A 30 L electrical radiator containing heating oil is placed in a 50 m3room. Both the roomand the oil in the radiator are initially at 10◦C. The radiator with a rating of 1.8 kW is nowturned on. At the same time, heat is lost from the room at an average rate of 0.35 kJ/s.After some time, the average temperature is measured to be 20◦C for the air in the room,and 50◦C for the oil in the radiator. Taking the density and the specific heat of the oil to be950 kg/m3and 2.2 kJ/kg◦C, respectively, determine how long the heater is kept on. Assumethe room is well sealed so that there are no air leaks.

Answer:

The temperature of silver at this given resistivity is 2971.1 ⁰C

Explanation:

The resistivity of silver is calculated as follows;

[tex]R_t = R_o[1 + \alpha(T-T_o)]\\\\[/tex]

where;

Rt is the resistivity of silver at the given temperature

Ro is the resistivity of silver at room temperature

α is the temperature coefficient of resistance

To is the room temperature

T is the temperature at which the resistivity of silver will be two times the resistivity of iron at room temperature

[tex]R_t = R_o[1 + \alpha(T-T_o)]\\\\\R_t = 1.59*10^{-8}[1 + 0.0038(T-20)][/tex]

Resistivity of iron at room temperature = 9.71 x 10⁻⁸ ohm.m

When silver's resistivity becomes 2 times the resistivity of iron, we will have the following equations;

[tex]R_t,_{silver} = 2R_o,_{iron}\\\\1.59*10^{-8}[1 + 0.0038(T-20)] =(2 *9.71*10^{-8})\\\\\ \ (divide \ through \ by \ 1.59*10^{-8})\\\\1 + 0.0038(T-20) = 12.214\\\\1 + 0.0038T - 0.076 = 12.214\\\\0.0038T +0.924 = 12.214\\\\0.0038T = 12.214 - 0.924\\\\0.0038T = 11.29\\\\T = \frac{11.29}{0.0038} \\\\T = 2971.1 \ ^0C[/tex]

Therefore, the temperature of silver at this given resistivity is 2971.1 ⁰C

Answer:

The net charge on the shell is 30x10^-9C

Explanation:

Pls see attached file

Answer:

m = 0.0125 kg

Explanation:

Let us apply the formula for the speed of a wave on a string that is under tension:

[tex]v = \sqrt{\frac{F}{\mu} }[/tex]

where F = tension force

μ = mass per unit length

Mass per unit length is given as:

μ  = m / l

where m = mass of the string

l = length of the string

This implies that:

[tex]v = \sqrt{\frac{F}{m/l} }\\ \\v = \sqrt{\frac{F * l}{m} }[/tex]

Let us make mass, m, the subject of the formula:

[tex]v^2 = \frac{F * l}{m}\\\\m = \frac{F * l}{v^2}[/tex]

From the question:

F = 20 N

l = 4.50 m

v = 85 m/s

Therefore:

[tex]m = \frac{20 * 4.5}{85^2}\\\\m = \frac{90}{7225}\\ \\m = 0.0125 kg[/tex]

Answer:

The  angular speed is [tex]w = 5.89 \ rad/s[/tex]

Explanation:

From the question we are told that

    The time taken is  [tex]t = 1.6 s[/tex]

    The number of somersaults  is n  =  1.5

The total angular displacement during the somersault is mathematically represented as

         [tex]\theta = n * 2 * \pi[/tex]

substituting values

        [tex]\theta = 1.5 * 2 * 3.142[/tex]

       [tex]\theta = 9.426 \ rad[/tex]

 The angular speed is mathematically represented as

         [tex]w = \frac{\theta }{t}[/tex]

substituting values

         [tex]w = \frac{9.426}{1.6}[/tex]

          [tex]w = 5.89 \ rad/s[/tex]

Answer:

C You should deflect the ball back toward your friend.

