Answer:
Muscular endurance is how many times you can move a weight without getting tired.
Muscular strength is the amount force you can put out.
when a stationary rugby ball is kicked, it is contact with a player's about for 0.05 s. during this short time, the ball accelerates at 600m/s/s.calculate the speed at which the ball leaves the player's boot
Answer:
30 m/s
Explanation:
Applying,
v = u+at................ Equation 1
Where v = final speed of the ball, u = initial speed of the ball, a = acceleration, t = time.
From the question,
Given: u = 0 m/s (stationary), a = 600 m/s², t = 0.05 s
Substitute these values into equation 5
v = 0+(600×0.05)
v = 30 m/s
Hence the speed at which the ball leaves the player's boot is 30 m/s
80 grams of iron at 100°C is dropped into 200 of water at 20°C contained in an iron vessel of mass 50 gram find the resulting temperature.
Answer:
the resulting temperature is 23.37 ⁰C
Explanation:
Given;
mass of the iron, m₁ = 80 g = 0.08 kg
mass of the water, m₂ = 200 g = 0.2 kg
mass of the iron vessel, m₃ = 50 g = 0.05 kg
initial temperature of the iron, t₁ = 100 ⁰C
initial temperature of the water, t₂ = 20 ⁰C
specific heat capacity of iron, c₁ = 462 J/kg⁰C
specific heat capacity of water, c₂ = 4,200 J/kg⁰C
let the temperature of the resulting mixture = T
Apply the principle of conservation of energy;
heat lost by the hot iron = heat gained by the water
[tex]m_1c_1 \Delta t_1 = m_2c_2\Delta t_2\\\\m_1c_1 (100 - T) = m_2c_2 (T- 20)\\\\0.08 \times 462 (100-T) = 0.2 \times 4,200 (T-20)\\\\36.96 (100-T) = 840 (T-20) \\\\100 - T = 22.72 (T-20)\\\\100-T = 22.72 T - 454.4 \\\\554.4 = 23.72T\\\\T = \frac{554.4}{23.72} \\\\T = 23.37 \ ^0C[/tex]
Therefore, the resulting temperature is 23.37 ⁰C
A solid conducting sphere of radius ra is placed concentrically inside a conducting spherical shell of inner radius rb1 and outer radius rb2. The inner sphere carries a charge Q while the outer sphere does not carry any net charge. The potential for rb1 < r < rb2 is:________
Answer:
The right answer is "[tex]\frac{KQ}{r_b_2}[/tex]".
Explanation:
As the outer spherical shell is conducting, so there is no electric field in side from
⇒ [tex]r_b_1 < r < r_b_2[/tex].
So the electric potential at all points inside the conducting shell that from
⇒ [tex]r_b_1<r<r_b_2[/tex]
and will be similar as well as equivalent to the potential on the outer surface of the shell that will be:
⇒ [tex]v=\frac{KQ}{r_b_2}[/tex]
Thus the above is the right solution.
A rigid, insulated tank whose volume is 10 L is initially evacuated. A pinhole leak develops and air from the surroundings at 1 bar, 25 C enters the tank until the pressure in the tank becomes 1 bar (assume ideal gas model k=1.4 for the air). Find:
A) final temperature in tank.
B) amount of air that leaks into tank in grams.
C) amount of entropy produced in J/K.
Answer:
The answer is "[tex]143.74^{\circ} \ C , 8.36\ g, and \ 2.77\ \frac{K}{J}[/tex]"
Explanation:
For point a:
Energy balance equation:
[tex]\frac{dU}{dt}= Q-Wm_ih_i-m_eh_e\\\\[/tex]
[tex]W=0\\\\Q=0\\\\m_e=0[/tex]
From the above equation:
[tex]\frac{dU}{dt}=0-0+m_ih_i-0\\\\\Delta U=\int^{2}_{1}m_ih_idt\\\\[/tex]
because the rate of air entering the tank that is [tex]h_i[/tex] constant.
