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3 years ago

Astronomers initially had difficulty identifying the emission lines in quasar spectra at optical wavelengths because

Physics

1 answer:

Rus_ich [418]3 years ago

6 0

No one expected violet & ultraviolet spectral lines to be shifted towards the red.

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Friction in a car's engine produces some wasted _____ energy.

alexira [117]

The answer to this is B.

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3 years ago

Read 2 more answers

A grandfather clock keeps time using a pendulum consisting of a light rod connected to a small heavy mass. With a rod of length

german

Answer:

The length of the rod should be

$\frac{L}{4} \\$

Explanation:

Period of simple pendulum is given by

$T=2\pi\sqrt{\frac{l}{g}} \\$

We have

$\frac{T_1^2}{T_2^2}=\frac{l_1}{l_2}\\\\\frac{2^2}{1^2}=\frac{L}{l_2}\\\\l_2=\frac{L}{4} \\$

The length of the rod should be

$\frac{L}{4} \\$

5 0

3 years ago

You need eye drops for your dry eyes. You brought only a few dollars with you to the store. Which is the best method of choosing

Mnenie [13.5K]

Answer:

Compare brands first, but purchase the one that has lots of commercial advertisements, Compare ingredients and purchase the brand that is safe to use with soft contact lenses, Compare prices and purchase the most inexpensive brand.

Explanation:

4 0

2 years ago

1. A 25.35-g piece of iron absorbs 1562.75 Joules of heat energy, and its temperature

Lera25 [3.4K]

Answer:

Q = C M T where C is the specific, M the mass, T the temperature change

Note 1 cal = 4.19 Joules

1562.75 J / (4.19 J/cal) = 378 cal

C = Q / (M * T) = 378 cal / (25.35 g * 155 deg C)

C = .096 cal / g deg C

8 0

1 year ago

A 7.5-cmcm-diameter horizontal pipe gradually narrows to 4.5 cmcm . When water flows through this pipe at a certain rate, the ga

tino4ka555 [31]

Answer

given,

diameter,d₁ = 7.5 cm

d₂ = 4.5 cm

P₁ = 32 kPa

P₂ = 25 kPa

Assuming, we have calculation of flow in the pipe

using continuity equation

A₁ v₁ = A₂ v₂

π r₁² v₁ = π r₂² v₂

$v_1= \dfrac{r_2^2}{r_1^2} v_2$

$v_1= \dfrac{2.25^2}{3.75^2} v_2$

$v_1= 0.36 v_2$

Applying Bernoulli's equation

$\Delta P = \dfrac{1}{2}\rho (v_2^2-v_1^2)$

$P_1-P_2 = \dfrac{1}{2}\rho (v_2^2-(0.36 v_2)^2)$

$32-25 = \dfrac{1}{2}1000\times v_2^2 (1 - 0.1269)$

$v_2=\sqrt{\dfrac{2\times 7\times 10^3}{1000\times (0.8704)}}$

$v_2=\sqrt{16.084}$

v₂ = 4.01 m/s

fluid flow rate

Q = A₂ V₂

Q = π (0.0225)² x 4.01

Q = 6.38 x 10⁻³ m³/s

flow in the pipe is equal to 6.38 x 10⁻³ m³/s

4 0

2 years ago