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2 years ago

At what wavelength would a star radiate the greatest amount of energy if the star has a surface temperature of 60,000 K?

Physics

1 answer:

kompoz [17]2 years ago

7 0

Answer:

$\lambda=4.81\times 10^{-8}\ m$

Explanation:

We have,

The surface temperature of the star is 60,000 K

It is required to find the wavelength of a star that radiated greatest amount of energy. Wein's displacement law gives the relation between wavelength and temperature such that :

$\lambda T=2.89\times 10^{-3}$

Here,

$\lambda$ = wavelength

$\lambda=\dfrac{2.89\times 10^{-3}}{60000}\\\\\lambda=4.81\times 10^{-8}\ m$

So, the wavelength of the star is $4.81\times 10^{-8}\ m$ .

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2 years ago

To collect quantitative data, scientists use all of the following except _____.

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2 years ago

Read 2 more answers

Two charged objects are 1 meter apart. Calculate the magnitude of the electric force between them if the two charges are +1.0 μC

Solnce55 [7]

Answer:

0.00899 N

Explanation:

The magnitude of the electrostatic force between two charges is given by the equation:

$F=k\frac{q_1 q_2}{r^2}$

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the charges

r is the distance between the two charges

And the force is:

- Repulsive if the two charges have same sign

- Attractive if the two charges have opposite sign

In this problem we have:

$q_1=1.0\mu C = 1.0\cdot 10^{6}C$ (charge of object 1)

$q_2=1.0\mu C = 1.0\cdot 10^{6}C$ (charge of object 2)

r = 1 m (separation between the objects)

So, the electric force is

$F=(8.99\cdot 10^9)\frac{(1.0\cdot 10^{-6})^2}{1^2}=0.00899 N$

5 0

2 years ago

A 15 m uniform ladder weighing 500 N rests against a frictionless wall. The ladder makes a 60° angle with horizontal. (a) Find t

scoray [572]

Answer:

a) F₁ = 267.3 N, N₁ = 1300 N, b) μ = 0.324

Explanation:

For this exercise we use the rotational equilibrium condition, we have a reference system is the floor and the anticlockwise rotations as positive, in the adjoint we can see a diagram of the forces

let's use subscript 1 for the ladder and 2 for the firefighter

∑ τ = 0

-W₁ x₁ - W₂ x₂ + N₁ y = 0

N₁ = $\frac{W_1 x_1 + W_2 x_2}{y}$ (1)

the center of mass of the ladder is at its geometric center,

d = L / 2 = 15/2 = 7.5 m

cos 60 = x₁ / d₁

x₁ = d₁ cos 60

x₁ = 7.5 cos 60

x₁ = 3.75 m

for the firefighter d₂ = 4 m

cos 60 = x₂ / d₂

x₂ = d₂ cos 60

x₂ = 4 cos 60 = 2 m

for the fulcrum d₃ = 15 m

sin 60 = y / d₃

y = d₃ sin 60

y = 15 sin 60

y = 13 m

we look for the Normal by substituting in equation 1

N₂ = $\frac{500 \ 3.75 \ + 800 \ 2}{13}$

N₂ = 267.3 N

now let's use the translational equilibrium relations

X axis

F₁ - N₂ = 0

F₁ = N₂

F₁ = 267.3 N

Axis y

N₁ - W₁ -W₂ = 0

N₁ = W₁ + W₂

N₁ = 500 + 800

N₁ = 1300 N

b) for this case change the firefighter's distance d₂ = 9 m

x₂ = 9 cos 60

x₂ = 4.5 m

we substitute in 1

N₂ = \frac{500 \ 3.75 \ + 800 \ 4.5}{13}

N₂ = 421.15 N

of the translational equilibrium equation on the x-axis

fr = F₁ = N₂

fr = 421.15 N

friction force has the expression

fr = μ N

in this case the reaction of the Earth to the support of the ladder is N1 = 1300N

μ = fr / N₁

μ = 421.15 / 1300

μ = 0.324

8 0

2 years ago

Umar has two copper pans, each containing 500cm3 of water. Pan A has a mass of 750g and pan B has a mass of 1.5kg. Which pan wil

Olin [163]

Answer:

heat required in pan B is more than pan A

Explanation:

Heat required to raise the temperature of the substance is given by the formula

$Q = ms\Delta T$

now we know that both pan contains same volume of water while the mass of pan is different

So here heat required to raise the temperature of water in Pan A is given as

$Q_1 = (m_w s_w + m_ps_p)\delta T$

$Q_1 = (0.5(4186) + 0.750(s))\Delta T$

Now similarly for other pan we have

$Q_2 = (m_w s_w + m_ps_p)\delta T$

$Q_2 = (0.5(4186) + 1.50(s))\Delta T$

So here by comparing the two equations we can say that heat required in pan B is more than pan A

3 0

2 years ago