Explanation:

This is because it would result in a completely inelastic collision, and the final velocity of me would be found using,

with m= mass, V=velocity, i=initial, f=final:

mV(me,i) +mV(ball,i) = [m(me)+m(b)]V(f)

So V(f) would be just the momentum of the ball divided by just MV mass of the ball and it will be higher resulting in inelastic collision

Answer:

A. You should catch the ball.

Explanation:

Catching the ball maximizes your speed by converting most of the momentum of the flying ball into the momentum of you and the ball. Since the ice is smooth, the friction between your feet and the ice is almost negligible, meaning less energy is needed to set your body in motion. Catching the ball means that you and the ball undergoes an inelastic collision, and part of the kinetic energy of the ball is transferred to you, setting you in motion. Deflecting the ball will only give you a relatively small speed compared to catching the ball.

Answer:

The object has an excess of [tex]10^{13}[/tex] electrons.

Explanation:

When an object has a negative charge he has an excess of electrons in its body. We can calculate the number of excessive electrons by dividing the charge of the body by the charge of one electron. This is done below:

[tex]n = \frac{\text{object charge}}{\text{electron charge}}\\n = \frac{-1.6*10^{-6}}{-1.6*10^{-19}} = 1*10^{-6 + 19} = 10^{13}[/tex]

The object has an excess of [tex]10^{13}[/tex] electrons.

Answer:

The major difference is the capacity of both batteries. The AA battery has a higher capacity (a higher current) than the AAA battery.

Explanation:

The AA batteries and the AAA batteries are very similar in their voltage; both of them have 1.5 V.

The difference between these two batteries is their size and also the current that they have. The AAA battery is smaller than the AA battery, which means that the amount of electrochemical material is lower, so the AA battery has a higher capacity (a higher current) than the AAA battery. Generally, AA battery has 2400 mAh capacity and AAA battery has a capacity of 1000mAh; this means that AA battery has almost three times the capacity of an AAA battery.      

Furthermore, the size of the AA battery makes it more common than the AAA battery and therefore has higher commercial demand.                                  

I hope it helps you!

Answer:

is this a company name.? java is a computer software right..

Answer:

A

Explanation:

since the wooden bat is an opaque object placed after a translucent object, light will come through the plastic sheet but will be unable to go through the bat. hence the dark shadow of the bat on a lit sheet

Answer:

  θ₁ = 85.5º       θ₂ = 12.98º

Explanation:

Let's analyze this projectile launch problem, the catapults are 100 m from the wall 15 m high, the objective is for the walls, let's look for the angles for which the rock stops touching the wall.

Let's write the equations for motion for this point

X axis

          x = v₀ₓ t

          x = v₀ cos θ t

Y axis

         y = [tex]v_{oy}[/tex] t - ½ g t2

         y = v_{o} sin θ t - ½ g t²

let's substitute the values

         100 = 80 cos θ t

           15 = 80 sin θ t - ½ 9.8 t²

we have two equations with two unknowns, so the system can be solved

let's clear the time in the first equation

           t = 100/80 cos θ

         15 = 80 sin θ (10/8 cos θ) - 4.9 (10/8 cos θ)²

         15 = 100  tan θ - 7.656 sec² θ

we can use the trigonometric relationship

         sec² θ = 1- tan² θ

we substitute

       15 = 100 tan θ - 7,656 (1- tan² θ)

       15 = 100 tan θ - 7,656 + 7,656 tan² θ

        7,656 tan² θ + 100 tan θ -22,656=0

let's change variables

       tan θ = u

        u² + 13.06 u + 2,959 = 0

let's solve the quadratic equation

       u = [-13.06 ±√(13.06² - 4  2,959)] / 2

       u = [13.06 ± 12.599] / 2

        u₁ = 12.8295

        u₂ = 0.2305

now we can find the angles

         u = tan θ

         θ = tan⁻¹ u

        θ₁ = 85.5º

         θ₂ = 12.98º

Answer:

