Answer:
3.6×10⁻⁶ Nm
Explanation:
From the question,
The expression for maximum torque is given as
τ = BANI.................Equation 1
where τ = maximum torque, B = magnetic field, A = Area of the coil, N = number of turns, I = current carried by the coil.
Given: B = 0.1 T, A = (5×12) = 60 cm² = 0.006 m², N = 600 turns, I = 10⁻⁵ A.
Substitute these values into equation 1
τ = 0.1(0.006)(600)(10⁻⁵)
τ = 3.6×10⁻⁶ Nm
(a) According to Hooke's Law, the force required to hold any spring stretched x meters beyond its natural length is f(x)=kx. Suppose a spring has a natural length of 20 cm. If a 25-N force is required to keep it stretched to a length of 30 cm, how much work is required to stretch it from 20 cm to 25 cm?
(b) Find the area of the region enclosed by one loop of the curve r=2sin(5θ).
Answer:
a) The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules, b) The area of the region enclosed by one loop of the curve [tex]r(\theta) = 2\cdot \sin 5\theta[/tex] is [tex]4\pi[/tex].
Explanation:
a) The work, measured in joules, is a physical variable represented by the following integral:
[tex]W = \int\limits^{x_{f}}_{x_{o}} {F(x)} \, dx[/tex]
Where
[tex]x_{o}[/tex], [tex]x_{f}[/tex] - Initial and final position, respectively, measured in meters.
[tex]F(x)[/tex] - Force as a function of position, measured in newtons.
Given that [tex]F = k\cdot x[/tex] and the fact that [tex]F = 25\,N[/tex] when [tex]x = 0.3\,m - 0.2\,m[/tex], the spring constant ([tex]k[/tex]), measured in newtons per meter, is:
[tex]k = \frac{F}{x}[/tex]
[tex]k = \frac{25\,N}{0.3\,m-0.2\,m}[/tex]
[tex]k = 250\,\frac{N}{m}[/tex]
Now, the work function is obtained:
[tex]W = \left(250\,\frac{N}{m} \right)\int\limits^{0.05\,m}_{0\,m} {x} \, dx[/tex]
[tex]W = \frac{1}{2}\cdot \left(250\,\frac{N}{m} \right)\cdot [(0.05\,m)^{2}-(0.00\,m)^{2}][/tex]
[tex]W = 0.313\,J[/tex]
The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules.
b) Let be [tex]r(\theta) = 2\cdot \sin 5\theta[/tex]. The area of the region enclosed by one loop of the curve is given by the following integral:
[tex]A = \int\limits^{2\pi}_0 {[r(\theta)]^{2}} \, d\theta[/tex]
[tex]A = 4\int\limits^{2\pi}_{0} {\sin^{2}5\theta} \, d\theta[/tex]
By using trigonometrical identities, the integral is further simplified:
[tex]A = 4\int\limits^{2\pi}_{0} {\frac{1-\cos 10\theta}{2} } \, d\theta[/tex]
[tex]A = 2 \int\limits^{2\pi}_{0} {(1-\cos 10\theta)} \, d\theta[/tex]
[tex]A = 2\int\limits^{2\pi}_{0}\, d\theta - 2\int\limits^{2\pi}_{0} {\cos10\theta} \, d\theta[/tex]
[tex]A = 2\cdot (2\pi - 0) - \frac{1}{5}\cdot (\sin 20\pi-\sin 0)[/tex]
[tex]A = 4\pi[/tex]
The area of the region enclosed by one loop of the curve [tex]r(\theta) = 2\cdot \sin 5\theta[/tex] is [tex]4\pi[/tex].
A spring attached to the ceiling is stretched one foot by a four pound weight. The mass is set in motion by pulling down 2 feet and then released in a medium with a damping force numerically equal to the velocity.
a. Find the Hook's law spring constant k.
b. Find its natural frequency ω.
c. Form a differential equation for the position x(t) of the mass.
d. Determine the solution for the position (in alternate form).
e. Find the times at which the mass passes the equilibrium second time heading up.
Given that,
Weight = 4 pound
[tex]W=4\ lb[/tex]
Stretch = 2 feet
Let the force be F.
