Answer:
From the initial height h
Explanation:
When a material or substance is drop from a height h, it possesses potential energy, immediately it is dropped from that height, the potential energy is gradually converted to kinetic energy, it gets to a point where the potential energy equals the kinetic energy, as the material touches the ground, all potential energy has been converted to kinetic energy already
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
5) What is the weight of a body in earth. if its weight is 5Newton
in moon?
Answer:
8.167
Explanation:
How do I find an apparent weight in N for a metal connected to a string submerged in water if a scale shows the mass 29.52 g when it is submerged ? Also how do I measure its density
The Tension of the string is going to be less when submerged in water by a value called the buoyancy force, so below in the attached file is explanation on how to calculate the apparent weight and density of the submerged object
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)
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
In a uniform electric field, the magnitude of torque is given by:-
Answer:
Electric dipole
Explanation:
the dipole axis makes an angle with the electric field. depending on direction (clockwise/aniclockwise) you get the torque
Hope this helps
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
Consider the Earth and the Moon as a two-particle system.
Find an expression for the gravitational field g of this two-particle system as a function of the distance r from the center of the Earth. (Do not worry about points inside either the Earth or the Moon. Assume the Moon lies on the +r-axis. Give the scalar component of the gravitational field. Do not substitute numerical values; use variables only. Use the following as necessary: G, Mm, Me, r, and d for the distance from the center of Earth to the center of the Moon.)"
sorry but I don't understand
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
The frequency of a physical pendulum comprising a nonuniform rod of mass 1.15 kg pivoted at one end is observed to be 0.658 Hz. The center of mass of the rod is 42.5 cm below the pivot point. What is the rotational inertia of the pendulum around its pivot point
Answer:
The rotational inertia of the pendulum around its pivot point is [tex]0.280\,kg\cdot m^{2}[/tex].
Explanation:
The angular frequency of a physical pendulum is measured by the following expression:
[tex]\omega = \sqrt{\frac{m\cdot g \cdot d}{I_{o}} }[/tex]
Where:
[tex]\omega[/tex] - Angular frequency, measured in radians per second.
[tex]m[/tex] - Mass of the physical pendulum, measured in kilograms.
[tex]g[/tex] - Gravitational constant, measured in meters per square second.
[tex]d[/tex] - Straight line distance between the center of mass and the pivot point of the pendulum, measured in meters.
[tex]I_{O}[/tex] - Moment of inertia with respect to pivot point, measured in [tex]kg\cdot m^{2}[/tex].
In addition, frequency and angular frequency are both related by the following formula:
[tex]\omega =2\pi\cdot f[/tex]
Where:
[tex]f[/tex] - Frequency, measured in hertz.
If [tex]f = 0.658\,hz[/tex], then angular frequency of the physical pendulum is:
[tex]\omega = 2\pi \cdot (0.658\,hz)[/tex]
[tex]\omega = 4.134\,\frac{rad}{s}[/tex]
From the formula for the physical pendulum's angular frequency, the moment of inertia is therefore cleared:
[tex]\omega^{2} = \frac{m\cdot g \cdot d}{I_{o}}[/tex]
[tex]I_{o} = \frac{m\cdot g \cdot d}{\omega^{2}}[/tex]
Given that [tex]m = 1.15\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]d = 0.425\,m[/tex] and [tex]\omega = 4.134\,\frac{rad}{s}[/tex], the moment of inertia associated with the physical pendulum is:
[tex]I_{o} = \frac{(1.15\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.425\,m)}{\left(4.134\,\frac{rad}{s} \right)^{2}}[/tex]
[tex]I_{o} = 0.280\,kg\cdot m^{2}[/tex]
The rotational inertia of the pendulum around its pivot point is [tex]0.280\,kg\cdot m^{2}[/tex].
Charge of uniform surface density (0.20 nC/m2) is distributed over the entire xy plane. Determine the magnitude of the electric field at any point having z
The question is not complete, the value of z is not given.
