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Brilliant_brown [7]
3 years ago
8

Do any of the galaxies appear to be closer to each other as your universe expands? What does this say about the possibility of g

alaxies colliding with each other?
Physics
1 answer:
Luda [366]3 years ago
8 0

Explanation:

In local galactic group the force of expansion of universe is overcome by the force of attraction due  to gravity. Best example is our own galaxy milky way and another giant galaxy in our local group Andromeda. Andromeda having enormous gravity is pulling milky way towards itself, overcoming the force of expansion.

So, there  are possibilities of collision despite the expansion of universe at a rapid pace. It is estimated that the milky way and Andromeda will collide each other after about 50 billion years from now.  

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You are pushing a 60 kg block of ice across the ground. You exert a constant force of 9 N on the block of ice. You let go after
kotykmax [81]

Answer: d = 33 cm or 0.33 m

Explanation: In physics, Work is the amount of energy transferred to an object to make it move. It can be expressed by:

W = F.d.cosθ

F is the force applied to the object, d is the displacement and θ is the angle formed between the force and the displacement.

For the ice block, the angle is 0, i.e., force and distance are at the same direction, so:

W = F.d.cos(0)

W = F.d

To determine d:

d = \frac{W}{F}

d = \frac{3}{9}

d = 0.33 m

The distance d the block ice moved is 33 cm.

7 0
3 years ago
A baton twirler is twirling her aluminum baton in a horizontal circle at a rate of 2.33 revolutions per second. A baton held hor
Nata [24]

Answer:

Explanation:

Given that;

horizontal circle at a rate of 2.33 revolutions per second

the magnetic field of the Earth is 0.500 gauss

the baton is 60.1 cm in length.

the magnetic field  is oriented at 14.42°

we wil get the area due to rotation of radius of baton is

\Delta A = \frac{1}{2} \Delta \theta R^2

The  formula for the induced emf is

E = \frac{\Delta  \phi}{\Delta  t}

\phi  = \texttt {magnetic flux}

E=\frac{\Delta (BA) }{\Delta  t}

=B\frac{\Delta  A}{\Delta  t}

B is the magnetic field strength

substitute

\texttt {substitute}\  \frac{1}{2} \Delta \theta R^2 \ \ for \Delta  A

E=B\frac{(\Delta  \theta R^3/2)}{\Delta  t} \\\\=\frac{1}{2} BR^2\omega

The magnetic field of the earth is oriented at 14.42

\omega =2.33\\\\L=60.1c,\\\\\theta=14.42\\\\B=0.5

we plug in the values in the equation above

so, the induce EMF will be

E=\frac{1}{2} \times (B\sin \theta)R^2\omega\\\\E=\frac{1}{2} \times (B\sin \theta)(\frac{L}{2} )\omega

=\frac{1}{2} \times0.5gauss\times\frac{0.0001T}{1gauss} \times\sin 14.42\times(\frac{60.1\times10^-^2m}{2} )^2(2.33rev/s)(\frac{2\pi rad}{1rev} )\\\\=2.5\times10^-^5\times0.2490\times0.0903\times14.63982\\\\=2.5\times10^-^5\times0.32917\\\\=8.229\times10^-^6V

6 0
3 years ago
Atomic math challenge will give brainly and thanks
nevsk [136]

Answer:

1. Hydrogen

Atomic # = 1

Atomic Mass = 1.00794  ( If you round it it's 1.008 )

# of protons = 1

# of neutrons = none

# of electrons = 1

8 0
3 years ago
Read 2 more answers
If the magnitude of the magnetic field is 6.50 mT at a distance of 12.8 cm from a long straight current carrying wire, what is t
Lunna [17]

Answer: magnitude of the magnetic field at a distance of 19.4 cm from the wire=4.29mT

Explanation:

According to  Biot-Savart law, A magnetic field generated by a current  carrying wire at a distance is represented as

B=μ₀I/ 2πr

B = magnetic field intensity 1000 mT =1T, 6.50mT = 6.50 X 10^-3T

 μ₀ =permeability of free space  4π × 10−7 H/m

I = current intensity

r = radius, 100cm = 1m, 12.8 cm= 12.8 x 10^-2m

6.50 X 10^-3 =  μ₀ x I/ 2 π X 12.8 X 10^-2

I =6.50  X 10 ^-3 X 2π  X  X 12.8 X 10^-2/  4π × 10−7 H/m

I= 4160 A

when the magnetic field is at 19.4 cm from the wire

B=μ₀I/ 2πr

= 4π × 10−7 H/m x4160/ 2π x 19.4 x 10^-2

=0.004288

= 4.29x 10 ^-3T

= 4.29mT

7 0
3 years ago
find the potential energy of an aircraft weighing 10000 bs at 5000 ft true altitude and 125 kts true air speed
Cloud [144]

Answer:

U=5*10^7ft.Ib

Explanation:

From the question we are told that

Weight  W= 10000bs

Altitude H=5000ft

Speed    V=125kts\\1kts=0.514m/s\\V=125*0.514=>64.25m/s

Generally the equation for Potential energy ids mathematically given as

Potential\ Energy\ U=mgh

U=Wh

U=10000*5000

U=5*10^7ft.Ib

6 0
3 years ago
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