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3 years ago

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In each case, three charged particles are fixed in place at the vertices of an equilateral triangle. The triangles are all the s

ame size. Rank the magnitude of the net electric force on the lower-left particle. Greatest ______ _______ _______ _______ Least Explain your reasoning.

1 answer:

spayn [35]3 years ago

5 0

Answer:

hello your question is incomplete attached below is the complete question

answer : $F_{E} > F_{F} > F_{A} = F_{B} > F_{C} =F_{D}$

Explanation:

The explanation behind the reasoning is attached below

Ranking the magnitude of the net electric force on the lower-left particle from the greatest to the least

$F_{E} > F_{F} > F_{A} = F_{B} > F_{C} =F_{D}$

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Water flows at 0.65 m/s through a 3.0 cm diameter hose that terminates in a 0.3 cm diameter nozzle. (Recall, the density of wate

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Answer:

v2 = 65 m/s

the speed of the water leaving the nozzle is 65 m/s

Explanation:

Given;

Water flows at 0.65 m/s through a 3.0 cm diameter hose that terminates in a 0.3 cm diameter nozzle

Initial speed v1 = 0.65 m/s

diameter d1 = 3.0 cm

diameter (nozzle) d2 = 0.3 cm

The volumetric flow rates in both the hose and the nozzle are the same.

V1 = V2 ........1

Volumetric flow rate V = cross sectional area × speed of flow

V = Av

Area = (πd^2)/4

V = v(πd^2)/4 ....2

Substituting equation 2 to 1;

v1(πd1^2)/4 = v2(πd2^2)/4

v1d1^2 = v2d2^2

v2 = (v1d1^2)/d2^2

Substituting the given values;

v2 = (0.65 × 3^2)/0.3^2

v2 = 65 m/s

the speed of the water leaving the nozzle is 65 m/s

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4 years ago

For each question, select the right answer from the choices below:

Leni [432]

Option (ii) B is the correct option. The object on the moon has greater mass.

To resolve this, utilize the formulas Force = Mass * Acceleration.

The equation can be used to find the mass given the force in Newtons, using 9.8 m/s² for the acceleration of gravity of the earth and 1.6 m/s² for the moon.

Calculating the mass on earth:

30 N = 9.8 m/s² * mass

This results in a mass of 3.0 kg for the object on Earth.

Calculating the mass of the moon:

30 N = 1.6 m/s²2 * mass

Thus, the moon's object has a mass of 19. kg.

This can be explained by the fact that the earth has a stronger gravitational pull than the moon, producing more force per kilogram of mass. As a result, the moon's mass must be bigger to produce the same amount of force at a lower acceleration from gravity (1.6 m/s² vs. 9.8 m/s²).

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1 year ago

A piano of mass 852 kg is lifted to a height of 3.5 m. How much gravitational potential energy is added to the piano? Accelerati

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$m=852 \ kg \\ h=3,5 \ m \\ g=9,8 \ m/s^2 \\ \boxed{P_e-?} \\ \bold{Solving:} \\ \boxed{P_e=m \cdot g \cdot h} \\ P_e=852 \ kg \cdot 9,8 \ m/s^2 \cdot 3,5 \ m =8 \ 349,6 \ N \cdot 3,5 \ m \\ \Rightarrow \boxed{P_e=29 \ 223,6 \ J}$

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3 years ago

A ball on a string makes 25.0 revolutions in 9.37 s, in a circle radius 0.450 m. What is it’s velocity?

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The velocity of the ball is 12.5 m/s

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The velocity of the ball is given by the ratio between the distance covered by the ball and the time taken:

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First, we calculate the distance covered. We know that the radius of the circle is

r = 0.450 m

And the length of the circumference is

$L=2\pi r = 2\pi(0.750)=4.7 m$

The ball makes 25.0 revolutions, so a total distance of

$d=(25.0)L=(25.0)(4.7)=117.5 m$

In a time of

t = 9.37 s

So, its velocity is

$v=\frac{117.5}{9.37}=12.5 m/s$

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3 years ago

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. A block of mass 4.0 kg on a spring has displacement as a function of time given by, x (t) = (1.0 cm) cos [(2.0 rad/s) t + 0.25

svlad2 [7]

Answer and Explanation:

1. Evaluate the function x(t) at t=0.5

$x(t)=1cos(2t+0.25)\\x(0.5)=cos(2*0.5+0.25)=cos(1+0.25)=cos(1.25)=0.3153223624\approx0.31cm$

2. The period of motion T can be calculated as:

$T=\frac{2\pi}{\omega}$

Where:

$\omega=2rad/s$

So:

$T=\frac{2\pi}{2}=\pi\approx3.14s$

3. The angular frequency can be expressed as:

$\omega=\sqrt{\frac{k}{m} }$

Solving for k:

$k=(\omega)^2*m=(2)^2*4=4*4=16\frac{N}{m}$

4. Find the derivate of x(t):

$\frac{dx}{dt} =v(t)=-2sin(2t+0.25)$

Now, the sine function reach its maximum value at π/2 so:

$2t+025=\frac{\pi}{2}$

Solving for t:

$t=\frac{\frac{\pi}{2} -0.25}{2} =0.6603981634s$

Evaluating v(t) for 0.6603981634:

$v(0.6603981634)=-2sin(2*0.6603981634+0.25)=-2sin(\frac{\pi}{2} )=-2*1=2$

So the maximum speed of the block is:

$v(0.6603981634)=2cm/s=0.02m/s$

In the negative direction of x-axis

5. The force is given by:

$F=kx$

The cosine function reach its maximum value at 2π so:

$2t+0.25=2\pi$

Solving for t:

$t=\frac{2\pi-0.25}{2} =3.016592654s$

Evaluating x(t) for 3.016592654:

$x(3.016592654)=cos(2*3.016592654+0.25)=cos(2\pi)=1cm=0.01m$

Therefore the the maximum force on the block is:

$F=16*0.01=0.16N$

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3 years ago