Assuming the value of z = 4.0m
Answer:
the magnitude of the electric field at any point having z(4.0 m) =
E = 5.65 N/C
Explanation:
given
σ(surface density) = 0.20 nC/m² = 0.20 × 10⁻⁹C/m²
z = 4.0 m
Recall
E =F/q (coulumb's law)
E = kQ/r²
σ = Q/A
A = 4πr²
∴ The electric field at point z =
E = σ/zε₀
E = 0.20 × 10⁻⁹C/m²/(4 × 8.85 × 10⁻¹²C²/N.m²)
E = 5.65 N/C
Cathode ray tubes in old television sets worked by accelerating electrons and then deflecting them with magnetic fields onto a phosphor screen. The magnetic fields were created by coils of wire on either side of the tube carrying large currents. In one such TV set, the phosphor screen is 51.2 cm wide, and is 11.1 cm away from the center of the magnetic deflection coils (that is, the center of the region of magnetic field). The electron beam is first accelerated through a 22,000 V potential difference before it enters the magnetic field region, which is 1.00 cm wide. The field is approximately uniform and perpendicular to the velocity of the electrons. If the field were turned off, the electrons would hit the center of the screen. What magnitude of magnetic field (in mT) is needed to deflect the electrons so that they hit the far edge of the screen
Answer:
B = 0.046T
Explanation:
given
size of the screen = 51.2cm
distance from center = 11.1cm
region of magnetic field = 1.00cm
V= 22000V= 22kV
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:
at the temperature at which we live, earth's core is solid or liquid?
Explanation:
The Earth has a solid inner core
From largest to smallest, rank the gravitational force that each planet exerts on the star with mass M. Distances are as follows:_________.a) Planet A is 3d from M b) Planet B is 1d from M c) Planet C is 1d from M d) Planet D is 2d from M e) Planet E is 3d from M
Answer:
Planet A = planet B > planet D > planet A = planet E
Explanation:
Gravitational force obeys the inverse square law. That is,
Gravitation force F is inversely proportional to the square of the distance between the two masses.
The larger the distance, the weaker the gravitational force F.
From largest to smallest, the rank of gravitational force that each planet exerts on the star with mass M. Distances are as follows:
Planet B, planet C, planet D, planet planet A, planet E.
Planet B and C may experience different the same gravitational force depending on their masses. This is also applicable to planet A and E.
Therefore,
Planet A = planet B > planet D > planet A = planet E
A hollow sphere of radius 1.66 m is in a region where the electric field is radial and directed toward the center of the sphere. If the magnitude of the field at the surface of the sphere is 21.5 N/C, what is the net electric flux through the spherical surface?
Answer:
The value of flux will be "744.1 N.m²/C".
Explanation:
The given values are:
Magnitude,
E = 21.5 N/C
Radius,
R = 1.66 m
As we know,
⇒ [tex]Flux = Area\times E[/tex]
On putting the estimated values, we get
⇒ [tex]=-21.5\tines (4\times \pi\times 1.66^2 )[/tex]
⇒ [tex]=744.1 \ N.m^2/C[/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.
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.
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)
g The radius of a spherical ball increases at a rate of 3 m/s. At what rate is the volume changing when the radius is equal to 2 meters
Answer:
dV/dt = 150.79 m^3/s
Explanation:
In order to calculate the rate of change of the volume, you calculate the derivative, respect to the radius of the sphere, of the volume of the sphere, as follow:
[tex]\frac{dV}{dt}=\frac{d}{dt}(\frac{4}{3}\pi r^3)[/tex] (1)
r: radius of the sphere
You calculate the derivative of the equation (1):
[tex]\frac{dV}{dt}=\frac{d}{dt}(\frac{4}{3}\pi r^3)=3\frac{4}{3}\pi r^2\frac{dr}{dt}=4\pi r^2\frac{dr}{dt}\\\\\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}[/tex](2)
where dr/dt = 3m/s
You replace the values of dr/dt and r=2m in the equation (2):
[tex]\frac{dV}{dt}=4\pi (2m)^2(3\frac{m}{s})=150.79\frac{m^3}{s}[/tex]
The rate of change of the sphere, when it has a radius of 2m, is 150.79m^3/s
A pickup truck moves at 25 m/s toward the east. Ahmed is standing in the back and throws a baseball in what to him is the southwest direction at 28 m/s (with respect to the truck). A person at rest on the ground would see the ball moving how fast in what direction? HTML EditorKeyboard Shortcuts
Answer:
Speed = 20 m/sec at 75 deg South of East = 20 m/sec at 15 deg East of South
Explanation:
given data
truck moves = 25 m/s toward the east.
