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Lana71 [14]
3 years ago
15

APEX What is one advantage of using primary sources when doing research on an

Physics
1 answer:
MissTica3 years ago
4 0

Answer:

C

Explanation:

They are first hand sources so they are more reliable and detailed...

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Describe what would happen if you rubbed a mineral with a Mohs hardness value of 7 against a mineral with a value of 5?
chubhunter [2.5K]
The mineral with Mohs hardness would be scratched because the mineral with Mohs 7 hardness is stronger than the Mohs 5 mineral. Eventually, that mineral would turn into dust if you kept rubbing it.
8 0
3 years ago
A bike rider pedals with constant acceleration to reach a velocity of 7.8 m/s over a time of 4.2 s. during the period of acceler
Artyom0805 [142]

To calculate the initial velocity of the bike, we use the following equation

d=\frac{1}{2} (u+v)t.

or

u=\frac{2d}{t} -v

Here, u is initial velocity, v is final velocity, t is the time and d is the distance covered by bike.

Given, u =7.8 m/s,d= 19 m and t=4.2 s.

Substituting these values in above equation, we get

u = \frac{2 \times 19}{4.2 \ s} -7.8 m/s = 9.05 \ m/s - 7.8 \ m/s \\\\ u= 1.2 m/s.

Thus, the initial velocity of the bike is 1.2 m/s.

3 0
3 years ago
In Ch. 1.6, the authors point out that interstellar space is not actually as empty as it seems. There is actually a lot of diffu
wel

Answer:

very small solid particles called interstellar dust.

Explanation:

In the space between the stars there is gas and dust, which represent at least 20% of the mass of our galaxy. In the Milky Way it is considered that there is a gas density of approximately 0.2 to 0.5 atoms / cm3 in the surroundings of the Sun; with respect to the dust an average of 1 g / cm3 is estimated.

Gas is about atoms and molecules, mainly hydrogen; In order of abundance, helium, carbon, oxygen, nitrogen and iron follow. On the other hand, the dust is tiny particles, generally smaller than 10 microns; the dust does not shine and therefore it is only distinguished when it is projected on bright regions (nebulae or clusters).

Interstellar matter is mainly concentrated towards the plane of the galaxy, in the strip corresponding to the Milky Way; there you can see bright nebulas of diffuse character called nebulas. These nebulae are classified according to three types: (a) bright or emission nebulae, (b) reflection nebulae and (c) planetary nebulae.

Hydrogen appears both ionized and neutral; The bright nebulae are composed of ionized hydrogen and other ionized elements. Non-ionized (neutral) hydrogen is found in the spiral arms of the Milky Way and can be detected through radio waves.

6 0
3 years ago
A 0.0250-kg bullet is accelerated from rest to a speed of 550 m/s in a 3.00-kg rifle. The pain of the rifle’s kick is much worse
kondaur [170]

Answer:

a) 4.583 m/s

b) 31.505 J

c) 0.491 m/s

d) 3.375 J

e)

   p_player = (110 kg)(8 m/s) = 880 kg m/s

   p_ball = (0.41 kg)(25 m/s) = 10.25 kg m/s

Explanation:

HI!

a)

We can calculate the recoil velocity by conservation of momentum, remember that p=mv.

The momentum of the bullet is:

p_b = (0.0250 kg)*(550 m/s )

The momentum of the rifle is:

p_r = (3 kg) * v

Since the total initial momentum is zero:

p_b = p_r

That is:

v = (550 m/s ) (0.0250 kg/ 3 kg ) = 4.583 m/s

b)

The kinetic energy gained by the rifle is:

K = (1/2) m v^2 = (1/2) *(3 kg) *(4.583 m/s)^2 = 31.505 J

c)

We use the same formula as in a), but with m=28kg instead of 3 kg

v = (550 m/s ) (0.0250 kg/ 28 kg ) = 0.491 m/s

d)

Again, the same formula as b, but with m=28 and v=0.491 m/s

K = 3.375 J

e)

p_player = (110 kg)(8 m/s) = 880 kg m/s

p_ball = (0.41 kg)(25 m/s) = 10.25 kg m/s

I believe that the kinetic energy is more related to the problem than the momentum. The relation between these two quantities is:

K = p^2/(2m)

usiing this relation, we get:

K_player = 3520 J

K_ball =  128.125 J

Therefore the kinetic energy of the player is around 27 time larger than the kinetic energy of the ball, that being said, the pain of being tackled by that player is around 27 times greater that being hit by the ball!

4 0
3 years ago
Three long parallel wires each carry 2.0-A currents in the same direction. The wires are oriented vertically, and they pass thro
blagie [28]

Answer:

21.2\times 10^{-6} T

Explanation:

i  = magnitude of current in each wire = 2.0 A

a  = length of the side of the square = 4 cm = 0.04 m

r  = length of the diagonal of the square = \sqrt{2} a = \sqrt{2} (0.04) = 0.057 m

B = magnitude of magnetic field by wires at A and C

B = \left ( \frac{\mu _{o}}{4\pi } \right )\left ( \frac{2i}{a} \right )

B = (10^{-7}) \left ( \frac{2(2)}{0.04} \right )

B = 10\times 10^{-6} T

B' = magnitude of magnetic field by wire at B

B' = \left ( \frac{\mu _{o}}{4\pi } \right )\left ( \frac{2i}{r} \right )

B' = (10^{-7}) \left ( \frac{2(2)}{0.057} \right )

B' = 7.02\times 10^{-6} T

Net magnitude of the magnetic field at D is given as

B_{net} = \sqrt{B^{2}+B^{2}} + B'

B_{net} = \sqrt{2} B + B'

B_{net} = \sqrt{2} (10\times 10^{-6}) + (7.02\times 10^{-6})

B_{net} = 21.2\times 10^{-6} T

8 0
3 years ago
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