The elongation of the spring after the mass attached is
[tex]x=2-1=1\ feet[/tex]
(a). We need to calculate the value of spring constant
Using Hooke's law
[tex]F=kx[/tex]
[tex]k=\dfrac{F}{x}[/tex]
Where, F = force
k = spring constant
x = elongation
Put the value into the formula
[tex]k=\dfrac{4}{1}[/tex]
[tex]k=4[/tex]
(b). We need to calculate the mass
Using the formula
[tex]F=mg[/tex]
[tex]m=\dfrac{F}{g}[/tex]
Where, F = force
g = acceleration due to gravity
Put the value into the formula
[tex]m=\dfrac{4}{32}[/tex]
[tex]m=\dfrac{1}{8}\ lb[/tex]
We need to calculate the natural frequency
Using formula of natural frequency
[tex]\omega=\sqrt{\dfrac{k}{m}}[/tex]
Where, k = spring constant
m = mass
Put the value into the formula
[tex]\omega=\sqrt{\dfrac{4}{\dfrac{1}{8}}}[/tex]
[tex]\omega=\sqrt{32}[/tex]
[tex]\omega=4\sqrt{2}[/tex]
(c). We need to write the differential equation
Using differential equation
[tex]m\dfrac{d^2x}{dt^2}+kx=0[/tex]
Put the value in the equation
[tex]\dfrac{1}{8}\dfrac{d^2x}{dt^2}+4x=0[/tex]
[tex]\dfrac{d^2x}{dt^2}+32x=0[/tex]
(d). We need to find the solution for the position
Using auxiliary equation
[tex]m^2+32=0[/tex]
[tex]m=\pm i\sqrt{32}[/tex]
We know that,
The general equation is
[tex]x(t)=A\cos(\sqrt{32t})+B\sin(\sqrt{32t})[/tex]
Using initial conditions
(I). [tex]x(0)=2[/tex]
Then, [tex]x(0)=A\cos(\sqrt{32\times0})+B\sin(\sqrt{32\times0})[/tex]
Put the value in equation
[tex]2=A+0[/tex]
[tex]A=2[/tex].....(I)
Now, on differentiating of general equation
[tex]x'(t)=-\sqrt{32}A\sin(\sqrt{32t})+\sqrt{32}B\cos(\sqrt{32t})[/tex]
Using condition
(II). [tex]x'(0)=0[/tex]
Then, [tex]x'(0)=-\sqrt{32}A\sin(\sqrt{32\times0})+\sqrt{32}B\cos(\sqrt{32\times0})[/tex]
Put the value in the equation
[tex]0=0+\sqrt{32}B[/tex]
So, B = 0
Now, put the value in general equation from equation (I) and (II)
So, The general solution is
[tex] x(t)=2\cos\sqrt{32t}[/tex]
(e). We need to calculate the time
Using formula of time
[tex]T=\dfrac{2\pi}{\omega}[/tex]
Put the value into the formula
[tex]T=\dfrac{2\pi}{4\sqrt{2}}[/tex]
[tex]T=1.11\ sec[/tex]
Hence, (a). The value of spring constant is 4.
(b). The natural frequency is 4√2.
(c). The differential equation is [tex]\dfrac{d^2x}{dt^2}+32x=0[/tex]
(d). The solution for the position is [tex] x(t)=2\cos\sqrt{32t}[/tex]
(e). The time period is 1.11 sec.
A 100-m long transmission cable is suspended between two towers. If the mass density is 2.01 kg/m and the tension in the cable is 3.00 x 104 N, what is the speed of transverse waves on the cable
Answer:
The speed is [tex]v =122.2 \ m/s[/tex]
Explanation:
From the question we are told that
The length of the wire is [tex]L = 100 \ m[/tex]
The mass density is [tex]\mu = 2.01 \ kg/m[/tex]
The tension is [tex]T = 3.00 *10^{4} \ N[/tex]
Generally the speed of the transverse cable is mathematically represented as
[tex]v = \sqrt{\frac{T}{\mu} }[/tex]
substituting values
[tex]v = \sqrt{\frac{3.0 *10^{4}}{2.01} }[/tex]
[tex]v =122.2 \ m/s[/tex]
A parallel-plate capacitor with circular plates of radius R is being discharged. The displacement current through a central circular area, parallel to the plates and with radius R/2, is 9.2 A. What is the discharging current?
Answer:
The discharging current is [tex]I_d = 36.8 \ A[/tex]
Explanation:
From the question we are told that
The radius of each circular plates is R
The displacement current is [tex]I = 9.2 \ A[/tex]
The radius of the central circular area is [tex]\frac{R}{2}[/tex]
The discharging current is mathematically represented as
[tex]I_d = \frac{A}{k} * I[/tex]
where A is the area of each plate which is mathematically represented as
[tex]A = \pi R ^2[/tex]
and k is central circular area which is mathematically represented as
[tex]k = \pi [\frac{R}{2} ]^2[/tex]
So
[tex]I_d = \frac{\pi R^2 }{\pi * [ \frac{R}{2}]^2 } * I[/tex]
[tex]I_d = \frac{\pi R^2 }{\pi * \frac{R^2}{4} } * I[/tex]
[tex]I_d = 4 * I[/tex]
substituting values
[tex]I_d = 4 * 9.2[/tex]
[tex]I_d = 36.8 \ A[/tex]
The electric field must be zero inside a conductor in electrostatic equilibrium, but not inside an insulator. It turns out that we can still apply Gauss's law to a Gaussian surface that is entirely within an insulator by replacing the right-hand side of Gauss's law, Qin/eo, with Qin/e, where ε is the permittivity of the material. (Technically, Eo is called the vacuum permittivity.) Suppose that a 70 nC point charge is surrounded by a thin, 32-cm-diameter spherical rubber shell and that the electric field strength inside the rubber shell is 2500 N/C.
What is the permittivity of rubber?
