Optical astronomers need a clear, dark sky to collect good data. Explain why radio astronomers have no problem observing in the UK where it is often very cloudy.

Answers

Answer 1

Answer:

Answer:

In the clarification portion elsewhere here, the definition of the concern is mentioned.

Explanation:

So like optical telescopes capture light waves, introduce it to concentrate, enhance it, as well as make it usable through different instruments via study, so radio telescopes accumulate weak signal light waves, introduce that one to focus, enhance it, as well as make this information available during research. To research naturally produced radio illumination from stars, galaxies, dark matter, as well as other natural phenomena, we utilize telescopes.

Optical telescopes detect space-borne visible light. There are some drawbacks of optical telescopes mostly on the surface:

Mostly at night would they have been seen.
Unless the weather gets cloudy, bad, or gloomy, they shouldn't be seen.

Although radio telescopes monitor space-coming radio waves. Those other telescopes, when they are already typically very massive as well as costly, have such an improvement surrounded by optical telescopes. They should be included in poor weather and, when they travel through the surrounding air, the radio waves aren't obscured by clouds. Throughout the afternoon and also some at night, radio telescopes are sometimes used.

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In an experiment to measure the acceleration of g due to gravity, two values, 9.96m/s (s is squared) and 9.72m/s (s is squared), are determined. Find (1) the percent difference of the measurements, (2) the percent error of each measurement, and (3) the percent error of their mean. (Accepted value:g=9.80m/s (s is squared))

Answers

Answer:

a)2.46 %

b)For 1 :101.52 %

For 2 : 99.08 %

c)100..4 %

Explanation:

Given that

g₁ = 9.96 m/s²

g₂ = 9.72 m/s²

The actual value of g = 9.8 m/s²

The difference Δ g = 9.96 -9.72 =0.24 m/s²

$The\ percentage\ difference=(0.24)/(9.72)* 100=2.46\ percentage\n$

For first one :

$Error\ in\ the\ percentage =(9.96)/(9.81)* 100 =101.52\ perncetage$

For second :

$Error\ in\ the\ percentage =(9.72)/(9.81)* 100 =99.08\ perncetage$

The mean g(mean )

$g(mean )=(9.96+9.72)/(2)\ m/s^2\ng(mean)=9.84\ m/s^2$

$The\ percentage=(9.84)/(9.8)* 100=100.40\ percentage$

a)2.46 %

b)For 1 :101.52 %

For 2 : 99.08 %

c)100..4 %

Final answer:

The percent difference between the two measurements is 2.44%. The percent error for the first measurement is 1.63%, and for the second measurement is 0.82%. The percent error of their mean is 0.41%.

Explanation:

In physics, the percent difference is calculated by subtracting the two values, taking the absolute value, dividing by the average of the two values, and then multiplying by 100. Therefore, the percent difference between the two measurements 9.96m/s² and 9.72m/s² is:

|(9.96-9.72)|/((9.96+9.72)/2)*100 = 2.44%

The percent error involves taking the absolute difference between the experimental value and the accepted value, divided by the accepted value, then multiplied by 100. So, the percent error for the two measurements with accepted value of 9.80m/s² are:

For 9.96m/s²: |(9.96-9.80)|/9.80*100 = 1.63%

For 9.72m/s²: |(9.72-9.8)|/9.8*100 = 0.82%

The percent error of the mean involves doing the above but using the mean of the experimental measurements:

|(Mean of measurements - Accepted value)|/Accepted value * 100 |(9.96+9.72)/2-9.8|/9.8*100 = 0.41%

Learn more about Percent Difference and Error here:

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A sphere has a charge of −84.0 nC and a radius of 5.00 cm. What is the magnitude of its electric field 3.90 cm from its surface?

Answers

Answer:

$E = -9.5* 10^4~N/C$

Explanation:

Gauss' Law should be applied to find the E-field 3.9 cm from the surface of the sphere.

In order to apply Gauss' Law, an imaginary spherical shell (Gaussian surface) should be placed around the original sphere. The exact position of the shell must be 3.9 cm from the surface of the original sphere.

Gauss' Law states that

$\int {\vec{E}d\vec{a}} = (Q_(enc))/(\epsilon_0)$

Here, the integral in the left-hand side is equal to the area of the imaginary surface. After all, the reason behind choosing the imaginary surface a spherical shell is to avoid this integral. The enclosed charge in the right-hand side is equal to the charge of the sphere, -84.0 nC. The radius of the imaginary surface must be 5 + 3.9 = 8.9 cm.

So,

$E4\pi r^2 = (-84* 10^(-9))/(8.8* 10^(-12))\nE4\pi (8.9 * 10^(-2))^2 = (-84* 10^(-9))/(8.8* 10^(-12))\n\nE = -9.5* 10^4~N/C$

Information that is easily converted into numbers and is stored as a on and off signals is _____ information.

Answers

"Binary" information

A uniform 190 g rod with length 43 cm rotates in a horizontal plane about a fixed, vertical, frictionless pin through its center. Two small 38 g beads are mounted on the rod such that they are able to slide without friction along its length. Initially the beads are held by catches at positions 10 cm on each sides of the center, at which time the system rotates at an angular speed of 12 rad/s. Suddenly, the catches are released and the small beads slide outward along the rod. Find the angular speed of the system at the instant the beads reach the ends of the rod. Answer in units of rad/s.

Answers

Answer:

The angular speed of the system at the instant the beads reach the ends of the rod is 14.87 rad/s

Explanation:

Moment of inertia is given as;

I = ¹/₁₂×ML² + 2mr²

where;

I is the moment of inertia

M is the mass of the rod = 0.19 kg

L is the length of the rod = 0.43 m

m is the mass of the bead = 0.038 kg

r is the distance of one bead

Initial moment of inertial is given as;

$I_i = (1)/(12)ML^2 +2mr_1^2$

Final moment of inertia is also given as

$I_f= (1)/(12)ML^2 +2mr_2^2$

Angular momentum is the product of angular speed and moment of inertia;

= Iω

From the principle of conservation of angular momentum;

$I_i \omega_i = I__f } \omega_f$

$((1)/(12)ML^2 +2mr_1^2) \omega_i = ((1)/(12)ML^2 +2mr_2^2) \omega_f$

Given;

ωi = 12 rad/s

r₁ = 10.0 cm = 0.1 m

r₂ = 10.0cm/4 = 2.5 cm = 0.025 m

Substitute these values in the above equation, we will have;

$((1)/(12)*0.19*(0.43)^2 +2*0.038(0.1)^2) 12 = ((1)/(12)*0.19*(0.43)^2 +2*0.038*(0.025)^2) \omega_f\n\n0.04425 =0.002975\ \omega_f\n\n\omega_f = (0.04425)/(0.002975) = 14.87\ rad/s$

Therefore, the angular speed of the system at the instant the beads reach the ends of the rod is 14.87 rad/s

Consider a proton travelling due west at a velocity of 5×10^5m/s. Assuming that the rth magnetic field has a strength of 5x10^-5Tand is directed due south calculate li) the magnitude of the force on the proton (q= 1.6x10^-9C)

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Answer:

Explanation:

Label the longitudinal wave

Answers

Answer:

???

Explanation:

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