Density = 10,933.93 kg/m^3

the density of the material is 10,933.93 kg/m^3

Explanation:

Density is the mass per unit volume

Density = mass/volume = m/V

Volume of a cylinder V = πr^2 h

Given;

Height h = 48.5mm = 0.0485 m

Radius r = diameter/2 = 49mm÷2 = 24.5mm = 0.0245m

Substituting the values;

Volume V = π×(0.0245^2)×0.0485

V = 0.000091458438030 m^3

V = 0.000091458 m^3

The mass is given as;

Mass = 1 kg

So, the density can be calculated as;

Density = 1/0.000091458

Density = 10933.92825785 kg/m^3

Density = 10,933.93 kg/m^3

the density of the material is 10,933.93 kg/m^3

Answer:

T₂ = 95.56°C

Explanation:

The final resistance of a material after being heated is given by the relation:

R' = R(1 + αΔT)

where,

R' = Final Resistance = 207.4 Ω

R = Initial Resistance = 154.9 Ω

α = Temperature Coefficient of Resistance of Tungsten = 0.0045 °C⁻¹

ΔT = Change in Temperature = ?

Therefore,

207.4 Ω = 154.9 Ω[1 + (0.0045°C⁻¹)ΔT]

207.4 Ω/154.9 Ω = 1 + (0.0045°C⁻¹)ΔT

1.34 - 1 = (0.0045°C⁻¹)ΔT

ΔT = 0.34/0.0045°C⁻¹

ΔT = 75.56°C

but,

ΔT = Final Temperature - Initial Temperature

ΔT = T₂ - T₁ = T₂ - 20°C

T₂ - 20°C = 75.56°C

T₂ = 75.56°C + 20°C

T₂ = 95.56°C

Answer:

the resistance of the longer one is twice as big as the resistance of the shorter one.

Explanation:

Given that :

For the shorter cylindrical resistor

Length = L

Diameter = D

Resistance = R1

For the longer cylindrical resistor

Length = 8L

Diameter = 4D

Resistance = R2

So;

We all know that the resistance of a given material can be determined by using the formula :

[tex]R = \dfrac{\rho L }{A}[/tex]

where;

A = πr²

[tex]R = \dfrac{\rho L }{\pi r ^2}[/tex]

For the shorter cylindrical resistor ; we have:

[tex]R = \dfrac{\rho L }{\pi r ^2}[/tex]

since 2 r = D

[tex]R = \dfrac{\rho L }{\pi (\frac{2}{2 \ r}) ^2}[/tex]

[tex]R = \dfrac{ 4 \rho L }{\pi \ D ^2}[/tex]

For the longer cylindrical resistor ; we have:

[tex]R = \dfrac{\rho L }{\pi r ^2}[/tex]

since 2 r = D

[tex]R = \dfrac{ \rho (8 ) L }{\pi (\frac{2}{2 \ r}) ^2}[/tex]

[tex]R = \dfrac{32\rho L }{\pi \ (4 D) ^2}[/tex]

[tex]R = \dfrac{2\rho L }{\pi \ (D) ^2}[/tex]

Sp;we can equate the shorter cylindrical resistor to the longer cylindrical resistor as shown below :

[tex]\dfrac{R_s}{R_L} = \dfrac{ \dfrac{ 4 \rho L }{\pi \ D ^2}}{ \dfrac{2\rho L }{\pi \ (D) ^2}}[/tex]

[tex]\dfrac{R_s}{R_L} ={ \dfrac{ 4 \rho L }{\pi \ D ^2}}* { \dfrac {\pi \ (D) ^2} {2\rho L}}[/tex]

[tex]\dfrac{R_s}{R_L} =2[/tex]

[tex]{R_s}=2{R_L}[/tex]

Thus; the resistance of the longer one is twice as big as the resistance of the shorter one.

