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lilavasa [31]
3 years ago
6

Help me with this question ​

Physics
1 answer:
Roman55 [17]3 years ago
4 0

Explanation:

Let's set the x-axis to be parallel to the and positive up the plane. Likewise, the y-axis will be positive upwards and perpendicular to the plane. As the problem stated, we are going to assume that m1 will move downwards so its acceleration is negative while m2 moves up so its acceleration is positive. There are two weight components pointing down the plane, m_1g \sin \theta and m_2g \sin \theta and two others pointing up the plane, the two tensions <em>T</em> along the strings. There is a normal force <em>N</em> pointing up from the plane and two pointing down, m_1g \sin \theta and m_2g \sin \theta. Now let's apply Newton's 2nd law to this problem:

<u>x-axis</u><u>:</u>

m1:\:\:\:\displaystyle \sum_i F_i = T - m_1g \sin \theta = - m_1a\:\:\:\:(1)

m2:\:\:\:\displaystyle \sum_i F_i = T - m_2g \sin \theta = m_2a\:\:\:\:(2)

<u>y-axis</u><u>:</u>

\:\:\:\displaystyle \sum_i F_i = N - m_1g \cos \theta - m_2g \cos \theta = 0

Use Eqn 1 to solve for T,

T = m_1(g \sin \theta - a)

Substitute this expression for T into Eqn 2,

m_1g \sin \theta - m_1a - m_2g \sin \theta = m_2a

Collecting all similar terms, we get

(m_1 + m_2)a = (m_1 - m_2)g \sin \theta

or

a = \left(\dfrac{m_1 - m_2}{m_1 + m_2} \right)g \sin \theta

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À stone is thrown
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A = -9.8

v = -9.8t -8

s = -4.9 t2 -8t +25

So… -5t^2 -8t + 25 =0, we’ll rearrange to 5t^2 + 8t - 25. We get two roots, one is positive and is 1.59 seconds

V = -9.8(1.59) - 8 = -23.6

So… it takes 1.59 seconds to hit the ground at -23.6 m/s.
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3 years ago
Ishani and John now try a problem involving a charging capacitor. An uncharged capacitor with C = 6.81 μF and a resistor with R
DENIUS [597]

Answer:

Q=81.72\times10^{-6}C

I=2.1\times10^{-5}A

Explanation:

The maximum charge on the capacitor will be, at the end of the process, given by the formula (and for our values):

Q=CV=(6.81\times10^{-6}F)(12V)=81.72\times10^{-6}C

The maximum current on the resistor will be, at the beginning of the process, given by the formula (and for our values):

I=\frac{V}{R}=\frac{12V}{5.8\times10^{5}\Omega}=2.1\times10^{-5}A

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3 years ago
What's a refractive medium that focuses light to a single point?
ss7ja [257]

Answer:

The convex lens is shaped so that all light rays that enter it parallel to its axis cross one another at a single point on the opposite side of the lens.

Explanation:

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How a measuring scale measure our weight
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Scales measure how much something weighs—and they do it by measuring how much force exists between the object you're weighing and planet Earth. Although scales measure force, they give you measurements of mass in kilograms, grams, pounds, or whatever.

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Learning Goal:
enot [183]

Answer:

A. U_0 = \dfrac{\epsilon_0 A V^2}{2d}

B. U_1 = \dfrac{\epsilon_0 A V^2}{6d}

C. U_2 = \dfrac{K\epsilon_0 A V^2}{2d}

Explanation:

The capacitance of a capacitor is its ability to store charges. For parallel-plate capacitors, this ability depends the material between the plates, the common plate area and the plate separation. The relationship is

C=\dfrac{\epsilon A}{d}

C is the capacitance, A is the common plate area, d is the plate separation and \epsilon is the permittivity of the material between the plates.

For air or free space, \epsilon is \epsilon_0 called the permittivity of free space. In general, \epsilon=\epsilon_r \epsilon_0 where \epsilon_r is the relative permittivity or dielectric constant of the material between the plates. It is a factor that determines the strength of the material compared to air. In fact, for air or vacuum, \epsilon_r=1.

The energy stored in a capacitor is the average of the product of its charge and voltage.

U = \dfrac{QV}{2}

Its charge, Q, is related to its capacitance by Q=CV (this is the electrical definition of capacitance, a ratio of the charge to its voltage; the previous formula is the geometric definition). Substituting this in the formula for U,

U = \dfrac{CV^2}{2}

A. Substituting for C in U,

U_0 = \dfrac{\epsilon_0 A V^2}{2d}

B. When the distance is 3d,

U_1 = \dfrac{\epsilon_0 A V^2}{2\times3d}

U_1 = \dfrac{\epsilon_0 A V^2}{6d}

C. When the distance is restored but with a dielectric material of dielectric constant, K, inserted, we have

U_2 = \dfrac{K\epsilon_0 A V^2}{2d}

6 0
3 years ago
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