If two charged objects each have 2.5 C of charge on them and are located 100 m apart, how strong is the electrostatic force between them?

Answer:

The coefficient of kinetic friction on the floor is 0.138

Explanation:

Given;

mass of the crate, m = 450 kg

force applied by the first worker, F₁ = 380 N

force applied by the second worker in the same direction as the first worker, F₁ = 230 N

frictional force opposing the motion of the box = -[tex]F_k[/tex]

Apply Newton's second law of motion;

∑F = ma

[tex]F_1 + F_2 - F_k = ma[/tex]

If the crate slides with constant speed, acceleration is zero (0).

[tex]F_1 + F_2 - F_k = ma = 0\\\\F_1 + F_2 - F_k = 0\\\\F_k = F_1 + F_2\\\\\mu _kmg= F_1 + F_2\\\\\mu _k = \frac{F_1 + F_2}{mg} \\\\\mu _k = \frac{380 + 230}{450 \times 9.8} \\\\\mu _k = 0.138[/tex]

Therefore, the coefficient of kinetic friction on the floor is 0.138

Answer:

a) Ink X is likely to be pure because it only contain 1 spot.

b) The chromatography tell us about ink Y that it is a mixture as it contain more than 1 spot.

c) The three different spots are separated out from ink Y at different heights beacaus different substance have different solubility.

The different spots from Y are found at various heights because they represent different compounds.

The term chromatography has to do with a method of separating the component of a substance. The term chromatography originally means color writing.

We can see that the pure ink is the ink marked X. We can see from the chromatogram that Y is a mixture of colors. The different spots from Y are found at various heights because they represent different compounds.

Learn more about chromatography:https://brainly.com/question/26491567?

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Answer:

doubled

Explanation:

When thrust is double so will the pressure I hope this helps

enjoy

Answer: b force

Explanation:

yes because the world comin g up with more technique

Solution :

The conditions for the maximum in the Young's experiment is :

d sin θ = m λ,     where m = 0, 1, 2, 3, .....

The angle between the central maximum and the 1st order maximum can be determined by setting the m = 1. So,

d sin θ =  λ

[tex]$\theta = \sin^{-1}\left(\frac{\lambda}{d}\right)$[/tex]

Given : d = 100 λ

[tex]$\theta = \sin^{-1}\left(\frac{\lambda}{100 \lambda}\right)$[/tex]

[tex]$\theta = \sin^{-1}\left(\frac{1}{100}\right)$[/tex]

 [tex]$=0.573^\circ$[/tex]

  = 0.01 rad

Answer:

the pearls have an electrical charge induced by contact with the ions of the solution and these charges are attracted by the electrode by a force electric

Explanation:

The pearls are suspended in a solution, when connecting the power source, it is subjected to an electric shock, the pearls have an electrical charge induced by contact with the ions of the solution and these charges are attracted by the electrode by a force electric

            F = q E

Answer:

The resulting force on the first object is 800 N.

Explanation:

Given;

force exerted on one of the objects, F₁ = 400 N

let the first charge = q₁

let the second charge = q₂

The force of repulsion between the objects is calculated using Coulomb's law;

[tex]F =\frac{kq_1q_2}{r^2} \\\\\frac{k}{r^2} = \frac{F}{q_1q_2} = \frac{F_1}{q_1\times2q_2} \\\\F_1 = \frac{F(q_1\times2q_2)}{q_1q_2} \\\\F_1 = 2F\\\\F_1 = 2(400 \ N)\\\\F_1 = 800 \ N[/tex]

Therefore, the resulting force on the first object is 800 N.

Answer:

F = 312 N

Explanation:

Given that,

The mass of a crate, m = 40 kg

Acceleration of the crate, a = 2 m/s²

As the carte is falling downward, the net force exerted by the rope on the carte is given by :

F = m(g-a)

Put all the values,

F = 40(9.8-2)

F = 312 N

Hence, the required force exerted by the rope on the crate is equal to 312 N.

Answer: See explanation

Explanation:

The evolutionary stages for the formation of planets from earliest to latest will be:

1. Dust keeps matter inside the disk cool enough for planet formation to start

2. Dust grains form condensation nuclei on which surrounding atoms condense to form small clumps of matter.

3. Small clumps of matter stick together via the process of accretion to form planetesimals a few hundred kilometers in diameter.

4. Planetesimals begin to accrete, forming protoplanets.

5. A collection of a few planet-sized protoplanets remain in a fairly cleared out disk around the star

Given :

Sally travels by car from one city to another. She drives for 30 min at 80 km/hr, 12.0 min at 105 km/hr, and 45 min at 40.0 km/hr.

To Find :

The total distance traveled.

Solution :

Total distance traveled is given as sum :

Distance = sum of  product of speed and time

D = (v₁ × t₁) + (v₂ × t₂) + (v₃ × t₃)

Putting all given value in this we get :

[tex]D = 80\times \dfrac{30}{60} + 105 \times \dfrac{12}{60} + 40\times \dfrac{45}{60}\\\\D = 91\ km[/tex]

Hence, this is the required solution.

c. a natural process

It is a natural process

Answer:

Option A (69.56 newtons) is the appropriate solution.

