1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Lelechka [254]
2 years ago
6

A star has a spectrum shift exhibiting a red shift. What can you conclude about the distance from earth to that star

Physics
2 answers:
Akimi4 [234]2 years ago
6 0
Red shift in the spectrum of a star is always taken to mean that the source of the light is moving away from us.

That's the whole basis for the notion of the expanding universe, since we don't know of any other mechanism that can shift light toward longer wavelengths.
sergejj [24]2 years ago
5 0
Its going away cuz red shift goes away (somehow i think of hell so think that u wanna go away from it)
You might be interested in
Jill can use a force of 12.0 N to lift a single box up 5.00 m. How many of these boxes must she lift it in a minute to use 60.0
Mama L [17]

Answer:

60 boxes

Explanation:

The work done by lifting a single box is equal to the force applied (the weight of the box) times the displacement of the box:

W_1 = Fd=(12.0 N)(5.00 m)=60 J

Power is related to the work done by the equation:

P=\frac{W}{t}

where W is the work done and t is the time. In this problem, we are told that the power used is P=60.0 W, while the time taken is t = 1 min = 60 s, so the total work done must be

W=Pt=(60.0 W)(60 s)=3600 J

Therefore, the number of boxes that she has to lift in order to use this power is the total work divided by the work done in lifting each box:

N=\frac{W}{W_1}=\frac{3,600 J}{60 J}=60

5 0
3 years ago
You are the juror of a case involving a drunken driver whose 1041 kg sports car ran into a stationary 1928 kg station wagon stop
mina [271]
1. Find the force of friction between the sports car and the station wagon stuck together and the road. The total mass m = 1928kg + 1041kg = 2969kg. The only force in the x-direction is friction: F = μ*N = μ * m * g 
2. Find the acceleration due to friction: 
F = m*a =  μ * m * g => a = μ * g = 0.6 * 9.81
3. Find the time it took the two cars stuck together to slide 12m:
x = 0.5*a*t² 
t = sqrt(2*x / a) = sqrt(2 * x / (μ * g) )
4. Find the initial velocity of the two cars:
v = a*t = μ * g * sqrt(2 * x / (μ * g) ) = sqrt( 2 * x * μ * g)
5. Use the initial velocity of the two cars combined to find the velocity of the sports car. Momentum must be conserved:

m₁ mass of sports car
v₁ velocity of sports car before the crash
m₂ mass of station wagon
v₂ velocity of station wagon before the crash = 0
v velocity after the crash

m₁*v₁ + m₂*v₂ = (m₁+m₂) * v = m₁*v₁ 
v₁ = (m₁+m₂) * v / m₁ = (m₁+m₂) * sqrt( 2 * x * μ * g) / m₁
v₁ = 33.9 m/s


7 0
3 years ago
The half-life of Iodine-131 is 8.0252 days. If 14.2 grams of I-131 is released in Japan and takes 31.8 days to travel across the
MakcuM [25]

Answer:

Explanation:

Half-life problems are modeled as exponential equations.  The half-life formula is P=P_o\left (\dfrac{1}{2} \right)^{\frac{t}{k}} where P_o is the initial amount, k is the length of the half-life, t is the amount of time that has elapsed since the initial measurement was taken, and P is the amount that remains at time t.

P=14.2\left (\dfrac{1}{2} \right)^{\frac{t}{8.0252}}

<u>Deriving the half-life formula</u>

If one forgets the half-life formula, one can derive an equivalent equation by recalling the basic an exponential equation, y=a b^{t}, where t is still the amount of time, and y is the amount remaining at time t.  The constants a and b can be solved for as follows:

Knowing that amount initially is 14.2g, we let this be time zero:

y=a b^{t}

(14.2)=ab^{(0)}

14.2=a *1

14.2=a

So, a=14.2, which represents out initial amount of the substance, and our equation becomes: y=14.2 b^{t}

