Question 1 of 10
What is the electric force acting between two charges of -0.0085 C and
-0.0025 C that are 0.0020 m apart?
Use F.- and k - 900-10° N m.cº.
kq,92
A. -4.8 - 1010N
B. -9.6x 107N
C. 9.6 x 10°N
D. 4.8 - 1010 N

The three statements that are correct about the mechanical waves are options B, C, and D.

Mechanical waves are produced due to disturbances in any matter. Which clearly indicates that the mechanical waves cannot travel through empty space.

The mechanical waves require matter to transfer energy through oscillation in the matter. So it is true that the waves transfer energy by causing particles of matter to move.

Also, the mechanical waves need matter to transfer energy, the waves can transfer energy through solids, liquids, and gases.

An example of a mechanical wave is the sound produced by the guitar strings.

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Answer:

quantity of charge that passes through Anita's window AC, Q = 316800 C

Explanation:

given  data

current I = 11 amps

runs t = 8.0 hours

solution

we get here quantity of charge that passes through Anita's window AC

the quantity of charge passing Q = I × t     .................1

Q =  11 A × ( 8 hr )×  ( 3600 s/hr)

so

quantity of charge that passes through Anita's window AC, Q = 316800 C

The quantity of charge that passes through Anita's window AC, will be

Q = 316800 C

The current is defined as the quantity of charge flowing from the wire per unit time. Also defined as the rate of flow of the charge through any circuit. The formula will be given as:

[tex]\rm I=\dfrac{Q}{t}[/tex]

Now it is given that:

current I = 11 amps

runs t = 8.0 hours

The quantity of charge that passes through Anita's window AC during these 8.0 hours. Will be calculated as:

[tex]Q=I\times t[/tex]

[tex]Q=11\times 8\times 3600[/tex]

[tex]Q=316800 \ C[/tex]

Thus the quantity of charge that passes through Anita's window AC, Q = 316800 C

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Answer:

a = 9 [m/s²]

Explanation:

To solve this problem we must use the following equation of kinematics.

[tex]v_{f}^{2}=v_{o}^{2} +2*a*x[/tex]

where:

Vf = final velocity = 30 [m/s]

Vo = initial velocity = 0

a = acceleration [m/s²]

x = displacement = 50 [m]

Now replacing:

[tex]30^{2}=0 +2*a*50\\100*a=900\\a=9[m/s^{2} ][/tex]

Answer:

1.7791 ft^3

Explanation:

The computation of the extra water is as follows;

Here we use the following formula

Volume of the water = ÷ Weight of concrete canoe ÷ value of the weight of the specific water

= 147 lb ÷ 62.4 lb /ft^3

= 2.356 ft^3

Now the volume of water occupied in ultra lightweight

= 36 lb ÷ 62.4 lb /ft^3

= 0.5769 ft^3

Now the water volume displaced is

= 2.356 - 0.5769

= 1.7791 ft^3

Answer:

0.135m/s

Explanation:

Using the law of conservation of momentum

m1u1+m2u2 = m1v1 +m2v2

m1 and m2 are the masses

u1 and u2 are the initial velocities

v1 and v2 are thee final velocities

Substitute the given parameters and get v2

0.015(0.225)+0.03(0.18) = 0.015(0.315)+0.03v2

0.003375+0.0054 = 0.004725 + 0.03v2

0.008775 - 0.004725 = 0.03v2

0.00405 = 0.03v2

v2 = 0.00405/0.03

v2 = 0.135m/s

Hence the final velocity of the 0.03kg marble after collision is 0.135m/s

Answer:

312,497.5Joules

Explanation:

Work done = force × distance

W = FS

Get the force

F = ma

F = 1250×9.8

F = 12250N

Get the distances using the equation of motion

v² = u² +2gS

30² =20²+2(9.8)S

900 =400+19.6S

900-400 =19.6S

500 = 19.6S

S = 500/19.6

S = 25.51m

Get the work done

Work done = 12250×25.51

Workdone = 312,497.5Joules

Answer:

3.13 x 10^5 J

Explanation:

The work done increases the car's kinetic energy so that its new speed is 30 m/s. So, we set t, the net work required to increase the car's speed equal to the change in the car's kinetic energy.

Wnet = △K = Kf - Ki = 1/2mvf^2 - 1/2mvi^2 = 1/2m(vf^2-vi^2) = 1/2(1250kg)((30m/s)^2 - (20m/s)^2) = 3.13x10^5 J

Answer:

x = 50 N

Explanation:

Given that we have a net force, a mass, and acceleration, we can use the fundamental formula for force found in newton's second law which is F = m × a.