[tex]\Delta U = h_i \int^{2}_{1} m_i dt \\\\= h_i(m_2 -m_1)\\\\m_2u_2-m_1u_2=h_i(M_2-m_1)\\\\[/tex]
Since the tank was initially empty and the inlet is constant hence, [tex]m_2u-0=h_1(m_2-0)\\\\m_2u_2=h_1m_2\\\\u_2=h_1\\\\[/tex]
Interpolate the enthalpy between [tex]T = 300 \ K \ and\ T=295\ K[/tex]. The surrounding air
temperature:
[tex]T_1= 25^{\circ}\ C\ (298.15 \ K)\\\\\frac{h_{300 \ K}-h_{295\ K}}{300-295}= \frac{h_{300 \ K}-h_{1}}{300-295.15}[/tex]
Substituting the value from ideal gas:
[tex]\frac{300.19-295.17}{300-295}=\frac{300.19-h_{i}}{300-298.15}\\\\h_i= 298.332 \ \frac{kJ}{kg}\\\\Now,\\\\h_i=u_2\\\\u_2=h_i=298.33\ \frac{kJ}{kg}[/tex]
Follow the ideal gas table.
The [tex]u_2= 298.33\ \frac{kJ}{kg}[/tex] and between temperature [tex]T =410 \ K \ and\ T=240\ K.[/tex]
Interpolate
[tex]\frac{420-410}{u_{240\ k} -u_{410\ k}}=\frac{420-T_2}{u_{420 k}-u_2}[/tex]
Substitute values from the table.
[tex]\frac{420-410}{300.69-293.43}=\frac{420-T_2}{{u_{420 k}-u_2}}\\\\T_2=416.74\ K\\\\=143.74^{\circ} \ C\\\\[/tex]
For point b:
Consider the ideal gas equation. therefore, p is pressure, V is the volume, m is mass of gas. [tex]\bar{R} \ is\ \frac{R}{M}[/tex] (M is the molar mass of the gas that is [tex]28.97 \ \frac{kg}{mol}[/tex] and R is gas constant), and T is the temperature.
[tex]n=\frac{pV}{TR}\\\\[/tex]
[tex]=\frac{(1.01 \times 10^5 \ Pa) \times (10\ L) (\frac{10^{-3} \ m^3}{1\ L})}{(416.74 K) (\frac{8.314 \frac{J}{mol.k} }{2897\ \frac{kg}{mol})}}\\\\=8.36\ g\\\\[/tex]
For point c:
Entropy is given by the following formula:
[tex]\Delta S = mC_v \In \frac{T_2}{T_1}\\\\=0.00836 \ kg \times 1.005 \times 10^{3} \In (\frac{416.74\ K}{298.15\ K})\\\\=2.77 \ \frac{J}{K}[/tex]
During a particular thunderstorm, the electric potential difference between a cloud and the ground is Vcloud - Vground = 4.20 108 V, with the cloud being at the higher potential. What is the change in an electron's electric potential energy when the electron moves from the ground to the cloud?
Answer:
The electric potential energy is 6.72 x 10^-11 J.
Explanation:
Potential difference, V = 4.2 x 10^8 V
charge of electron, q = - 1.6 x 10^-19 C
Let the potential energy is U.
U = q V
U = 1.6 x 10^-19 x 4.2 x 10^8
U = 6.72 x 10^-11 J
A ten loop coil of area 0.23 m2 is in a 0.047 T uniform magnetic field oriented so that the maximum flux goes through the coil. The average emf induced in the coil is
Answer:
Explanation:
From the question we are told that:
Number of turns [tex]N=10[/tex]
Area [tex]a=0.23m^2[/tex]
Magnetic field [tex]B=0.947T[/tex]
Generally the equation for maximum flux is mathematically given by
[tex]\phi=NBa[/tex]
[tex]\phi=10*0.047*0.23[/tex]
[tex]\phi=0.1081wbi[/tex]
Therefore induced emf
[tex]e= \frac{d\phi}{dt}[/tex]
Since
[tex]t=0[/tex]
Therefore
[tex]e=0[/tex]
the coefficient of linear expansion of aluminum is 24.0 * 10^-6K^-1 and th density of aluminum at ) C is 2.7 *10^3kg/m^3. what is the density of aluminum at 300 C
Answer:
The density at 300 C is [tex]2641.68 kg/m^3[/tex].