Answer:
The permittivity of rubber is [tex]\epsilon = 8.703 *10^{-11}[/tex]
Explanation:
From the question we are told that
The magnitude of the point charge is [tex]q_1 = 70 \ nC = 70 *10^{-9} \ C[/tex]
The diameter of the rubber shell is [tex]d = 32 \ cm = 0.32 \ m[/tex]
The Electric field inside the rubber shell is [tex]E = 2500 \ N/ C[/tex]
The radius of the rubber is mathematically evaluated as
[tex]r = \frac{d}{2} = \frac{0.32}{2} = 0.16 \ m[/tex]
Generally the electric field for a point is in an insulator(rubber) is mathematically represented as
[tex]E = \frac{Q}{ \epsilon } * \frac{1}{4 * \pi r^2}[/tex]
Where [tex]\epsilon[/tex] is the permittivity of rubber
=> [tex]E * \epsilon * 4 * \pi * r^2 = Q[/tex]
=> [tex]\epsilon = \frac{Q}{E * 4 * \pi * r^2}[/tex]
substituting values
[tex]\epsilon = \frac{70 *10^{-9}}{2500 * 4 * 3.142 * (0.16)^2}[/tex]
[tex]\epsilon = 8.703 *10^{-11}[/tex]
An electron has an initial velocity of (17.1 + 12.7) km/s, and a constant acceleration of (1.60 × 1012 m/s2) in the positive x direction in a region in which uniform electric and magnetic fields are present. If = (529 µT) find the electric field .
Answer:
Explanation:
Since B is perpendicular, it does no work on the electron but instead deflects it in a circular path.
q = 1.6 x 10-19 C
v = (17.1j + 12.7k) km/s = square root(17.1² + 12.7²) = 2.13 x 10⁴ m/s
the force acting on electron is
F= qvBsinΦ
F= (1.6 x 10⁻¹⁹C)(2.13.x 10⁴ m/s)(526 x 10⁻⁶ T)(sin90º)
F = 1.793x 10⁻¹⁸ N
The net force acting on electron is
F = e ( E+ ( vXB)
= ( - 1.6 × 10⁻¹⁹) ( E + ( 17.1 × 10³j + 12.7 × 10³ k)X( 529 × 10⁻⁶ ) (i)
= ( -1.6 × 10⁻¹⁹ ) ( E- 6.7k + 9.0j)
a= F/m
1.60 × 10¹² i = ( -1.6 × 10⁻¹⁹ ) ( E- 6.9 k + 7.56 j)/9.11 × 10⁻³¹
9.11 i = - ( E- 6.7 k + 9.0 j)
E = -9.11i + 6.7k - 9.0j
The ability to distinguish between acceleration and velocity will be critical to your understanding of many other concepts in this course. Some of the most prevalent issues arise in interpreting the sign of both the velocity and acceleration of an object. I would recommend reading through the section "The Sign of the Acceleration" carefully. An object moves with a positive acceleration. Could the object be moving with increasing speed, decreasing speed or constant speed?
Answer:
Object could only be moving with increasing speed.
Explanation:
Let us consider the general formula of acceleration:
a = (Vf - Vi)/t
Vf = Vi + at -------- equation 1
where,
Vf = Final Velocity
Vi = Initial Velocity
a = acceleration
t = time
FOR POSITIVE ACCELERATION:
Vf = Vi + at
since, both acceleration and time are positive quantities. Hence, it means that the final velocity of the object shall be greater than the initial velocity of the object.
Vf > Vi
It clearly shows that if an object moves with positive acceleration. It could only be moving with increasing speed.
Solving the same equation for negative acceleration shows that the final velocity will be less than initial velocity and object will be moving with decreasing speed.
And for the constant velocity final and initial velocities are equal and thus, acceleration will be zero.
Legacy issues $570,000 of 8.5%, four-year bonds dated January 1, 2019, that pay interest semiannually on June 30 and December 31. They are issued at $508,050 when the market rate is 12%.
1. Determine the total bond interest expense to be recognized.
Total bond interest expense over life of bonds:
Amount repaid:
8 payments of $24,225 $193,800
Par value at maturity 570,000
Total repaid 763,800
Less amount borrowed 645 669
Total bond interest expense $118.131
2. Prepare a straight-line amortization table for the bonds' first two years.
Semiannual Period End Unamortized Discount Carrying Value
01/01/2019
06/30/2019
12/31/2019
06/30/2020
12/31/2020
3. Record the interest payment and amortization on June 30. Note:
Date General Journal Debit Credit
June 30
4. Record the interest payment and amortization on December 31.
Date General Journal Debit Credit
December 31
Answer:
1) Determine the total bond interest expense to be recognized.
Total bond interest expense over life of bonds:
Amount repaid:
8 payments of $24,225: $193,800
Par value at maturity: $570,000
Total repaid: $763800 (193,800 + 570,000)
Less amount borrowed: $508050
Total bond interest expense: $255750 (763800 - 508,050)
2)Prepare a straight-line amortization table for the bonds' first two years.
