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Lesechka [4]
2 years ago
12

What type of Galaxy is considered the most common?

Physics
2 answers:
Semmy [17]2 years ago
4 0
Spiral galaxies are most common
lubasha [3.4K]2 years ago
4 0

Answer:

Spiral Galaxies

Explanation:

The most common type galaxy found throughout the universe is the spiral galaxy!

Happy to help. Lol! Good luck!

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A 4.9-MeV (kinetic energy) proton enters a 0.28-T field, in a plane perpendicular to the field. Part APart complete What is the
BartSMP [9]

Answer:

r=1.14m

Explanation:

\theta is the angle between the velocity and the magnetic field. So, the magnetic force on the proton is:

F_m=qvBsen\theta\\F_m=qvBsen(90^\circ)\\F_m=qvB

A charged particle describes a semicircle in a uniform magnetic field. Therefore, applying Newton's second law to uniform circular motion:

F_m=F_c\\qvB=F_c(1)

F_c is the centripetal force and is defined as:

F_c=m\frac{v^2}{r}

Here v is the proton's speed and r is the radius of the circular motion. Replacing this in (1) and solving for r:

qvB=\frac{mv^2}{r}\\r=\frac{mv^2}{qvB}\\r=\frac{mv}{qB}

Recall that 1 J is equal to 6.242*10^{12}MeV, so:

4.9MeV*\frac{1J}{6.242*10^{12}MeV}=7.85*10^{-13}J

We can calculate v from the kinetic energy of the proton:

K=\frac{mv^2}{2}\\\\v=\sqrt{\frac{2K}{m}}\\v=\sqrt{\frac{2(7.85*10^{-13}J)}{1.67*10^{-27}kg}}\\v=3.06*10^{7}\frac{m}{s}

Finally, we calculate the radius of the proton path:

r=\frac{mv}{qB}\\r=\frac{1.67*10^{-27}kg(3.06*10^{7}\frac{m}{s})}{1.6*10^{-19}C(0.28T)}\\r=1.14m

8 0
3 years ago
An umbrella tends to move upward on a windy day because _____.
masha68 [24]
E. all of the above

An umbrella tends to move upward on a windy day because _<span>A. buoyancy increases with increasing wind speed </span>
<span>B. air gets trapped under the umbrella and pushes it up </span>
<span>C. the wind pushes it up </span>
<span>D. a low-pressure area is created on top of the umbrella </span>

3 0
3 years ago
A shopper does 157 J of work pushing a cart with 10.9 N force
Tanzania [10]

The cart travelled a distance of 14.4 m

Explanation:

The work done by a force when pushing an object is given by:

W=Fd cos \theta

where:

F is the magnitude of the force

d is the displacement

\theta is the angle between the direction of the force and the displacement

In this problem we have:

W = 157 J is the work done on the cart

F = 10.9 N is the magnitude of the force

\theta=0^{\circ}, assuming the force is applied parallel to the motion of the cart

Therefore we can solve for d to find the distance travelled by the cart:

d=\frac{W}{F cos \theta}=\frac{157}{(10.9)(cos 0)}=14.4 m

Learn more about work:

brainly.com/question/6763771  

brainly.com/question/6443626  

#LearnwithBrainly

4 0
2 years ago
You plan to take a trip to the moon. Since you do not have a traditional spaceship with rockets, you will need to leave the eart
hichkok12 [17]

Answer:

v = 3.5 \times 10^5 m/s

Explanation:

At some distance from the Earth the force of attraction due to moon is balanced by the force due to Moon

so we will have

\frac{GM_em}{r^2} = \frac{GM_m}{(d-r)^2}

now we have

\frac{d - r}{r} = \sqrt{\frac{M_m}{M_e}}

\frac{3.844\times 10^8 - r}{r} = \sqrt{\frac{7.36 \times 10^{22}}{5.9742\times 10^{24}}}

so we will have

r = 3.46 \times 10^8 m

Now by energy conservation

-\frac{GM_e}{R_e} - \frac{GM_m}{d - (R_e + R_m)} + \frac{1}{2}v^2 = -\frac{GM_e}{r} - \frac{GM_m}{d - r}

-6.26 \times 10^{8} - 13046 + \frac{1}{2}v^2 = -1.15 \times 10^6 - 1.28 \times 10^5

v = 3.5 \times 10^5 m/s

7 0
3 years ago
How much energy (in Joules) is released when 12.0 g of water cools from 20.0 °C to 11.0 °C? This is a grade 10 question from the
KATRIN_1 [288]

Answer: - 452.088joule

Explanation:

Given the following :

Mass of water = 12g

Change in temperature(Dt) = (11 - 20)°C = - 9°C

Specific heats capacity of water(c) = 4.186j/g°C

Q = mcDt

Where Q = quantity of heat

Q = 12g × 4.186j/g°C × - 9°C

Q = - 452.088joule

7 0
3 years ago
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