throws a baseball = 28 m/s southwest
solution
first we take here Speed of truck w.r.to ground i.e. V(p/g) = 25 m/sec toward the east so we can say
V(p/g) = (25 i) m/sec ........................1
and
Speed of baseball w.r.t. pickup i.e. V(b/p) = 28 m/sec toward the South West and we know that south west direction is in third quadrant
and here both component (x and y) are negative
So that we can say it
V(b/p) = -28 × cos(45) i - 28 × sin(45) j = -19.8 i - 19.8 j
and
now we use here relative motion velocity for ball w.r.t ground
V(b/g) = V(b/p) + V(p/g ) ..........................2
put here value and we get
V(b/g) = (-19.8 i - 19.8 j) + 25 i = 5.2 i - 19.8 j
so
Magnitude of that velocity
| V(b/g) | = [tex]\sqrt{(5.2^2 + 19.8^2)}[/tex]
| V(b/g) | = 20.47 m/sec
so that Direction will be here
Direction = arctan (19.8 ÷ 5.2)
Direction = 75.3° South of East
so that
Speed = 20.47 m/sec at 75.3 deg South of East
and 2 significant
Speed = 20 m/sec at 75 deg South of East = 20 m/sec at 15 deg East of South
A car moving at a speed of 25 m/s enters a curve that traces a circular quarter turn of radius 129 m. The driver gently applies the brakes, slowing the car with a constant tangential acceleration of magnitude 1.2 m/s2.a) Just before emerging from the turn, what is the magnitudeof the car's acceleration?
b) At that same moment, what is the angle q between the velocity vector and theacceleration vector?
I am having trouble because this problem seems to have bothradial and tangential accleration. I tried finding the velocityusing V^2/R, but then that didnt take into account thedeceleration. Any help would be great.
Answer:
8.7 m/s^2
82.15°
Explanation:
Given:-
- The initial speed of the car, vi = 25 m/s
- The radius of track, r = 129 m
- Car makes a circular " quarter turn "
- The constant tangential acceleration, at = 1.2 m/s^2
Solution:-
- We will solve the problem using rotational kinematics. Determine the initial angular velocity of car ( wi ) as follows:
[tex]w_i = \frac{v_i}{r} \\\\w_i = \frac{25}{129}\\\\w_i = 0.19379 \frac{rad}{s}[/tex]
- Now use the constant tangential acceleration ( at ) and determine the constant angular acceleration ( α ) for the rotational motion as follows:
at = r*α
α = ( 1.2 / 129 )
α = 0.00930 rad/s^2
- We know that the angular displacement from the initial entry to the exit of the turn is quarter of a turn. The angular displacement would be ( θ = π/2 ).
- Now we will use the third rotational kinematic equation of motion to determine the angular velocity at the exit of the turn (wf) as follows:
[tex]w_f^2 = w_i^2 + 2\alpha*theta\\\\w_f = \sqrt{0.19379^2 + 0.00930\pi } \\\\w_f = 0.25840 \frac{rad}{s}[/tex]
- We will use the evaluated final velocity ( wf ) and determine the corresponding velocity ( vf ) as follows:
[tex]v_f = r*w_f\\\\v_f = 129*0.2584\\\\v_f = 33.33380 \frac{x}{y}[/tex]
- Now use the formulation to determine the centripetal acceleration ( ac ) at this point as follows:
[tex]a_c = \frac{v_f^2}{r} \\\\a_c = \frac{33.3338^2}{129} \\\\a_c = 8.6135 \frac{m}{s^2}[/tex]
- To determine the magnitude of acceleration we will use find the resultant of the constant tangential acceleration ( at ) and the calculated centripetal acceleration at the exit of turn ( ac ) as follows:
[tex]|a| = \sqrt{a^2_t + a_c^2} \\\\|a| = \sqrt{1.2^2 + 8.6135^2} \\\\|a| = 8.7 \frac{m}{s^2}[/tex]
- To determine the angle between the velocity vector and the acceleration vector. We need to recall that the velocity vector only has one component and always tangential to the curved path. Hence, the velocity vector is parallel to the tangential acceleration vector ( at ). We can use the tangential acceleration ( at ) component of acceleration ( a ) and the centripetal acceleration ( ac ) component of the acceleration and apply trigonometric ratio as follows:
[tex]q = arctan \frac{a_c}{a_t} = arctan \frac{8.7}{1.2} \\\\q = 82.15 ^.[/tex]
Answer: The angle ( q ) between acceleration vector ( a ) and the velocity vector ( v ) at the exit of the turn is 82.15° .