Explanation:
density at 0 degree, d = 2700 kg/m3
coefficient of linear expansion, = 24 x 10^{-6} /K
Let the density at 300 C is d'.
Use the formula of the cubical expansion
[tex]d'= d (1-3\alpha \Delta T)\\\\d'= 2700(1- 3 \times 24\times 10^{-6}\times 300)\\\\d' = 2641.68 kg/m^3[/tex]
In a large chemical factory, a feed pipe carries a liquid at a speed of 5.5 m/s. A pump pushes the liquid along at a gauge pressure of 140,000 Pa. The liquid travels upward 6.0 m and enters a tank at a gauge pressure of 2,000 Pa. The diameter of the pipe remains constant. At what speed does the liquid enter the tank
Answer:
v₂ = 15.24 m / s
Explanation:
This is an exercise in fluid mechanics
Let's write Bernoulli's equation, where the subscript 1 is for the factory pipe and the subscript 2 is for the tank.
P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂
They indicate the pressure in the factory P₁ = 140000 Pa, the velocity
v₁ = 5.5 m / s and the initial height is zero y₁ = 0
the tank is at a pressure of P2 = 2000 Pa and a height of y₂ = 6.0 m
P₁ -P₂ + ρ g (y₁ -y₂) + ½ ρ v₁² = ½ ρ v₂²
let's calculate
140,000 - 2000 + ρ 9.8 (0- 6) + ½ ρ 5.5² = ½ ρ v₂²
138000 - ρ 58.8 + ρ 15.125 = ½ ρ v2²
v₂² = 2 (138000 /ρ - 58.8 + 15.125)
v₂ = [tex]\sqrt{\frac{276000}{\rho } - 43.675 }[/tex]
In the exercise they do not indicate what type of liquid is being used, suppose it is water with
ρ = 1000 kg / m³
v₂ = [tex]\sqrt{\frac{276000}{1000} - 43.675}[/tex]
v₂ = 15.24 m / s
Which quantities below of a solid object on this planet are NOT the same as on Earth?
Choose all
possible answers.
Weight
Mass
Volume
Density
Acceleration when it falls vertically.
Color
Answer:
Weight, acceleration when it falls vertically, are not same as that of earth.
Explanation:
Weight of the object is given by the product of mass of the object and the acceleration due to gravity of the planet.
So, the weight of object is not same as that on earth.
The mass is defined as the amount of matter contained in the object.
So, the mass of the object is same as that of earth.
The volume of the object is defined as the space occupied by the object.
So, the volume of the object is same as that of earth.
The density is defined as the ratio of mass of the object to its volume.
So, the density of the object is same as that of earth.
The acceleration due to gravity on a planet depends on the mass of planet and radius of planet.
So, the acceleration is not same as that of earth.
The color of the object is its characteristic.
It is same as that of earth.
An electric field E⃗ =5.00×105ı^N/C causes the point charge in the figure to hang at an angle. What is θ?
We have that the angle is
[tex]\theta=32.53[/tex]
From the Question we are told that
E⃗ =5.00×105ı^N/C
Generally the equation for Tension is mathematically given
[tex]W=Tcos\theta[/tex]
Where
[tex]tan\theta=\frac{2.5*10^{-9}(5*10{5})}{2*10^{-3}(9.8)}[/tex]
[tex]\theta=32.53[/tex]
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Picture isn’t the best but could use the help
Answer:
60 deg from Y-axis is 150 deg from X-axis
x = L cos theta = 23,2 cos 150 = -.867 * 23.2 = -20.1
what is the prefix notation of 0.0000738?
