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Pie
3 years ago
12

Complete the sentence below using the words provided in parentheses (). For two universes that are the same size, the universe w

ith the faster expansion rate must be _______ (younger/older) than the universe with the slower expansion rate. The slope of the line in the Hubble plot of the ______ (younger/older) universe will be ______ (steeper/flatter)
Physics
1 answer:
miss Akunina [59]3 years ago
4 0

Answer:

For two universe that is the same size, the universe with the faster expansion rate must be <u>younger</u> than the universe with a slower expansion rate. The slope in the line of the Hubble plot of the <u>older</u> universe will be <u>flatter</u>.

Explanation:

Hubble's law establishes a direct relationship between a galaxy's distance and its redshift-determined recessional velocity. The redshifts of galaxy clusters and distant galaxies are equivalent and proportional to their distances from us, according to Hubble's theorem.

Thus, to copy and complete the given question.

For two universe that is the same size, the universe with the faster expansion rate must be <u>younger</u> than the universe with a slower expansion rate. The slope in the line of the Hubble plot of the <u>older</u> universe will be <u>flatter</u>.

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Calculate the acceleration if you push with a 20-N horizontal force on a 2-kg block on a horizontal friction-free air table
Afina-wow [57]
Acceleration = force / mass = 20 / 2 = 10 m/s^2
3 0
3 years ago
A dog barks to alert his owner. What is the medium for the sound waves produced by the dog barking?
777dan777 [17]

Answer:

The medium for the sound waves produced by the dog barking is the air

Explanation:

Sound waves are a form of mechanical waves such that it moves from one point to another by the vibration, back and forth of the medium particles from one layer to the next, thereby the particles of the medium remain in the same location and only the wave energy is propagated

The medium which has particles that vibrate back and forth in order to transmit the barking of the dog is the air.

6 0
3 years ago
A helicopter goes straight up 500m from a landing pad. It then goes north 20m. Then it goes down 452m. a) What is the displaceme
gavmur [86]

Answer:

a

  x-component             20 \  m

y-component     500 -  452 =  48 \  m

b

 Magnitude d  =  52 \  m

direction is  \theta  = 67.4^o

Explanation:

From the question we are told that

   The first  vertical distance is  y_1   = 500 \  m

    The  first horizontal distance  is  x =  20 \  m

    The  second vertical distance is  y_2  =  452 \  m

Generally the displacement is  

x-component             20 \  m

y-component     500 -  452 =  48 \  m

Generally the helicopters displacement is mathematically evaluated as  

       d  =  \sqrt{ x- component ^2  +  y- component ^2 }

      d  =  \sqrt{ 20t ^2  + 48 ^2 }

      d  =  52 \  m

The  direction is the angle the displacement of the helicopter makes with the horizontal which is mathematically evaluated as

         \theta  = tan ^{-1}[ \frac{48}{20}]

=>       \theta  = tan ^{-1}[ 2.4 ]

=>      \theta  = 67.4^o

   

3 0
3 years ago
A heavy mirror that has a width of 2 m is to be hung on a wall as shown in the figure below. The mirror weighs 700 N and the wir
loris [4]

The translational equilibrium condition allows finding that the response for cable length with a maximum tension is

      L = 2.56 m

Newton's second law says that the force is proportional to the mass and the acceleration of the body, in the special case that the acceleration is zero, the relationship is called the translational equilibrium condition.

              ∑ F = 0

 

Where the bold indicates vectors, F is the force and the sum is for all external forces.

The reference systems are coordinate systems with respect to which the decomposition of the vectors is carried out and the measurements are made, in this case we will use a system with the horizontal x axis and the vertical y axis.

