Two particles are 15 meters apart.
Particle A has a charge of 6.0*10^-4 C

Particle B has a charge of 5.0*10^-4 C.

The resulting Coulomb force is 12 N. At the same distance, what combination of charges would yield the same Coulomb force?

A. Particle A: 8.0*10^-4 C, and Particle B: 3.0*10^-4 C

B. Particle A: 12.0*10^-4 C, and Particle B: 2.5*10^-4 C

C. Particle A: 9.0*10^-4 C, and Particle B: 8.0*10^-4 C

D. Particle A: 9.0*10^-4 C, and Particle B: 3.0*10^-4 C

Answer:All magnets have north and south poles. Opposite poles are attracted to each other, while the same poles repel each other. When you rub a piece of iron along a magnet, the north-seeking poles of the atoms in the iron line up in the same direction. The force generated by the aligned atoms creates a magnetic field.

Explanation: I did this exact question before

Answer:

As the cars ascend the next hill, some kinetic energy is transformed back into potential energy. Then, when the cars descend this hill, potential energy is again changed to kinetic energy. This conversion between potential and kinetic energy continues throughout the ride.

Explanation:

hope it helps U

Answer:

because when you shape glass it's not a shiny as diamond

Answer:

300 m/s

Explanation:

2d = vt

v = 2d/t

v = 2×90/.3

v=300 m/s

d = distance

t = time

v = velocity/speed of sound

Explanation:

for part A:

work done is equal to change in potential energy which is

MgH

so the answer is 1.9 × 1 × 9.81

ANSWER FOR PART A: 18.639

FOR PART B:

power = flow rate x gh

= 5×9.81×1.9

93.195 watts

Answer: I am so sorry i hadn't learned this Yet~!

Explanation:

The electric potential V(z) on the z-axis is :  V = [tex](\frac{Q}{a^2} ) [ (a^2 + z^2)^{\frac{1}{2} } -z[/tex]

The magnitude of the electric field on the z axis is : E = kб 2[tex]\pi[/tex]( 1 - [z / √(z² + a² ) ] )

Given data :

V(z) =2kQ / a²(v(a² + z²) ) -z  

Considering a disk with radius R

Charge = dq

Also the distance from the edge to the point on the z-axis = √ [R² + z²].

The surface charge density of the disk ( б ) = dq / dA

Small element charge dq =  б( 2πR ) dr

dV  [tex]\frac{k.dq}{\sqrt{R^2+z^2} } \\\\= \frac{k(\alpha (2\pi R)dR}{\sqrt{R^2+z^2} }[/tex]  ----- ( 1 )

Integrating equation ( 1 ) over for full radius of a

∫dv = [tex]\int\limits^a_o {\frac{k(\alpha (2\pi R)dR)}{\sqrt{R^2+z^2} } } \,[/tex]

 V = [tex]\pi k\alpha [ (a^2+z^2)^\frac{1}{2} -z ][/tex]

     = [tex]\pi k (\frac{Q}{\pi \alpha ^2})[(a^2 +z^2)^{\frac{1}{2} } -z ][/tex]

Therefore the electric potential V(z) = [tex](\frac{Q}{a^2} ) [ (a^2 + z^2)^{\frac{1}{2} } -z[/tex]

Also

The magnitude of the electric field on the z axis is : E = kб 2[tex]\pi[/tex]( 1 - [z / √(z² + a² ) ] )

Hence we can conclude that the answers to your question are as listed above.

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When you rub a dry cloth across glass, it creates charged static electricity, which attracts little non-charged dust particles.

Explanation: First fix your units. acceleration is in m/s/s.

T-Mg=Ma

M=348.1+169.2 =517.3 kg

T= 517.3 (9.81+4.11)

I hope this helps

Hypotenuse be H

Apply Pythagorean theorem

[tex]\\ \sf\Rrightarrow H^2=P^2+B^2[/tex]

[tex]\\ \sf\Rrightarrow H^2=5^2+9^2[/tex]

[tex]\\ \sf\Rrightarrow H^2=25+81[/tex]

[tex]\\ \sf\Rrightarrow H^2=106[/tex]

[tex]\\ \sf\Rrightarrow H=\sqrt{106}cm[/tex]

The athlete can improve performance by building strength, coordination and balance.