Answer:
7.38 × 10-5
Explanation:
All numbers in scientific notation or standard form are written in the form m × 10n, where m is a number between 1 and 10 ( 1 ≤ |m| < 10 ) and the exponent n is a positive or negative integer.
To convert 0.0000738 into scientific notation, follow these steps:
Move the decimal 5 times to the right in the number so that the resulting number, m = 7.38, is greater than or equal to 1 but less than 10
Since we moved the decimal to the right the exponent n is negative
n = -5
Write in the scientific notation form, m × 10n
= 7.38 × 10-5
Therefore, the decimal number 0.0000738 written in scientific notation is 7.38 × 10-5 and it has 3 significant figures.
Answer = 7.38 × 10-5
The top of a swimming pool is at ground level. If the pool is 2.60 m deep, how far below ground level does the bottom of the pool appear to be located for the following conditions?
a. The pool is completely filled with water.
______m below ground level
b. The pool is filled halfway with water.
______m below ground level
Answer:
a) [tex]d_g=1.95m[/tex]
b) [tex]d_g'=2.3m[/tex]
Explanation:
From the question we are told that:
Depth [tex]d=2.60[/tex]
a)
Generally the equation for distance to ground is mathematically given by
[tex]\frac{d_g}{d}=\frac{1}{n}[/tex]
Where
[tex]n=\frac{4}{3}[/tex]
Therefore
[tex]d_g=\frac{d}{n}[/tex]
[tex]d_g=\frac{2.6}{4/3}[/tex]
[tex]d_g=1.95m[/tex]
b)
For when The pool is filled halfway with water
[tex]\frac{d_g}{d}=\frac{1}{n}[/tex]
[tex]d_g'=\frac{1.3}{4/3}[/tex]
[tex]d_g'=0.98m[/tex]
Therefore
[tex]d_g'=(1.3+0.98)[/tex]
[tex]d_g'=2.3m[/tex]
A hollow steel ball weighing 4 pounds is suspended from a spring. This stretches the spring 1515 feet. The ball is started in motion from the equilibrium position with a downward velocity of 88 feet per second. The air resistance (in pounds) of the moving ball numerically equals 4 times its velocity (in feet per second) . Suppose that after t seconds the ball is y feet below its rest position. Find y in terms of t. (Note that the positive direction is down.) Take as the gravitational acceleration 32 feet per second per second.
Answer:
[tex]\mathbf{y(t) =0.408e^{-6.20t }-0.408e^{-25.80t}}}[/tex]
Explanation:
Given that:
The weight of ball = 4 pounds
The spring stretch x = 1/15 feet
Using the relation of weight on an object:
W = mg
m = W/g
m = 4 / 32
m = 1/8
Now, from Hooke's law:
F = kx
4 =k(1/5)
k = 20 lb/ft
However, since the air resistance is 4 times the velocity;
Then, we can say:
C = 4
Now, for the damped vibration in the spring-mass system, we have:
[tex]m\dfrac{d^2 y}{dx^2}+ c\dfrac{dy}{dt}+ky = 0[/tex]
[tex](\dfrac{1}{8})\dfrac{d^2 y}{dx^2}+ 4\dfrac{dy}{dt}+20y = 0[/tex]
[tex]\dfrac{d^2 y}{dx^2}+ 32\dfrac{dy}{dt}+160y = 0[/tex]
Solving the differential equation:
m² + 32m + 160 = 0
Solving the equation:
m = -25.80 or m = -6.20
So, the general solution for the equation is:
[tex]y (t)= c_1 e^{-6.20t}+c_2e^{-25.80t}[/tex]
[tex]y '(t)=-6.20 c_1 e^{-6.20t}-25.80c_2e^{-25.80t}[/tex]
y(0) = 0 ; y'(0) = 8
[tex]y (0)= c_1 e^{-6.20(0)}+c_2e^{-25.80(0)}[/tex]
[tex]c_1 +c_2 = 0 ---(1)[/tex]
At y'(0) = 8
[tex]y '(0)=-6.20 c_1 e^{-6.20(0)}-25.80c_2e^{-25.80(0)} \\ \\ 8=-6.20 c_1 e^{0}-25.80c_2e^{0} \\ \\ 8=-6.20 c_1 -25.80c_2--- (2)[/tex]
From (1), let [tex]c_1 = -c_2[/tex], then replace the value of c_1 into equation (2)
[tex]8=-6.20 (-c_2)-25.80c_2[/tex]
[tex]8=6.20c_2-25.80c_2[/tex]
[tex]8=-19.60c_2[/tex]
[tex]c_2=\dfrac{ 8}{-19.60}[/tex]
[tex]c_2 = -0.408[/tex]
From [tex]c_1 = -c_2[/tex]
[tex]c_1 = -(-0.408)[/tex]
[tex]c_1 = 0.408[/tex]
∴
The required solution in terms of t is:
[tex]\mathbf{y(t) =0.408e^{-6.20t }-0.408e^{-25.80t}}}[/tex]
Suppose you want to design an air bag system that can protect the driver at a speed 100 km/h (60 mph) if the car hits a brick wall.
