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2 years ago

I need help please i really do appreciate it

Physics

2 answers:

SVETLANKA909090 [29]2 years ago

5 0

Answer:

Inter molecular forces

Explanation:

cluponka [151]2 years ago

3 0

Answer:Substances

Explanation:

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Does a break in one branch affect another branch in a parallel circuit

SashulF [63]

No. Parallel branches have the same potential difference across them. A break in one does not affect the other.

8 0

2 years ago

A nerve signal travels 150 meters per second. Determine the number of kilometers that the nerve signal will travel in the same t

joja [24]

Explanation:

It is given that,

A nerve signal travels 150 meters per second. It is the speed of the nerve signal. We need to convert the number of kilometers that the nerve signal will travel in the same time.

We know that,

1 kilometer = 1000 meter

1 hour = 3600 seconds

$150\ \dfrac{m}{s}=150\times \dfrac{1/1000\ km}{1/3600\ h}$

$150\ m/s=540\ km/h$

So, the nerve signal will travel at the rate of 540 km/h. Hence, this is the required solution.

6 0

3 years ago

A photon is absorbed by an electron that is in the n = 3 state of a hydrogen atom, causing the hydrogen atom to become ionized.

Wittaler [7]

Answer:

$\lambda=3.99*10^{-7}m$

Explanation:

According to the law of conservation of energy, the energy of the absorbed photon must be equal to the binding energy of the electron plus the energy of the released electron:

$E_p=E_b+E_r\\\frac{hc}{\lambda}=\frac{13.6eV}{n^2}+\frac{m_ev^2}{2}$

1 eV is equal to $1.6*10^{-19}J$ , so:

$13.6eV*\frac{1.6*10^{-19}J}{1eV}=2.18*10^{-18}J$

Solving for $\lambda$ and replacing the given values:

$\lambda=\frac{hc}{\frac{2.18*10^{-18}J}{n^2}+\frac{m_ev^2}{2}}\\\lambda=\frac{6.63*10^{-134}J\cdot s(3*10^8\frac{m}{s})}{\frac{2.18*10^{-18}J}{3^2}+\frac{(9.11*10^{-31}kg)(7.5*10^5\frac{m}{s})^2}{2}}\\\lambda=3.99*10^{-7}m$

4 0

3 years ago

What would Hubble's constant be if we found one galaxy moving away at 30,000 km/s at a distance of 600 Mpc?

lora16 [44]

Answer:

H₀ = 1.6 x 10⁻¹⁸ s⁻¹

Explanation:

The Hubble's Constant can be found by the following formula:

$v = H_o D\\\\H_o = \frac{v}{D}$

where,

H₀ = Hubble's Constant = ?

v = speed of galaxy = 30000 km/s = 3 x 10⁷ m/s

D = Distacance = 600 Mpc = (6 x 10⁸ pc)(3.086 x 10¹⁶ m/1 pc)

D = 18.52 x 10²⁴ m

Therefore,

$H_o = \frac{3\ x\ 10^7\ m/s}{18.52\ x\ 10^{24}\ m}$

3 0

2 years ago

N experiment is performed in deep space with two uniform spheres, one with mass 27.0 and the other with mass 107.0 . They have e

Reptile [31]

Answer:

Explanation:

Apply the law of conservation of energy

$KE_i+PE_i=KE_f+PE_f$

$Gm_1m_2[\frac{1}{r_f} -\frac{1}{r_1} ]=\frac{1}{2} (m_1v_1^2+m_2v_2^2)$

from the law of conservation of the linear momentum

$m_1v_1=m_2v_2$

Therefore,

$Gm_1m_2[\frac{1}{r_f} -\frac{1}{r_1} ]=\frac{1}{2} (m_1v_1^2+m_2v_2^2)$

$=\frac{1}{2} [m_1v_1^2+m_2[\frac{m_1v_1}{m_2} ]^2]\\\\=\frac{1}{2} [m_1v_1^2+\frac{m_1^2v_1^2}{m_2} ]\\\\=\frac{m_1v_1^2}{2} [\frac{m_1+m_2}{m_2} ]$

$v_1^2=[\frac{2Gm_2^2}{m_1+m_2} ][\frac{1}{r_f} -\frac{1}{r_1} ]$

Substitute the values in the above result

$v_1^2=[\frac{2Gm_2^2}{m_1+m_2} ][\frac{1}{r_f} -\frac{1}{r_1} ]$

$=[\frac{2(6.67\times 10^-^1^1)(107)^2}{27+107} ][\frac{1}{26} -\frac{1}{41}] \\\\=1.6038\times 10^-^1^0\\\\v_1=\sqrt{1.6038\times 106-^1^0} \\\\=1.2664 \times 10^-^5m/s$