Semiannual Interest Period End; Unamortized Discount; Carrying Value
01/01/2019 61,950 508,050
06/30/2019 54,206 515,794
12/31/2019 46,462 523,538
06/30/2020 38,718 531,282
12/31/2020 30,974 539,026
3) Record the interest payment and amortization on June 30:
June 30 Bond interest expense, dr 31969
Discount on bonds payable, Cr (61950/8) 7743.75
Cash, Cr ( 570000*8.5%/2) 24225
4) Record the interest payment and amortization on December 31:
Dec 31 Bond interest expense, Dr 31969
Discount on bonds payable, Cr 7744
Cash, Cr 24225
A soft tennis ball is dropped onto a hard floor from a height of 1.50 m and rebounds to a height of 1.10 m. (a) Calculate its velocity just before it strikes the floor. (b) Calculate its velocity just after it leaves the floor on its way back up. (c) Calculate its acceleration during contact with the floor if that contact lasts 3.50 ms (3.50×10−3s). (d) How much did the ball compress during its collision with the floor, assuming the floor is absolutely rigid
Answer:
(a) v = 5.42m/s
(b) vo = 4.64m/s
(c) a = 2874.28m/s^2
(d) Δy = 5.11*10^-3m
Explanation:
(a) The velocity of the ball before it hits the floor is given by:
[tex]v=\sqrt{2gh}[/tex] (1)
g: gravitational acceleration = 9.8m/s^2
h: height where the ball falls down = 1.50m
[tex]v=\sqrt{2(9.8m/s^2)(1.50m)}=5.42\frac{m}{s}[/tex]
The speed of the ball is 5.42m/s
(b) To calculate the velocity of the ball, after it leaves the floor, you use the information of the maximum height reached by the ball after it leaves the floor.
You use the following formula:
[tex]h_{max}=\frac{v_o^2}{2g}[/tex] (2)
vo: velocity of the ball where it starts its motion upward
You solve for vo and replace the values of the parameters:
[tex]v_o=\sqrt{2gh_{max}}=\sqrt{2(9.8m/s^2)(1.10m)}=4.64\frac{m}{s}[/tex]
The velocity of the ball is 4.64m/s
(c) The acceleration is given by:
[tex]a=\frac{\Delta v}{\Delta t}=\frac{v_o-v}{3.50*10^{-3}s}=\frac{4.64m/s-(-5.42m)/s}{3.50*10^{-3}s}=2874.285\frac{m}{s^2}[/tex]
[tex]a=\frac{\Delta v}{\Delta t}=\frac{v_o-v}{3.50*10^{-3}s}=\frac{4.64m/s-5.42m/s}{3.50*10^{-3}s}=-222.85\frac{m}{s^2}[/tex]
The acceleration of the ball is 2874.28/s^2
(d) The compression of the ball is:
[tex]\Deta y=\frac{v^2}{2(a)}=\frac{(5.42m/s)^2}{2(2874.28m/s^2)}=5.11*10^{-3}m[/tex]
THe compression of the ball when it strikes the floor is 5.11*10^-3m
A Ferris wheel starts at rest and builds up to a final angular speed of 0.70 rad/s while rotating through an angular displacement of 4.9 rad. What is its average angular acceleration
Answer:
The average angular acceleration is 0.05 radians per square second.
Explanation:
Let suppose that Ferris wheel accelerates at constant rate, the angular acceleration as a function of change in angular position and the squared final and initial angular velocities can be clear from the following expression:
[tex]\omega^{2} = \omega_{o}^{2} + 2 \cdot \alpha\cdot (\theta-\theta_{o})[/tex]
Where:
[tex]\omega_{o}[/tex], [tex]\omega[/tex] - Initial and final angular velocities, measured in radians per second.
[tex]\alpha[/tex] - Angular acceleration, measured in radians per square second.
[tex]\theta_{o}[/tex], [tex]\theta[/tex] - Initial and final angular position, measured in radians.
Then,
[tex]\alpha = \frac{\omega^{2}-\omega_{o}^{2}}{2\cdot (\theta-\theta_{o})}[/tex]
Given that [tex]\omega_{o} = 0\,\frac{rad}{s}[/tex], [tex]\omega = 0.70\,\frac{rad}{s}[/tex] and [tex]\theta-\theta_{o} = 4.9\,rad[/tex], the angular acceleration is:
[tex]\alpha = \frac{\left(0.70\,\frac{rad}{s} \right)^{2}-\left(0\,\frac{rad}{s} \right)^{2}}{2\cdot \left(4.9\,rad\right)}[/tex]
[tex]\alpha = 0.05\,\frac{rad}{s^{2}}[/tex]
Now, the time needed to accelerate the Ferris wheel uniformly is described by this kinematic equation:
[tex]\omega = \omega_{o} + \alpha \cdot t[/tex]
Where [tex]t[/tex] is the time measured in seconds.
The time is cleared and obtain after replacing every value:
[tex]t = \frac{\omega-\omega_{o}}{\alpha}[/tex]
If [tex]\omega_{o} = 0\,\frac{rad}{s}[/tex], [tex]\omega = 0.70\,\frac{rad}{s}[/tex] and [tex]\alpha = 0.05\,\frac{rad}{s^{2}}[/tex], the required time is:
[tex]t = \frac{0.70\,\frac{rad}{s} - 0\,\frac{rad}{s} }{0.05\,\frac{rad}{s^{2}} }[/tex]
[tex]t = 14\,s[/tex]
Average angular acceleration is obtained by dividing the difference between final and initial angular velocities by the time found in the previous step. That is:
[tex]\bar \alpha = \frac{\omega-\omega_{o}}{t}[/tex]
If [tex]\omega_{o} = 0\,\frac{rad}{s}[/tex], [tex]\omega = 0.70\,\frac{rad}{s}[/tex] and [tex]t = 14\,s[/tex], the average angular acceleration is:
[tex]\bar \alpha = \frac{0.70\,\frac{rad}{s} - 0\,\frac{rad}{s} }{14\,s}[/tex]
[tex]\bar \alpha = 0.05\,\frac{rad}{s^{2}}[/tex]
The average angular acceleration is 0.05 radians per square second.
