Since when was the light we see now emanating from the quasar? Note that the distance between the Earth and the quasar is 598 Mpc ​

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Answer 1
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The blade of a windshield wiper moves through an angle of 90.08 in 0.40 s. The tip of the blade moves on the arc of a circle that has a radius of 0.45 m. What is the magnitude of the centripetal acceleration of the tip of the blade

Answers

Answer:

The centripetal acceleration is 6.95 m/s²

Explanation:

Given;

angular displacement of the blade, θ = 90.08⁰

duration of motion of the blade, t = 0.4 s

radius of the circle moved by the blade, r = 0.45 m

The angular speed of the blade in radian is calculated as;

[tex]\omega = \frac{\theta}{t} \times \frac{\pi \ radian}{180^0} \\\\\omega = \frac{90.08 ^0}{0.4 \ s} \times \frac{\pi \ radian}{180^0} \\\\\omega = 3.93 \ rad/s[/tex]

The centripetal acceleration is calculated as;

a = ω²r

a = (3.93)² x 0.45

a = 6.95 m/s²

A car travels at a constant speed around a circular track whose radiu is 2.6 km. The goes once arond the track in 360s . What is the magnitude

Answers

Answer:

Centripetal acceleration = 0.79 m/s²

Explanation:

Given the following data;

Radius, r = 2.6 km

Time = 360 seconds

Conversion:

2.6 km to meters = 2.6 * 1000 = 2600 meters

To find the magnitude of centripetal acceleration;

First of all, we would determine the circular speed of the car using the formula;

[tex] Circular \; speed (V) = \frac {2 \pi r}{t}[/tex]

Where;

r represents the radius and t is the time.

Substituting into the formula, we have;

[tex] Circular \; speed (V) = \frac {2*3.142*2600}{360} [/tex]

[tex] Circular \; speed (V) = \frac {16338.4}{360} [/tex]

Circular speed, V = 45.38 m/s

Next, we find the centripetal acceleration;

Mathematically, centripetal acceleration is given by the formula;

[tex] Centripetal \; acceleration = \frac {V^{2}}{r}[/tex]

Where;

V is the circular speed (velocity) of an object.r is the radius of circular path.

Substituting into the formula, we have;

[tex] Centripetal \; acceleration = \frac {45.38^{2}}{2.6}[/tex]

[tex] Centripetal \; acceleration = \frac {2059.34}{2600}[/tex]

Centripetal acceleration = 0.79 m/s²

The maximum amount of pulling force a truck can apply when driving on
concrete is 8760 N. If the coefficient of static friction between a trailer and
concrete is 0.8, what is the heaviest that the trailer can be and still be pulled
by the truck?
O A. 8760 N
O B. 12,680 N
O C. 10,950 N
O D. 7240 N

Answers

Answer:

8760 N

Explanation:

think this is the right answer :)

5N
5 N
19 N
19 N

Pls help look at the pic

Answers

Answer:

b. is the correct answer ....

consider a circular loop of wire carrying a counterclockwise current as shown. Indicate the direction of the magnetic field at points both inside and outside of the loop.

Answers

Answer:

in this case around the loop the field points downwards on the outside and upwards on the inside

Explanation:

To find the direction of the magnetic field in a wire you must use the right hand rule.

The thumb points in the direction of the current flow and the other created fingers point in the direction of the magnetic field.

Therefore in this case around the loop the field points downwards on the outside and upwards on the inside

If a second ball were dropped from rest from height ymax, how long would it take to reach the ground

Answers

Answer:

[tex](b)\ t_1 - t_0[/tex]

[tex](d)\ t_2 - t_1[/tex]

[tex](e)\ \frac{t_2 - t_0}{2}[/tex]

Explanation:

Given

See attachment for complete question

Required

How long to reach the ground from the maximum height

First, calculate the time of flight (T)

[tex]T =t_2 - t_0[/tex]

The time taken (t) from maximum height to the ground is:

[tex]t = \frac{1}{2}T[/tex]

So, we have:

[tex]t = \frac{t_2 - t_0}{2}[/tex]

Another representation is:

At ymax, the time is: t1

On the ground, the time is t2

The difference between these times is the time taken.

