Two particles are separated by 0.38 m and have charges of -6.25 x 10-9C
and 2.91 x 10-9 C. Use Coulomb's law to predict the force between the
particles if the distance is doubled. The equation for Coulomb's law is
Fe = kq92, and the constant, k, equals 9.00 x 10°N-m/c2.
A. -2.83 x 10-7N
B. 2.83 x 10-7N
C. -1.13 x 10-6N
D. 1.13 x 10-6N

Answer:

Salt melts ice and helps keep water from re-freezing by lowering the freezing point of water. This phenomenon is called freezing point depression. Salt only helps if there is a little bit of liquid water available. The salt has to dissolve into its ions in order to work.

Explanation:

Answer:

The head-to-tail method is a graphical way to add vectors, The tail of the vector is the starting point of the vector, and the head (or tip) of a vector is the final, pointed end of the arrow.

Explanation:

Answer:

B

Explanation:

To solve this we must be knowing each and every concept related to  electrical energy.  Therefore, the correct option is option B among all the given options.        

The work done by an electric charge is referred to as electrical energy. For time t seconds, if electricity I ampere passes throughout a conductor or any other conducting element with a potential differential v volts across it.

The kilowatt hour is both a practical as well as an economic unit of electrical energy. The basic commercial unit is the watt-hour, one and kilowatt hour equals 1000 watt hours. Electric supply providers charge their customers every kilowatt hour unit of electricity used. This kilowatt hour seems to be a BOT unit, or board of trade unit.  The electrical energy that does not become light energy, the lightbulb transforms it into thermal energy.

Therefore, the correct option is option B among all the given options.        

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Answer:

[tex]F = 5.4*10^{-8}\ N[/tex]

Explanation:

Given

Represent the mass of the student with M and the mass of the slice of pizza with m

[tex]M = 92kg[/tex]

[tex]m = 550g[/tex]

[tex]d = 25cm[/tex]

Required

Determine the force of attraction

This is calculated as:

[tex]F = \frac{GMm}{d^2}[/tex]

Where G = gravitational constant

[tex]G = 6.67408 * 10^{-11}\ m^3 kg^{-1} s^{-2}[/tex]

Convert both mass to kilogram and distance to metre

[tex]m = 550g[/tex]

[tex]m = 550kg/1000[/tex]

[tex]m = 0.55kg[/tex]

[tex]d = 25cm[/tex]

[tex]d = 25m/100[/tex]

[tex]d = 0.25m[/tex]

Substitute these values in [tex]F = \frac{GMm}{d^2}[/tex]

[tex]F = \frac{6.67408 * 10^{-11} * 92 * 0.55}{0.25^2}[/tex]

[tex]F = \frac{6.67408 * 92 * 0.55* 10^{-11} }{0.25^2}[/tex]

[tex]F = \frac{337.708448* 10^{-11} }{0.0625}[/tex]

[tex]F = 5403.335168* 10^{-11}[/tex]

[tex]F = 5.403335168* 10^3*10^{-11}[/tex]

[tex]F = 5.403335168*10^{3-11}[/tex]

[tex]F = 5.403335168*10^{-8}[/tex]

[tex]F = 5.4*10^{-8}\ N[/tex]

62×3.0

think so not sure

Answer:

When traveling behind other vehicles, there should be at least a four second space between your vehicles. When the car in front of you passes a stationary object, slowly count to yourself. If you pass the object before the allotted time, you should back off. When traveling at night or inclement weather, these times should be doubled.

Don't talk on a cell phone while driving. Phones detract from your ability to concentrate on the road and increase your chance of a collision by nearly 400%. If you must use the phone, pull over to a safe, well-lit parking lot and place your call there. After completing your call you may continue on your way.bey all speed limits and signs.

Answer:

Explanation:

horizontal component of velocity of throw = 20 cos20 = 18.8 m /s

vertical downwards component = 20 sin20 = 6.84 m /s

time to displace by height 30 m = t  , initial velocity u = 6.84 m /s

h = ut + 1/2 gt²

30 = 6.84 t + .5 x 9.8 t²

4.9 t² + 6.84 t - 30 = 0

t = - 6.84 ±√( 6.84² + 4 x 4.9 x 30 ) / 2x 4.9

=  - 6.84 ±√( 46.78 + 588 ) / 9.8

=  - 6.84 ±√(634.78 ) / 9.8

= - 6.84 ±25.2 / 9.8

= 1.87 s

horizontal displacement in 1.87 s

= 18.8 x 1.87

= 35.15 m .

Answer:

a.  -4.166 J/K

b. 8.37 J/K

c. 4.21 J/K

d. entropy always increases.