1. A ski-plane with a total mass of 1200 kg lands towards the west on a frozen lake at 30.0
m/s. The coefficient of kinetic friction between the skis and the ice is 0.200. How far does
the plane slide before coming to a stop?
Answer:
d = 229.5 m
Explanation:
It is given that,
Total mass of a ski-plane is 1200 kg
It lands towards the west on a frozen lake at 30.0 m/s.
The coefficient of kinetic friction between the skis and the ice is 0.200.
We need to find the distance covered by the plane before coming to rest. In this case,
[tex]\mu mg=ma\\\\a=\mu g\\\\a=0.2\times 9.8\\\\a=1.96\ m/s^2[/tex]
It is decelerating, a = -1.96 m/s²
Now using the third equation of motion to find the distance covered by the plane such that :
[tex]v^2-u^2=2ad\\\\d=\dfrac{-u^2}{2a}\\\\d=\dfrac{-(30)^2}{2\times -1.96}\\\\d=229.59\ m[/tex]
So, the plane slide a distance of 229.5 m.
Oceanographers often express the density of sea water in units of kilograms per cubic meter. If the density of sea water is 1.025 g/cm3 at 15ºC, what is its density in kilograms per cubic meter?
Answer: The density in kilograms per cubic meter is 1025
Explanation:
Density is defined as mass contained per unit volume.
Given : Density of sea water = [tex]1.025g/cm^3[/tex]
Conversion : [tex]1.025g/cm^3=?kg/m^3[/tex]
As 1 g = 0.001 kg
Thus 1.025 g =[tex]\frac{0.001}{1}\times 1.025=0.001025kg[/tex]
Also [tex]1cm^3=10^{-6}m^3[/tex]
Thus [tex]1.025g/cm^3=\frac{0.001025}{10^{-6}kg/m^3}=1025kg/m^3[/tex]
Thus density in kilograms per cubic meter is 1025
In a double-slit interference experiment you are asked to use laser light of different wavelengths and determine the separation between adjacent maxima. You observe that this separation is greatest when you illuminate the double slit with In a double-slit interference experiment you are asked to use laser light of different wavelengths and determine the separation between adjacent maxima. You observe that this separation is greatest when you illuminate the double slit with:_________.
1. yellow light.
2. red light.
3. blue light.
4. green light.
5. The separation is the same for all wavelengths.
Answer:
Red light
Explanation:
This because All interference or diffraction patterns depend upon the wavelength of the light (or whatever wave) involved. Red light has the longest wavelength (about 700 nm)
In a uniform electric field, the magnitude of torque is given by:-
Answer:
Electric dipole
Explanation:
the axis of a dipole makes an angle with the electric field. depending on the direction (clockwise/anticlockwise) we can get the torque (positive/negative).
hope this helps :D
A cylinder initially contains 1.5 kg of air at 100 kPa and 17 C. The air is compressed polytropically until the volume is reduced by one-half. The polytropic exponent is 1.3. Determine the work done (absolute value) and heat transfer for this process.
Answer:
Work done = 96.15 KJ
Heat transferred = 24 KJ
Explanation:
We are given;
Mass of air;m = 1.5 kg
Initial pressure;P1 = 100 KPa
Initial temperature;T1 = 17°C = 273 + 17 K = 290 K
Final volume;V2 = 0.5V1
Since the polytropic exponent is 1.3,thus it means;
P2 = P1[V1/V2]^(1.3)
So,P2 = 100(V1/0.5V1)^(1.3)
P2 = 100(2)^(1.3)
P2 = 246.2 KPa
To find the final temperature, we will make use of combined gas law.
So,
(P1×V1)/T1 = (P2×V2)/T2
T2 = (P2×V2×T1)/(P1×V1)
Plugging in the known values;
T2 = (246.2×0.5V1×290)/(100×V1)
V1 cancels out to give;
T2 = (246.2×0.5×290)/100
T2 = 356.99 K
The boundary work for this polytropic process is given by;
W_b = - ∫P. dv between the initial boundary and final boundary.
Thus,
W_b = - (P2.V2 - P1.V1)/(1 - n) = -mR(T2 - T1)/(1 - n)
R is gas constant for air = 0.28705 KPa.m³/Kg.K
n is the polytropic exponent which is 1.3
Thus;
W_b = -1.5 × 0.28705(356.99 - 290)/(1 - 1.3)
W_b = 96.15 KJ
The formula for the heat transfer is given as;
Q_out = W_b - m.c_v(T2 - T1)
c_v for air = 0.718 KJ/Kg.k
Q_out = 96.15 - (1.5×0.718(356.99 - 290))
Q_out = 24 KJ
Lightbulbs are typically rated by their power dissipation when operated at a given voltage. Which of the following lightbulbs has the largest resistance when operated at the voltage for which it's rated?
