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3 years ago

I need help in physics please help me

Physics

1 answer:

artcher [175]3 years ago

3 0

The first step is to represent the vectors shown in the image in Cartesian coordinates.

For the vector C we have a magnitude of 4.8 and an angle 22 ° with the axis -y (direction j)

To write this vector in Cartesian coordinates we must find its component in x (address i) and in the y axis.

$x = 4.8sin(22)i\\\\y = 4.8cos(22)(-j)$

So:

$C = 4.8sin(22) i + 4.8cos(22)(-j)\\\\C = 1.798\ i - 4.450\ j$

For Vector B we have a magnitude of 5.6 and an angle of 33 with the -x axis (-i direction)

So:

$x = 5.6cos(33)(-i)\\\\y = 5.6 sin(33)(-j)$

So:

$B = 5.6cos(33)(-i) + 5.6sin(33)(-j)\\\\B = -4.696\ i - 3.05\ j$

Finally the sum of B + C is made component by component in the following way:

$F = (-4.696 +1.798)i + (-4.450 - 3.05)j\\\\F = -2.898\ i - 7.5\ j$

Finally the magnitude of f is:

$|F| = \sqrt{(-2,898)^2 + (-7.5)^2}$

| F | = 8.04

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2 years ago

g a small smetal sphere, carrying a net charge is held stationarry. what is the speed are 0.4 m apart

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Complete Question

A small metal sphere, carrying a net charge q1=−2μC, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q2= -8μC and mass 1.50g, is projected toward q1. When the two spheres are 0.80m apart, q2 is moving toward q1 with speed 20ms−1. Assume that the two spheres can be treated as point charges. You can ignore the force of gravity.The speed of q2 when the spheres are 0.400m apart is.

Answer:

The value $v_2 = 4 \sqrt{10} \ m/s$

Explanation:

From the question we are told that

The charge on the first sphere is $q_1 = 2\mu C = 2*10^{-6} \ C$

The charge on the second sphere is $q_2 = 8 \mu C = 8*10^{-6} \ C$

The mass of the second charge is $m = 1.50 \ g = 1.50 *10^{-3} \ kg$

The distance apart is $d = 0.4 \ m$

The speed of the second sphere is $v_1 = 20 \ ms^{-1}$

Generally the total energy possessed by when $q_2$ and $q_1$ are separated by $0.8 \ m$ is mathematically represented

$Q = KE + U$

Here KE is the kinetic energy which is mathematically represented as

$KE = \frac{1 }{2} m (v_1)^2$

substituting value

$KE = \frac{1 }{2} * ( 1.50 *10^{-3}) (20 )^2$

$KE = 0.3 \ J$

And U is the potential energy which is mathematically represented as

$U = \frac{k * q_1 * q_2 }{d }$

substituting values

$U = \frac{9*10^9 * 2*10^{-6} * 8*10^{-6} }{0.8 }$

$U = 0.18 \ J$

$Q = 0.3 + 0.18$

$Q = 0.48 \ J$

Generally the total energy possessed by when $q_2$ and $q_1$ are separated by $0.4 \ m$ is mathematically represented

$Q_f = KE_f + U_f$

Here $KE_f$ is the kinetic energy which is mathematically represented as

$KE_f = \frac{1 }{2} m (v_2^2$

substituting value

$KE_f = \frac{1 }{2} * ( 1.50 *10^{-3}) (v_2 )^2$

$KE_f = 7.50 *10^{ -4} (v_2 )^2$

And $U_f$ is the potential energy which is mathematically represented as

$U_f = \frac{k * q_1 * q_2 }{d }$

substituting values

$U_f = \frac{9*10^9 * 2*10^{-6} * 8*10^{-6} }{0.4 }$

$U_f = 0.36 \ J$

From the law of energy conservation

$Q = Q_f$

$0.48 = 0.36 +(7.50 *10^{-4} v_2^2)$

$v_2 = 4 \sqrt{10} \ m/s$

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Answer:

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Explanation:

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E = 1/2ke²................ Equation 1

Where E = elastic potential energy of the spring, k = force constant of the spring, e = extension

Given: K = 10 N/m, e = 2.00 m

Substituting these values into Equation 1

E = 1/2(10)(2)²

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