So;

[tex]t = t_2 - t_1[/tex]

Since air resistance is to be ignored, then

[tex]t_2 - t_1 = t_1 - t_0[/tex] --- i.e. time to reach the maximum height from the ground equals time to reach the ground from the maximum height

Which of the following would likely happen if a person’s lactic acid system had difficulty breaking down glycogen in the muscles?
The person would have difficulty swimming across a lake.
The person would have difficulty sprinting in a race.
The person would have difficulty cycling down a hill.
The person would have difficulty running a marathon.

Answers

Answer:

The last one

Answer:  I think that its b, they would have difficulty sprinting in a race

Explanation:

A weightlifter presses a 200 N weight 0.5 m over his head in 2 s. What is the power of the weightlifter

Answers

Answer:

50 watts

Explanation:

Applying,

Power (P) = Workdone (W)/Time(t)

But,

Work done (W) = Force (F)×distance(d)

Therefore,

P = Fd/t..................... Equation 1

Where P =  power of the weightlifter, F = Force applied, d = distance, t = time.

From the question,

Given: F = 200 N, d = 0.5 m, t = 2 s

Substitute these values into equation 1

P = (200×0.5)/2

P = 100/2

P = 50 watts

A ball is launched from the ground with a horizontal speed of 30 m/s and a vertical speed of 30 m/s. What will the vertical speed be right before it hits the ground?
A. 0 m/s
B. 15 m/s
C. 40 m/s
D. 30 m/s

Answers

Answer:

Explanation:

The nice thing about parabolic motion is that the object launched from a certain height will have the same velocity coming down when it reaches that height again, just in the opposite direction. For us, that means if the velocity of the ball right off the ground is 30 m/s, then right before it hits the ground again it will be -30 m/s (the negative just means that the direction is the opposite). Your choice is D.

What are the relationships between the temperature scales of Fahrenheit, Kelvin, Celsius, and Rankine

Answers

Fahrenheit scale is called the Rankine (°R) scale. These scales are related by the equations K = °C + 273.15, °R = °F + 459.67, and °R = 1.8 K. Zero in both the Kelvin and Rankine scales is at absolute zero.

If you move 10 times farther away from a source of light, then how will the
apparent brightness of that source change?
it will become 10 times less bright
it will become 2 times less bright
its brightness will not change
O it will become 100 times less bright

Answers

It will become 100 times less bright

please help! will mark brainliest

Answers

Answer:

27.D.28.A

Explanation:

THE ANSWERME MY ANSWERWAH IS ME AND MY ANSWER WAHHHWALA NA SIYA WALA NA SIYA MASAYA MASAYA MASAYA WAHHtinapon na siya tipon na siya wahhsa basurahan wahh wahh ...............

You’re working with a patient who suddenly falls. You should?

Answers

help them up and make sure there ok

light travels approximately 982,080,000 ft/s, and one year has approximately 32,000,000 seconds. A light year is the distance light travels in one year.

Answers

Answer:

The distance traveled in 1 year is: [tex]3.143*10^{16}ft[/tex]

Explanation:

Given

[tex]s = 982,080,000 ft/s[/tex] --- speed

[tex]t = 32,000,000 s[/tex] --- time

Required

The distance traveled

This is calculated as:

[tex]Speed = \frac{Distance}{Time}[/tex]

So, we have:

[tex]Distance = Speed * Time[/tex]

This gives:

[tex]Distance = 982,080,000 ft/s * 32,000,000 s[/tex]

[tex]Distance = 982,080,000 * 32,000,000ft[/tex]

[tex]Distance = 3.143*10^{16}ft[/tex] -- approximated

Which of these is a source of thermal energy inside earth

Answers

There's no multiple answers that you added if that's what you meant but it possibly could be Magma or radioactive decay of particles from the earths core if those two are any of the options

After your school's team wins the regional championship, students go to the dorm roof and start setting off fireworks rockets. The rockets explode high in the air and the sound travels out uniformly in all directions. If the sound intensity is 1.62 10-6 W/m2 at a distance of 165 m from the explosion, at what distance from the explosion is the sound intensity half this value

Answers

Answer:

required distance is 233.35 m

Explanation:

Given the data in the question;

Sound intensity [tex]I[/tex] = 1.62 × 10⁻⁶ W/m²

distance r = 165 m

at what distance from the explosion is the sound intensity half this value?

we know that;

Sound intensity [tex]I[/tex] is proportional to 1/(distance)²

i.e

[tex]I[/tex] ∝ 1/r²

Now, let r² be the distance where sound intensity is half, i.e [tex]I[/tex]₂ = [tex]I[/tex]₁/2

Hence,

[tex]I[/tex]₂/[tex]I[/tex]₁ = r₁²/r₂²

1/2 = (165)²/ r₂²

r₂² = 2 × (165)²

r₂² = 2 × 27225

r₂² = 54450

r₂ = √54450

r₂ = 233.35 m

Therefore, required distance is 233.35 m

If the child has a mass of 13.9 kg, calculate the magnitude of the force in newtons the mother exerts on the child under the following conditions. (b) The elevator accelerates upward at 0.898 m/s2. 148.702 N

Answers

The elevator accelerates upward at an acceleration, then the magnitude of the force is 148.84 N.