A. 0.8 W, 1.5 V
B. 6 W 3 V
C. 4 W, 4.5 V
D. 8 W, 6 V
Answer:
The arrangement with the greatest resistance is the light bulb of option C. 4 W, 4.5 V
Explanation:
The equation for electric power is
power P = IV
also, I = V/R,
substituting into the equation, we have
[tex]P = \frac{V^{2} }{R}[/tex]
[tex]R = \frac{V^{2} }{P}[/tex]
a) [tex]R = \frac{1.5^{2} }{0.8}[/tex] = 2.8 Ω
b) [tex]R = \frac{3^{2} }{6}[/tex] = 1.5 Ω
c) [tex]R = \frac{4.5^{2} }{4}[/tex] 5.06 Ω
d) [tex]R = \frac{6^{2} }{8}[/tex] = 4.5 Ω
from the calculations, one can see that the lightbulb with te greates resistance is
C. 4 W, 4.5 V
Calculate the angular momentum of a solid uniform sphere with a radius of 0.150 m and a mass of 13.0 kg if it is rotating at 5.70 rad/s about an axis through its center.
Answer:
The angular momentum of the solid sphere is 0.667 kgm²/s
Explanation:
Given;
radius of the solid sphere, r = 0.15 m
mass of the sphere, m = 13 kg
angular speed of the sphere, ω = 5.70 rad/s
The angular momentum of the solid sphere is given;
L = Iω
Where;
I is the moment of inertia of the solid sphere
ω is the angular speed of the solid sphere
The moment of inertia of solid sphere is given by;
I = ²/₅mr²
I = ²/₅ x (13 x 0.15²)
I = 0.117 kg.m²
The angular momentum of the solid sphere is calculated as;
L = Iω
L = 0.117 x 5.7
L = 0.667 kgm²/s
Therefore, the angular momentum of the solid sphere is 0.667 kgm²/s
what are the rays that come of the sun called? A. Ultraviolent rays B. Gamma rays C. soundwaves D. sonic rays
Answer:
The answer is ultraviolet rays...
Explanation:
...because the ozone layer protects us from the UV rays of the sun.
5) What is the weight of a body in earth. if its weight is 5Newton
in moon?
Answer:
8.167
Explanation:
what is mean by the terminal velocity
Terminal Velocity is the constant speed that a falling thing reaches when the resistence of a medium prevents the thing to reach any further speed.
Best of Luck!
Given small samples of three liquids, you are asked to determine their refractive indexes. However, you do not have enough of each liquid to measure the angle of refraction for light retracting from air into the liquid. Instead, for each liquid, you take a rectangular block of glass (n= 1.52) and Place a drop of the liquid on the top surface f the block. you shine a laser beam with wavelength 638 nm in vacuum at one Side of the block and measure the largest angle of incidence for which there is total internal reflection at the interface between the glass and the liquid. Your results are given in the table.
Liquid A B C
θ 52.0 44.3 36.3
Required:
a. What is the refractive index of liquid A at this wavelength?
b. What is the refractive index of liquid B at this wavelength?
c. What is the refractive index of liquid C at this wavelength?
Answer:
A — 1.198B — 1.062C — 0.900Explanation:
The index of refraction of the liquid can be computed from ...
[tex]n_i\sin{(\theta_t)}=n_t[/tex]
where ni is the index of refraction of the glass block (1.52) and θt is the angle at which there is total internal refraction. nt is the index of refraction of the liquid.
For the given incidence angles, the computed indices of refraction are ...
A: n = 1.52sin(52.0°) = 1.198
B: n = 1.52sin(44.3°) = 1.062
C: n = 1.52sin(36.3°) = 0.900
A stretch spring has an elastic potential energy of 35 J when it is stretched 0.54m. What is the spring constant of the spring?
Answer:
240.1 N/m
Explanation:
Applying the formula for the potential energy stored in a stretched spring,
E = 1/2ke².................... Equation 1
Where E = potential energy, k = spring constant, e = extension.
make k the subject of the equation,
k = 2E/e².................. Equation 2
Given: E = 35 J, e = 0.54 m
Substitute into equation 2
k = 2(35)/0.54²
k = 70/0.2916
k = 240.1 N/m.
Hence the spring constant of the spring is 240.1 N/m
Answer:
240 N/m
Explanation:
edge2o2o
Problem 2: A 21-W horizontal beam of light of wavelength 430 nm, travelling at speed c, passes through a rectangular opening of width 0.0048 m and height 0.011 m. The light then strikes a screen at a distance 0.36 m behind the opening.
a) E = 4.62E-19 Joules
b) N = 4.543E+19 # of photons emitted
c) If the beam of light emitted by the source has a constant circular cross section whose radius is twice the height of the opening the beam is approaching, find the flow density of photons as the number of photons passing through a square meter of cross-sectional area per second.
d) Calculate the number of photons that pass through the rectangular opening per second.
e) The quantity NO that you found in part (d) gives the rate at which photons enter the region between the opening and the screen. Assuming no reflection from the screen, find the number of photons in that region at any time.
f) How would the number of photons in the region between the opening and the screen change, if the photons traveled more slowly? Assume no change in any other quantity, including the speed of light.
Answer:
a. E = 4.62 × 10⁻¹⁹J
b. n = 4.54 × 10¹⁹photons
c. 2.99 × 10²²photons/m²
d. 1.58 ×10¹⁸photons/seconds
e. n = 1.896 × 10 ⁹ photons
f. the number of photons will be more if it travels slowly
Explanation:
Three sleds (30kg sled connected by tension rope B to 20kg sled connected by tension rope A to 10kg sled) are being pulled horizontally on frictionless horizontal ice using horizontal ropes. The pull is horizontal and of magnitude 143N . Required:a. Find the acceleration of the system. b. Find the tension in rope A. c. Find the tension in rope B.