Given a mass of 150 kg, and an acceleration 3.0m/s². We can substitute these two values in our formula to calculate the magnitude of these forces or it's net force to identify the unknown force acting on our known force for this situation to work.

_______

F (Net force) = F2 (Second force which we are given) - F1 (First force) = m × a

m (mass which we are given) = 150 kg

a (acceleration which we are given) = 3.0m/s

________

So F = m × a → F2 - F1 = m × a →

500 - F1 = 150 × 3.0 → 500 - F1 = 450 →

-F1 = -50 → F1 = 50

Answer:

11.6km

4.4km in the negative direction

Explanation:

Distance is the total length of path covered and traveled by a body.

So, for this car on a straight line;

  Total distance  = 8km + 3.6km  = 11.6km

Displacement is the distance traveled along a path and the direction it takes.

It is a vector quantity with magnitude and directional attributes.

For this journey;

 Displacement  = 8km  - 3.6km  = 4.4km in the negative direction.

The question is incomplete, here is the complete question:

Which is greater, the energy of one photon of orange light or the energy of one quantum of radiation having a wavelength of [tex]3.36\times 10^{-9}m[/tex]

Answer: The energy of one quantum of radiation having wavelength [tex]3.36\times 10^{-9}m[/tex] is greater than the energy of 1 photon of orange light.

Explanation:

To calculate the energy of one photon, we use the Planck's equation:

[tex]E=\frac{N_Ahc}{\lambda}[/tex]

where,

E = energy of radiation

[tex]N_A[/tex] = Avogadro's number = [tex]6.022\times 10^{23}mol^{-1}[/tex]

h = Planck's constant = [tex]6.626\times 10^{-34}Js[/tex]

c = speed of light = [tex]3\times 10^8 m/s[/tex]

[tex]\lambda}[/tex] = wavelength of radiation

For 1 photon, the term [tex]N_A[/tex] does not appear

[tex]\lambda}[/tex] = 620 nm = [tex]620\times 10^{-9}m[/tex]             (Conversion factor: [tex]1nm=10^{-9}m[/tex] )

Putting values in above equation, we get:

[tex]E=\frac{6.626\times 10^{-34}Js\times 3\times 10^8m/s}{620\times 10^{-9}m}\\\\E=3.206\times 10^{-19}J[/tex]

[tex]\lambda}[/tex] = [tex]3.36\times 10^{-9}m[/tex]

Putting values in above equation, we get:

[tex]E=\frac{6.022\times 10^{23}mol^{-1}\times 6.626\times 10^{-34}Js\times 3\times 10^8m/s}{3.36\times 10^{-9}m}\\\\E=3.56\times 10^{7}J/mol[/tex]

Hence, the energy of one quantum of radiation having wavelength [tex]3.36\times 10^{-9}m[/tex] is greater than the energy of 1 photon of orange light.

Answer:

171.5m

Explanation:

The velocity of sound in water = 343m/s

Time taken = 1.00secs

using the formula to calculate the distance

2x = vt

x is the distance

v is the speed of sound

t is the time

x = vt/2

x = 343(1)/2

x = 171.5m

hence their separation 1.00 s after the second object is released is 171.5m

Answer:

[tex]PK_T = constant[/tex]  

Explanation:

[tex]PK_T[/tex]  = constant

terminal velocity is the maximum const. velocity to achieved by the partical when particle is free fall with the effect of gravity

so that here

eq of motion is

Fnet = mg - kv     ................1

here a = 0

v = terminal velocity

0 = mg - kV

v = [tex]\frac{mg}{k}[/tex]  m/s

Answer:

The force remains the same.

Explanation:

Let the magnitude of the forces in each case be F1 and F2 respectively.

The charges are Q1 and Q2.

The distance of separation between them is R.

Hence, for F1;

F1 = KQ1Q2/R^2

For F2:

F2 = K * 3Q1 * 3Q2/(3R)^2

F2 = 9KQ1Q2/9R^2

F2 =KQ1Q2/R^2

Hence F1=F2

The force is the same in both cases!

Answer:

true

true

false

true

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Answer:

154° at 195 km/h

Explanation:

The helicopter is moving south at 175 km/h, relative to the wind.

But the wind is moving east at 85 km/h, relative to the ground.

This means that the helicopter is moving south east relative to the ground.

Every hour, the helicopter will move 175 km to the south and 85 km to the east, relative to the ground.

This means that we can determine the speed and direction of the helicopter using a right triangle and simple trigonometry.

Refer to the triangle b1.

The distance traveled by the helicopter in 1 hour is denoted by d.

d is the hypotenuse of the right triangle.

Using the Pythagorean Theorem, we can calculate d to be 195 km (rounded to 3 s. f.)