Centripetal acceleration = (speed squared) / (radius)

Centripetal acceleration = (10 m/s)² / (1.0 m)

Centripetal acceleration = (100 m²/s²) / (1.0 m)

Centripetal acceleration = 100 m/s²

Answer:

Towards the west.

Explanation:

The direction of a magnetic field lines is the direction north end of a compass needle points. The magnetic field exert force on positive charge.

Using the magnetic rule,which indicate that in order to find the direction of magnetic force on a moving charge, the thumb of the right hand point in the direction of force, the index finger in the direction of velocity charge and the middle finger in the direction of magnetic field.

According to the right hand rule, the electron moving moving west which is the thumb, the direction of the electron is west which is the middle finger and it is upward

Answer:

i

  [tex]v = 142.595 \ m/s[/tex]

ii

  [tex]f = 109.69 \ Hz[/tex]

iii1 )

  [tex]f_2 =219.4 Hz[/tex]

iii2)

   [tex]f_3 =329.1 Hz[/tex]

iii3)

    [tex]f_4 =438.8 Hz[/tex]

Explanation:

From the question we are told that

    The length of the string is  [tex]l = 0.65 \ m[/tex]

     The tension on the string is  [tex]T = 61 \ N[/tex]

     The mass per unit length is  [tex]m = 3.0 \ g/m = 3.0 * \frac{1}{1000} = 3 *10^{-3 } \ kg /m[/tex]

The speed of wave on the string is mathematically represented as

       [tex]v = \sqrt{\frac{T}{m} }[/tex]

substituting values

      [tex]v = \sqrt{\frac{61}{3*10^{-3}} }[/tex]

     [tex]v = 142.595 \ m/s[/tex]

generally the  string's  frequency is mathematically represented as

         [tex]f = \frac{nv}{2l}[/tex]

n = 1  given that the frequency we are to find is the fundamental frequency

So

      substituting values

       [tex]f = \frac{142.595 * 1 }{2 * 0.65}[/tex]

       [tex]f = 109.69 \ Hz[/tex]

The  frequencies at which the string would vibrate include

1       [tex]f_2 = 2 * f[/tex]

Here [tex]f_2[/tex] is  know as the second harmonic and the value is  

      [tex]f_2 = 2 * 109.69[/tex]

      [tex]f_2 =219.4 Hz[/tex]

2

[tex]f_3 = 3 * f[/tex]

Here [tex]f_3[/tex] is  know as the third harmonic and the value is  

      [tex]f_3 = 3 * 109.69[/tex]

     [tex]f_3 =329.1 Hz[/tex]

3

     [tex]f_3 = 4 * f[/tex]

Here [tex]f_4[/tex] is  know as the fourth harmonic and the value is  

      [tex]f_3 = 4 * 109.69[/tex]

     [tex]f_4 =438.8 Hz[/tex]

Answer:

30.67m

Explanation:

Using one of the equations of motion as follows, we can describe the path of the projectile in its horizontal or vertical displacement;

s = ut ± [tex]\frac{1}{2} at^2[/tex]               ------------(i)

Where;

s = horizontal/vertical displacement

u = initial horizontal/vertical component of the velocity

a = acceleration of the projectile

t = time taken for the projectile to reach a certain horizontal or vertical position.

Since the question requires that we find the vertical distance from where the projectile was launched to where it hit the target, equation (i) can be made more specific as follows;

h = vt ± [tex]\frac{1}{2} at^2[/tex]               ------------(ii)

Where;

h = vertical displacement

v = initial vertical component of the velocity = usinθ

a = acceleration due to gravity (since vertical motion is considered)

t = time taken for the projectile to hit the target

From the question;

u = 47m/s, θ = 0.6rads

=> usinθ = 47 sin 0.6

=> usinθ = 47 x 0.5646 = 26.54m/s

t = 1.7s

Take a = -g = -10.0m/s   (since motion is upwards against gravity)