Answer:
a) a = 2.383 m / s², b) T₂ = 120,617 N , c) T₃ = 72,957 N
Explanation:
This is an exercise of Newton's second law let's fix a horizontal frame of reference
in this case the mass of the sleds is 30, 20 10 kg from the last to the first, in the first the horizontal force is applied.
a) request the acceleration of the system
we can take the sledges together and write Newton's second law
T = (m₁ + m₂ + m₃) a
a = T / (m₁ + m₂ + m₃)
a = 143 / (10 +20 +30)
a = 2.383 m / s²
b) the tension of the cables we think through cable A between the sledges of 1 and 20 kg
on the sled of m₁ = 10 kg
T - T₂ = m₁ a
in this case T₂ is the cable tension
T₂ = T - m₁ a
T₂ = 143 - 10 2,383
T₂ = 120,617 N
c) The cable tension between the masses of 20 and 30 kg
T₂ - T₃ = m₂ a
T₃ = T₂ -m₂ a
T₃ = 120,617 - 20 2,383
T₃ = 72,957 N
PLS HELP ILL MARK U BRAINLIEST I DONT HAVE MUCH TIME!!
A football player of mass 103 kg running with a velocity of 2.0 m/s [E] collides head-
on with a 110 kg player on the opposing team travelling with a velocity of 3.2 m/s
[W]. Immediately after the collision the two players move in the same direction.
Calculate the final velocity of the two players.
Answer:
The final velocity of the two players is 0.69 m/s in the direction of the opposing player.
Explanation:
Since the players are moving in opposite directions, from the principle of conservation of linear momentum;
[tex]m_{1} u_{1}[/tex] - [tex]m_{2}u_{2}[/tex] = [tex](m_{1} + m_{2} )[/tex] v
Where: [tex]m_{1}[/tex] is the mass of the first player, [tex]u_{1}[/tex] is the initial velocity of the first player, [tex]m_{2}[/tex] is the mass of the second player, [tex]u_{2}[/tex] is the initial velocity of the second player and v is the final common velocity of the two players after collision.
[tex]m_{1}[/tex] = 103 kg, [tex]u_{1}[/tex] = 2.0 m/s, [tex]m_{2}[/tex] = 110 kg, [tex]u_{2}[/tex] = 3.2 m/s. Thus;
103 × 2.0 - 110 × 3.2 = (103 + 110)v
206 - 352 = 213 v
-146 = 213 v
v = [tex]\frac{-146}{213}[/tex]
v = -0.69 m/s
The final velocity of the two players is 0.69 m/s in the direction of the opposing player.
A crate of mass 9.2 kg is pulled up a rough incline with an initial speed of 1.58 m/s. The pulling force is 110 N parallel to the incline, which makes an angle of 20.2° with the horizontal. The coefficient of kinetic friction is 0.400, and the crate is pulled 5.10 m.A) How much work is done by the gravitational force on thecrate?
B) Determine the increase in internal energy of the crate-inclinesystem owing to friction.
C) How much work is done by the 100N force on the crate?
D) What is the change in kinetic energy of the crate?
E) What is the speed of the crate after being pulled 5.00m?
Given that,
Mass = 9.2 kg
Force = 110 N
Angle = 20.2°
Distance = 5.10 m
Speed = 1.58 m/s
(A). We need to calculate the work done by the gravitational force
Using formula of work done
[tex]W_{g}=mgd\sin\theta[/tex]
Where, w = work
m = mass
g = acceleration due to gravity
d = distance
Put the value into the formula
[tex]W_{g}=9.2\times(-9.8)\times5.10\sin20.2[/tex]
[tex]W_{g}=-158.8\ J[/tex]
(B). We need to calculate the increase in internal energy of the crate-incline system owing to friction
Using formula of potential energy
[tex]\Delta U=-W[/tex]
Put the value into the formula
[tex]\Delta U=-(-158.8)\ J[/tex]
[tex]\Delta U=158.8\ J[/tex]
(C). We need to calculate the work done by 100 N force on the crate
Using formula of work done
[tex]W=F\times d[/tex]
Put the value into the formula
[tex]W=100\times5.10[/tex]
[tex]W=510\ J[/tex]
We need to calculate the work done by frictional force
Using formula of work done
[tex]W=-f\times d[/tex]
[tex]W=-\mu mg\cos\theta\times d[/tex]
Put the value into the formula
[tex]W=-0.4\times9.2\times9.8\cos20.2\times5.10[/tex]
[tex]W=-172.5\ J[/tex]
We need to calculate the change in kinetic energy of the crate
Using formula for change in kinetic energy
[tex]\Delta k=W_{g}+W_{f}+W_{F}[/tex]
Put the value into the formula
[tex]\Delta k=-158.8-172.5+510[/tex]
[tex]\Delta k=178.7\ J[/tex]
(E). We need to calculate the speed of the crate after being pulled 5.00m
Using formula of change in kinetic energy
[tex]\Delta k=\dfrac{1}{2}m(v_{2}^2-v_{1}^{2})[/tex]
[tex]v_{2}^2=\dfrac{2\times\Delta k}{m}+v_{1}^2[/tex]
Put the value into the formula
[tex]v_{2}^2=\dfrac{2\times178.7}{9.2}+1.58[/tex]
[tex]v_{2}=\sqrt{\dfrac{2\times178.7}{9.2}+1.58}[/tex]
[tex]v_{2}=6.35\ m/s[/tex]
Hence, (A). The work done by the gravitational force is -158.8 J.
