A total charge of 62 nC is uniformly distributed throughout a non-conducting sphere with a radius of 5.00 cm. The electric potential at r = 15.0 cm , relative to the potential far away, is:________

Answer:

the final temperature is [tex]\mathbf{T_f = -377.2^0 C}[/tex]

Explanation:

The change in length of a bar can be expressed with the relation;

[tex]\Delta L = L_f - L_i[/tex]   ---- (1)

Also ; the relative or fractional increase in length  is proportional to the change in temperature.

Mathematically;

ΔL/L_i ∝ kΔT

where;

k is replaced with ∝ (the proportionality constant )

[tex]\dfrac{ \Delta L}{L_i}=\alpha \Delta T[/tex]    ---- (2)

From (1) ;

[tex]L_f = \Delta L + L_i[/tex]  ---  (3)

from (2)

[tex]{ \Delta L}=\alpha \Delta T*{L_i}[/tex]  ---- (4)

replacing (4) into (3);we have;

[tex]L_f =(\alpha \Delta T*{L_i} ) + L_i[/tex]

On re-arrangement; we have

[tex]L_f = L_i + \alpha L_i (\Delta T )[/tex]

from the given question; we can say that :

[tex](L_f)_{brass}}} = (L_f)_{Al}[/tex]

So;

[tex]L_{brass} + \alpha _{brass} L_{brass}(\Delta T) = L_{Al} + \alpha _{Al} L_{Al}(\Delta T)[/tex]

Making the change in temperature the subject of the formula; we have:

[tex]\Delta T = \dfrac{L_{Al}-L_{brass}}{\alpha _ {brass} L_{brass}-\alpha _{Al}L_{Al}}[/tex]

where;

[tex]L_{Al}[/tex] = 10.02 cm

[tex]L_{brass}[/tex] = 10.00 cm

[tex]\alpha _{brass}[/tex] = 19 × 10⁻⁶ °C ⁻¹

[tex]\alpha_{Al}[/tex] = 24  × 10⁻⁶ °C ⁻¹

[tex]\Delta T = \dfrac{10.02-10.00}{19*10^{-6} \ \ {^0}C^{-1} *10.00 -24*10^{-6} \ \ {^0}C^{-1} *10.02}[/tex]

[tex]\Delta T[/tex] = −396.1965135 ° C

[tex]\Delta T[/tex] ≅ −396.20  °C

Given that the initial temperature [tex]T_i = 19^0 C[/tex]

Then ;

[tex]\Delta T = T_f - T_i[/tex]

[tex]T_f = \Delta T + T_I[/tex]

Thus;

[tex]T_f =(-396.20 + 19.0)^0 C[/tex]

[tex]\mathbf{T_f = -377.2^0 C}[/tex]

Thus; the final temperature is [tex]\mathbf{T_f = -377.2^0 C}[/tex]

Answer:

E = V/5 x10⁻³

Explanation:

if the potential difference is V

then electric field E is given by

E = V/d

d = 0.5cm = 5 x 10⁻³m

E = V/5 x10⁻³

Answer:

the acceleration [tex]a^{\to} = (0.0159 \ \ m/s^2 )i[/tex]

Explanation:

Given that:

the initial speed v₁ = 0 m/s i.e starting from rest ; since the car accelerates at a distance Δx = 6 miles in order to teach that final speed v₂ of 63.15 km/h.

So;  the acceleration for the first 6 miles can be calculated by using the formula:

v₂² = v₁² + 2a (Δx)

Making acceleration  a the subject of the formula in the above expression ; we have:

v₂² - v₁² = 2a (Δx)

[tex]a = \dfrac{v_2^2 - v_1^2 }{2 \Delta x}[/tex]

[tex]a = \dfrac{(63.15 \ km/s)^2 - (0 \ m/s)^2 }{2 (6 \ miles)}[/tex]

[tex]a = \dfrac{(17.54 \ m/s)^2 - (0 \ m/s)^2 }{2 (9.65*10^3 \ m)}[/tex]

[tex]a =0.0159 \ m/s^2[/tex]

Thus;

Assume the car moves in the +x direction;

the acceleration [tex]a^{\to} = (0.0159 \ \ m/s^2 )i[/tex]

Answer:

a.     F = 2.32*10^-18 N

b.     The force F is 2.59*10^11 times the weight of the electron

Explanation:

a. In order to calculate the magnitude of the force exerted on the electron you first calculate the acceleration of the electron, by using the following formula:

[tex]v^2=v_o^2+2ax[/tex]         (1)

v: final speed of the electron = 6.60*10^5 m/s

vo: initial speed of the electron = 4.00*10^5 m/s

a: acceleration of the electron = ?

x: distance traveled by the electron = 5.40cm = 0.054m

you solve the equation (2) for a and replace the values of the parameters:

[tex]a=\frac{v^2-v_o^2}{2x}=\frac{(6.60*10^5m/s)^2-(4.00*10^5m/s)^2}{2(0.054m)}\\\\a=2.55*10^{12}\frac{m}{s^2}[/tex]

Next, you use the second Newton law to calculate the force:

[tex]F=ma[/tex]

m: mass of the electron = 9.11*10^-31kg

[tex]F=(9.11*10^{-31}kg)(2.55*10^{12}m/s^2)=2.32*10^{-18}N[/tex]

The magnitude of the force exerted on the electron is 2.32*10^-18 N

b. The weight of the electron is given by:

[tex]F_g=mg=(9.11*10^{-31}kg)(9.8m/s^2)=8.92*10^{-30}N[/tex]

The quotient between the weight of the electron and the force F is:

[tex]\frac{F}{F_g}=\frac{2.32*10^{-18}N}{8.92*10^{-30}N}=2.59*10^{11}[/tex]

The force F is 2.59*10^11 times the weight of the electron

Answer:

More current will be loss through the metal wire strands if the force on them was repulsive, and more stress will be induced on the wire strands due to internal and external flexing.

Explanation:

A wire bundle is made up of wire strands bunched together to increase flexibility that is not always possible in a single solid metal wire conductor. In the strands of wire carrying a high voltage power, each strand carries a certain amount of current, and the current through the strands all travel in the same direction. It is know that for two conductors or wire, separated by a certain distance, that carries current flowing through them in the same direction, an attractive force is produced on these wires, one on the other. This effect is due to the magnetic induction of a current carrying conductor. The forces between these strands of the high voltage wire bundle, pulls the wire strands closer, creating more bond between these wire strands and reducing internal flex induced stresses.

If the case was the opposite, and the wires opposed themselves, the effect would be that a lot of cost will be expended in holding these wire strands together. Also, stress within the strands due to the repulsion, will couple with external stress from the flexing of the wire, resulting in the weakening of the material.

The biggest problem will be that more current will be lost in the wire due to increased surface area caused by the repulsive forces opening spaces between the strand. This loss is a s a result of the 'skin effect' in wire transmission, in which current tends to flow close to the surface of the metal wire. The skin effect generates power loss as heat through the exposed surface area.

Answer:

a) correct answer is b higher , b) correct answer is b higher , c) correct answer is b faster , d)  traveling wave , e)

Explanation:

A traveling wave is described by the expression

            y = A sin (kx - wt)

where k is the wave vector and w is the angular velocity

let's examine every situation presented

a) a new faster pulse is generated

A faster pulse should have a higher angular velocity

equal speed is related to the period and frequency

            w = 2π f = 2π / T

therefore in this case the period must decrease so that the angular velocity increases

the correct answer is c narrower

b) Generate a pulse, but move your hand more.

Moving the hand increases the amplitude (A) of the pulse

the correct answer is b higher

c) generates a pulse but the force is tightened

Set means that more tension force is applied to the string, so the velicate changes

       v = √ (T /μ)

the correct answer is b faster

d) move your hand back and forth

in this case you would see a pulse series whose sum corresponds to a traveling wave

e) Advance a frame the movie

in this case the wave will be displaced a whole period to the right

the correct answer is b

f) move your hand faster

the waves will have a maximum fast, so they are closer

answer C

g) decrease wave frequency

Since the speed of the wave is a constant m ak, decreasing the frequency must increase the wavelength to keep the velocity constant.

the correct answer is b increases its wavelength

Answer:

The two value of the wavelength for the out of tune guitar is  

[tex]\lambda _2 = (6.48,6.52) \ cm[/tex]

Explanation:

From the question we are told that

     The wavelength of the note is [tex]\lambda = 6.50 \ cm = 0.065 \ m[/tex]

     The difference in beat frequency is [tex]\Delta f = 17.0 \ Hz[/tex]

