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3 years ago

5

The average radial velocity of galaxies in the Hercules cluster is 10,800 km/s. (a) using H0 = 73 km/s/Mpc, find the distance to

the Hercules cluster in Mpc. (b) How would your answer to (a) differ if the Hubble constant had a smaller value?

1 answer:

ArbitrLikvidat [17]3 years ago

4 0

Answer:

a) 147.95 Mpc

Explanation:

Using Hubble's formula

$v = H_od$

where;

v = radical velocity

$H_o$ = Hubble's constant

d = distance

Given that:

The average radial velocity of galaxies in the Hercules cluster v = 10,800 km/s

Also using $H_o$ = 73 km/s/ Mpc ; we make distance d the subject of the formula:

Then distance d can be written as:

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

$d = \frac{10,800 \ km/s}{73 \ km/s/Mpc}$

d = 147.95 Mpc

b)

Now, if the Hubble constant had a smaller value, then for a given velocity the distance to the galaxy will increase because distance d is inversely proportional to $H_o$

i.e

d ∝ $\frac{1}{H_0}$

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A copper cable needs to carry a current of 160 A with a power loss of only 2.0 W/m. What is the required radius of the copper ca

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Answer:

The radius of the cable is 0.0083 m or 8.3 mm.

Explanation:

The resistance of copper cable of 1 meter length will be given by

$R_{cable} = \frac{\rho \times l}{a}$ .... (i)

where the resistivity of copper is $\rho = 1.7 \times 10^{-8} \Omega.m$ , and l is the length of the wire which is considered to be 1m, and a is the cross sectional area of the wire in $m^{2}$ .

From the formula of power we know that, $P = I^2 R$ .... (ii)

Therefore 2 W/m = $(160)^2 \times R$ .... (iii)

where the resistance,R, actually means the resistance of the cable per meter.

Therefore R ( resistance of cable per meter)

= $\frac{2}{160^2} = 7.812 \times 10^{-5}$ ohms / meter. .... (iv)

Therefore from (i)

$7.812 \times 10^{-5}$ = $\frac{1.7 \times 10^{-8} \times 1}{a} = \frac{1.7 \times 10^{-8} \times 1}{\pi r^{2} }$ ..... (v)

where cross sectional area of the cable, a = $\pi r^2$ ,

where r is the radius of the cable, and length of cable,l = 1m.

Therefore r = $\sqrt{\frac{ 1.7 \times 10^{-8}}{\pi \times 7.812 \times 10^{-5} } } = 0.0083m = 8.3 mm$

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3 years ago

Two electrons are at rest and separated by a distance of 4.32 × 10-10 m. When they are released they accelerate away from each o

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Answer:

Speed of electron when their separation increased by a factor of 4.10 is 9.41 x 10⁵ m/s .

Explanation:

The electric potential energy is given by the relation :

$U = \frac{kq_{1}q_{2} }{r}$

Here q₁ and q₂ are the two charge particles and r is the distance between them and k is electric constant.

In this case, there are two electrons which are separated by the distance 4.32 x 10⁻¹⁰ m.

Let e be the electron charge and r₁ be the distance between them. Then, the initial electric potential energy is :

$U_{1} = \frac{ke^{2} }{r_{1} }$

Now, the distance between the electrons increases by the factor of 4.10. Let r₂ be the new distance between them i.e. r₂ = 4.10 r₁.

Thus, the new electric potential energy is :

$U_{2} = \frac{ke^{2} }{r_{2} }=\frac{ke^{2} }{4.10r_{1} }$

Applying law of conservation of energy :

ΔU = ΔK

Here ΔU is change in electric potential energy and ΔK is change in kinetic energy.

( U₁ - U₂ ) = ( K₂ - K₁ )

Here K₂ and K₁ are initial and final kinetic energy of electron.

Since, the electron initially is at rest, so its initial kinetic energy is zero. Thus, the above equation becomes:

K₂ = U₁ - U₂

$\frac{1}{2}mv^{2}=\frac{ke^{2} }{r_{1} }- \frac{ke^{2} }{4.10r_{1} }$

Here m and v are the mass and final speed of electron respectively.

$v^{2}=\frac{2}{m} \frac{ke^{2} }{r_{1} }(1- \frac{1 }{4.10 })$

Substitute 9.1 x 10⁻³¹ kg for m, 9 x 10⁹ N m² C⁻² for k, 1.6 x 10⁻¹⁹ C for e and 4.32 x 10⁻¹⁰ m for r₁ in the above equation.

$v^{2}=\frac{2}{9.1\times10^{-31} } \frac{9\times10^{9}\times(1.6\times10^{-19})^{2} }{4.32\times10^{-10} }(1- \frac{1 }{4.10 })$

$v^{2}=8.86\times10^{11}$

v = 9.41 x 10⁵ m/s

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3 years ago

What two simple machines are found in a bike?

omeli [17]

I think it's a pulley and a lever.

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3 years ago

You stand on a frictional platform that is rotating at 1.8 rev/s. Your arms are outstretched, and you hold a heavy weight in eac

dusya [7]

Answer:

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489.81804 J

Explanation:

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$I_f$ = Final moment of inertia = 5.1 kgm²

$\omega_i$ = Initial angular speed = 1.8 rev/s

$\omega_f$ = Final angular speed

As the angular momentum of the system is conserved

$I_i\omega_i=I_f\omega_f\\\Rightarrow \omega_f=\dfrac{I_i\omega_i}{I_f}\\\Rightarrow \omega_f=\dfrac{9.3\times 1.8}{5.1}\\\Rightarrow \omega_f=3.28235\ rev/s=3.28235\times 2\pi=20.62361\ rad/s$

The resulting angular speed of the platform is 20.62361 rad/s

Change in kinetic energy is given by

$\Delta K=\dfrac{1}{2}(I_f\omega_f^2-I_i\omega_i^2)\\\Rightarrow \Delta K=\dfrac{1}{2}(5.1\times (20.62361)^2-9.3\times (1.8\times 2\pi)^2)\\\Rightarrow \Delta K=489.81804\ J$

The change in kinetic energy of the system is 489.81804 J

As the work was done to move the weight in there was an increase in kinetic energy

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3 years ago

2. Tomas is hanging from a tree limb, that is inclined at a 65° angle. The force

LUCKY_DIMON [66]

Answer:

57 N

Explanation:

Were are told that the force

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F_net = √(51.66² + 24.09²)

F_net = √3249.0837

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3 years ago