Hence the helicopter is traveling at 195 km/h relative to the ground.

To calculate the direction we use,

tan (x) = opposite/adjacent = 85/175

So the angle x is,

[tex]x = arctan (\frac{85}{175} )[/tex] = 25.9°

Relative to the North, the helicopter is moving at 180° - 25.9° = 154° (rounded to 3 s. f.)

The work done is 480 Joules.

Option 3 is correct.

The work done  is computed as multiplication of force and displacement.

Given that,  Force = 20 N

Displacement = 4 meter per second

For 6 second,

Displacement [tex]=4*6=24m[/tex]

                  [tex]Workdone=Force*displacement\\\\Workdone=20*24=480J[/tex]

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Answer: See explanation

Explanation:

The density would like be calculated by dividing the mass by volume. This will be:

Density = Mass / Volume

Density = 390 / 50

Density = 7.8g/cm

Answer:

my assuytyyhyyyyyy gftdrrdtiifyb tvhyvth rv b yy

Answer:

There is no work done.

Explanation:

Given the following data;

Mass = 10 kg

To find the work done?

In Physics, work done can be defined as the amount of energy transfered when an object or body is moved over a distance due to the action of an external force.

Mathematically, work done is given by the formula;

Work done = force * distance

[tex] W = F * d[/tex]

Where,

However, the weight suspended in the air by a strong cable does not move or experience any form of displacement. Therefore, there is no work done.

Answer:

The combination of Earth's elliptical orbit and the tilt of its axis results in the Sun taking different paths across the sky at slightly different speeds each day. This gives us different sunrise and sunset times each day.

Explanation:

Answer:

The answer is below

Explanation:

a) What are the three longest wavelengths for standing waves on a 270-cm-long string that is fixed at both ends? b. If the frequency of the second-largest wavelength is 50.0 Hz, what is the frequency of the third-longest wave length?

Solution:

a) The wavelengths (λ) for standing waves is given by the formula:

[tex]\lambda_m=\frac{2*length\ of\ string}{m}\\\\Where\ m=1,2,3,.\ .\ .\\\\Given\ that\ length\ of\ string = 270\ cm=2.7\ m,\ m=1,2,3(three\ longest\ wavelengths)\\\\Hence:\\\\\lambda_1=\frac{2(2.7)}{1}=5.4\ m\\\\\lambda_2=\frac{2(2.7)}{2}=2.7\ m \\\\\lambda_3=\frac{2(2.7)}{3}=1.8\ m[/tex]

b) The frequency (f) and wavelength (λ) is given by:

fλ = constant

Hence:

[tex]f_2\lambda_2=f_3\lambda_3\\\\f_2=50\ Hz\\\\2.7*50=f_3(1.8)\\\\f_3=\frac{2.7*50}{1.8} \\\\f_3=75\ Hz[/tex]

The three longest wavelengths for the standing waves on a 270-cm long string that is fixed at both ends are:

1. 5.4 meters.

2. 2.7 meters.

3. 1.8 meters.

Given the following data:

To determine the three (3) longest wavelengths for these standing waves:

Mathematically, the wavelength for standing waves is given by the formula:

[tex]\lambda_n = \frac{2L}{n}[/tex]

Where:

Note: n = 1, 2, and 3.

When n = 1:

[tex]\lambda_1 = \frac{2\times 2.7}{1} \\\\\lambda_1 = 5.4 \;meters[/tex]

When n = 2:

[tex]\lambda_2 = \frac{2\times 2.7}{2} \\\\\lambda_2 = 2.7 \;meters[/tex]

When n = 3:

[tex]\lambda_3 = \frac{2\times 2.7}{3} \\\\\lambda_3 =\frac{5.4}{3} \\\\\lambda_3 = 1.8 \;meters[/tex]

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Answer:

2.87m

Explanation:

Using the law of gravitation to solve this question

F = GMm/r²

G is the gravitational constant

M and m are the masses

r is the distance between the masses

Substitute the given values

G = 6.67×10^-11 m³/kgs²

M =8.8 x 10^6 kg

m = 5.6 x 10^5 kg

F =440N

400 = 6.67×10^-11×8.8 x 10^6 ×5.6 x 10^5/r²

400r² = 328.698×10

400r² = 3286.98

r² = 3286.98/400

r² = 8.21745

r = √8.21745

r = 2.87m

Hence the distance of separation is 2.87m

The distance of separation will be "0.86 m".

