All categories

3 years ago

What would Hubble's constant be if we found one galaxy moving away at 30,000 km/s at a distance of 600 Mpc?

Physics

1 answer:

lora16 [44]3 years ago

3 0

Answer:

H₀ = 1.6 x 10⁻¹⁸ s⁻¹

Explanation:

The Hubble's Constant can be found by the following formula:

$v = H_o D\\\\H_o = \frac{v}{D}$

where,

H₀ = Hubble's Constant = ?

v = speed of galaxy = 30000 km/s = 3 x 10⁷ m/s

D = Distacance = 600 Mpc = (6 x 10⁸ pc)(3.086 x 10¹⁶ m/1 pc)

D = 18.52 x 10²⁴ m

Therefore,

$H_o = \frac{3\ x\ 10^7\ m/s}{18.52\ x\ 10^{24}\ m}$

You might be interested in

64 g of sulfur dioxide (SO2) contains 32 g of oxygen. Calculate how much sulfur it contains.

Ghella [55]

In sulfur dioxide, there are 2 oxygen atoms and 1 sulfur atom. As there are 32g of sulfur and 32g of oxygen, that would mean that each oxygen atom would weigh about 16g. Given that, the mass of a single sulfur atom is twice that of a single oxygen atom.

5 0

3 years ago

Read 2 more answers

8-14 A Cu-30% Zn alloy tensile bar has a strain-hardening coefficient of 0.50. The bar, which has an initial diameter of 1 cm an

77julia77 [94]

Answer:

The true stress at true strain 0.05cm/cm is 80MPa

Explanation:

Given that

the strength coefficient is K

true strain is ε

strain hardening exponent is n

initial diameter of bar is d = 1cm, (10mm)

tensile force is F

engineering stress(S) = 120

the engineering stress(S) = $\frac{F}{\frac{\pi }{4}(d^2) }$

To find force (F) =

120 = $\frac{F}{\frac{\pi }{4}(100^{2} )}$

F = 120 * (π/4) * (100)

F = 9425N

Calculate the true strain (ε) = In (l₀ / l₁)

where

l₀ = initial length of the metallic bar = 3cm

l₁ = final length of metallic bar = 3.5cm

ε = In (3.5 / 3)

= In 1.1667

= 0.154cm/cm

Calculate the true stress (σ) at fracture point

= $\frac{F}{\frac{\pi }{4}(d^2) }}$

tensile force is F and final diameter of bar is d₁ (d in the eqn)

Substitute 9425 N for F and 0.926 cm (9.26mm) for d₁ (d in the eqn)

σ = $\frac{9425}{\frac{\pi }{4}(9.26^2) }}$

= 140MPa

To find the strength coefficient (K) of the material bar

K = $\frac{140}{\sqrt{0.154} }$

K = $\frac{140}{0.3925}$

= 356.75MPa

To calculate the true stress σ true strain of 0.05cm/cm

K = 356.75MPa

σ = $356.75(0.05)^0^.^5$

= $356.75 ( 0.2236)$

= 80MPa

The true stress at true strain 0.05cm/cm is 80MPa

6 0

3 years ago

What is the volume in liters of the basketball?

babymother [125]

he basket ball diameter used by NBA men players is around 9.55 inches

So diameter = 9.55 inch = 0.243 m

So radius of basket ball = 0.1215 m

Volume , $V=\frac{4}{3} \pi r^3$

$V = \frac{4}{3}* \pi *0.1215^3=0.0075 m^3=7.5 L$

So volume of basket ball used by men NBA players = 7.5 L

5 0

3 years ago

Read 2 more answers

What<br> A moving object always has energy in its

elena-14-01-66 [18.8K]

The answer is a cell I think hope it helps sorry if wrong

8 0

3 years ago

............................................................??????????????

den301095 [7]

Answer:

$I_{1}$ = 0.7 A, $I_{2}$ = 1.3 A and ε = 7.4 V

Explanation:

From the given circuit, applying Kirchhoff's rule;

Ammeter reading, $I_{0}$ = 2 A

⇒ $I_{0}$ = $I_{1}$ + $I_{2}$ = 2 A

Dividing the circuit to loops 1 and 2.

a. From loop 1,

15 + 7 $I_{1}$ - 5 $I_{0}$ = 0

15 + 7 $I_{1}$ - 10 = 0 (since $I_{0}$ = 2 A)

7 $I_{1}$ - 5 = 0

$I_{1}$ = 0.7 A

But, $I_{0}$ = $I_{1}$ + $I_{2}$

⇒ 2 = 0.7 + $I_{2}$

$I_{2}$ = 1.3 A

b. From loop 2,

ε + 2 $I_{2}$ - 5 $I_{0}$ = 0

ε + 2 $I_{2}$ - 10 = 0

ε + 2.6 - 10 = 0

ε - 7.4 = 0

ε = 7.4 V

Therefore, $I_{1}$ = 0.7 A, $I_{2}$ = 1.3 A and ε = 7.4 V.

8 0

3 years ago