rod B

6 0

3 years ago

What is the transmitted intensity of light if an additional polarizer is added perpendicular to the first polarizer in the setup

Fantom [35]

Answer:

3) Transmitted intensity of light if unpolarized light passes through a single polarizing filter = 40 W/m²

- Transmitted intensity of light if an additional polarizer is added perpendicular to the first polarizer in the setup described = 7.5 W/m²

Explanation:

Complete Question

3) What is the transmitted intensity of light if unpolarized light passes through a single polarizing filter and the initial intensity is 80 W/m²?

- What is the transmitted intensity of light if an additional polarizer is added perpendicular to the first polarizer in the setup described in Question 3 (the setup)? Show all work in your answer.

The image of this setup attached to this question as obtained from online is attached to this solution.

Solution

3) When unpolarized light passes through a single polarizer, the intensity of the light is cut in half.

Hence, if the initial intensity of unpolarized light is I₀ = 80 W/m²

The intensity of the light rays thay pass through the first single polarizer = I₁ = (I₀/2) = (80/2) = 40 W/m²

- According to Malus' law, the intensity of transmitted light through a polarizer is related to the intensity of the incident light and the angle at which the polarizer is placed with respect to the major axis of the polarizer before the current polarizer of concern.

I₂ = I₁ cos² θ

where

I₂ = intensity of light that passes through the second polarizer = ?

I₁ = Intensity of light from the first polarizer that is incident upon the second polarizer = 40 W/m²

θ = angle between the major axis of the first and second polarizer = 30°

I₂ = 40 (cos² 30°) = 40 (0.8660)² = 30 W/m²

In the same vein, the intensity of light that passes through the third/additional polarizer is related to the intensity of light that passes through the second polarizer and is incident upon this third/additional polarizer through

I₃ = I₂ cos² θ

I₃ = intensity of light that passes through the third/additional polarizer = ?

I₂ = Intensity of light from the second polarizer that is incident upon the third/additional polarizer = 30 W/m²

θ = angle between the major axis of the second and third/additional polarizer = 60° (although, it is 90° with respect to the first polarizer, it is the angle it makes with the major axis of the second polarizer, 60°, that matters)

I₃ = 30 (cos² 60°) = 30 (0.5)² = 7.5 W/m²

Hope this Helps!!!

5 0

3 years ago

A sprinter with a mass of 70 kg accelerates at a rate of 5 m/

Crank

F = ma

= 70kg × 5ms⁻²

F =350N

8 0

3 years ago

A 0.145 kg baseball is flying at a height of 2.50 m. How much potential energy

Studentka2010 [4]

Gravitational potential energy is the energy stored in an object as the result of its vertical position or height. It can be calculated as its weight times the height. This is:

$\boxed{U_g=W\cdot h=mgh}$

Where:

W is the weight of the object
m is the mass
g is gravitational acceleration
h is the height

Evaluating:

$U_g = (0.145\;kg)(9.8\;m/s^2)(2.50\;m)\\\\\boxed{U_g=3.5525\;J}$

R/ The ball has 3.5525 J of potential energy.

3 0

3 years ago