Substitute these values into equation (ii) as follows;

h = vt - [tex]\frac{1}{2} at^2[/tex]

h = 26.54(1.7) - [tex]\frac{1}{2} (10)(1.7)^2[/tex]

h = 45.118 - 14.45

h = 30.67m

Therefore, the vertical distance is 30.67m        

Answer:

The moment of inertia is  [tex]I =0.14 \ kg \cdot m^2[/tex]

Explanation:

From the question we are told that

    The length of the rod is  [tex]l = 0.900 \ m[/tex]

     The mass of the rod is  [tex]m = 0.500 \ kg[/tex]

      The distance of the center of mass from the pivot is  [tex]d = 0.500 \ m[/tex]

      The period of the rod's motion is  [tex]T = 1.49 \ s[/tex]

Generally the period of the motion is mathematically represented as

       [tex]T = 2 \pi * \sqrt{\frac{I}{m* g * d} }[/tex]

Where [tex]I[/tex] is the moment of inertia about the pivot so making [tex]I[/tex] the subject of formula

      [tex]I = [\frac{T}{2\pi } ]^2 * m * g * d[/tex]

substituting values

        [tex]I = [\frac{1.49}{2* 3.142 } ]^2 * 0.5 * 9.8 * 0.5[/tex]

       [tex]I =0.14 \ kg \cdot m^2[/tex]

Answer:

17.94 kN/C is the electric field intensity at the origin due to the charges.

Explanation:

From the question, we are told that

The distance of 1 μC from origin = 1 m

The distance of 2 μC from origin = 2 m

The distance of 3 μC from origin = 3 m

The distance of 4 μC from origin = 5 m

Therefore, for us to find the electric field intensity, we'll solve below:

The formula for Electric field intensity = ( k * q ) / ( r * r )

where , r is distance ,

k = 9 * 10^9 ,

and , q is charge .

now ,

electric field intensity at the origin = [ k * 10^(-6) / 1 * 1 ] +[ k * 2 * 10^(-6) / 2 * 2 ] + [ k * 3 * 10^(-6) / 3 * 3 ] + [ k * 4 * 10^(-6) / 5 * 5 ]

=> electric field intensity at the origin = k * 10^(-6) [ 1 + 1/2 + 1/3 + 4/25 ] N/C

=> electric field intensity at the origin = 9 * 10^9 * 10^(-6) * 1.99 N/C

=> electric field intensity at the origin = 17.94 kN/C

Answer:

D

Explanation:

The power equation is P= V^2/R

Please let me know if this helped! Please rate it the brainlist if possible!

As a result of the given scenario, power delivered from the battery to combination decreases. The correct option is D.

A resistor is a two-terminal passive electrical component that uses electrical resistance as a circuit element.

Resistors are used in electronic circuits to reduce current flow, adjust signal levels, divide voltages, and bias active elements.

A resistor is a component of an electronic circuit that limits or regulates the flow of electrical current. Resistors can also be used to supply a fixed voltage to an active device such as a transistor.

The current through resistors is the same when they are connected in series. The battery voltage is divided among resistors.

Adding more resistors to a series circuit increases total resistance and thus lowers current. However, in a parallel circuit, adding more resistors in parallel creates more options while decreasing total resistance.

Thus, the correct option is D.

For more details regarding resistors, visit:

https://brainly.com/question/24297401

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Answer:

the other force= (16.3i + 14.6j)N

EXPLANATION:

Given:

Mass=2.80-kg

t= 1.2s

Since the object started from rest, the origin is (0,0) which symbolize the the object's initial position.