Explanation:

Given :

Temperature at hot reservoir , [tex]$T_h$[/tex] = 720 K

Temperature at cold reservoir , [tex]$T_c$[/tex] = 358 K

Transfer of heat, dQ = 3.00 kJ = 3000 J

(a). In the hot reservoir, the change of entropy is given by:

[tex]$dS_h= -\frac{dQ}{t_h}$[/tex]              (the negative sign shows the loss of heat)

[tex]$dS_h= -\frac{3000}{720}$[/tex]

      =  -4.166 J/K

(b)  In the cold reservoir, the change of entropy is given by:

[tex]$dS_c= \frac{dQ}{t_c}$[/tex]              

[tex]$dS_c= \frac{3000}{358}$[/tex]

      =  8.37 J/K

(c). The entropy change in the universe is given by:

[tex]$dS=dS_h+dS_c$[/tex]

    = -4.16+8.37

   = 4.21 J/K

(d). According to the concept of entropy, the entropy of the universe is always increasing and never decreasing for an irreversible process. If the entropy of universe decreases, it violates the laws of thermodynamics. Hence, in part (c), the result have to be positive.

Answer:

The ratio is  [tex]\frac{B_1}{B_2} = 1.265[/tex]

Explanation:

From the question we are told that

   The density of fresh water is  [tex]\rho__{f}} = 998 \ kg/m^3[/tex]

     The density of ethanol is  [tex]\rho_{e} = 789 \ kg /m^3[/tex]

Generally speed of a wave in a substance is mathematically represented as

    [tex]v = \sqrt{\frac{B}{\rho} }[/tex]

Here B is the adiabatic bulk modulus of the substance  while [tex]\rho[/tex]  is the density of the substance

So at constant wave speed

     [tex]\sqrt{\frac{B_1}{\rho_1} } = \sqrt{\frac{B_2}{\rho_2} }[/tex]

=>   [tex]\frac{B_1}{\rho_1} = \frac{B_2}{\rho_2}[/tex]

=>   [tex]B_1 \rho_2 = B_2\rho_1[/tex]    

=>  [tex]\frac{B_1}{B_2} = \frac{\rho_1}{\rho_2}[/tex]

Here  [tex]\rho_1 =\rho__{f}} = 998 \ kg/m^3[/tex]   and  [tex]\rho_2 = \rho_{e} = 789 \ kg /m^3[/tex]

So

  =>  [tex]\frac{B_1}{B_2} = \frac{998}{789}[/tex]    

  =>  [tex]\frac{B_1}{B_2} = 1.265[/tex]    

Answer:

Explanation:

Step one:

given data

mass = 7kg

acceleration =2m/s^2

time= 9seconds

acceleration = velocity/time

velocity= acceleration *time

velocity=2*9

velocity= 18m/s

distance moved= velocity* time

distance= 18*9

distance=162m

we also know that the force on impulse is given as

Ft=mv

F=mv/t

F=7*18/9

F=126/9

F=14N

work done = Force* distance

work done=14*162

work=2268Joules

work= 2.27kJ

Answer:

However, those astronauts would experience a second spectacle: A solar eclipse caused by the Earth – the Sun disappearing behind the dark disc of the Earth. When Earth inhabitants witness a lunar eclipse, Moon inhabitants would, simultaneously be witnessing a solar eclipse.

Answer:

a = 11.03 [m/s²]

Explanation:

To solve this problem we must use Newton's second law which tells us that the sum of forces is equal to the product of mass by acceleration.

ΣF = m*a

where:

F = force = 160000 [N]

m = mass = 14500 [kg]

a = acceleration [m/s²]

160000 = 14500*a

a = 11.03 [m/s²]

the emitter of x-rays rotates around the patient and the detector, placed in diametrically opposite side, picks up the image of a body section (beam and detector move in synchrony)

i believe that should be ur answer =) good luck !

Answer:

Two ways to determine ones muscular strength is to either lift the heaviest weight possible or by doing one repetition. The advantages these two methods is that it is easy to determine and does not require calculations or estimating.

Explanation:

Answer:

Two ways to determine ones muscular strength is to either lift the heaviest weight possible or by doing one repetition. The advantages these two methods is that it is easy to determine and does not require calculations or estimating.

Explanation:

Answer:

h' = 55.3 m

Explanation:

First, we analyze the horizontal motion of the projectile, to find the time taken by the arrow to reach the orange. Since, air friction is negligible, therefore, the motion shall be uniform:

s = vt

where,

s = horizontal distance between arrow and orange = 60 m

v = initial horizontal speed of the arrow = v₀ Cos θ

θ = launch angle = 30°

v₀ = launch speed = 35 m/s

Therefore,

60 m = (35 m/s)Cos 30° t

t = 60 m/30.31 m/s

t = 1.98 s

Now, we analyze the vertical motion to find the height if arrow at this time. Using second equation of motion:

h = Vi t + (1/2)gt²

where,

Vi = Vertical Component of initial Velocity = v₀ Sin θ = (35 m/s)Sin 30°

Vi = 17.5 m/s

Therefore,

h = (17.5 m/s)(1.98 s) + (1/2)(9.81 m/s²)(1.98 s)²

h = 34.6 m + 19.2 m

h = 53.8 m

since, the arrow initially had a height of y = 1.5 m. Therefore, its final height will be:

h' = h + y

h' = 53.8 m + 1.5 m

h' = 55.3 m

Answer:

