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Naddik [55]
3 years ago
6

Match each item to its best description.

Physics
2 answers:
Mashcka [7]3 years ago
5 0
A solar system is a group of planets which orbit a star. Our solar system is comprised of planets like Mercury, Earth, and Jupiter. A solar nebula is a large cloud of gas and dust. An accretion disk occurs when gas and dust spin, typically the main body of this disk is a star.
andre [41]3 years ago
5 0

Solar nebula --> "gas and dust spins"

Accretion disk --> "large cloud of gas and dust"

Solar system --> "planets orbit a star"

(Apex class ;)

You might be interested in
A small, 150 g cart is moving at 1.60 m/s on a frictionless track when it collides with a larger, 5.00 kg cart at rest. After th
NemiM [27]

Answer:

Mass of larger cart will be 0.0213 m /sec

Explanation:

We have given mass of small cart m_1=150gram=0.150kg

Initial velocity of small cart v_{1i}=1.6m/sec

Mass of larger block m_2=5kg

Initial velocity of larger cart v_{2i}=0m/sec

After the collision velocity of smaller cart v_{1f}=-0.890m/sec

Now according conservation of momentum

m_1v_{1i}+m_2v_{2i}=m_1v{1f}+m_2v_{2f}

0.150\times 1.6+5\times 0=0.150\times -0.890+5\times v_{2f}

5\times v_{2f}=0.1065

v_{2f}=0.0213m/sec

5 0
3 years ago
Two blocks can collide in a one-dimensional collision. The block on the left hass a mass of 0.30 kg and is initially moving to t
snow_lady [41]

Answer:

a) 3.632 m/s

b) 0.462 m/s

Explanation:

Using the law of conservation of momentum:

m_{1} u_{1} + m_{2} u_{2}= m_{1} V_{1} + m_{2} V_{2}..........(1)

m_{1} = 0.30 kg\\u_{1} = 2.4 m/s\\m_{2} = 0.80 kg\\u_{2} = 0 m/s

Substituting the above values into equation (1) and make V2 the subject of the formula:

0.3(2.4) + 0.80(0)= 0.3 V_{1} + 0.8 V_{2}\\

V_{2} = \frac{0.72 - 0.3 V_{1}}{0.8}..................(2)

Using the law of conservation of kinetic energy:

0.5m_{1} u_{1} ^{2} + 1.2 = 0.5m_{1} V_{1} ^{2} + 0.5m_{2} V_{2} ^{2}\\0.5(0.3) (2.4) ^{2} + 1.2 = 0.5(0.3) V_{1} ^{2} + 0.5(0.8)V_{2} ^{2}\\

2.064 = 0.15 V_{1} ^{2} + 0.4V_{2} ^{2}.......(3)

Substitute equation (2) into equation (3)

2.064 = 0.15 V_{1} ^{2} + 0.4(\frac{0.72 - 0.3V_{1} }{0.8})  ^{2}\\2.064 = 0.15 V_{1} ^{2} + 0.4(\frac{0.5184 - 0.432V_{1} + 0.09V_{1} ^{2}  }{0.64}) \\1.32096 = 0.096 V_{1} ^{2} + 0.20736 - 0.1728V_{1} + 0.036V_{1} ^{2} \\0.132 V_{1} ^{2} - 0.1728V_{1} - 1.1136 = 0\\V_{1} = 3.632 m/s

Substituting V_{1} into equation(2)

V_{2} = \frac{0.72 - 0.3 *3.632}{0.8}\\V_{2} = \frac{0.72 - 0.3 *(3.632)}{0.8}\\V_{2} = 0.462 m/s

8 0
3 years ago
A very long nonconducting cylinder of diameter 10.0 cm carries charge distributed uniformly over its surface. Each meter of leng
Murrr4er [49]

Answer:

Explanation:

The concept of electric field, force acting on proton is applied and appropriate derivations were made to calculate the distance from the surface as shown in the attached file.

7 0
3 years ago
What is pmolecule and hmolecules
Sedbober [7]

Answer:

Compound Basics

Explanation:

Molecule is the general term used to describe any atoms that are connected by chemical bonds. ... Hydrogen gas (H2) is a molecule, but not a compound because it is made of only one element. Water (H2O) can be called a molecule or a compound because it is made of hydrogen (H) and oxygen (O) atoms.

3 0
3 years ago
The electric field in a region of space increases from 0 to 2150 N/C in 5.00 s. What is the magnitude of the induced magnetic fi
Feliz [49]

To solve this problem we will use the Ampere-Maxwell law, which   describes the magnetic fields that result from a transmitter wire or loop in electromagnetic surveys. According to Ampere-Maxwell law:

\oint \vec{B}\vec{dl} = \mu_0 \epsilon_0 \frac{d\Phi_E}{dt}

Where,

B= Magnetic Field

l = length

\mu_0 = Vacuum permeability

\epsilon_0 = Vacuum permittivity

Since the change in length (dl) by which the magnetic field moves is equivalent to the perimeter of the circumference and that the electric flow is the rate of change of the electric field by the area, we have to

B(2\pi r) = \mu_0 \epsilon_0 \frac{d(EA)}{dt}

Recall that the speed of light is equivalent to

c^2 = \frac{1}{\mu_0 \epsilon_0}

Then replacing,

B(2\pi r) = \frac{1}{C^2} (\pi r^2) \frac{d(E)}{dt}

B = \frac{r}{2C^2} \frac{dE}{dt}

Our values are given as

dE = 2150N/C

dt = 5s

C = 3*10^8m/s

D = 0.440m \rightarrow r = 0.220m

Replacing we have,

B = \frac{r}{2C^2} \frac{dE}{dt}

B = \frac{0.220}{2(3*10^8)^2} \frac{2150}{5}

B =5.25*10^{-16}T

Therefore the magnetic field around this circular area is B =5.25*10^{-16}T

3 0
3 years ago
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