Given:

Mass,

Force,

As we know,

→   [tex]F = \frac{Gm_1 m_2}{r^2}[/tex]

By putting the values, we get

→ [tex]440=\frac{6.67\times 10^{-11}\times 5.6\times 10^5\times 8.8\times 10^6}{r^2}[/tex]

→   [tex]r^2 = \frac{6.67\times 5.6\times 8.8}{440}[/tex]

→   [tex]r^2 = 0.74704[/tex]

→     [tex]r = \sqrt{0.74704}[/tex]

→        [tex]= 0.86 \ m[/tex]

Thus above approach is correct.        

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Answer:

The resistance of a wire is directly proportional to its length and inversely proportional to its cross-sectional area

Explanation:

Answer:

The shortest time in which the driver can complete the operation is approximately 6.32456 seconds

Explanation:

The given parameters are;

The speed of the truck and the car = 35 mi/h ≈ 51.33 ft./s

Let "t" represent the time it takes the car to pass the truck by 40 ft., we have;

The distance covered by the truck = 35 mi/h × t

The distance covered by the car = 35 mi/h × t + 80 ft = 51.33

Let the distance over which the car accelerates = d

We have;

d = 51.33×t₁ + 1/2×5×t₁²

51.33·t + 80 - d = (51.33 + 5·t₁)·(t - t₁) - 1/2·20·(t - t₁)²

51.33 = 51.33 + 5·t₁ - 20·(t - t₁)

∴ 5·t₁ = 20·(t - t₁)

5·t₁ = 20·t - 20·t₁

25·t₁ = 20·t

t₁ = 4·t/5

(51.33 + 5·t₁)·(t - t₁) - 1/2·20·(t - t₁)² + 51.33×t₁ + 1/2×5×t₁² = 51.33·t + 80

We get;

(51.33 + 5·(4·t/5))·(t - (4·t/5)) - 1/2·20·(t - (4·t/5))² + 51.33×(4·t/5) + 1/2×5×(4·t/5)² = 51.33·t + 80

(51.33 + 4·t)·(t/5) - 2·t²/5 + 51.33 × (4·t/5) + 8·t²/5 = 51.33·t + 80

51.33×t/5 + 51.33×4·t/5 + 4·t²/5 - 2·t²/5 + 8·t²/5 = 51.33·t + 80

51.33·t + 10·t²/5 = 51.33·t + 80

2·t² = 51.33·t + 80 - 51.33·t = 80

t² = 80/2 = 40

t = √40 = 2·√10 ≈ 6.32456

The shortest time in which the driver can complete the operation is "t" ≈ 6.32456 seconds

Answer:heat brings it up then down

Explanation:

Answer:

1.625g

Explanation:

The half - life of a radioactive substance is the time taken for half of it to decay to half of its original composition.

 For the given specie, the original composition is 26g  and the half life is 15hrs:

                At time 0hrs                    Mass  = 26g

For the first half life at time 15hrs         Mass  = 13g

                                             30hrs        Mass  = 6.5g

                                              45hrs        Mass = 3.25g

                                              60hrs        Mass  = 1.625g

Answer:

All objects, irrespective of their mass, experience the same acceleration g when falling freely under the influence of gravity at the same point on the Earth. If gravity is the only force acting on an object, the sum of kinetic energy and gravitational energy is constant. ...

Answer:

Gravity, also called gravitation, in mechanics, the universal force of attraction acting between all matter. ... On Earth all bodies have a weight, or downward force of gravity, proportional to their mass, which Earth's mass exerts on them. Gravity is measured by the acceleration that it gives to freely falling objects.The Earth's gravitational force accelerates objects when they fall. It constantly pulls, and the objects constantly speed up.Gravity is the force of attraction between two objects with mass and is dependent on the distance between these objects. Gravity is the force that repels two objects that have opposite charges. It is dependent upon the charges of the objects.

Explanation:

Answer:

680 Kg.m/s

Explanation:

Mass of player; m_p = 105 kg.

Speed of player before Collision; v_pi = 8.5 m/s

Mass of referee; m_r = 85 kg

Speed of referee before collision; v_ri = 3.5 m/s

Speed of referee after collision; v_rf = 6 m/s

From conservation of momentum,

Initial momentum = final momentum

Thus;

(m_p × v_pi) + (m_r × v_ri) = (m_p × v_pf) + (m_r × v_rf)

Where (m_p × v_pf) is the momentum of the player after collision.

Thus, Plugging in the relevant values, we have;

(105 × 8.5) + (85 × 3.5) = (m_p × v_pf) + (85 × 6)

(m_p × v_pf) = (105 × 8.5) + (85 × 3.5) - (85 × 6)

(m_p × v_pf) = 680 Kg.m/s

Answer:

C. 36J

Explanation:

KE = ½ m V²

= ½ × 8 × 3²

= 4 × 9

= 36 J