We will need to calculate the magnitude of the displacement using the below formula;

d = (1/2)at2 + v0t + d0

But note that

d0 = 0,( initial position)

v0 = 0( initial position)

a is the net acceleration

d = √[4.202 + (-3.30)2] m = 5.34 m

Hence, the magnitude of the displacement is 5.34 m, then we can make 'a' the subject of formula in the above expression in order to calculate the value for acceleration, note that d0 = 0,( initial position) and v0 = 0( initial position)

d = (1/2)at2

a = 2d/t2 = 2(5.34)/(1.20)2 m/s2 = 7.42 m/s2

the net acceleration is 7.42 m/s2

Acceleration in terms of the vector can be calculated as

a=2(ri - r0)/t^2

Where t =1.2s which is the time

a= 2(4.2i - 3.30j)/ 1.2^2

a=( 5.83i - 4.58j)m/s

now the net force can now be calculated since we have known the value of acceleration, using the formula below;

F(x) = ma - mg

Where a = 5.83i - 4.58j)m/s and g= 9.8m/s

2.8(5.83i - 4.58j)m/s - (2.80 × 9.8)m/s^2

Therefore, the other force= (16.3i + 14.6j)N

Answer:

no the only things that can travel at the speed of light are waves in the electromagnetic spectrum

Answer:

The air-water interface is an example of boundary. The transmitted portion of the initial wave energy is way smaller than the reflected portion. This makes the boundary  wave hard to hear.

When both the source of the sound and your ears are located underwater, the sound is louder because the sound waves can travel directly to your ear.

Explanation:

The air-to-water sound wave transmission is inhibited because more of reflection than transmission of the wave occurs at the boundary. In the end, only about 30% of the sound wave eventually reaches underwater. For sound generated underwater, all the wave energy is transmitted directly to the observer. Sound wave travel faster in water than in air because, the molecules of water are more densely packed together, and hence can easily transmit their vibration to their neighboring molecules, when compared to air.

Answer: The air-water interface is an example of boundary. The (transmitted) portion of the initial wave is way smaller than the (reflected) portion. This makes the (transmitted) wave hard to hear.

When both the source of the sound and your ears are located underwater, the sound is louder because the sound waves can (travel directly to your ears.)

Explanation:

The part of the sound wave that is transmitted across the boundary between air and water is much smaller than the part of the wave that is reflected. This is what makes it hard to hear your friend knocking two rocks together above the surface.

When you and the rocks are underwater, the sound that comes from knocking the rocks together can travel directly to your ears rather than having to be transmitted across mediums.

Answer:

0.85c

Explanation:

Rest mass of Kaon [tex]M_{0K}[/tex] = 494 MeV/c²

Rest mass of proton [tex]M_{0P}[/tex]  = 938 MeV/c²

The rest energy is gotten by multiplying the rest mass by the square of the speed of light c²

for the kaon, rest energy [tex]E_{0K}[/tex] = 494c² MeV

for the proton, rest energy [tex]E_{0P}[/tex] = 938c² MeV

Recall that the rest energy, and the total energy are related by..

[tex]E[/tex] = γ[tex]E_{0}[/tex]

which can be written in this case as

[tex]E_{K}[/tex] = γ[tex]E_{0K}[/tex] ...... equ 1

where [tex]E[/tex] = total energy of the kaon, and

[tex]E_{0}[/tex] = rest energy of the kaon

γ = relativistic factor = [tex]\frac{1}{\sqrt{1 - \beta ^{2} } }[/tex]

where [tex]\beta = \frac{v}{c}[/tex]

But, it is stated that the total energy of the kaon is equal to the rest mass of the proton or its equivalent rest energy, therefore...

[tex]E_{K}[/tex] = [tex]E_{0P}[/tex] ......equ 2

where [tex]E_{K}[/tex] is the total energy of the kaon, and

[tex]E_{0P}[/tex] is the rest energy of the proton.