In the attachment we can see a free body diagram of the system, let's write the equilibrium condition for each axis.

x-axis  

         Tₓ -Tₓ = 0

y-axis

         T_y + T_y - W =0

         2T_y - W = 0

Let's use trigonometry to decompose the tension, we can see from the graph and the adjoint that each string is half the length, let's call the angle θ

          cos θ = \frac{T_x}{T}

          sin θ = \frac{T_y} { T}

          Tₓ = T cos θ

          T_y = T sin θ

We substitute

          2 T sin θ = W          (1)

The text indicates that the length of the block is 2 m, so the distance to the midpoint is

        x = 1 m

Let's use the Pythagoras' Theorem            

            H² = CA² + CO²

           CO = \sqrt{H^2 - CA^2}

           CA = x

           CO = \sqrt{(\frac{L}{2} )^2  - 1 }

Where CO is the opposite leg,  CA is the adjacent leg and H is the hypotenuse indicating H = L / 2,

Let's write the trigonometry functions

           sin θ = \frac{CO}{H}

Let's substitute        

            sin θ = \frac{2 \sqrt{\frac{L^2}{4} -1 } }{L}

Let's subtitute in the equation  1

          2 T  ( \frac{2 \sqrt{\frac{L^2}{4} -1 } }{L}   ) = W

          \sqrt{\frac{L^2}{4}-1  }  = \frac{1}{4}  \ \frac{W}{T} \ L

Let's solve by squaring

         \frac{L^2}{4}  -1 = \frac{1}{4} \ (\frac{W}{T} )^2 \ \frac{L^2}{4}  

         \frac{L^2}{4} ( 1 - \frac{1}{4} (\frac{W}{T})^2  ) -1 =0

They indicate that the maximum tension of the cable is T = 700N and the weight is worth W = 700N, we substitute the values

        \frac{L^2}{4} ( 1- \frac{1}{4}) - 1 =0 \\\frac{3 L^2}{16} = 1 \\ L^2 = \frac{16}{3} \\L = \sqrt{\frac{16}{3} }

        L =   2.31 m    

In conclusion using the translational equilibrium condition we can find that the response for cable length with maximum tension is

      L = 2.31 m

Learn more about translational equilibrium here:

brainly.com/question/12966823

7 0
3 years ago
Read 2 more answers
A 0.500-mol sample of an ideal monatomic gas at 400 kPa and 300 K, expands quasi-statically until the pressure decreases to 160
Marianna [84]

Answer:

(a). The final volume is 2.5 times the initial volume.

The work done is 1143 J.

(b). The final temperature is 207.9 K

The work done is 574 J.

Explanation:

Given that,

Sample of an ideal gas = 0.500 mol

Initial pressure = 400 kPa

Final pressure = 160 kPa

Temperature = 300 K

(a) for isothermal,

Temperature will be same.

We need to calculate the volume of gas

P_{f}V_{f}=P_{i}V_{i}

V_{f}=(\dfrac{P_{i}}{P_{f}})V_{i}

Put the value into the fomrula

V_{f}=(\dfrac{400}{160})V_{i}

V_{f}=2.5 V_{i}

We need to calculate the work done

Using equation of energy

dQ=dW

dQ=nRTln(\dfrac{P_{in}}{P_{f}})

dQ=0.500\times8.314\times300\times ln(\dfrac{400}{160})

dQ=1143\ J

(b). For adiabatic,

No transfer of heat between system and surroundings

We need to calculate the final temperature

Using formula of gas

P_{f}^{1-\gamma}T_{f}^{\gamma}=P_{i}^{1-\gamma}T^{\gamma}

T_{f}=(\dfrac{P_{i}}{P_{f}})^{\frac{1-\gamma}{\gamma}}T_{i}

Put the value into the formula

T_{f}=(\dfrac{400}{160})^{\frac{1-\frac{5}{3}}{\frac{5}{3}}}\times300

T_{f}=207.9\ K

We need to calculate the wok done in adiabatic

Using formula of work done

W=\dfrac{nR(T_{i}-T_{f})}{\gamma-1}

W=\dfrac{0.500\times8.314\times(300-207.9)}{\dfrac{5}{3}-1}

W=574\ J

Hence, (a). The final volume is 2.5 times the initial volume.

The work done is 1143 J.

(b). The final temperature is 207.9 K

The work done is 574 J.

6 0
3 years ago
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