How much heat does it take to raise the temperature of 7.0 kg of water from
25-C to 46-C? The specific heat of water is 4.18 kJ/(kg.-C).
Use Q = mcTr-T)
A. 148 kJ
B. 176 kJ
C. 610 kJ
D. 320 kJ
Answer:
non of the above
Explanation:
Quantity of heat = mass× specific heat× change in temperature
m= 7kg c= 4.18 temp= 46-25=21°
.......H= 7×4.18×21= 614.46kJ
Answer:610 KJ
Explanation:A P E X answers
Which of the following emissions is associated with burning coal? a. sulfur dioxide b. carbon dioxide c. nitrous oxides d. all of the above
Answer:
all of the above
Explanation:
because it is.
To get an idea of the order of magnitude of inductance, calculate the self-inductance in henries for a solenoid with 900 loops of wire wound on a rod 6 cm long with radius 1 cm?
Answer:
The self-inductance is [tex]L = 0.0053 \ H[/tex]
Explanation:
From the question we are told that
The number of loops is [tex]N = 900[/tex]
The length of the rod is [tex]l =6 \ cm = 0.06 \ m[/tex]
The radius of the rod is [tex]r = 1 \ cm = 0.01 \ m[/tex]
The self-inductance for the solenoid is mathematically represented as
[tex]L = \frac{\mu_o * A * N^2 }{l}[/tex]
Now the cross-sectional of the solenoid is mathematically evaluated as
[tex]A = \pi r^2[/tex]
substituting values
[tex]A =3.142 * 0.01 ^2[/tex]
[tex]A = 3.142 *10^{-4} \ m^2[/tex]
and [tex]\mu_o[/tex] is the permeability of free space with a value [tex]\mu_o = 4\pi * 10^{-7} N/A^2[/tex]
substituting values into above equation
[tex]L = \frac{ 4\pi * 10^{-7} ^2* 3.142*10^{-4} * 900^2 }{0.06}[/tex]
[tex]L = 0.0053 \ H[/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]
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.
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.
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
An interference pattern is produced by light with a wavelength 550 nm from a distant source incident on two identical parallel slits separated by a distance (between centers) of 0.500 mm .
a. If the slits are very narrow, what would be the angular position of the second- order, two-slit interference maxima?
b. Let the slits have a width 0.300 mm. In terms of the intensity lo at the center of the central maximum, what is the intensity at the angular position in part "a"?
Answer:
a
[tex]\theta = 0.0022 rad[/tex]
b
[tex]I = 0.000304 I_o[/tex]
Explanation:
From the question we are told that
The wavelength of the light is [tex]\lambda = 550 \ nm = 550 *10^{-9} \ m[/tex]
The distance of the slit separation is [tex]d = 0.500 \ mm = 5.0 *10^{-4} \ m[/tex]
Generally the condition for two slit interference is
[tex]dsin \theta = m \lambda[/tex]
Where m is the order which is given from the question as m = 2
=> [tex]\theta = sin ^{-1} [\frac{m \lambda}{d} ][/tex]
substituting values
[tex]\theta = 0.0022 rad[/tex]
Now on the second question
The distance of separation of the slit is
[tex]d = 0.300 \ mm = 3.0 *10^{-4} \ m[/tex]
The intensity at the the angular position in part "a" is mathematically evaluated as
[tex]I = I_o [\frac{sin \beta}{\beta} ]^2[/tex]
Where [tex]\beta[/tex] is mathematically evaluated as
[tex]\beta = \frac{\pi * d * sin(\theta )}{\lambda }[/tex]
substituting values
[tex]\beta = \frac{3.142 * 3*10^{-4} * sin(0.0022 )}{550 *10^{-9} }[/tex]
[tex]\beta = 0.06581[/tex]
So the intensity is
[tex]I = I_o [\frac{sin (0.06581)}{0.06581} ]^2[/tex]
[tex]I = 0.000304 I_o[/tex]