Knowing that the "half-life" is 8.0252 days (note that the unit here is "days", so times for all future uses of this equation must be in "days"), we know that the amount remaining after that time will be one-half of what we started with:

\left(\frac{1}{2} *14.2 \right)=14.2 b^{(8.0252)}

\dfrac{7.1}{14.2}=\dfrac{14.2 b^{8.0252}}{14.2}

0.5=b^{8.0252}

\sqrt[8.0252]{\frac{1}{2}}=\sqrt[8.0252]{b^{8.0252}}

\sqrt[8.0252]{\frac{1}{2}}=b

Recalling exponent properties, one could find that  \left ( \frac{1}{2} \right )^{\frac{1}{8.0252}}=b, which will give the equation identical to the half-life formula.  However, recalling this trivia about exponent properties is not necessary to solve this problem.  One can just evaluate the radical in a calculator:

b=0.9172535661...

Using this decimal approximation has advantages (don't have to remember the half-life formula & don't have to remember as many exponent properties), but one minor disadvantage (need to keep more decimal places to reduce rounding error).

So, our general equation derived from the basic exponential function is:

y=14.2* (0.9172535661)^t  or y=14.2*(0.5)^{\frac{t}{8.0252}} where y represents the amount remaining at time t.

<u>Solving for the amount remaining</u>

With the equation set up, substitute the amount of time it takes to cross the Pacific to solve for the amount remaining:

y=14.2* (0.9172535661)^{(31.8)}          y=14.2*(0.5)^{\frac{(31.8)}{8.0252}}

y=14.2* 0.0641450581                    y=14.2*(0.5)^{3.962518068}

y=0.9108598257                              y=14.2* 0.0641450581

                                                        y=0.9108598257

Since both the initial amount of Iodine, and the amount of time were given to 3 significant figures, the amount remaining after 31.8days is 0.911g.

8 0
1 year ago
You are designing a delivery ramp for crates containing exercise equipment. The 1890 N crates will move at 1.8 m/s at the top of
Mashcka [7]

Answer:

K = 588.3 N/m

Explanation:

From a forces diagram, and knowing that for the maximum value of K, the crate will try to rebound back up (Friction force will point downward):

Fe - Ff - W*sin(22) = 0    Replacing Fe = K*X   and then solving for X:

X = \frac{Ff + W*sin(22)}{K}=\frac{1223}{K}

By conservation of energy:

\frac{K*X^{2}}{2}-mg*d*sin(22)-\frac{m*V^{2}}{2}=-Ff*d

Replacing our previous value for X and solving the equation for K, we get maximum value to prevent the crate from rebound:

K = 588.3 N/m

6 0
3 years ago
Which has a charge with the greatest magnitude?
pishuonlain [190]

Answer:

an object with charge -3

3 0
2 years ago
Other questions:
  • A lever and fulcrum are used to lift a fallen tree, which has a weight of 480N. if the lever has a mechanical advantage of 5.5,
    5·1 answer
  • An object with 274 J of GPE is 140cm above the ground. What is its mass?
    12·1 answer
  • Johannes Kepler, an apprentice of Brahe, believed in the heliocentric universe but rejected past astronomers' belief in
    12·2 answers
  • 2. Montesquieu's view of the separation of powers was later expressed in the United States Government through which document? A.
    11·1 answer
  • A hollow spherical shell has mass 8.05kg and radius 0.215m . It is initially at rest and then rotates about a stationary axis th
    13·2 answers
  • The current in a toaster is 20 amps, and its voltage is 200 volts. What is the resistance in the toaster?
    7·2 answers
  • An electron (q=-1.602×10-19C) is placed .03m away from spherical object with a net charge of -7.2 C.
    12·1 answer
  • If the roller coaster car in the above problem were moving at twice the speed, then what would be its new kinetic energy
    7·1 answer
  • why are we all cheating if we payed attention in what they trying to tell us we wouldn't have to cheat but we are not paying att
    9·2 answers
  • Convert 800 cm to meters.
    11·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!