Generally the frequency of the note played by the guitar that is in tune is  

        [tex]f_1 = \frac{v_s}{\lambda}[/tex]

Where [tex]v_s[/tex] is the speed of sound with a constant value [tex]v_s = 343 \ m/s[/tex]

       [tex]f_1 = \frac{343}{0.0065}[/tex]

      [tex]f_1 = 5276.9 \ Hz[/tex]

The difference in beat is mathematically represented as

       [tex]\Delta f = |f_1 - f_2|[/tex]

Where [tex]f_2[/tex] is the frequency of the sound from the out of tune guitar

     [tex]f_2 =f_1 \pm \Delta f[/tex]

substituting values

      [tex]f_2 =f_1 + \Delta f[/tex]

      [tex]f_2 = 5276.9 + 17.0[/tex]  

     [tex]f_2 = 5293.9 \ Hz[/tex]

The wavelength for this frequency is

      [tex]\lambda_2 = \frac{343 }{5293.9}[/tex]

     [tex]\lambda_2 = 0.0648 \ m[/tex]

    [tex]\lambda_2 = 6.48 \ cm[/tex]

For the second value of the second frequency

     [tex]f_2 = f_1 - \Delta f[/tex]

     [tex]f_2 = 5276.9 -17[/tex]

      [tex]f_2 = 5259.9 Hz[/tex]

The wavelength for this frequency is

   [tex]\lambda _2 = \frac{343}{5259.9}[/tex]

   [tex]\lambda _2 = 0.0652 \ m[/tex]

   [tex]\lambda _2 = 6.52 \ cm[/tex]

This question involves the concepts of beat frequency and wavelength.

The possible wavelengths of the out-of-tune guitar are "6.48 cm" and "6.52 cm".

The beat frequency is given by the following formula:

[tex]f_b=|f_1-f_2|\\\\[/tex]

f₂ = [tex]f_b[/tex] ± f₁

where,

f₂ = frequency of the out-of-tune guitar = ?

[tex]f_b[/tex] = beat frequency = 17 Hz

f₁ = frequency of in-tune guitar = [tex]\frac{speed\ of\ sound\ in\ air}{\lambda_1}=\frac{343\ m/s}{0.065\ m}=5276.9\ Hz[/tex]

Therefore,

f₂ = 5276.9 Hz ± 17 HZ

f₂ = 5293.9 Hz (OR) 5259.9 Hz

Now, calculating the possible wavelengths:

[tex]\lambda_2=\frac{speed\ of\ sound}{f_2}\\\\\lambda_2 = \frac{343\ m/s}{5293.9\ Hz}\ (OR)\ \frac{343\ m/s}{5259.9\ Hz}\\\\[/tex]

λ₂ = 6.48 cm (OR) 6.52 cm

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Answer:

force on larger piston = [tex]\frac{fA}{a}[/tex]

Explanation:

we label the pistons as piston A and piston B

small piston A:

area = a

force = f

large piston B:

area = A

force  = ?

Pascal's law of pressure state that the pressure delivered to a liquid is transmitted undiminished to every portion of the fluid.

we know that pressure = force ÷ area

pressure of piston A = [tex]\frac{f}{a}[/tex]

pressure of piston B = [tex]\frac{(force on piston B)}{A}[/tex]

obeying Pascal's law, the system pressures must be equal. Therefore

[tex]\frac{f}{a} = \frac{(force on piston B)}{A}[/tex]

force on large piston (B) = F = [tex]\frac{fA}{a}[/tex]

Answer:

a. 94.54 N

b. 0.356 m/s^2

Explanation:

Given:-

- The mass of the inclined block, M = 100 kg

- The mass of the vertically hanging block, m = 10 kg

- The angle of inclination, θ = 20°

- The coefficient of friction of inclined surface, u = 0.3

Find:-

a) The magnitude of tension in the cable

b) The acceleration of the system

Solution:-

- We will first draw a free body diagram for both the blocks. The vertically hanging block of mass m = 10 kg tends to move "upward" when the system is released.

- The block experiences a tension force ( T ) in the upward direction due the attached cable. The tension in the cable is combated with the weight of the vertically hanging block.

- We will employ the use of Newton's second law of motion to express the dynamics of the vertically hanging block as follows:

                        [tex]T - m*g = m*a\\\\[/tex]  ... Eq 1

Where,

              a: The acceleration of the system

- Similarly, we will construct a free body diagram for the inclined block of mass M = 100 kg. The Tension ( T ) pulls onto the block; however, the weight of the block is greater and tends down the slope.