The value is  [tex]\alpha = 24.5 \ rad/s^2[/tex]

Explanation:

From the question we are told that

  The weight of the elevator is [tex]W = 7300 \ N[/tex]

  The acceleration it is to be given is  [tex]a = 0.150 \ g = 0.150 * 9.8 = 1.47 \ m/s^2[/tex]

  The diameter of the shaft is [tex]d = 12.0 \ cm = 0.12 \ m[/tex]

Generally the radius is mathematically represented as

      [tex]r = \frac{d}{2}[/tex]

=>   [tex]r = \frac{0.12}{2}[/tex]

=>   [tex]r = 0.06 \ m[/tex]

Generally the minimum angular acceleration is mathematically represented as

       [tex]\alpha = \frac{a}{r}[/tex]

=>    [tex]\alpha = \frac{1.47}{0.06}[/tex]

=>    [tex]\alpha = 24.5 \ rad/s^2[/tex]

Answer:hat are some examples of energy transformation?

The Sun transforms nuclear energy into heat and light energy.

Our bodies convert chemical energy in our food into mechanical energy for us to move.

An electric fan transforms electrical energy into kinetic energy.

Explanation:

Answer:

C. 4.3 seconds

Explanation:

B 0.56 s is the time period of a twirlers baton.

Centripetal acceleration is defined as the property of the motion of an object which traversing a circular path.

Any object that is moving in a circle and has an acceleration vector pointed towards the center of that circle is known as Centripetal acceleration.

The centripetal acceleration is given by:

a = 4π²R/T²

Given values are:

a = 47.8 m/s²

D = 0.76 m  so , R = 0.76/2 = 0.38m

Using this formula,

47.8*T² = 4π² x0.38

T² = [tex]\frac{4*3.14^2*0.38}{47.8}[/tex]

T = 0.56 s

Therefore,

A twirlers baton is 0.76 m long and spins around its center. The end of the baton has a centripetal acceleration of 47.8 m/s2 which have  time period of 0.56 s.

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.

Answer:

1.

Explanation:

Hello!

In this case, for such mathematical operations, we can wee that the slash represents a fraction or a division, say 8 ÷ 4 = 2, 6 ÷ 3 = 2, 20 ÷ 4 = 5, etc. In such a way, since the operation 2/2, represents 2 ÷ 2, it is clear that two is once in 2, therefore, the result is:

2 ÷ 2 = 1.

Best regards!

Answer:

a) The potential energy of the system is 1/9 of the total mechanical energy, when [tex]x= \frac{1}{3}\cdot A[/tex].

b) The fraction of the total mechanical energy that is kinetic if the displacement is 1/2 the amplitude is 1/2.

c) The maximum kinetic energy is increased by a factor of 9.

Explanation:

a) From Mechanical Physics, we remember that the mechanical energy of mass-spring system ([tex]E[/tex]), measured in joules, is the sum of the translational kinetic energy ([tex]K[/tex]), measured in joules, and elastic potential energy ([tex]U[/tex]), measured in joules. That is:

[tex]E = K + U[/tex] (1)

By definitions of translational kinetic energy and elastic potential energy, we have the following expressions:

[tex]K = \frac{1}{2}\cdot m \cdot v^{2}[/tex] (2)

[tex]U = \frac{1}{2}\cdot k\cdot x^{2}[/tex] (3)

Where:

[tex]m[/tex] - Mass, measured in kilograms.

[tex]v[/tex] - Velocity of the mass, measured in meters per second.

[tex]k[/tex] - Spring constant, measured in newtons per meter.

[tex]x[/tex] - Elongation of the spring, measured in meters.

If we know that [tex]U = \frac{1}{9}\cdot E[/tex], [tex]k = k[/tex] and [tex]E = \frac{1}{2}\cdot k \cdot A^{2}[/tex], then:

[tex]\frac{1}{18}\cdot k\cdot A^{2} = \frac{1}{2}\cdot k\cdot x^{2}[/tex]

[tex]\frac{1}{9}\cdot A^{2} = x^{2}[/tex]

[tex]x= \frac{1}{3}\cdot A[/tex]

The potential energy of the system is 1/9 of the total mechanical energy, when [tex]x= \frac{1}{3}\cdot A[/tex].

b) If we know that [tex]k = k[/tex], [tex]x = \frac{1}{2}\cdot A[/tex] and [tex]E = \frac{1}{2}\cdot k \cdot A^{2}[/tex], then the equation of energy conservation associated with the system is:

[tex]\frac{1}{2}\cdot k\cdot A^{2} = \frac{1}{4}\cdot k\cdot A^{2}+K[/tex]

[tex]K = \frac{1}{4}\cdot k\cdot A^{2}[/tex]

The fraction of the total mechanical energy that is kinetic if the displacement is 1/2 the amplitude is 1/2.

c) From the Energy Conservation equation associated with the system, we know that increasing the amplitude by a factor of 3 represents an increase in the elastic potential energy by a factor of 9. Then, the maximum kinetic energy is increased by a factor of 9.