This is an individual who is proficient in sports and other forms of physical exercise.

Improvement of performance based on the kinematic variable of speed and acceleration can be achieved by building strength, coordination and balance by performing plyometric exercises etc.

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Answer:

If the lifting force varies jointly with wing surface area and the square of the plane's velocity, then:

F = kAv²

If the lift is 27,800 pounds when the area is 190 sq. ft and the velocity is 230 mph, then the constant of proportionality is:

27,800 = k(190)(230)²

k = (27,800) / (190)(230)² = 139/50255

The lifting force the plane slows down to 200 mph is:

F = (139/50255)(190)(200)²

F = 21,020.79

Rounded to the nearest pound, the lifting force is 21,021 pounds.

(a) The time of flight be for a tennis ball launched by the ball machine is 0.19 s.

(b) The range of the tennis ball be is 1.01 m.

(c)  The final velocity of the ball with which it reaches the ground is 7.16 m/s.

The time of flight of the tennis ball is calculated as follows;

h = vt + ¹/₂gt²

1.2 = 5.3t + 0.5(9.8)t²

1.2 = 5.3t + 4.9t²

4.9t² + 5.3t - 1.2 = 0

a = 4.9, b = 5.3, c = 1.2

solve using quadratic formula

t = 0.19 s

Thus, the time of flight be for a tennis ball launched by the ball machine is 0.19 s.

The range of the tennis ball is calculated as follows;

R = vt

R = 5.3 x 0.19

R = 1.01 m

The final velocity of the ball with which it reaches the ground is calculated as follows;

vf = vo + gt

vf = 5.3 + 9.8(0.19)

vf = 7.16 m/s

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Now

[tex]\\ \sf\rightarrowtail Power=\dfrac{Work\:done}{Time\:taken}[/tex]

[tex]\\ \sf\rightarrowtail Power=\dfrac{250}{1}[/tex]

[tex]\\ \sf\rightarrowtail Power=250W[/tex]

Answer:

Explanation:

The hot exhaust from a rocket travels in one direction, and the rocket travels in the opposite direction. This is an example of?

A. Equal and opposite forces.

B. Friction.

C. Balanced forces.

D. Inertia.

Answer:

Because you have a greater understanding of the mentality and conduct of individuals you care about, you are better able to understand their perspective and are more likely to recognize possible situational factors for their behavior.

The magnitude of magnitude force on the proton is 2.4 x 10⁻¹² N.

The magnitude of magnitude force on the proton is calculated as follows;

F = qvB sinθ

where;

Substitute the given parameters and solve for the magnetic force.

F = (1.6 x 10⁻¹⁹) x (3 x 10⁶) x (5) X(sin90)

F = 2.4 x 10⁻¹² N

Thus, the magnitude of magnitude force on the proton is 2.4 x 10⁻¹² N.

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The  force required to pull one of the microscope sliding at a constant speed of 0.28 m/s relative to the other is zero.

The force required to pull one of the microscope sliding at a constant speed of 0.28 m/s relative to the other is determined by applying Newton's second law of motion as shown below;

F = ma

where;

At a constant speed, the acceleration of the object will be zero.

F = m x 0

F = 0

Thus, the  force required to pull one of the microscope sliding at a constant speed of 0.28 m/s relative to the other is zero.

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Answer:

[tex]\displaystyle \sqrt{\frac{(5/2)\, (g - a)\, m\, R^{2}}{M^{2}\, a}}[/tex], assuming that the tension in the rope is the only tangential force on the sphere ([tex]g[/tex] denote the gravitational acceleration.)

Explanation:

The forces on the bucket are:

Since the weight of the bucket and the tension from the rope are in opposite directions, the magnitude of the net force would be:

[tex]\begin{aligned} \|\text{Net Force}\| =\; & \|\text{Weight}\| - \|\text{Tension}\| \end{aligned}[/tex].