- As the block moves down the slope it experiences frictional force ( F ) that acts up the slope due to the contact force ( N ) between the block and the plane.

- We will employ the static equilibrium of the inclined block in the normal direction and we have:

                        [tex]N - M*g*cos ( Q )= 0\\\\N = M*g*cos ( Q )[/tex]

- The frictional force ( F ) is proportional to the contact force ( N ) as follows:

                        [tex]F = u*N\\\\F = u*M*g *cos ( Q )[/tex]

- Now we will apply the Newton's second law of motion parallel to the plane as follows:

                       [tex]M*g*sin(Q) - T - F = M*a\\\\M*g*sin(Q) - T -u*M*g*cos(Q) = M*a\\[/tex] .. Eq2

- Add the two equation, Eq 1 and Eq 2:

                      [tex]M*g*sin ( Q ) - u*M*g*cos ( Q ) - m*g = a* ( M + m )\\\\a = \frac{M*g*sin ( Q ) - u*M*g*cos ( Q ) - m*g}{M + m} \\\\a = \frac{100*9.81*sin ( 20 ) - 0.3*100*9.81*cos ( 20 ) - 10*9.81}{100 + 10}\\\\a = \frac{-39.12977}{110} = -0.35572 \frac{m}{s^2}[/tex]

- The inclined block moves up ( the acceleration is in the opposite direction than assumed ).

- Using equation 1, we determine the tension ( T ) in the cable as follows:

                     [tex]T = m* ( a + g )\\\\T = 10*( -0.35572 + 9.81 )\\\\T = 94.54 N[/tex]

Answer:

-17.8 V

Explanation:

The induced emf in a coil is given as:

[tex]E = \frac{-NdB\pi r^2}{dt}[/tex]

where N = number of loops

dB = change in magnetic field

r = radius of coil

dt = elapsed time

From the question:

N = 50

dB = final magnetic field - initial magnetic field

dB = 0.35 - 0.10 = 0.25 T

r = 3 cm

dt = 2 ms = 0.002 secs

Therefore, the induced emf is:

[tex]E = \frac{-50 * 0.25 * \pi * 0.03^2}{0.002} \\E = -17.8 V[/tex]

Note: The negative sign implies that the EMf acts in an opposite direction to the change in magnetic flux.

Answer:

the two elements are resistor R and inductor L

answers in two significant figures

R = 9.6Ω

L = 54mH

Explanation:

Answer: No, we can have a displacement equal to 0 while the distance traveled is different than zero.

Explanation:

Ok, let's write the definitions:

Displacement: The displacement is equal to the difference between the final position and the initial position.

Distance traveled: Total distance that you moved.

So, for example, if at t = 0s, you are in your house, then you go to the store, and then you return to your house, we have:

The displacement is equal to zero, because the initial position is your house and the final position is also your house, so the displacement is zero.

But the distance traveled is not zero, because you went from you traveled the distance from your house to the store two times.

So no, we can have a displacement equal to zero, but a distance traveled different than zero.

Answer:

The found acceleration in terms of h and t is:

[tex]a=\frac{h}{5(t_1)^2}[/tex]

Explanation:

(The complete question is given in the attached picture. We need to find the acceleration in terms of h and t in this question)

We are given 3 stages of movement of elevator. We'll first model them each of the stage one by one to find the height covered in each stage. After that we'll find the total height covered by adding heights covered in each stage, and equate it to Total height h. From that we can find the formula for acceleration.

Constant acceleration, starts from rest.

Distance = [tex]y = \frac{1}{2}a(t_1)^2[/tex]

Velocity = [tex]v_1=at_1[/tex]

Constant velocity where

Velocity = [tex]v_o=v_1=at_1[/tex]

Distance =

Constant deceleration where

Velocity = [tex]v_0=v_1=at_1[/tex]

Distance =

[tex]y_3=v_1t_3-\frac{1}{2}a(t_3)^2\\\text{Where}~t_3=t_1\\y_3=v_1t_1-\frac{1}{2}a(t_1)^2\\\text{Where}~ v_1t_1=a(t_1)^2\\y_3=a(t_1)^2-\frac{1}{2}a(t_1)^2\\\text{Subtracting both terms:}\\y_3=\frac{1}{2}a(t_1)^2[/tex]

Total height = y₁ + y₂ + y₃

Total height = [tex]\frac{1}{2}a(t_1)^2+4a(t_1)^2+\frac{1}{2}a(t_1)^2 = 5a(t_1)^2[/tex]

Find acceleration by rearranging the found equation of total height.

