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2 years ago

8

Which of the following is supportive evidence of the big bang theory? pulsars the different shapes of galaxies cosmic microwave

background radiation galactic clusters

2 answers:

Juli2301 [7.4K]2 years ago

6 0

Cosmic microwave background radiation

Lunna [17]2 years ago

3 0

Cosmic microwave background radiation

You might be interested in

Heated lithium atoms emit photons of light with an energy of 2.961 × 10−19 J. Calculate the frequency and wavelength of one of t

tatiyna

Answer:

4.5 x 10¹⁴ Hz

666.7 nm

1.8 x 10⁵ J

The color of the emitted light is red

Explanation:

E = energy of photons of light = 2.961 x 10⁻¹⁹ J

f = frequency of the photon

Energy of photons is given as

E = h f

2.961 x 10⁻¹⁹ = (6.63 x 10⁻³⁴) f

f = 4.5 x 10¹⁴ Hz

c = speed of light = 3 x 10⁸ m/s

λ = wavelength of photon

Using the equation

c = f λ

3 x 10⁸ = (4.5 x 10¹⁴) λ

λ = 0.6667 x 10⁻⁶ m

λ = 666.7 x 10⁻⁹ m

λ = 666.7 nm

n = number of photons in 1 mole = 6.023 x 10²³

U = energy of 1 mole of photons

Energy of 1 mole of photons is given as

U = n E

U = (6.023 x 10²³) (2.961 x 10⁻¹⁹)

U = 1.8 x 10⁵ J

The color of the emitted light is red

6 0

3 years ago

jonny [76]

Answer:

about half the people who develop the condition survive.

Explanation:

Cardiogenic shock is a condition in which your heart suddenly can't pump enough blood to meet your body's needs. ... Cardiogenic shock is rare, but it's often fatal if not treated immediately.

7 0

3 years ago

A tree loses water to the air by the process of transpiration at the rate of 110 g/h. This water is replaced by the upward flow

Ksivusya [100]

Answer:

The answer is 1.87nm/s.

Explanation:

The $110g/hr$ water loss must be replaced by $110g/hr$ of sap. 110g of sap corresponds to a volume of

$110g \div \dfrac{1040*10^3g}{1*10^6cm^3} = 106cm^3$

thus rate of sap replacement is

$106cm^3/hr = 106*10^{-6}m^3/3600s = 2.94*10^{-8}m^3/s$

The volume of sap in the vessel of length $x$ is

$V = Ax$ ,

where $A$ is the cross sectional area of the vessel.

For 2000 such vessels, the volume is

$V = 2000Ax$

taking the derivative of both sides we get:

$\dfrac{dV}{dt} = 2000A \dfrac{dx}{dt}$

on the left-hand-side $\dfrac{dx}{dt}$ is the velocity $v$ of the sap, and on right-hand-side $\dfrac{dV}{dt} = 2.94*10^{-8}m^3/s$ ; therefore,

$2.94*10^{-8}m^3/s=2000Av$

and since the cross-sectional area is

$A = \pi (\dfrac{100*10^{-3}m}{2} )^2 = 7.85*10^{-3}m^2$ ;

therefore,

$2.94*10^{-8}m^3/s =2000(7.85*10^{-3}m^2)v$

solving for $v$ we get:

$v = \dfrac{2.94*10^{-8}m^3/s}{2000(7.85*10^{-3}m^2)}$

$\boxed{v =1.875*10^{-9}m/s = 1.875nm/s}$

which is the upward speed of the sap in each vessel.

8 0

3 years ago

Milk with a density of 1020 kg/m3 is transported on a level road in a 9-m-long, 3-m-diameter cylindrical tanker. The tanker is c

jeka57 [31]

Solution :

Given data is :

Density of the milk in the tank, $\rho = 1020 \ kg/m^3$

Length of the tank, x = 9 m

Height of the tank, z = 3 m

Acceleration of the tank, $a_x = 2.5 \ m/s^2$

Therefore, the pressure difference between the two points is given by :

$P_2-P_1 = -\rho a_x x - \rho(g+a)z$

Since the tank is completely filled with milk, the vertical acceleration is $a_z = 0$

$P_2-P_1 = -\rho a_x x- \rho g z$

Therefore substituting, we get

$P_2-P_1=-(1020 \times 2.5 \times 7) - (1020 \times 9.81 \times 3)$

$=-17850 - 30018.6$

$=-47868.6 \ Pa$

$=-47.868 \ kPa$

Therefore the maximum pressure difference in the tank is Δp = 47.87 kPa and is located at the bottom of the tank.

4 0

2 years ago

A spring with spring constant 11.5 N/m hangs from the ceiling. A 490 g ball is attached to the spring and allowed to come to res

Natalija [7]

Answer:

The time constant is $\tau = 17.5 \ s$

Explanation:

From the question we are told that

The spring constant is $k = 11.5 \ N/m$

The mass of the ball is $m_b = 490 \ g = 0.49 \ kg$

The amplitude of the oscillation t the beginning is $x = 6.70 cm = 0.067 \ m$

The amplitude after time t is $x_t = 2.20 cm = 0.022 \ m$

The number of oscillation is $N = 30$

Generally the time taken to attain the second amplitude is mathematically represented as

$t = N * T$ Here T is the period of oscillation

$t = N * 2\pi \sqrt{\frac{m}{k} }$

=> $t = 30 * 2 * 3.142 * \sqrt{\frac{ 0.490}{11.5} }$

=> $t = 38.88 \ s$

Generally the amplitude at time t is mathematically represented as

$x(t) = x e^{-\frac{at}{2m} }$

Here a is the damping constant so

at $t = 38.88 \ s$ , $x_t = 2.20 cm = 0.022 \ m$

So

$0.022 = 0.067 e^{-\frac{a * 38.88}{2 * 0.490} }$

=> $0.3284 = e^{-\frac{a * 38.88}{2 * 0.490} }$

taking natural log of both sides

=> $ln(0.3284 ) = -\frac{a * 38.88}{2 * 0.490} }$

=> $a = 0.028$

Generally the time constant is mathematically represented as

$\tau = \frac{m}{a}$

=> $\tau = \frac{0.490}{ 0.028}$

=> $\tau = 17.5 \ s$

4 0

3 years ago