The upward tension in the rope prevents the bucket from accelerating at [tex]g[/tex] (free fall.) Rather, the bucket is accelerating at an acceleration of only [tex]a[/tex]. The net force on the bucket would be thus [tex]m\, a[/tex].

Rearrange the equation for the net force on the bucket to find the magnitude of the tension in the rope would be:

[tex]\begin{aligned} & \|\text{Tension}\| \\ =\; & \|\text{Weight}\| - \|\text{Net Force}\| \\ =\; & m\, g - m\, a \\ =\; & (g - a)\, m\end{aligned}[/tex].

At a distance of [tex]R[/tex] from the center of the sphere, the tension in the rope [tex](g - a)\, m[/tex] would exert a torque of [tex](g - a)\, m\, R[/tex] on the sphere. If this tension is the only tangential force on this sphere, the net torque on the sphere would be [tex](g - a)\, m\, R\![/tex].

Let [tex]M[/tex] denote the mass of this sphere. The moment of inertia of this filled sphere would be [tex]I = (2/5)\, M^{2}[/tex].

Therefore, the magnitude of the angular acceleration of this sphere would be:

[tex]\begin{aligned}& \|\text{Angular Acceleration}\| \\ =\; & \frac{\|\text{Net Torque}\|}{(\text{Moment of Inertia})} \\ =\; & \frac{(g - a)\, m\, R}{(2/5)\, M^{2}} \end{aligned}[/tex].

The bucket is accelerating at a magnutide of [tex]a[/tex] downwards. The rope around the sphere need to unroll at an acceleration of the same magnitude, [tex]a\![/tex]. The tangential acceleration of the sphere at the surface would also need to be [tex]\! a[/tex].

Since the surface of the sphere is at a distance of [tex]R[/tex] from the center, the angular acceleration of this sphere would be [tex](a / R)[/tex].

Hence the equation:

[tex]\begin{aligned}& \frac{(g - a)\, m\, R^{2}}{(2/5)\, M^{2}} = \|\text{Angular Acceleration}\| = \frac{a}{R} \end{aligned}[/tex].

Solve this equation for [tex]M[/tex], the mass of this sphere:

[tex]\begin{aligned}& \frac{(g - a)\, m\, R^{2}}{(2/5)\, M^{2}} = \frac{a}{R} \end{aligned}[/tex].

[tex]\begin{aligned}M^{2} &= \frac{(g - a)\, m\, R^{2}}{(2/5)\, a} \\ &= \frac{(5/2)\, (g - a)\, m\, R^{2}}{a}\end{aligned}[/tex].

[tex]\begin{aligned}M&= \sqrt{\frac{(5/2)\, (g - a)\, m\, R^{2}}{a}}\end{aligned}[/tex].

(a) The force the ground exerts on each set of rear wheels when the plane is at rest on the runway is 0.743 MN.

(b) The force the ground exerts on the front set of wheels is 0.239 MN.

The concept of center mass of an object can be used to dtermine the mass distribution of the airplane along the line through the center.

Position of the center mass of the plane is calculated as follows;

Let the position of the center mass, Xcm = y

the center mass is 3 m in front of rear wheels, that is

21.7 - y = 3

y = 21.7 - 3

y = 18.7 m

Xcm = 18.7 m

Let the mass of the plane at front wheels = M1

Let the mass of the plane at rear wheels = M2

[tex]X_{cm} = \frac{M_1x_1 + M_2x_2}{M_1 + M_2}[/tex]

[tex]18.7 = \frac{M_1(0) + M_2(21.7)}{177000} \\\\3,309,900 = 21.7M_2\\\\M_2 = \frac{3,309,900}{21.7} \\\\M_2 = 152,529.95 \ kg[/tex]

There are two rear wheels, and the force exerted on each wheel due to mass of the airplane at this position is calculated as follows;

[tex]W = mg\\\\W_1 = W_2 = \frac{1}{2} (mg) = \frac{1}{2} (152,529.95 \times 9.8) = 743,396.76 \ N= 0.743 \ MN[/tex]

M1 + M2 = 177,000

M1 = 177,000 - M2

M1 = 177,000 - 152,529.95

M1 = 24,470.05 kg

W = mg

W = 24,470.05 x 9.8

W = 239,806.5 N = 0.239 MN

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The lifting force when a child pushes a toy truck up an inclined plane is  equal to the total amount of work done.