Total Height = h

h = 5a(t₁)²

[tex]a=\frac{h}{5(t_1)^2}[/tex]

Answer:

0.15 m

Explanation:

First calculating the center of mass from the saw horse

[tex]\frac{3.15}{2} -1=0.575 m[/tex]

from the free body diagram we can write

Taking moment about the saw horse

55.9×9.81×y=14.5×0.575×9.81

y= 0.15 m

So, the painter walk from the saw horse on the right until the board begins to tip is 0.15 m far.

Answer:

the coefficient of volume expansion of the glass is [tex]\mathbf{ ( \beta_{glass} )= 1.333 *10^{-5} / K}[/tex]

Explanation:

Given that:

Initial volume of the glass flask = 1000 cm³ = 10⁻³ m³

temperature of the glass flask and mercury= 1.00° C

After heat is applied ; the final temperature = 52.00° C

Temperature change ΔT = 52.00° C - 1.00° C = 51.00° C

Volume of the mercury overflow = 8.50 cm^3 = 8.50 ×  10⁻⁶ m³

the coefficient of volume expansion of mercury is 1.80 × 10⁻⁴ / K

The increase in the volume of the mercury =  10⁻³ m³ ×  51.00 × 1.80 × 10⁻⁴

The increase in the volume of the mercury = [tex]9.18*10^{-6} \ m^3[/tex]

Increase in volume of the glass =  10⁻³ × 51.00 × [tex]\beta _{glass}[/tex]

Now; the mercury overflow = Increase in volume of the mercury - increase in the volume of the flask

the mercury overflow = [tex](9.18*10^{-6} - 51.00* \beta_{glass}*10^{-3})\ m^3[/tex]

[tex]8.50*10^{-6} = (9.18*10^{-6} -51.00* \beta_{glass}* 10^{-3} )\ m^3[/tex]

[tex]8.50*10^{-6} - 9.18*10^{-6} = ( -51.00* \beta_{glass}* 10^{-3} )\ m^3[/tex]

[tex]-6.8*10^{-7} = ( -51.00* \beta_{glass}* 10^{-3} )\ m^3[/tex]

[tex]6.8*10^{-7} = ( 51.00* \beta_{glass}* 10^{-3} )\ m^3[/tex]

[tex]\dfrac{6.8*10^{-7}}{51.00 * 10^{-3}}= ( \beta_{glass} )[/tex]

[tex]\mathbf{ ( \beta_{glass} )= 1.333 *10^{-5} / K}[/tex]

Thus; the coefficient of volume expansion of the glass is [tex]\mathbf{ ( \beta_{glass} )= 1.333 *10^{-5} / K}[/tex]

Answer:

Explanation:

each grid corresponding  0.1s⁻¹.

0.2grid unit = 0.2×0.1 =0.02s⁻¹

distance of the star from telescope

d = 1/p

d= 1/0.02= 50 par sec

1par sec = 3.26 light year

1 light year = 9.5×10¹²km

3.26ly=3.084×10¹³km

d= 50×3.084×10¹³=1.55×10¹⁵km

Answer:

Option B:

The normal force

Explanation:

The normal force does no work as the box slides down the ramp.

Work can only be done when the force succeeds in moving the object in the direction of the force.

All the other forces involved have a component that is moving the box in their direction.

However, the normal force does not, as it points downwards into the ramp. Since the normal force is pointing into the ramp, and the box is sliding down the ramp, we can say that no work is being done by the normal force because the box is not moving in its direction (which would have been the box moving into the ramp)

Sun's energy comes from the nuclear fusion taking place inside. In nuclear fusion two light nuclei fuses together to form a heavy nuclei with the release of greater amount of energy.

Nuclear fusion is the process of combining two light nuclei to form a heavy nuclei. In this nuclear process, tremendous energy is released. This is the source of heat and light in stars.

On the other hand, nuclear fission is the process of breaking of a heavy nuclei into two lighter nuclei. Fission also produces massive energy. But in comparison, more energy is produced by nuclear fusion.