Explanation:

According to the question,

On the X-axis,

⇒ [tex]T_1Cos30^{\circ}-T_2Cos60^{\circ}=0[/tex]

or,

    [tex]T_1Cos 30^{\circ}=T_2Cos60^{\circ}[/tex]

On substituting the values, we get

      [tex]T_1\times \frac{\sqrt{3} }{2}=T_2\times \frac{1}{2}[/tex]

      [tex]T_1\times \sqrt{3} =T_2[/tex]....(equation 1)

On the Y-axis,

⇒ [tex]T_1Sin30^{\circ}+T_2Sin60^{\circ}=139.3 \ N[/tex]

                        [tex]\frac{T_1}{2} +\frac{\sqrt{3} }{2} =139.2 \ N[/tex]

                    [tex]T_1+\sqrt{3}T_2=139.2\times 2[/tex]

From equation 1, we get

           [tex]T_1+\sqrt{3}\times \sqrt{3}T_1 =278.4 \ N[/tex]

                        [tex]T_1+3T_1=278.4 \ N[/tex]

                                [tex]4T_1=278.4 \ N[/tex]

                                  [tex]T_1=\frac{278.4}{4}[/tex]

                                       [tex]=69.6 \ N[/tex]  

Answer:

69.58

Explanation:

This question is incomplete, the complete question is;

People who do very detailed work close up, such as jewelers, often can see objects clearly at much closer distance than the normal 25 cm.

a) What is the power in D of the eyes of a woman who can see an object clearly at a distance of only 8.50 cm? (Assume the lens-to-retina distance is 2.00 cm.)

b) What is the size in mm of an image of a 8.00 mm object, such as lettering inside a ring, held at this distance? (Include the sign of the value in your answer.)  __ mm

Answer:

1) the power in D of the eyes of a woman is 61.7647 D

2) the size in mm of an image of a 8.00 mm object is -1.882 mm

Explanation:

Given the data in the question;

a) power in D of the eyes of woman who can see an object clearly at a distance of only 8.5 cm and the lens-to-retina distance is 2.00 cm,

so

u = 8.5 cm = ( 8.5 / 100 )m = 0.085 m

v = 2.00 cm = ( 2 / 100 )m =  0.02 m

Now, we know that power of lens p = 1 / u + 1 / v

so we substitute

p = ( 1 / 0.085 ) + ( 1 / 0.02 )

p = 11.7647 + 50

p = 61.7647 D

Therefore,  the power in D of the eyes of a woman is 61.7647 D

b) What is the size in mm of an image of a 8.00 mm object, such as lettering inside a ring, held at this distance? (Include the sign of the value in your answer.)

we know that;

m = -v / u

we substitute

m = -0.02 / 0.085

m = -0.2353

since H₀ = 8.0 mm

H[tex]_i[/tex] = m × H₀

H[tex]_i[/tex] = -0.2353 × 8.0

H[tex]_i[/tex] = -1.882 mm

the size in mm of an image of a 8.00 mm object is -1.882 mm

Answer:

I dont. understand the question, maybe insert the picture?

Answer:

spirit animal?

Explanation:

Answer:

a = -0.4m/s²

Explanation:

v_f = v_I + (a)(t)

a(t) = v_f-v_I

a = (v_f-v_I)/t

a = (0m/s-12m/s)/30s

a = -0.4m/s²

Answer:

 K = 7.56 10⁻⁴ J

Explanation:

The rotational kinetic energy is

        K = ½ I w²

They ask us for the kinetic energy of the wheel, which can be approximated as a thin ring, its moment of inertia is

         I = M r²

the linear speed of the gerbils is equal to the linear speed of the wheel. The linear and rotational variables are related

          v = w r

          w = v / r

   we substitute

         K = ½ (M r²) v² / r²

         K = ½ M v²

let's calculate

         K = ½ 5 10⁻³ 0.55²

         K = 7.56 10⁻⁴ J

SOLVED DOWN BELOW

Explanation:

In series the same current goes thru both resistors, equiv resistance is 200 ohms, then using ohms law

I = 25/200

I= .125 amps or 125 ma

__________

R= r1 * r2 / r1 +r2

R= 100 * 100 / 100 + 100

R= 10000 / 200

R= 50 ohms

By Newton's second law, the net force on the object is

∑ F = m a

∑ F = (2.00 kg) (8 i + 6 j ) m/s^2 = (16.0 i + 12.0 j ) N

Let f be the unknown force. Then

∑ F = (30.0 i + 16 j ) N + (-12.0 i + 8.0 j ) N + f

=>   f = (-2.0 i - 12.0 j ) N

Answer:

The tension in the horizontal string is 25.2 N.

Explanation:

mass, m = 1.2 kg

Angle, A = 65 degree

The tension in the string is T.

