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4 years ago

8

How many neutrons does carbon have

2 answers:

Sav [38]4 years ago

4 0

Carbon has 6 neutrons!

jek_recluse [69]4 years ago

4 0

Carbon has 6 neutrons

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Rosa eats a peanut butter sandwich for lunch. Peanut butter contains a lot

shusha [124]

Cmndkdkkdnd djnd dhdbhd s ahah who w baggage wbhebwnwh

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3 years ago

A 0.035 bullet strikes a 5kg wooden block that is stationary and gets stuck inside the block. The block and bullet move together

hjlf

Answer:

1240 m/s

Explanation:

[Do you mean 0.035 kg bullet?]

Since the momentum of the system is conserved,

Let u m/s be the initial velocity of the bullet

0.035u = 8.6(5+0.035)

u ~= 1237.1714 m/s

u = 1240 m/s (3 sig. fig.)

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3 years ago

Question 5 of 10

spayn [35]

Answer: A) 0.19 kg.m/s South just answered it right on Apex!

Explanation:

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3 years ago

A conducting rod (length = 2.0 m) spins at a constant rate of 2.0 revolutions per second about an axis that is perpendicular to

vladimir2022 [97]

Answer:

0.050V

Explanation:

To solve this problem it is necessary to use the concepts related to the potential between two objects that have a magnetic field, this concept is represented in the equation.

$\int dV = \int_{0}^{l/2} Bv (dl)$

Where,

v= tangencial velocity

B = Magnetic Field

We know for definition that,

$v= l\omega$

Where,

L = length

$\omega =$ Angular velocity

We can replace this values in our first equation then,

$\int dV = \int_{0}^{l/2} B (l\omega) (dl)$

Integrating we have,

$V = \frac{1}{8} Bl^2 \omega$

Replacing the values,

$V= \frac{1}{8} (8*10^{-3})(12.56)(4)$

$V = 0.050V$

Therefore the potential difference between the center of the rod and the other rod is 0.050V

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3 years ago

Plaskett's binary system consists of two stars that revolve In a circular orbit about a center of mass midway between them. This

Sergio039 [100]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The mass is $M =1.43 *10^{32} \ kg$

Explanation:

From the question we are told that

The mass of the stars are $m_1 = m_2 =M$

The orbital speed of each star is $v_s = 240 \ km/s =240000 \ m/s$

The orbital period is $T = 12.5 \ days = 12.5 * 2 4 * 60 *60 = 1080000\ s$

The centripetal force acting on these stars is mathematically represented as

$F_c = \frac{Mv^2}{r}$

The gravitational force acting on these stars is mathematically represented as

$F_g = \frac{GM^2 }{d^2}$

So $F_c = F_g$

=> $\frac{mv^2}{r} = \frac{Gm_1 * m_2 }{d^2}$

=> $\frac{v^2}{r} = \frac{GM}{(2r)^2}$

=> $\frac{v^2}{r} = \frac{GM}{4r^2}$

=> $M = \frac{v^2*4r}{G}$

The distance traveled by each sun in one cycle is mathematically represented as

$D = v * T$

$D = 240000 * 1080000$

$D = 2.592*10^{11} \ m$

Now this can also be represented as

$D = 2 \pi r$

Therefore

$2 \pi r= 2.592*10^{11} \ m$

=> $r= \frac{2.592*10^{11}}{2 \pi }$

=> $r= 4.124 *10^{10} \ m$

So

$M = \frac{v^2*4r}{G}$

=> $M = \frac{(240000)^2*4*(4.124*10^{10})}{6.67*10^{-11}}$

=> $M =1.43 *10^{32} \ kg$

3 0

3 years ago