Nuclear fission is used in nuclear power generators. The light energy and  heat energy comes form the nuclear fusion of hydrogens to form helium nuclei. Hence, option D is correct.

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Answer: [tex]\frac{P_B}{P_A}[/tex] = 8.5264

Explanation: Power is the rate of energy transferred per unit of time: P = [tex]\frac{E}{t}[/tex]

The energy from the engine is converted into kinetic energy, which is calculated as: [tex]KE = \frac{1}{2}.m.v^{2}[/tex]

To compare the power of the two cars, first find the Kinetic Energy each one has:

K.E. for Model A

[tex]KE_A = \frac{1}{2}.m.v^{2}[/tex]

K.E. for model B

[tex]KE_B = \frac{1}{2}.m.(2.92v)^{2}[/tex]

[tex]KE_B = \frac{1}{2}.m.8.5264v^{2}[/tex]

Now, determine Power for each model:

Power for model A

[tex]P_{A}[/tex] = [tex]\frac{m.v^{2} }{2.t}[/tex]

Power for model B

[tex]P_B = \frac{m.8.5264.v^{2} }{2.t}[/tex]

Comparing power of model B to power of model A:

[tex]\frac{P_B}{P_A} = \frac{m.8.5264.v^{2} }{2.t}.\frac{2.t}{m.v^{2} }[/tex]

[tex]\frac{P_B}{P_A} =[/tex] 8.5264

Comparing power for each model, power for model B is 8.5264 better than model A.

Given that,

Sun exposure = 30%, 45%, 60%, 75%, 90%

Stem mass (g) = 275, 415, 563, 815, 954

Stem volume (ml) = 1100, 1215, 1425, 1610, 1742

(a). We need to convert the mass measurements to kilograms (kg) and the volume measurements to cubic meters

Using conversion of mass

[tex]1\ g=0.001\ kg[/tex]

Conservation of volume

[tex]1\ Lt=0.001\ m^3[/tex]

[tex]1\ mL=1\times10^{-6}\ m^3[/tex]

So, mass in kg

Stem mass (kg) = 0.275, 0.415, 0.563, 0.815, 0.954

Volume in m³,

Stem volume (m³) = 0.0011, 0.001215, 0.001425, 0.001610, 0.001742

(b). We need to calculate the density of the samples

Using formula of density

[tex]\rho=\dfrac{m}{V}[/tex]

Where, m = mass

V = volume

If the m = 0.275 kg and V = 0.0011 m³

Put the value into the formula

[tex]\rho=\dfrac{0.275}{0.0011}[/tex]

[tex]\rho=250\ kg/m^3[/tex]

If the m = 0.415 kg and V = 0.001215 m³

Put the value into the formula

[tex]\rho=\dfrac{0.415}{0.001215}[/tex]

[tex]\rho=341.56\ kg/m^3[/tex]

[tex]\rho=342\ kg/m^3[/tex]

If the m = 0.563 kg and V = 0.001425 m³

Put the value into the formula

[tex]\rho=\dfrac{0.563}{0.001425}[/tex]

[tex]\rho=395.08\ kg/m^3[/tex]

If the m = 0.815 kg and V = 0.001610 m³

Put the value into the formula

[tex]\rho=\dfrac{0.815}{0.001610}[/tex]

[tex]\rho=506.21\ kg/m^3[/tex]

If the m = 0.954 kg and V = 0.001742 m³

Put the value into the formula

[tex]\rho=\dfrac{0.954}{0.001742}[/tex]

[tex]\rho=547.6\ kg/m^3[/tex]

[tex]\rho=548\ kg/m^3[/tex]

(c). We need to convert the density values to scientific notation

In scientific notation

The densities are

[tex]\rho\ (kg/m^3)= 2.50\times10^{2}, 3.42\times10^{2}, 3.95\times10^{2}, 5.06\times10^{2}, 5.48\times10^{2}[/tex]

Hence, This is required solution.

