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valina [46]
2 years ago
5

consider the hypothetical observation irregular galaxies outside the local group are moving toward us. from part a, this observa

tion would contradict the idea of an expanding universe. why?
Physics
1 answer:
alina1380 [7]2 years ago
3 0

Answer:

spiral galaxies move away from us 10% faster than elliptical galaxies at the same distances;  irregular galaxies outside the Local Group are moving toward us;  galaxy speeds are faster in summer than in winter

Explanation:

This may help I'm not sure.

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A long solenoid has 103 turns/cm and carries current i. An electron moves within the solenoid in a circle of radius 2.60 cm perp
Aneli [31]

Answer:

0.23348 A

Explanation:

B = Magnetic field

v = Velocity of electron = 1.38\times 10^7\ m/s

q = Charge of electron = 1.6\times 10^{-19}\ C

\mu_0 = Vacuum permeability = 4\pi \times 10^{-7}\ H/m

r = Radius of circle = 0.026 m

N = Number of turns = 103 turns/cm = 103\times 100\ turns/m

I = Current

The magnetic and centripetal force will be balanced

Bqv=m\frac{v^2}{r}\\\Rightarrow B=\frac{mv}{qr}

The magnetic field in solenoid is given by

B=N\mu_0 I\\\Rightarrow I=\frac{B}{N\mu_0}

From the first equation

I=\frac{\frac{mv}{qr}}{N\mu_0}\\\Rightarrow I=\frac{mv}{N\mu_0qr}\\\Rightarrow I=\frac{9.11\times 10^{-31}\times 1.38\times 10^7}{103\times 100\times 4\pi \times 10^{-7}\times 1.6\times 10^{-19}\times 0.026}\\\Rightarrow I=0.23348\ A

The current in the solenoid is 0.23348 A

8 0
2 years ago
CAN someone help ASAP?
lyudmila [28]

Answer:someone help me

Explanation:

7 0
2 years ago
Two runners start at the same point on a straight track. The first runs with constant acceleration so that he covers 98 yards in
charle [14.2K]

Answer:

94.13 ft/s

Explanation:

<u>Given:</u>

  • t = time interval in which the rock hits the opponent = 10 s - 5 s = 5 s
  • s = distance to be moved by the rock long the horizontal = 98 yards
  • y = displacement to be moved by the rock during the time of flight along the vertical = 0 yard

<u>Assume:</u>

  • u = magnitude of initial velocity of the rock
  • \theta = angle of the initial velocity with the horizontal.

For the motion of the rock along the vertical during the time of flight, the rock has a constant acceleration in the vertically downward direction.

\therefore y = u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow 0 = u\sin \theta 5 +\dfrac{1}{2}(-9.8)\times 5^2\\\Rightarrow u\sin \theta 5 =\dfrac{1}{2}(9.8)\times 5^2......(1)\\

Now the rock has zero acceleration along the horizontal. This means it has a constant velocity along the horizontal during the time of flight.

\therefore u\cos \theta t = s\\\Rightarrow u\cos \theta 5 = 98.....(2)\\

On dividing equation (1) by (2), we have

\tan \theta = \dfrac{25}{20}\\\Rightarrow \tan \theta = 1.25\\\Rightarrow \theta = \tan^{-1}1.25\\\Rightarrow \theta = 51.34^\circ

Now, putting this value in equation (2), we have

u\cos 51.34^\circ\times  5 = 98\\\Rightarrow u = \dfrac{98}{5\cos 51.34^\circ}\\\Rightarrow u =31.38\ yard/s\\\Rightarrow u =31.38\times 3\ ft/s\\\Rightarrow u =94.13\ ft/s

Hence, the initial velocity of the rock must a magnitude of 94.13 ft/s to hit the opponent exactly at 98 yards.

3 0
3 years ago
Lactic acid is produced during low-moderate exercise <br> True <br> False
Marizza181 [45]
The answer to your question is true.
3 0
3 years ago
Read 2 more answers
Vector A → has magnitude 8.78 m at 37.0 ∘ from the + x axis. Vector B → has magnitude 8.26 m at 135.0 ∘ from the + x axis. Vecto
kodGreya [7K]

Answer:

R = (- 3.72î + 8.29j)

Magnitude of R = 9.09 m

Explanation:

Let î and j represent unit vectors along the x and y axis respectively.

Vector A --> magnitude 8.78 m, direction 37.0° from the +x-axis

Let the x and y components of this vector be Aₓ and Aᵧ

A = (Aₓî + Aᵧj) m

The components given magnitude and direction from the +x-axis are calculated as

Aₓ = A cos θ and Aᵧ = A sin θ

Aₓ = (8.78 cos 37°) = 7.01 m

Aᵧ = (8.78 sin 37°) = 5.28 m

A = (7.01î + 5.28j) m

Vector B has magnitude 8.26 m and direction 135° from the +x-axis

B = (Bₓî + Bᵧj) m

Bₓ = (8.26 cos 135°) = - 5.84 m

Bᵧ = (8.26 sin 135°) = 5.84 m

B = (-5.84î + 5.84j) m

Vector C has magnitude 5.65 m and direction 210° from the +x-axis

C = (Cₓî + Cᵧj) m

Cₓ = (5.65 cos 210°) = - 4.89 m

Cᵧ = (5.65 sin 210°) = - 2.83 m

C = (- 4.89î - 2.83j) m

The resultant force is a vector sum of all the forces. Let the resultant force be R

R = (Rₓî + Rᵧj) m

R = A + B + C = (7.01î + 5.28j) + (-5.84î + 5.84j) + (- 4.89î - 2.83j)

Summing the î and j components seperately,

R = (- 3.72î + 8.29j) m

To get its magnitude,

Magnitude of R = √(Rₓ² + Rᵧ²) = √((-3.72)² + (8.29)²) = 9.09 m

8 0
3 years ago
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