What is Force?

The force is the action of push or pull which makes an object to move or stop.

Given the mass of child m =13.9 kg, acceleration a =0.898 m/s², then the force will be given by

F = m(g-a)

F = 13.9 x (9.81 - (-0.898))

F = 148.84 N

Thus, the magnitude of the force is  148.84 N.

Learn more about force.

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A hot-air balloon stays aloft because hot air at atmospheric pressure is less dense than cooler air at the same pressure. If the volume of the balloon is 500.0 m^3 and the surrounding air is at 15.0°C. What must the temperature of the air in the balloon be for it to lift a total load of 290 kg (in addition to the mass of the hot air)? The density of air at 15.0°C and atmospheric pressure is 1.23kg/m^3.

Answers

Answer:

272° C

Explanation:

Given :

Volume of the balloon, V = 500 [tex]m^3[/tex]

The temperature of the surrounding air, [tex]T_{air} = 15^\circ C[/tex]

Total load, [tex]m_{T}[/tex] = 290 kg

Density of the air, [tex]$\rho_{air} = 1.23 \ kg/m^3$[/tex]

We known buoyant force,

[tex]$F_B = \rho_{air} V$[/tex]

For a 290 kg lift,  [tex]$m_{hot} = \frac{F_B}{g} = 290 \ kg$[/tex]

[tex]$m=\rho V$[/tex]

∴ [tex]$m_{hot}=\rho_{hot} V ; \ \ \ \ \ \frac{F_B}{g}-m_{hot} = 290 \ kg$[/tex]

  [tex]$(\rho_{air} - \rho_{hot}) V= 290 \ kg$[/tex]

  [tex]$\rho_{hot} = \rho_{air}- \frac{290}{V} \ kg = 1.23 \ kg/m^3 - \frac{290 \ kg}{500 \cm^3}$[/tex]

  [tex]$\rho_{hot}= 0.65 \ kg/m^3 =\frac{\rho M}{R T_{hot}}$[/tex]

 ∴ [tex]$\rho_{hot} T_{hot}= \rho_{air} T_{air}$[/tex]

   [tex]$T_{hot}= T_{air}\left[\frac{\rho_{air}}{\rho_{hot}}\right]$[/tex]

          [tex]$=288 \ K \times \frac{1.23 \ kg/m^3}{0.65 \ kg/m^3}$[/tex]

         = 545 K

          [tex]$=272^\circ C$[/tex]

Therefore, temperature of the air in the balloon is 272 degree Celsius.

To lift a load more than the weight of the balloon, the temperature of the air in the balloon has to be higher than the air in the surrounding.

The temperature of the air in the balloon to lift a total load of 290 kg is approximately 272.12°C.

Reasons:

Given information are;

Volume of the balloon = 500.0 m³

Temperature of the surrounding air = 15.0°C

Density of air at 15.0°C = 1.23 kg/m³

Required:

The temperature required to lift 290kg.

Solution:

Let, [tex]\rho _{air , b}[/tex], represent the density of the air in the balloon, we have;

[tex]\rho _{air , b}[/tex] × 500.0 + 290 = 1.23 × 500

Therefore;

[tex]\displaystyle \rho _{air , b} = \frac{1.23 \times 500- 290}{500} = 0.65[/tex]

According to the Ideal Gas Law, we have;

ρ₁ × R × T₁ = ρ₂ × R × T₂

Therefore;

[tex]\displaystyle T_2 = \mathbf{\frac{\rho_1 \times T_1}{\rho_2}}[/tex]

Therefore;

[tex]\displaystyle T_2 = \frac{1.23\times288.15}{0.65} \approx 545.27[/tex]

The temperature of the balloon, T₂ ≈ 545.27 -  273.15 = 272.12

The temperature of the air in the balloon, T₂ ≈ 272.12 °C

Learn more hear:

https://brainly.com/question/11236279

What are impact and non-impact printers?​

Answers

Impact printers involve mechanical components for conducting printing. It is a type of printer that works by direct contact of an ink ribbon with paper.