The work-energy theroem states, the work done on an object is equal to the change in mechanical energy of the object.

The net force applied by the child on the truck, will create a lifting force on the object.

Fd - Ffd = k.E

Where;

Thus, the lifting force when a child pushes a toy truck up an inclined plane is  equal to the total amount of work done.

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Answer:

Newton

Explanation:

The SI unit for weight is newton.

(a) The initial energy of the mass is 76.485 J.

(b) The compression of the spring in the absence of friction is 21.2 cm.

(c)  The compression of the spring in the presence of friction is 19.4 cm.

The initial energy of the mass is determined as follows;

E = K.E + P.E

E = ¹/₂mv² + mgh

where;

E = ¹/₂mv²  + mg x Lsinθ

P.E = ¹/₂(9.4)(2.5)² +  (9.4)(9.8)(0.87)(sin36)

P.E = 76.485 J

The compression of the spring in the absence of friction is calculated as follows;

Ux = E

¹/₂kx² = 76.485

kx² = 2(76.485)

x² = (2 x 76.485)/k

x = √(2 x 76.485)/k

x = √(2 x 76.485 / 3413.7)

x = 0.212 m

x = 21.2 cm

The compression of the spring in the presence of friction is determined by applying the principle of conservation of energy.

E - Fd = Ux

E - μmgd = ¹/₂kx²

76.11 - (0.27 x 9.4 x 9.8 x 0.48) = ¹/₂(3413)x²

64.17 = 1706.5x²

x² = 0.0376

x = √0.0376

x = 0.194

x = 19.4 cm

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We need evidence to be able to continually test theories about the composition and origin of our solar system and galaxies in order to verify the theories and to make laws.

A theory is an explanation for observations of the natural world that has been carried by using the scientific method, and which brings together many facts.

So we can conclude that we need evidence to test theories about the composition and origin of our solar system in order to verify the theories and to make laws.

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Answer:

Gravitational force is the force that attracts objects towards each other. Two factors that affect the gravitational force between objects are the mass of the two objects and the distance between

Explanation:

Gravity is what pulls us towards the Earth if we were to jump into the air, so it is the force that pulls things towards other things. The bigger the objects are the more gravity they have, so a planet has more gravity than say, an apple. Distance between objects also makes their gravity change, so the Earth's pull on the moon is different than the Earth's pull on the sun.

Hopefully this helps- let me know if you have any questions!

Describe gravitational force in your own words. Which two factors affect the gravitational force between two objects?

[tex]F = \frac{G m{ \tiny1}m{ \tiny2} }{ {r}^{2} } [/tex]

where

Answer:

what is heat and transfer

Answer:

The articles appearing under "Milestones in Physics" will give an insight into special events or situations that have been decisive for the evolution of Physics

Answer:

We have invented many new ways to get energy, especially renewable means of energy, and we have made many new carbon-neutral devices more widespread and cheaper.

Modern physics milestones are getting harder and harder to get because almost every element has been discovered, many gaseous and liquid substances have been experimented with and dozens of other mid 20th century discoveries.

Answer:

16 reflection,incidence

If a person generates about 412005 joules of heat from the body,  the water temperature is mathematically given as

t=21.6296C

Question Parameter(s):

The person is submerged neck-deep into a tub containing 2124 kg of water at 20.9 °C

Generally, the equation for the Heat   is mathematically given as

Heat gained =Heat loess

Thereofore

mw*cw*(t-2160)=1.5*10^5

[tex]t=21.60+\frac{1.5*10^5}{mw*Cw}\\\\t=21.60+\frac{1.5*10^5}{1.2*10^3*4186}[/tex]

t=21.6296C

In conclusion, the tempreature

t=21.6296C

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