Answer:

Explanation:

Given that

The electric fields of strengths E = 187,500 V/m and

and The magnetic  fields of strengths B = 0.1250 T

The diameter d is 25.05 cm which is converted to 0.2505m

The radius is (d/2)

= 0.2505m / 2 = 0.12525m

The given formula to find the magnetic force is [tex]F_{ma}=BqV---(i)[/tex]

The given formula to find the electric force is [tex]F_{el}=qE---(ii)[/tex]

The velocity of electric field and magnetic field is said to be perpendicular

Electric field is equal to magnectic field

Equate equation (i) and equation (ii)

[tex]Bqv=qE\\\\v=\frac{E}{B}[/tex]

[tex]v=\frac{187500}{0.125} \\\\v=15\times10^5m/s[/tex]

It is said that the particles moves in semi circle, so we are going to consider using centripetal force

[tex]F_{ce}=\frac{mv^2}{r}---(iii)[/tex]

magnectic field is equal to centripetal force

Lets equate equation (i) and (iii)

[tex]Bqr=\frac{mv^2}{r} \\\\\frac{q}{m}=\frac{v}{Br} \\\\\frac{q}{m} =\frac{15\times 10^5}{0.125\times0.12525} \\\\=\frac{15\times10^5}{0.015656} \\\\=95808383.23\\\\=958.1\times10^5C/kg[/tex]

Therefore,  the particle's charge-to-mass ratio is [tex]958.1\times10^5C/kg[/tex]

b)

To identify the particle

Then 1/ 958.1 × 10⁵ C/kg

The charge to mass ratio is very close to that of a proton, which is about 1*10^8 C/kg

Therefore the particle is proton.

Answer:

If instead a charge 2q were placed at that same point the force will be 2F.

Explanation:

The electric force is equal to:

[tex] F = q*E [/tex]    (1)

Where:

F: is the electric force

q: is the charge

E: is the electric field

We can see that in equation (1) the electric force (F) is proportional to the charge q, thus, if now the charge it's the double (2q) then the force will be the double too:    

Initially:

[tex] F_{1} = q_{1}*E [/tex]

Now,

[tex](2q_{1}})*E = 2(q_{1}*E) = 2F_{1}[/tex]  

Therefore, if instead a charge 2q were placed at that same point the force will be 2F.

I hope it helps you!            

Answer:

f = 2200 Hz

Explanation:

It is given that,

Speed of a wave is 242 m/s

The distance between crests is 0.11 m

We need to find the frequency of the wave. The distance between crests is called wavelength of a wave. So,

[tex]v=f\lambda\\\\f=\dfrac{v}{\lambda}\\\\f=\dfrac{242}{0.11}\\\\f=2200\ Hz[/tex]

So, the frequency of the wave is 2200 Hz.

Answer:2200 hz

Explanation:

Answer:

The total internal energy change for the entire process is  -0.94 kJ

Explanation:

Process A to B is an isothermal process, therefore, [tex]u_A[/tex] - [tex]u_B[/tex] = 0

Process B to C

P = -mV + C

When P = 12, V = 0.12

When P = 4, V = 0.135

Therefore, we have;

12 = -m·0.12 + C

4 = -m·0.135 + C

Solving gives

m = 533.33

C = 76

[tex]T = \dfrac{1}{nR} \times (-533.33 \times V^2 + 76 \times V)[/tex]

p₂ = p₁V₁/V₂ = 12*0.1/0.12 = 10 kPa

The work done = 0.5*(0.135 - 0.12)*(4 - 10.0) = -0.045 kJ = -45 J

For heat supplied

Assuming an approximate polytropic process, we have;

Work done = (p₃×v₃ - p₂×v₂)/(n - 1)

Which gives;

-45 = (4*0.135 - 10*0.12)/(n -1)

∴ n -1 = (4*0.135 - 10*0.12)/-45 =   14.67

n = 15.67

Q = W×(n - γ)/(γ - 1)

Q = -45*(15.67 - 1.4)/(1.4 - 1) = -1,605.375 J

u₃ - u₂ = Q + W = -1,605.375 J - 45 J = -1650 J = -1.65 kJ

For the constant pressure process D to C, we have;

[tex]Q = c_p \times \dfrac{p}{R} \times (V_4 -V_3) = \dfrac{5}{2} \times p \times (V_4 -V_3)[/tex]

Q₄₋₃ = (0.1 - 0.135) * 4*5/2 = -0.35 kJ

W₄₋₃ = 4*(0.1 - 0.135) = -0.14 kJ

u₄ - u₃ = Q₄₋₃ + W₄₋₃ = -0.14 kJ + -0.35 kJ = -0.49 kJ

For the process D to A, we have a constant volume process

[tex]Q_{1-4} = \dfrac{c_v}{R} \times V \times (p_1 - p_4) = \dfrac{3}{2} \times 0.1 \times (12 - 4) = 1.2 \ kJ[/tex]

W₁₋₄ = 0 for constant volume process, therefore, u₁ - u₄ = 1.2 kJ

The total internal energy change Δ[tex]u_{process}[/tex] for the entire process is therefore;

Δ[tex]u_{process}[/tex] = u₂ - u₁ + u₃ - u₂ + u₄ - u₃ + u₁ - u₄ = 0  - 1.65 - 0.49 + 1.2 = -0.94 kJ.