In Non-Impact printers, no mechanical moving component is used.

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A tank containing a fluid is stirred by a paddle wheel. The work input to the paddle wheel is 5090 kJ. The heat transfer from the tank is 1500 kJ. Consider the tank and the fluid inside a control surface and determine the change in internal energy, in kJ, of this control mass.

Answers

Answer: [tex]3590\ kJ[/tex]

Explanation:

Given

Paddle wheel work is [tex]W=-5090\ kJ\quad \text{work is done on the system}[/tex]

Heat transfer from the tank is [tex]Q=-1500\ kJ\quad \text{heat taken from the system}[/tex]

From the first law of thermodynamics

Change in the internal energy of the system is equal to the difference of heat and work .

[tex]\Rightarrow \Delta U=Q-W\\\Rightarrow \Delta U=-1500-(-5090)\\\Rightarrow \Delta U=3590\ kJ[/tex]

Therefore, the change in internal energy is [tex]3590\ kJ[/tex]



А bus has started to move from
the rest with an acceleration of
0.25 m/s². find its final velocity

Answers

Thank lord for that please thank lord please thank

Convert 387.1 K to °C

Answers

387.1 kelvins = 113.95 degrees celsius

A car drives 110 km in 2 hours. Calculate the speed of the car

Answers

Answer: 55 kmph

Explanation: Divide 110 by 2

Please help! ❤️
I’ll make you the Brainlyest, I can’t get this one wrong.

Answers

The answer is A because i know that it is and you will get the answer correct

a) Find the current in the 1 Ω resistor.
b ) Find the current in the 8 Ω resistor.
c ) Find the current in the 5 Ω resistor.

PLEASE HELP I NEED THIS TODAY

Answers

Answer:

a)I=V/R

39.5 amp

Explanation:

because the voltage in serious with 1ohm resistor

At what speed was object A moving ?

Answers

Answer:

C

Explanation:

The answer is C because if you look at the 1 hour mark it shows 10km

Answer:It will be 10km/hour

Explanation:

Activity 1
The equation for the combustion of butane gas is given below.
1.1
1.2
AH < 0
butane(g) + 1302(g) → 8CO2(g) + 10H2O(g)
Define the term activation energy.
Is the combustion reaction of butane exothermic or endothermic? Give
reason for the answer.
Draw a sketch graph of potential energy versus course of reaction for
reaction above.
3
Clearly indicate the following on the graph:
o
Activation energy
Heat of reaction (AH)
Reactants and products
Determine the empirical formula of butane gas if it consists of 82,76%
and 17,24% hydrogen.​

Answers

Answer:

I don't know hhaha ammmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm I wish I could help

A 1.2 kg basketball is thrown upwards. What is the potential energy of the basketball at the top of its path if it reaches a height of 15.6 m?

Answers

Answer:

Answer is 183.6 J

Explanation:

Using the Physics reference sheet the formula for Potential energy is

(mass) x (gravity) x (height)

Mass= 1.2

Gravity I used is 9.81 (use 10 to get the answer most schools use)

Height= 15.6

A ballistic pendulum is a device for measuring the speed of a projectile. The projectile is launched horizontally and embeds in a stationary block on the end of a string. The block-projectile system swings upward after the collision, reaching a maximum height. Which of the following statements is correct about the collision between the projectile-block system?

a. Kinetic energy of the system is conserved.
b. Linear momentum of the system is conserved.
c. Linear momentum of the system is not conserved.
d. The total mechanical energy of the system is conserved

Answers

(B) linear momentum of the system is conserved

A charged particle enters into a uniform magnetic field such that its velocity vector is perpendicular to the magnetic field vector. Ignoring the particle's weight, what type of path will the particle follow

Answers

Answer:

a circular path

Explanation:

In a magnetism field if a charged particle having a charge of magnitude '' enters  such that its velocity vector V is 90° to the direction of the magnetic field "B'', then it will experience a force, called Lorentz force F

[tex]F = V\times B[/tex]

According to the property of cross-product, the Lorentz force (F) acting on the particle will be perpendicular to the instantaneous position of the particle, making the path of the particle to be a circular path.

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