Answer:

a. 250kg I think it's the right answer. hope it helps:)

Answer:

C.10

Explanation:

because when you divide 50 divided by 5 = 10

Answer:

b.dance

Explanation:

don't know the explanation

Answer:

The number is  [tex]N = 300[/tex]

Explanation:

From the question we are told that

   The  net charge is  [tex]Q = -4.8 *10^{-17 } \ C[/tex]

Generally the charge on a electron is [tex]e = - 1.60 *10^{-19 } \ C[/tex]

Generally the number of excess electrons is mathematically represented as

      [tex]N = \frac{Q}{e}[/tex]

=>  [tex]N = \frac{-4.8 *10^{-17}}{-1.60 *10^{-19}}[/tex]

=>  [tex]N = 300[/tex]

Answer:

F = - 3.56*10⁵ N

Explanation:

To attempt this question, we use the formula for the relationship between momentum and the amount of movement.

I = F t = Δp

Next, we try to find the time that the average speed in the contact is constant (v = 600m / s), so we say

v = d / t

t = d / v

Given that

m = 26 g = 26 10⁻³ kg

d = 50 mm = 50 10⁻³ m

t = d/v

t = 50 10⁻³ / 600

t = 8.33 10⁻⁵ s

F t = m v - m v₀

This is so, because the bullet bounces the speed sign after the crash is negative

F = m (v-vo) / t

F = 26*10⁻³ (-500 - 640) / 8.33*10⁻⁵

F = - 3.56*10⁵ N

The negative sign is as a result of the force exerted against the bullet

Answer:

Searching in google I found the total mass and the radius of the ball (m = 1.5 kg and r = 10 cm) which are needed to solve the problem!  

The ball rotates 6.78 revolutions.

Explanation:

Searching in google I found the total mass and the radius of the ball (m = 1.5 kg and r = 10 cm) which are needed to solve the problem!        

At the bottom the ball has the following angular speed:

[tex] \omega_{f} = \frac{v_{f}}{r} = \frac{4.9 m/s}{0.10 m} = 49 rad/s [/tex]

Now, we need to find the distance traveled by the ball (L) by using θ=28° and h(height) = 2 m:

[tex] sin(\theta) = \frac{h}{L} \rightarrow L = \frac{h}{sin(\theta)} = \frac{2 m}{sin(28)} = 4.26 m [/tex]

To find the revolutions we need the time, which can be found using the following equation:                

[tex] v_{f} = v_{0} + at [/tex]  

[tex] t = \frac{v_{f} - v_{0}}{a} [/tex] (1)

So first, we need to find the acceleration:

[tex] v_{f}^{2} = v_{0}^{2} + 2aL \rightarrow a = \frac{v_{f}^{2} - v_{0}^{2}}{2L} [/tex]    (2)  

By entering equation (2) into (1) we have:

[tex] t = \frac{v_{f} - v_{0}}{\frac{v_{f}^{2} - v_{0}^{2}}{2L}} [/tex]

Since it starts from rest (v₀ = 0):  

[tex] t = \frac{2L}{v_{f}} = \frac{2*4.26 m}{4.9 m/s} = 1.74 s [/tex]

Finally, we can find the revolutions:  

[tex] \theta_{f} = \frac{1}{2} \omega_{f}*t = \frac{1}{2}*49 rad/s*1.74 s = 42.63 rad*\frac{1 rev}{2\pi rad} = 6.78 rev [/tex]

Therefore, the ball rotates 6.78 revolutions.

I hope it helps you!                                                                                                                                                                                          

Answer:

The correct answer is option B

Explanation:

Step one:

given data

a. force F= 110N

distance s= 3meters

we know that work= Force* distance

work= 110*3

Work= 330Joules

Step two:

data

Force= 60 pounds

distance= 30 ft

convert pounds to Newton

1 pound= 4.44822N

60 pounds= 60*4.44822

=266.9N

convert ft to meteres

1 ft = 0.3048meter

30ft= 0.3048*30

=9.144N

we know that work= Force* distance

work= 266.9N*9.144N

Work= 2440.53Joules

Answer:the answer is D

Explanation: I just took the quiz