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inn [45]
4 years ago
8

The presence of dwarf galaxies around the Milky Way supports what picture of our galaxy’s formation?

Physics
1 answer:
elena-14-01-66 [18.8K]4 years ago
4 0

Answer:

The presence of dwarf galaxies around the Milky Way supports what picture that our galaxy was formed by a coming together or combination of smaller systems

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Four astronauts are in a spherical space station. (a) If, as is typical, each of them breathes about 500 cm
mario62 [17]

Answer:

A) if each astronaut breathes about 500 cm³, the total volume of air breathed in a year is 14716.8m³.

B) The Diameter of this spherical space station should be 30.4m

Explanation:

The breathing frequency (according to Rochester encyclopedia) is about 12-16 breath per minute. if we take the mean value (14 breath per minute), we can estimate the total breaths of a person along a year:

f_b=14\frac{br}{min} \cdot \frac{60min}{1hr} \frac{24hr}{1day}\frac{365day}{1year}=29433600\frac{br}{year}

If we multiply this for the number of people in the station and the volume each breath needs, we obtain the volume breathed in a year.

The volume of a sphere is:

V_{sph}=\frac{4\pi}{3}r^3

So the diameter is:

D=2r=2\sqrt[3]{\frac{3V_{sph}}{4\pi}} =30.4m

6 0
3 years ago
A charge q of magnitude 6.4 × 10^-19 coulombs moves from point A to point B in an electric field of 6.5 × 10^4 newtons/coulomb.
lana [24]

In this problem we have the electric field intensity E:

E = 6.5 × 10^4 newtons/coulomb

We have the magnitude of the load:

q = 6.4 × 10 ^{-19} coulombs

We also have the distance d that the load moved in a direction parallel to the field 1.2 × 10^{-2} meters.

We know that the electric potential energy (PE) is:

PE = qEd

So:

PE = (6.4 × 10^{-19})(6.5 × 10^4)(1.2 × 10^{-2})

PE = 5.0 x 10^{-16} joules

None of the options shown is correct.

6 0
3 years ago
A man does 500 j work pushing a car a distance of 2m how much force does he apply
olasank [31]

Answer: 250 N

Explanation:

Use equation for work

W=F*d

d=2m

W=500J

F=?

-----------------------

W=Fd

F=W/d

F=500J/2m

F=250N

6 0
4 years ago
Read 2 more answers
A 60-W light bulb radiates electromagnetic waves uniformly in all directions. At a distance of 1.0 mm from the bulb, the light i
slega [8]

Answer:

The appropriate solution is:

(a) \frac{1}{4}(I_o)

(b) \frac{1}{4} (u_o)

(c) \frac{1}{2}B_o

Explanation:

According to the question, the value is:

Power of bulb,

= 60 W

Distance,

= 1.0 mm

Now,

(a)

⇒  \frac{I}{I_o} =\frac{r_o_2}{r_2}

On applying cross-multiplication, we get

⇒  I=I_o\times \frac{1_2}{2^2}

⇒     =I_o\times \frac{1}{4}

⇒     =\frac{1}{4} (I_o)

(b)

As we know,

⇒ \frac{u}{u_o} =\frac{I}{I_o}

By putting the values, we get

⇒ u=\frac{1}{4}(u_o)

(c)

⇒ \frac{B^2}{B_o^2} =\frac{u}{u_o}

         =\frac{I}{I_o}

⇒ B=B_o\times \sqrt{\frac{1}{4} }

⇒     =\frac{1}{2}(B_o)

4 0
3 years ago
What are the three basic parts of a circuit?
Nat2105 [25]

Answer: energy source, path, and load

Explanation:

5 0
4 years ago
Read 2 more answers
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