Estimate how fast the air bag must inflate to effectively protect the driver. How does the use of a seat belt help the driver?
When solving question that contains equations and the use mathematical computations, It is always ideal to list the parameters given.
Now, given that:
the speed of the car which is the initial velocity (u) = 100 km/h before it hits the wall.after hitting the wall, the final velocity will be (v) = 0 km/hAssumptions:
Suppose we make an assumption that the distance traveled during the collision of the car with the brick wall (S) = 1 mThat the car's acceleration is also constant.∴
For a motion under constant acceleration, we can apply the kinematic equation:
[tex]\mathsf{v^2 = u^2 + 2as}[/tex]
where;
v = final velocity u = initial velocitya = accelerations = distanceFrom the above equation, making acceleration (a) the subject of the formula:
[tex]\mathsf{v^2 - u^2 =2as }[/tex]
[tex]\mathsf{a = \dfrac{v^2 - u^2 }{2s}}[/tex]
The initial velocity (u) is given in km/h, and we need to convert it to m/s as it has an effect on the unit of the acceleration.
since 1 km/h = 0.2778 m/s
100 km/h = 27.78 m/s
[tex]\mathsf{a = \dfrac{(0)^2 - (27.78)^2 }{2(1)}}[/tex]
[tex]\mathsf{a = \dfrac{- 771.7284 }{2}}[/tex]
a = - 385.86 m/s²
Similarly, from the kinematic equation of motion, the formula showing the relation between time, acceleration, and velocity is;
v = u + at
where;
v = 0
-u = at
[tex]\mathsf{t = \dfrac{-u}{a}}[/tex]
[tex]\mathsf{t = \dfrac{-27.78}{-385.86}}[/tex]
t = 0.07 seconds
An airbag is designed in such a way as to prevent the driver from hitting on the steering wheel or other hard substance that could damage the part of the body. The use of the seat belt is to keep the driver in shape and in a balanced position against the expansion that occurred by the airbag during the collision on the brick wall.
Thus, we can conclude that in order to estimate how fast the airbag must inflate to effectively protect the driver, the airbag must be inflated at 0.07 seconds faster before the collision to effectively protect the driver.
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A 12.0 g sample of gas occupies 19.2 L at STP. what is the of moles and molecular weight of this gas?
At STP, 1 mole of an ideal gas occupies a volume of about 22.4 L. So if n is the number of moles of this gas, then
n / (19.2 L) = (1 mole) / (22.4 L) ==> n = (19.2 L•mole) / (22.4 L) ≈ 0.857 mol
If the sample has a mass of 12.0 g, then its molecular weight is
(12.0 g) / n ≈ 14.0 g/mol
A car starting at rest accelerates at 3m/seconds square How far has the car travelled after 4s?
