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kondor19780726 [428]

3 years ago

6

Determine a formula for the acceleration of the system in terms of mA, mB, θ, and g. Ignore the mass of the cord and pulley. Exp

ress your answer in terms of the variables mA, mB, θ, and g.

1 answer:

jekas [21]3 years ago

8 0

Answer:

$a=\frac{mBg-mAgSin\theta}{mA+mB}$

Explanation:

Given two mass on an incline code $mA$ and $mB$ and an angle of inclination $\theta$ . $g$ . Assume that $mA$ is the weight being pulled up and $mB$ the hanging weight.

-The equations of motion from Newton's Second Law are:

$mBg-T=mBa$ where a is the acceleration.

#Substituting for $T$ (tension) gives:

$mBg-mAsin\theta-mAa=mBa$

#and solving for $a:$

$a=\frac{mBg-mAgSin\theta}{mA+mB}$ which is the system's acceleration.

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Water flows straight down from an open faucet. The cross-sectional area of the faucet is 2.4 × 10-4m2 and the speed of the water

Ksenya-84 [330]

To solve this problem it is necessary to apply the continuity equations in the fluid and the kinematic equation for the description of the displacement, velocity and acceleration.

By definition the movement of the Fluid under the terms of Speed, acceleration and displacement is,

$v_2^2 = v_1^2 + 2gh$

Where,

$V_i =$ Velocity in each state

g= Gravity

h = Height

Our values are given as,

$A_1 = 2.4*10^{-4} m^2$

$v_1 = 0.8 m/s$

$h = 0.11m$

Replacing at the kinetic equation to find $V_2$ we have,

$v_2 = \sqrt{v_1^2 + 2gh}$

$v_2 = \sqrt{(0.8 m/s)^2 + 2(9.80 m/s2)(0.11 m)}$

$v_2= 1.67 m/s$

Applying the concepts of continuity,

$A_1v_1 = A_2v_2$

We need to find A_2 then,

$A_2= \frac{A_1v_1 }{v_2}$

So the cross sectional area of the water stream at a point 0.11 m below the faucet is

$A_2= \frac{A_1v_1 }{v_2}$

$A_2= \frac{(2.4*10^{-4})(0.8)}{(1.67)}$

$A_2= 1.14*10^{-4} m2$

Therefore the cross-sectional area of the water stream at a point 0.11 m below the faucet is $1.14*10^{-4} m2$

8 0

3 years ago

How do particles move when a transverse wave passes through a medium?

bagirrra123 [75]

<span> In </span>transverse waves<span>, </span>particles<span> of the</span>medium<span> vibrate </span>to<span> and from in a direction perpendicular </span>to<span> the direction of energy transport. </span>

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3 years ago

A small sphere has a harge of 9uC and other small sphere has a charge of 4uC.

Helga [31]

Answer:

Electrical force, F = 90 N

Explanation:

It is given that,

Charge on sphere 1, $q_1=9\ \mu C=9\times 10^{-6}\ C$

Charge on sphere 2, $q_1=4\ \mu C=4\times 10^{-6}\ C$

Distance between two spheres, d = 6 cm = 0.06 m

Let F is the electrical force between them. It is given by the formula of electric force which is directly proportional to the product of charges and inversely proportional to the square of distance between them such that,

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

$F=9\times 10^9\times \dfrac{9\times 10^{-6}\times 4\times 10^{-6}}{(0.06)^2}$

F = 90 N

So, the electrical force between them is 90 N. Hence, this is the required solution.

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3 years ago

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Anastaziya [24]

Line up in a direction parallel to the magnetic field lines<span />

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3 years ago

Water falls without splashing at a rate of 0.370 l/s from a height of 2.90 m into a 0.690-kg bucket on a scale. if the bucket is

raketka [301]

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3 years ago