Answer:

By convention a negative torque leads to clockwise rotation and a positive torque leads to counterclockwise rotation.

here weight of the child =21kgx9.8m/s2 = 205.8N

the torque exerted by the child Tc = - (1.8)(205.8) = -370.44N-m ,negative sign is inserted because this torque is clockwise and is therefore negative by convention.

torque exerted by adult Ta = 3(151) = 453N , counterclockwise torque.

net torque Tnet = -370.44+453 =82.56N , which is positive means counterclockwise rotation.

b) Ta = 2.5x151 = 377.5N-m

Tnet = -370.44+377.5 = 7.06N-m , positive ,counterclockwise rotation.

c)Ta = 2x151 = 302N-m

Tnet = -370.44+302 = -68.44N-m, negative,clockwise rotation.

Answer:

a) The distance from the cornea vertex to the retina is 2.37 cm

b) This distance is shorter than for the normal eye.

Explanation:

a) Let refractive index of air,

n(air) = x = 1

Let refractive index of lens,

n(lens) = y = 1.4

Object distance, s = 36 cm

Radius of curvature, R = 0.65 cm

The distance from the cornea vertex to the retina is the image distance because image is formed in the retina.

Image distance, s' = ?

(x/s) + (y/s') = (y-x)/R

(1/36) + (1.4/s') = (1.4 - 1)/0.65

1.4/s' = 0.62 - 0.028

1.4/s' = 0.592

s' = 1.4/0.592

s' = 2.37 cm

Distance from the cornea vertex to the retina is 2.37 cm

(b) For a normal eye, the distance between the cornea vertex and the retina is 2.60 cm. Since 2.37 < 2.60, this distance is shorter than for normal eye.

Answer:

c. The astronaut does not need to worry: the charge will remain on the outside surface.

Explanation:

The astronaut need not worry because according to Gauss's law of electrostatic, a hollow charged surface will have a net zero charge on the inside. This is the case of a Gauss surface, and all the charges stay on the surface of the metal chamber. This same principle explains why passengers are safe from electrostatic charges, in an enclosed aircraft, high up in the atmosphere; all the charges stay on the surface of the aircraft.

Answer:

1.424 μC

Explanation:

I'm assuming here, that the charged ball is suspended by the string. If the string also is deflected by the angle α, then the forces acting on it would be: mg (acting downwards),

tension T (acting along the string - to the pivot point), and

F (electric force – acting along the line connecting the charges).

We then have something like this

x: T•sin α = F,

y: T•cosα = mg.

Dividing the first one by the second one we have

T•sin α/ T•cosα = F/mg, ultimately,

tan α = F/mg.

Since we already know that

q1=q2=q, and

r=2•L•sinα,

k=9•10^9 N•m²/C²

Remember,

F =k•q1•q2/r², if we substitute for r, we have

F = k•q²/(2•L•sinα)².

tan α = F/mg =

= k•q²/(2•L•sinα)² •mg.

q = (2•L•sinα) • √(m•g•tanα/k)=

=(2•0.5•0.486) • √(0.0142•9.8•0.557/9•10^9) =

q = 0.486 • √(8.61•10^-12)

q = 0.486 • 2.93•10^-6

q = 1.424•10^-6 C

q = 1.424 μC.

Answer:

1.35 kJ  

Explanation:

KE = ½mv² = ½ × 0.030  kg × (300 m·s⁻¹)² = 1350 J = 1.35 kJ

Given:-

To Find: Kinetic energy of the bullet.

We know,

Eₖ = ½mv²

where,

thus,

Eₖ = ½(30 g)(300 m/s)²

= (15 g)(90000 m²/s²)

= 1350000 g m²/s²

= 1350 kg m²/s²

= 1350 J

= 1.35 kJ (Ans.)