Answer:
24 meters
Explanation:
Find the final velocity. 12m/s
d=[final-initial]/2×time
D=(6m/s)×4=24 m/s
Electromagnetic Wave PenetrationRadio waves and mircowaves are used in therapy to provide "deep heating" of tissue because the waves penetrate beneth the surface of the body and deposit energy. We define the penetration depth as the depth at which the wave intensity has decreased to 37% of its value at the surface. The penetration depth is 15cm for 27MHz radio waves. For radio frequencies such as this, the penetration depth is proportional to (sqrt of lambda), which is the square root of the wavelength. What is the wavelength of 27 MHz radio waves? A. 11 m.B. 9.0 m.C. 0.011 m.D. 0.009 m.If the frequency of the radio waves is increased, the depth of penetration:_____.A. Increases. B. Does not change. C. Decreases. 3. For 27 MHzMHz radio waves, the wave intensity has been reduced by a factor of 3 at a depth of approximately 15 cmcm. At this point in the tissue, the electric field amplitude has decreased by a factor of:____.A. 9 77. B. 3√3. C. 3. D. √3.
Answer:
Hence the square root of the wavelength. What is the wavelength of 27 MHz radio waves 11 m; option A is the correct answer.
Hence, as the frequency of the radio waves increases the penetration depth decreases.
Option C is the correct answer.
Hence the electric field amplitude has decreased by a factor of √3.
Hence the correct option is D.
Explanation:
[tex]c=n\lambda[/tex], where the symbols have their usual meanings.
[tex]\lambda =c/n\\= 3\times10^{8}/27\times10^{6}\\\\=11.11 m[/tex]
Hence, wavelength = 11 m; option A is the correct answer.
As given that, the penetration depth is proportional to the square root of wavelength, and we know that,
wavelength and frequency are inversely proportional to each other.
Hence, as the frequency of the radio waves increases the penetration depth decreases.
Option C is the correct answer.
Electric field amplitude is directly proportional to the square root of the intensity. Therefore, the ratio between the electric field amplitude is,
[tex]\frac{E\sigma_{1} }{E\sigma_{2} } =\sqrt{\frac{I_{1} }{I_{2} } }[/tex]
Here to find the ratio of intensities,
[tex]{\frac{I_{1} }{I_{2} } } = 3[/tex]
Therefore,
[tex]\frac{E\sigma_{1} }{E\sigma_{2} } =\sqrt{3 }[/tex]
Hence the correct option is D.
A 1.8-m-tall diver is standing completely submerged on the bottom of a swimming pool, in 3.0 m of water. You are sitting on the end of the diving board, almost directly over her.
Required:
How tall does the diver appear to be?
Answer:
[tex]H=1.35m[/tex]
Explanation:
From the question we are told that:
Height [tex]h=1.8m[/tex]
Depth of Water [tex]d=3.0m[/tex]
Where
Refraction index of water [tex]n = 4 / 3[/tex]
Generally the equation for Apparent height of the diver is mathematically given by
[tex]H=h\frac{h}{n}[/tex]
[tex]H=\frac{1.8}{\frac{4}{3}}[/tex]
[tex]H=1.35m[/tex]
A cylindrical water tank has a height of 20cm and a radius of 14cm. If it is filled to 2/5 of its capacity, calculate.
I. Quantity of water in the tank
II. Quantity of water left to fill the tank to its capacity.
Answer:
4.926 L Y 7.389 L
Explanation:
first you calculate the tank volume
V = π[tex](14 cm)^{2}[/tex](10 cm = [tex]12315 cm^{3}[/tex]
then you convert to liters
[tex]12315 cm^{3}[/tex] = 12.315 l
then you calculate the liters of water
2/5(12.35 l) = 4.926 l
finally we calculate the amount without water
12.315 l - 4.926 l = 7.389 l
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6. Two astronauts of equal mass step from the top of a rocket on Venus. One slides down a 10- degree ramp while the other slides down a 75-degree ramp. If all friction is ignored, which astronaut reaches the surface of Venus with the lower kinetic energy
Answer:
Final kinetic energies of both astronauts will be the same.