From [tex]E_{K}[/tex] = [tex]E_{0P}[/tex] = 938c²    

equ 1 becomes

938c² = γ494c²

γ = 938c²/494c² = 1.89

γ = [tex]\frac{1}{\sqrt{1 - \beta ^{2} } }[/tex] = 1.89

1.89[tex]\sqrt{1 - \beta ^{2} }[/tex] = 1

squaring both sides, we get

3.57( 1 - [tex]\beta^{2}[/tex]) = 1

3.57 - 3.57[tex]\beta^{2}[/tex] = 1

2.57 = 3.57[tex]\beta^{2}[/tex]

[tex]\beta^{2}[/tex] = 2.57/3.57 = 0.72

[tex]\beta = \sqrt{0.72}[/tex] = 0.85

but, [tex]\beta = \frac{v}{c}[/tex]

v/c = 0.85

v = 0.85c

Answer:

The magnetic field at mid point between two parallel wires is 1.2 x 10⁻⁵ T

Explanation:

Given;

current in the first wire, I₁ = 10 A

current in the second wire, I₂ = 20 A

distance between the two wires, d = 1.0 m

Magnetic field at mid point between two parallel wires is calculated as;

[tex]B = \frac{\mu_o I_1}{2\pi r} + \frac{\mu_o I_2}{2\pi r} \\\\B = \frac{\mu_o }{2\pi r}(I_1 +I_2)[/tex]

where;

r is the midpoint between the wires, = 0.5 m

μ₀ is the permeability of free space, = 4π x 10⁻⁷

[tex]B = \frac{\mu_o }{2\pi r}(I_1 +I_2)\\\\B = \frac{4\pi*10^{-7} }{2\pi *0.5}(10 +20)\\\\B = \frac{4\pi*10^{-7} *30}{2\pi *0.5}\\\\B = 1.2 *10^{-5} \ T[/tex]

Therefore, the magnetic field at mid point between two parallel wires is 1.2 x 10⁻⁵ T

Answer:

94.248 g/sec

Explanation:

For solving the total current of the blood passing first we have to solve the cross sectional area which is given below:

[tex]A_1 = \pi R^2\\\\A_1 = \pi (1)^2\\\\A_1 = 3.1416 cm^2[/tex]

And, the velocity of blood pumping is 30 cm^2

Now apply the following formula to solve the total current

[tex]Q = \rho A_1V_1\\\\Q = (1)(3.1416)(30)\\\\[/tex]

Q =  94.248 g/sec

Basically we applied the above formula So, that the total current could come

Answer:

200volts

Explanation:

Pls see attached file

Answer:

100 V

Explanation:

Electric field E = 100 V/m

Potential at the origin = 300 V

Potential at point (-2m, 0) i.e 2 m behind the origin = ?

From the equation ΔV = EΔd,

ΔV = [tex]V_{0} - V_{x}[/tex]

where [tex]V_{0}[/tex] is the potential at origin,

and [tex]V_{x}[/tex] is the potential at point (-2, 0)

E = electric field

Δd = 0 - (-2) = 2 m

[tex]V_{0} - V_{x}[/tex] = 300 - [tex]x[/tex]

equating, we have

 300 - [tex]x[/tex] = 100 x 2

300 - [tex]x[/tex] = 200

[tex]x[/tex] = 100 V

Answer:

q/6Eo

Explanation:

See attached file pls

Answer:

Explanation:

Given data

mass of  load m= 425 kg

height/distance h=64 m

acceleration a= 1.8 m/s^2

The work done can be calculated using the expression

work done= force* distance

but force= mass *acceleration

hence work done= 425*1.8*64= 48,960 J

work done= 48.96 kJ

Answer:

The minimum speed required  is 2.62m/s

Explanation:

The value of  gravitational acceleration = g = 9.81 m/s^2

Radius of the vertical circle = R = 0.7 m

Given the mass of the pail of water = m

The speed at the highest point of the circle = V

The centripetal force will be needed must be more than the weight of the pail of water in order to not spill water.

Below is the calculation:

[tex]\frac{mV^{2}}{R} = mg[/tex]

[tex]V = \sqrt{gR}[/tex]

[tex]V = \sqrt{9.81 \times 0.7}[/tex]

[tex]V = 2.62 m/s[/tex]