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2 years ago

Help!!!! please!!! i’m not sure what the answer is for both questions please

Physics

1 answer:

n200080 [17]2 years ago

7 0

Answer:

a.) 600N

b.) 2 m/s^2

Explanation:

a.)

In order to solve for the needed force, you must use Newton's 2nd law which states that a force is equal to mass multiplied by acceleration.

Basically: Force = mass * acceleration ; F = ma

Since both mass and acceleration are already given, just plug them into the formula and solve for force:

Mass = 200kg

Acceleration = 3m/s^2

F = m * a = 200 * 3 = 600kg * m/s^2 = 600N

b.)

In order to find the acceleration, like the hint already says, you need to find the net force first and then use Newton's 2nd law.

To calculate net force:

Since both forces are parallel (affect the same axis) and in opposite directions, you can subtract the two in order to find resultant magnitude. The resultant direction is the direction of the force with the large magnitude.

Net Force: 8N - 4N = 4N

Net Force Direction: since 8N > 4N, the direction is in the same direciton as the 8N force = Left

Finally, you find acceleration using newton's 2nd law, F = ma.

Since you already know mass and net force, plug in to solve for acceleration:

4N = 2kg * a

a = 4 N / 2kg = 2m/s^2

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What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1

notka56 [123]

Complete Question

Part of the question is shown on the first uploaded image

The rest of the question

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1 and q2 at x3 = -1.220 m ? Your answer may be positive or negative, depending on the direction of the force. Express your answer numerically in newtons to three significant figures.

Answer:

The net force exerted on the third charge is $F_{net}= 3.22*10^{-5} \ J$

Explanation:

From the question we are told that

The third charge is $q_3 = 55 nC = 55 *10^{-9} C$

The position of the third charge is $x = -1.220 \ m$

The first charge is $q_1 = -16 nC = -16 *10^{-9} \ C$

The position of the first charge is $x_1 = -1.650m$

The second charge is $q_2 = 32 nC = 32 *10^{-9} C$

The position of the second charge is $x_2 = 0 \ m$

The distance between the first and the third charge is

$d_{1-3} = -1.650 -(-1.220)$

$d_{1-3} = -0.43 \ m$

The force exerted on the third charge by the first is

$F_{1-3} = \frac{k q_1 q_3}{d_{1-3}^2}$

Where k is the coulomb's constant with a value $9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.$

substituting values

$F_{1-3} = \frac{9*10^{9}* 16 *10^{-9} * (55*10^{-9})}{(-0.43)^2}$

$F_{1-3} = 4.28 *10^{-5} \ N$

The distance between the second and the third charge is

$d_{2-3} = 0- (-1.22)$

$d_{2-3} =1.220 \ m$

The force exerted on the third charge by the first is mathematically evaluated as

$F_{2-3} = \frac{k q_2 q_3}{d_{2-3}^2}$

substituting values

$F_{2-3} = \frac{9*10^{9} * (32*10^{-9}) *(55*10^{-9})}{(1.220)^2}$

$F_{2-3} = 1.06*10^{-5} N$

The net force is

$F_{net} = F_{1-3} -F_{2-3}$

substituting values

$F_{net} = 4.28 *10^{-5} - 1.06*10^{-5}$

$F_{net}= 3.22*10^{-5} \ J$

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3 years ago

A transformer has a primary coil with 375 turns of wire and a secondary coil with 1,875 turns. An AC voltage source connected ac

Sonbull [250]

Answer:

The rms voltage (in V) measured across the secondary coil is 459.62 V

Explanation:

Given;

number of turns in the primary coil, Np = 375 turns

number of turns in the secondary coil, Ns = 1875 turns

peak voltage across the primary coil, Ep = 130 V

peak voltage across the secondary coil, Es = ?

$\frac{N_P}{N_s} = \frac{E_p}{E_s} \\\\E_s = \frac{N_sE_p}{N_p} \\\\E_s = \frac{1875*130}{375} \\\\E_s = 650 \ V$

The rms voltage (in V) measured across the secondary coil is calculated as;

$V_{rms} = \frac{V_0}{\sqrt{2} } = \frac{E_s}{\sqrt{2} } \\\\V_{rms} = \frac{650}{\sqrt{2} } = 459.62 \ V$

Therefore, the rms voltage (in V) measured across the secondary coil is 459.62 V

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andrezito [222]

Answer:

R = 1.2295 10⁵ m

Explanation:

After reading your problem they give us the diameter of the lens d = 4.50 cm = 0.0450 m, therefore if we use the Rayleigh criterion for the resolution in the diffraction phenomenon, we have that the minimum separation occurs in the first minimum of diffraction of one of the bodies m = 1 coincides with the central maximum of the other body

θ = 1.22 λ / D

where the constant 1.22 leaves the resolution in polar coordinates and D is the lens aperture

how angles are measured in radians

θ = y / R

where y is the separation of the two bodies (bulbs) y = 2 m and R the distance from the bulbs to the lens

$\frac{y}{R} = 1.22 \frac{ \lambda}{D}$

R = $\frac{ y \ D}{1.22 \lambda}$

let's calculate

R = $\frac{ 2 \ 0.045}{ 1.22 \ 600 \ 10^{-9}}$

R = 1.2295 10⁵ m

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Answer:

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