So, T cos A = m g

T x cos 65 =  1.2 x 9.8

T x 0.4226 = 11.76

T = 27.83 N

Tension in the horizontal string, T' = T sin A = 27.83 x sin 65 = 25.22 N

Answer:

Explanation:

See the figure attached . T₁ and T₂ are tension in the inclined and horizontal string .

The vertical component of T₁ will balance weight and horizontal component will balance the tension T₂.

T₁ sin 65 = mg and T₁ cos 65 = T₂

Dividing ,

Tan 65 = mg / T₂

T₂ = mg / tan 65

= 1.2 x 9.8 / 2.1445

= 5.5 N

Answer:

E = 16581.6 J

Explanation:

Given that,

Current, I = 4.9 A

Time for which the current is set up, I = 4.7 min = 282 s

The voltage of the battery, V = 12 V

We need to find how much chemical energy of the battery reduced. Let It is E. We know that,

E = P t

Where

P is power of battery, P = VI

So,

[tex]E=VIt[/tex]

Put all the values,

[tex]E=12\times 4.9\times 282\\E=16581.6\ J[/tex]

So, 16581.6 J of chemical energy of the battery is reduced.

Answer:

Explanation:

From the given information:

Let's assume that the missing function is:

s(t) = t³ - 6t², t ≥ 0

From part (b), we are to find the given  required terms when time t = 2

So; from the function s(t) =  t³ - 6t², t ≥ 0

[tex]velocity \ v(t) \ = \dfrac{d}{dt}s(t)[/tex]

[tex]velocity \ v(t) \ = \dfrac{d}{dt}(t^3 - 6t^2)[/tex]

[tex]velocity \ v(t) \ = 3t^2 - 12t[/tex]

[tex]acceleration a(t) = \dfrac{d}{dt}*v(t)[/tex]

[tex]acceleration a(t) = \dfrac{d}{dt}(3t^2 - 12 t)[/tex]

[tex]acceleration\ a(t) = 6t - 12[/tex]

At time t = 2

The position; S(2) = (2)² - 6(2)²

S(2) = 8 - 6(4)

S(2) = 8 - 24

S(2) = - 16 ft

v(2) = 3(2)² - 12 (2)

v(2) = 3(4) - 24

v(2) = 12 - 24

v(2) = - 12 ft/s

speed = |v(2)|

|v(2)|  = |(-12)|

|v(2)| = 12 ft/s

acceleration = 6t - 12

acceleration = 6(2) - 12

acceleration =  12 - 12

acceleration =  0 ft/s²

Answer:

The sensitivity of the device = 1.234 Hz per km

Explanation:

Given  

Frequency (f) = 10 gHz

Speed of the car = 120 Km/h

As per the doppler’s effect

V = (change in frequency /frequency) *(c/2)

Substituting the given values, we get –  

Change in frequency = {(2*10^9*120)/(3*10^8)} * (1000/3600)

Change in frequency =  37.03 Hz

b) speed of light = wavelength * frequency

3*10^8 = wavelength * 10*10^9

Wavelength = 0.03 m

Sensitivity = change in frequency /wavelength = 37.03/0.03 = 1234 Hz/m

1.234 Hz per km

Answer:

The correct answer is "[tex]0.5\times 10^{-6} \ m[/tex]".

Explanation:

Given:

[tex]\frac{\lambda D}{d} =2.8\times 10^{-3}[/tex]

[tex]d = 0.5\times 10^{-3}[/tex]

[tex]D = 2.80[/tex]

Now,

The wavelength will be:

⇒ [tex]\lambda = 2.8\times 10^{-3}\times \frac{d}{D}[/tex]

By putting the values, we get

⇒    [tex]=\frac{2.8\times 10^{-3}\times 0.5\times 10^{-3}}{2.8}[/tex]

⇒    [tex]=\frac{1.4\times 10^{-6}}{2.8}[/tex]

⇒    [tex]=0.5\times 10^{-6} \ m[/tex]  

Complete Question

The accepted speed of sound at atmospheric pressure and 0 *C is 331.5 m/s. The speed of sound increases 0.607 m/s for every *C. Calculate the speed of sound at the temperature of your room(70F) and compare your measured value to the accepted value.

Answer:

[tex]V_{Tc}=344.314m/s[/tex]

Explanation:

From the question we are told that:

Speed of sound at Temperature [tex]0 \textdegree[/tex] [tex]V_0=331.5m/s[/tex]

Rate of Speed increase [tex]\triangle V_{infty}=0.607[/tex]

Generally the equation for Temperature in Celsius is mathematically given by

 [tex]Tc=\frac{100}{180}(T_f-32)[/tex]

 [tex]Tc=0.56*38[/tex]

 [tex]Tc=21.11 textdegree C[/tex]

Therefore speed at Tc

 [tex]V_{Tc}=V_0+(Tc)( V_{infty})[/tex]

 [tex]V_{Tc}=331.5+(21.11)(0.607)[/tex]

 [tex]V_{Tc}=344.314m/s[/tex]