Explanation:
If we ignore all the friction present between the ramp and the person. Then essentially there is no loss of energy in the system. Hence the initial potential energies of the astronauts must be equal to their final kinetic energies.
Now the potential energy depends upon mass, height and acceleration due to gravity. All these parameters for both the astronauts. Therefore,both astronauts have same initial potential energies.
Similarly, the final kinetic energies of astronauts will also be the same.
A technician builds an RLC series circuit which includes an AC source that operates at a fixed frequency and voltage. At the operating frequency, the resistance R is equal to the inductive reactance XL. The technician notices that when the plate separation of the parallel-plate capacitor is reduced to one-half its original value, the current in the circuit doubles. Determine the initial capacitive reactance in terms of the resistance R.
Answer:
Xc = (0.467 - 0.427j)R
Explanation:
Since the resistance in the circuit is R, the reactance of the inductor is XL and the reactance of the capacitor is XC, then the impedance of the circuit is
Z = √[R² + (XL - XC)²]
Since the inductive reactance XL equals the resistance R, we have that
Z = √[R² + (XL - XC)²]
Z = √[R² + (R - XC)²]
Thus, the current in the circuit is thus I = V/Z = V/√[R² + (R - XC)²]
Now, when the plate separation of the parallel plate capacitor is reduced to one-half its original value, the current doubles. Also, when the plate separation is reduced to half, the capacitance doubles since C ∝ 1/d where C is capacitance and d separation between the plates. Since the capacitance doubles, the new reactance XC' is twice the initial reactance XC. So, XC' = 2XC. Thus the new impedance is thus
Z' = √[R² + (R - XC')²]
Z' = √[R² + (R - 2XC)²]
The new current is I' = V/Z' = V/√[R² + (R - 2XC)²]
Since the current doubles, I' = 2I.
V/√[R² + (R - 2XC)²] = 2V/√[R² + (R - XC)²]
1/√[R² + (R - 2XC)²] = 2/√[R² + (R - XC)²]
√[R² + (R - XC)²] = 2√[R² + (R - 2XC)²]
squaring both sides, we have
[R² + (R - XC)²] = 4[R² + (R - 2XC)²]
expanding the brackets, we have
[R² + R² - 2RXC + XC²] = 4[R² + R² - 4RXC + 4XC²]
[2R² - 2RXC + XC²] = 4[2R² - 4RXC + 4XC²]
2R² - 2RXC + XC² = 8R² - 16RXC + 16XC²
collecting like terms, we have
16RXC - 2RXC + XC² - 16XC² = 8R² - 2R²
14RXC - 15XC² = 6R²
15XC² - 14RXC + 6R² = 0
Using the quadratic formula to find XC, we have
[tex]XC = \frac{-(-14R) +/- \sqrt{(-14R)^{2} - 4 X 15 X 6R^{2} } }{2 X 15}\\= \frac{-(-14R) +/- \sqrt{196R^{2} - 360R^{2} } }{30}\\ \\= \frac{14R +/- \sqrt{- 164R^{2} } }{30}\\ \\= \frac{14R +/- 12.81Ri }{30}\\\\= 0.467R +/- 0.427Ri[/tex]
Since it is capacitive, we take the negative part.
So, Xc = (0.467 - 0.427j)R
Which two factors does the power of a machine depend on? А. work and distance B.. force and distance C. work and time D. time and distance?
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The answer is...
C. Work and time.
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When a ray of light passes through a boundary between two transparent media, it changes direction. What is this phenomenon known as?
