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3 years ago

PLEASE I NEED HELP WITH 6 and 8 !!!! I will love u sm

Physics

2 answers:

likoan [24]3 years ago

4 0

Answer:

Explanation:

006

They are acting in opposite directions. Therefore the net force is found by subtraction. The sign is the same as the larger number.

Net Force = 99.6 - 52.8 = 46.8 N acting in the same direction as the 99.6 which is upward.

008

If the two forces act in the same direction, the net force is found by addition.

Net Force = 99.6 + 52.8 = 152.4 N downward.

Georgia [21]3 years ago

4 0

They’re acting in opposite directions o

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frozen [14]

Connexus Physics pre-ap semester A final exam, all correct answers. And y’all better be freaking grateful, because it took a long time to type all this out.

1) In the steps of the scientific method…
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D. y = 1/2x

4) In the graph, what is the correct description…
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5) The motion of a ball on an inclined plane is described…
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6) Acceleration due to gravity is also called
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7) Which of the following is a physical quantity…
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8) In a coordinate system, a vector is…
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9) Which of the following is an example of projectile motion?
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14) A passenger on a bus moving east sees a man standing on a curb…
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18) Which of the following are not units of power?
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3 years ago

Free charges do not remain stationary when close together. To illustrate this, calculate the magnitude of the instantaneous acce

ASHA 777 [7]

Answer:

a=2.304×10¹⁶m/s²

Explanation:

Given data

Distance d=2.5 nm=2,5×10⁻⁹m

Mass of proton m=1.6×10⁻²⁷kg

charge of proton q=1.6×10⁻¹⁹C

To find

acceleration a

Solution

Apply the Coulombs Law

$F=k\frac{q_{1}q_{2} }{r^{2} }$

Where k is coulombs constant (k=9×10⁹Nm²/C²)

q=q₁=q₂

r=d

$F=k\frac{|q^{2} |}{d^{2} }\\ as \\F=ma\\ma=k\frac{|q^{2} |}{d^{2} }\\a=\frac{k}{m} \frac{|q^{2} |}{d^{2} }\\a=\frac{(9*10^{9} )*(1.6*10^{-19} )^{2} }{(1.6*10^{-27} )*(2.5*10^{-9} )^{2} }\\ a=2.304*10^{16}m/s^{2}$

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4 years ago

According to the principle of segregation, _____.

Anastaziya [24]

Answer:

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3 years ago

A nonconducting ring with a radius of 11.5 cm is uniformly charged with a total positive charge of 10.0 µC. The ring rotates at

zhuklara [117]

Answer:

$B=1.21*10^{-10}T$

Explanation:

The magnitude of the magnetic field on the axis of the ring is given by:

$B=\frac{\mu_0 IR^2}{2(r^2+R^2)^{\frac{3}{2}}}(1)$

$\mu_0$ is the permeability of free space, $I$ is the flowing current through the ring, $R$ is the ring's radius and $r$ is the distance to the center of the ring.

The flowing current through the ring is defined as the ring's charge divided into the time taken by the charge to complete one revolution, that is, the period $T=\frac{2\pi}{\omega}$ . So, we have:

$I=\frac{q}{T}\\I=\frac{q}{\frac{2\pi}{\omega}}\\\\I=\frac{\omega q}{2\pi}\\I=\frac{18\frac{rad}{s}(10*10^{-6}C)}{2\pi}\\I=2.87*10^{-5}A$

Now, replacing in (1):

$B=\frac{(4\pi*10^{-7}\frac{T\cdot m}{A})(2.87*10^{-5}A)(0.115m)^2}{2((0.05m)^2+(0.115m)^2)^{\frac{3}{2}}}\\B=1.21*10^{-10}T$

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4 years ago