(B). The increase in internal energy of the crate-incline system owing to friction is 158.8 J.
(C). The work done by 100 N force on the crate is 510 J.
(D). The change in kinetic energy of the crate is 178.7 J.
(E). The speed of the crate after being pulled 5.00m is 6.35 m/s
The average density of the body of a fish is 1080kg/m^3 . To keep from sinking, the fish increases its volume by inflating an internal air bladder, known as a swim bladder, with air.
By what percent must the fish increase its volume to be neutrally buoyant in fresh water? Use 1.28kg/m^3 for the density of air at 20 degrees Celsius. (change in V/V)
Answer:
Increase of volume (F) = 8.01%
Explanation:
Given:
Density of fish = 1,080 kg/m³
Density of water = 1,000 kg/m³
density of air = 1.28 kg/m³
Find:
Increase of volume (F)
Computation:
1,080 kg/m³ + [F × 1.28 kg/m³ ] = (1+F) × 1,000 kg/m³
1,080 + 1.28 F =1,000 F + 1,000
80 = 998.72 F
F = 0.0801 (Approx)
F = 8.01% (Approx)
If the frequency of a periodic wave is cut in half while the speed remains the same, what happens to the wavelength
Answer:
The wavelength becomes twice the original wavelength
Explanation:
Recall that for regular waves, the relationship between wavelength, velocity (i.e speed) and frequency is given by
v = fλ
where,
v = velocity,
f = frequency
λ = wavelength
Before a change was made to the frequency, we have: v₁ = f₁ λ₁
After a change was made to the frequency, we have: v₂ = f₂ λ₂
We are told that the speed remains the same, so
v₁ = v₂
f₁ λ₁ = f₂ λ₂ (rearranging this)
f₁ / f₂ = λ₂/λ₁ --------(1)
we are given that the frequency is cut in half.
f₂ = (1/2) f₁ (rearranging this)
f₁/f₂ = 2 -------------(2)
if we substitute equation (2) into equation (1):
f₁ / f₂ = λ₂/λ₁
2 = λ₂/λ₁
λ₂ = 2λ₁
Hence we can see that the wavelength after the change becomes twice (i.e doubles) the initial wavelength.
Use Kepler's third law to determine how many days it takes a spacecraft to travel in an elliptical orbit from a point 6 590 km from the Earth's center to the Moon, 385 000 km from the Earth's center.
Answer:
1.363×10^15 seconds
Explanation:
The spaceship travels an elliptical orbit from a point of 6590km from the earth center to the moon and 38500km from the earth center.
To calculate the time taken from Kepler's third Law :
T^2 = ( 4π^2/GMe ) r^3
Where Me is the mass of the earth
r is the average distance travel
G is the universal gravitational constant. = 6.67×10-11 m3 kg-1 s-2
π = 3.14
Me = mass of earth = 5.972×10^24kg
r =( r minimum + r maximum)/2 ......1
rmin = 6590km
rmax = 385000km
From equation 1
r = (6590+385000)/2
r = 391590/2
r = 195795km
From T^2 = ( 4π^2/GMe ) r^3
T^2 = (4 × 3.14^2/ 6.67×10-11 × 5.972×10^24) × 195795^3
= ( 4×9.8596/ 3.983×10^14 ) × 7.5059×10^15
= 39.4384/ 3.983×10^14 ) × 7.5059×10^15
= (9.901×10^14) × 7.5059×10^15
T^2 = 7.4321× 10^30
T =√7.4321× 10^30
T = 2.726×10^15 seconds
The time for one way trip from Earth to the moon is :
∆T = T/2
= 2.726×10^15 /2
= 1.363×10^15 secs
A circular coil of wire of 200 turns and diameter 2.0 cm carries a current of 4.0 A. It is placed in a magnetic field of 0.70 T with the plane of the coil making an angle of 30° with the magnetic field. What is the magnetic torque on the coil?
Answer:
0.087976 Nm
Explanation:
The magnetic torque (τ) on a current-carrying loop in a magnetic field is given by;
τ = NIAB sinθ --------- (i)
Where;
N = number of turns of the loop
I = current in the loop
A = area of each of the turns
B = magnetic field
θ = angle the loop makes with the magnetic field
From the question;
N = 200
I = 4.0A
B = 0.70T
θ = 30°
A = π d² / 4 [d = diameter of the coil = 2.0cm = 0.02m]
A = π x 0.02² / 4 = 0.0003142m² [taking π = 3.142]
Substitute these values into equation (i) as follows;
τ = 200 x 4.0 x 0.0003142 x 0.70 sin30°
τ = 200 x 4.0 x 0.0003142 x 0.70 x 0.5
τ = 200 x 4.0 x 0.0003142 x 0.70
τ = 0.087976 Nm
Therefore, the torque on the coil is 0.087976 Nm