Choose the best answer from the choices provided.
a. reflection
b. refraction
c. interference
d. diffraction
e. dispersion
Answer:
I think Refraction
Explanation:
I think refraction is light which passes through a boundary between 2 transparent media.
monochromatic light of wavelength 500 nm is incident normally on a diffraction grating. if the third order maximum is 32. how many total number of maximuima can be seen
Answer:
The total number of maxima that can be seen is 11
Explanation:
Given the data in the question
wavelength λ = 500 nm = 5 × 10⁻⁷ m
if the third order maximum is 32
i.e m = 3 and θ = 32°
Now, we know that condition for diffraction maximum is as follows;
d × sinθ = m × λ
so we substitute in our given values
d × sin( 32° ) = 3 × 5 × 10⁻⁷ m
d × sin( 32° ) = 1.5 × 10⁻⁶ m
d = [ 1.5 × 10⁻⁶ m ] / sin( 32° )
d = 2.83 × 10⁻⁶ m
Now, maxima n when θ = 90° will be;
sin( 90° ) = nλ / d
1 = nλ / d
d = nλ
n = d / λ
we substitute
n = [ 2.83 × 10⁻⁶ m ] / [ 5 × 10⁻⁷ m ]
n = 5.66
so 5 is the max value
hence, total maxima value is;
⇒ 2n + 1 = 2( 5 ) + 1 = 10 + 1 = 11
Therefore, total number of maxima that can be seen is 11
The carpenter question plz ASAP
Answer:
d. The hammer falls with a constant acceleration
Explanation:
Since gravity is the only thing that is acting on the hammer as it falls and gravity is a form of acceleration then acceleration of 9.81m/s² which is gravity is the correct answer.
If the Velocity of the body
is increased to 3v, determine the kinetic energy
ANSWER; KE=5mv^2 so it is proportional to v^2.
Explanation:So if you triple the velocity you are replacing v with 3v. Then you get (3v)^2=9v^2.
A source of emf is connected by wires to a resistor and electrons flow in the circuit the wire diameter is teh same throughout teh circuit compared to the drift speed of the electrons before entering the source of emf, the drift speed ot eh electrons afte rleaving the source of emf is: ___________
a. faster.
b. slower.
c. the same.
d. either A or B depending on circumstances.
e. any of A, B , or C depending on circumstances.
A solid object is made of two materials, one material having density of 2 000 kg/m3 and the other having density of 6 000 kg/m3. If the object contains equal masses of the materials, what is its average density
Answer:
[tex]\rho_{avg}=4000kg/m^3[/tex]
Explanation:
From the question we are told that:
Density of Material 1 [tex]\rho_1=2000kg/m^3[/tex]
Density of Material 2 [tex]\rho_2=6000kg/m^3[/tex]
Generally the equation for Average density is mathematically given by
[tex]\rho_{avg}=frac{\rho _1+rho _2}{2}[/tex]
[tex]\rho_{avg}=\frac{2000+6000}{2}[/tex]
[tex]\rho_{avg}=4000kg/m^3[/tex]
A T-shirt cannon launches a shirt at 5.30 m/s from a platform height of 4.00 m from ground level. How fast (in m/s) will the shirt be traveling if it is caught by someone whose hands are at 5.20 m from ground level (b) 4.00 m from ground level?
Answer:
(a) the velocity of the shirt is 2.14 m/s
(b) the velocity of the shirt is 5.3 m/s
Explanation:
Given;
initial velocity of the shirt, u = 5.3 m/s
height of the platform above the ground, h = 4.00 m
(a) When the shirt is caught by someone whose hand is 5.20 m from the ground level, the height traveled by the shirt = 5.2 m - 4.0 m = 1.2 m
The velocity at this position is calculated as;
[tex]v^2 = u^2 + 2(-g)h\\\\v^2 = u^2 - 2gh\\\\v^2 = 5.3^2 - (2\times 9.8 \times 1.2)\\\\v^2 = 4.57\\\\v= \sqrt{4.57} \\\\v = 2.14 \ m/s[/tex]
(b) When the shirt is caught by someone whose hand is 4.00 m from the ground level, the height traveled by the shirt = 4.00 m - 4.00 m = 0 m
The velocity at this position is calculated as;
[tex]v^2 = u^2 + 2(-g)h\\\\v^2 = u^2 - 2gh\\\\v^2 = 5.3^2 - (2\times 9.8 \times 0)\\\\v^2 = 28.09\\\\v= \sqrt{28.09} \\\\v = 5.3 \ m/s[/tex]
