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2 years ago

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Astronomers were at first surprised to find complicated molecules in the interstellar medium. They thought ultra-violet light fr

om stars would break apart such molecules. What protects the molecules we observe from being broken apart

1 answer:

jeka57 [31]2 years ago

8 0

Answer:

The dust present in the clouds.

Explanation:

The complicated composition molecules that can be found in space are generally associated with clouds of dust. The significant amount of dust in these clouds provides protection not only for these molecules, but for any body that makes up or is associated with dust clouds.

It is exactly this dust that protects the molecules against the action of ultraviolet rays.

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This is for physical science, need help :/

marusya05 [52]

1). From the frame of reference of a passenger on the airplane looking out of his window, the tree appears to be moving, at roughly 300 miles per hour toward the left of the picture.

2). The SI unit best suited to measuring the height of a building is the meter.

3). 'Displacement' is the straight-line distance and direction from the start-point to the end-point, regardless of the path that was followed to get there.

The ball started out in the child's hand, and it ended up 2 meters away from her in the direction of the wall. So the displacement of the ball from the beginning to the end of the story is: 2 meters toward the wall.

6 0

2 years ago

A closed container initially holds 50 monatomic Aparticles that have a combined energy of 480 units. After 100 monatomic B parti

Molodets [167]

Answer:

"8 units" is the appropriate answer.

Explanation:

According to the question,

Throughout equilibrium all particles are of equivalent intensity, and as such the integrated platform's total energy has been uniformly divided across all individuals.

Now,

The total energy will be:

= $480+720$

= $1200 \ units$

The total number of particles will be:

= $50+100$

= $150$

hence,

Energy of each A particle or each B particle will be:

= $\frac{1200}{150}$

= $8 \ units$

5 0

2 years ago

The speed of light in a vacuum is 2.99x10^8 m/s. calculate its speed in miles per hour.

34kurt

You'll never get the correct answer without the correct conversion factor. Note carefully that you have no decimal. It should be
<span>1 km = 0.6214 miles </span>
<span>1000 m = 1 km </span>
<span>60 seconds = 1 minute </span>
<span>60 minutes = 1 hour. </span>
<span>2.998E8 m/s x (1 km/1000m) x (0.6214 miles/km) x (60 sec/min) x (60 min/hr) = ?</span>

6 0

3 years ago

Two 0.2304 cm x 0.2304 cm square aluminum electrodes, spaced 0.5974 mm apart are connected to a 61 V battery. What is the charge

Arisa [49]

Answer:

The charge on each plate is 0.0048 nC

Explanation:

for the distance between the plates d and given the area of plates, A, and ε = 8.85×10^-12 C^2/N.m^2, the capacitance of the plates is given by:

C = (A×ε)/d

=[(0.2304×10^-2)(0.2304×10^-2)×(8.85×10^-12))/(0.5974×10^-3)

= 7.86×10^-14 F

then if the plates are connected to a battery of voltage V = 61 V, the charge on the plates is given by:

q = C×V

= (7.86×10^-14)×(61)

= 4.80×10^-14 C

≈ 0.0048 nC

Therefore, the charge on each plate is 0.0048 nC.

3 0

2 years ago

Block A has a mass of 0.5kg, and block B has a mass of 2kg. Block is is released at a height of 0.75 meters above B. The coeffic

VikaD [51]

Answer:

0.075 m

Explanation:

The picture of the problem is missing: find it in attachment.

At first, block A is released at a distance of

h = 0.75 m

above block B. According to the law of conservation of energy, its initial potential energy is converted into kinetic energy, so we can write:

$m_Agh=\frac{1}{2}m_Av_A^2$

where

$g=9.8 m/s^2$ is the acceleration due to gravity

$m_A=0.5 kg$ is the mass of the block

$v_A$ is the speed of the block A just before touching block B

Solving for the speed,

$v_A=\sqrt{2gh}=\sqrt{2(9.8)(0.75)}=3.83 m/s$

Then, block A collides with block B. The coefficient of restitution in the collision is given by:

$e=\frac{v'_B-v'_A}{v_A-v_B}$

where:

e = 0.7 is the coefficient of restitution in this case

$v_B'$ is the final velocity of block B

$v_A'$ is the final velocity of block A

$v_A=3.83 m/s$

$v_B=0$ is the initial velocity of block B

Solving,

$v_B'-v_A'=e(v_A-v_B)=0.7(3.83)=2.68 m/s$

Re-arranging it,

$v_A'=v_B'-2.68$ (1)

Also, the total momentum must be conserved, so we can write:

$m_A v_A + m_B v_B = m_A v'_A + m_B v'_B$

where

$m_B=2 kg$

And substituting (1) and all the other values,

$m_A v_A = m_A (v_B'-2.68) + m_B v_B'\\v_B' = \frac{m_A v_A +2.68 m_A}{m_A + m_B}=1.30 m/s$

This is the velocity of block B after the collision. Then, its kinetic energy is converted into elastic potential energy of the spring when it comes to rest, according to

$\frac{1}{2}m_B v_B'^2 = \frac{1}{2}kx^2$

where

k = 600 N/m is the spring constant

x is the compression of the spring

And solving for x,

$x=\sqrt{\frac{mv^2}{k}}=\sqrt{\frac{(2)(1.30)^2}{600}}=0.075 m$

5 0

2 years ago