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3 years ago

Which type of galaxy has arms that contain sites of active star formation and start close to a bulge in the center?

Physics

2 answers:

Sergeeva-Olga [200]3 years ago

8 0

Barred spiral is correct answer

Serhud [2]3 years ago

5 0

Answer:

normal spiral I think. I hope this helps

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A 30kg boxed is pushed with a force of 20N. What is the boxes acceleration. Please show work

kozerog [31]

Answer:

<h3>The answer is 0.67 m/s²</h3>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

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

f is the force

m is the mass

From the question we have

$a = \frac{20}{30} = \frac{2}{3} \\ = 0.6666666...$

We have the final answer as

Hope this helps you

8 0

3 years ago

The magnetic field at the center of a 1.40-cm-diameter loop is 2.50 mT . PART A) What is the current in the loop?

NISA [10]

Explanation:

It is given that,

Diameter of loop, d = 1.4 cm

Radius of loop, r = 0.7 cm = 0.007 m

Magnetic field, $B=2.5\ mT=2.5\times 10^{-3}\ T$

(A) Magnetic field of a current loop is given by :

$B=\dfrac{\mu_oI}{2r}$

I is the current in the loop

$I=\dfrac{2Br}{\mu_o}$

$I=\dfrac{2\times 2.5\times 10^{-3}\times 0.007}{4\pi \times 10^{-7}}$

I = 27.85 A

(B) Magnetic field at a distance r from a wire is given by :

$B=\dfrac{\mu_o I}{2\pi r}$

$r=\dfrac{\mu_o I}{2\pi B}$

$r=\dfrac{4\pi \times 10^{-7}\times 27.85}{2\pi \times 2.5\times 10^{-3}}$

r = 0.00222 m

$r=2.2\times 10^{-3}\ m$

Hence, this is the required solution.

3 0

3 years ago

A uniformly charged ring of radius 10.0 cm has a total charge of 75.0 mC. Find the electric field on the axis of the ring at (a)

wlad13 [49]

Answer:

(a) 6650246.305 N/C

(b) 24150268.34 N/C

(c) 6408227.848 N/C

(d) 665024.6305 N/C

Explanation:

Given:

Radius of the ring (r) = 10.0 cm = 0.10 m [1 cm = 0.01 m]

Total charge of the ring (Q) = 75.0 μC = $75\times 10^{-6}\ \mu C$ [1 μC = 10⁻⁶ C]

Electric field on the axis of the ring of radius 'r' at a distance of 'x' from the center of the ring is given as:

$E_x=\dfrac{kQx}{(x^2+r^2)^\frac{3}{2}}$

Plug in the given values for each point and solve.

(a)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=1.00\ cm=0.01\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.01)}{((0.01)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{6750}{1.015\times 10^{-3}}\\\\E_x=6650246. 305\ N/C$

(b)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=5.00\ cm=0.05\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.05)}{((0.05)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{33750}{1.3975\times 10^{-3}}\\\\E_x=24150268.34\ N/C$