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3 years ago

Https://brainly.com/question/14419503?referrer=searchResults

Physics

2 answers:

Naily [24]3 years ago

8 0

Answer:

Explanation:

Ludmilka [50]3 years ago

5 0

Answer:

what

Explanation:

Is every thing okay

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What is the molecular formula for the hydrocarbon in the mass spectrum above?

PtichkaEL [24]

What is the structure of a hydrocarbon that has $\mathrm{M}^{+}=120$ in its mass spectrum and has the following $1 \mathrm{H}$ NMR spectrum? 7.25 $\delta(5 \mathrm{H}, \text { broad singlet); } 2.90 \delta(1 \mathrm{H}, \text { septet, } J=7 \mathrm{Hz}) ; 1.22 \delta(6 \mathrm{H},\text { doublet, }$ $J=7 \mathrm{Hz})$

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3 years ago

Two particles, one with charge -6.29 × 10^-6 C and one with charge 5.23 × 10^-6 C, are 0.0359 meters apart. What is the magnitud

navik [9.2K]

Answer:

Force, F = −229.72 N

Explanation:

Given that,

First charge particle, $q_1=-6.29\times 10^{-6}\ C$

Second charged particle, $q_2=5.23\times 10^{-6}\ C$

Distance between charges, d = 0.0359 m

The electric force between the two charged particles is given by :

$F=k\dfrac{q_1q_2}{d^2}$

$F=9\times 10^9\times \dfrac{-6.29\times 10^{-6}\times 5.23\times 10^{-6}}{(0.0359)^2}$

F = −229.72 N

So, the magnitude of force that one particle exerts on the other is 229.72 N. Hence, this is the required solution.

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3 years ago

The gravitational force exerted by a proton on an electron is 2x1039 times weaker than the electric force that the proton exerts

pickupchik [31]

To solve this problem we will rely on the theorems announced by Newton and Coulomb about the Gravitational Force and the Electrostatic Force respectively.

In the case of the Force of gravity we have to,

$F_g = G\frac{m_pm_e}{d^2}$

Here,

G = Gravitational Universal Constant

$m_p$ = Mass of Proton

$m_e$ = Mass of Electron

d = Distance between them.

$F_g = (6.673*10^{-11} kg^{-1} \cdot m^3 \cdot s^{-2}) (\frac{(1.672*10^{-27}kg)(9.109*10^{-31})}{(52.9pm)^2})$

$F_g = 3.631*10^{-47}N$

In the case of the Electric Force we have,

$F_e = k\frac{q_pq_e}{d^2}$

k = Coulomb's constant

$q_p$ = Charge of proton

$q_e$ = Charge of electron

d = Distance between them

$F_e = (9*10^9N\cdot m^2 \cdot C^{-2})(\frac{(1.602*10^{-19}C)(1.602*10^{-19}C)}{(52.9pm)^2})$

$F_e = 82.446*10^{-9}N$

Therefore

$\frac{F_e}{F_g} = 2.270*10^{39}$

We can here prove that the statement is True

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3 years ago

What happens to waves near the shore?

Lady_Fox [76]

Answer As a wave travels across the open ocean, it gains speed. When a wave reaches a shallow coastline, the wave begins to slow down due to the friction caused by the approaching shallow bottom. ... Think of it like driving a car at high speed and then slamming on the breaks. Everything is going to fly to the front.:Waves at the Shoreline: As a wave approaches the shore it slows down from drag on the bottom when water depth is less than half the wavelength (L/2). The waves get closer together and taller. ... Eventually the bottom of the wave slows drastically and the wave topples over as a breaker. hope this helps have a nice night❤️❤️❤️

Explanation:

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3 years ago

A student pulls a block over a rough surface with a constant force FP that is at an angle θ above the horizontal, as shown above

gizmo_the_mogwai [7]

Answer:

B.The force of friction between the block and surface will decrease.

Explanation:

The force of friction is given by

$F_s = \mu N$

where $\mu$ is the coefficient of friction and $N$ is the normal force.

When the student pulls on the block with force $F_p$ at an angle $\theta$ , the normal force on the block becomes

$N = Mg- F_psin(\theta)$

and hence the frictional force becomes

$F_s = \mu (Mg- F_psin(\theta))$ .

Now, as we increase $\theta$ , $sin(\theta)$ increases which as a result decreases the normal force $Mg- F_psin(\theta)$ , which also means the frictional force decreases; Hence choice B stands true.

<em>P.S: Choice D is tempting but incorrect since the weight </em> $W=mg$ <em> is independent of the external forces on the block. </em>

6 0

4 years ago

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