Answer:
the final angular velocity of the particle is approximately 38.18 Rad/s
Explanation:
To start with, let's make sure that units of angle measure are the same, converting everything into radians:
[tex]4500^o\, \frac{\pi}{180^o}= 25\,\pi[/tex]
And now we can use the kinematic formulas for rotational motion:
[tex]\theta-\theta_0=\omega_0\,t+\frac{1}{2} \alpha\,t^2[/tex]
Therefore we can find the initial angular velocity [tex]\omega_0[/tex] of the particle:
[tex]\theta-\theta_0=\omega_0\,t+\frac{1}{2} \alpha\,t^2\\25\,\pi=\omega_0\,(3)+\frac{1}{2} (8)\,(3)^2\\25\,\pi-36=\omega_0\,(3)\\\omega_0=\frac{25\,\pi-36}{3} \\\omega_0\approx 14.18\,\,\,rad/s[/tex]
and now we can estimate the final angular velocity using the kinematic equation for angular velocity;
[tex]\omega=\omega_0\,+\alpha\,t\\\omega=14.18+8\,(3)\\\omega=38.18\,\,\,rad/s[/tex]
A small wave pulse and a large wave pulse approach each other on a string; the large pulse is moving to the right.
Sometime after the pulses have met and passed each other, which of the following statements is correct? (More than one answer may be correct)
- the large pulse continues moving to the right
- the large pulse continues unchanged, moving to the right
- the small pulse is reflected and moves off to the right with a smaller amplitude
- the small pulse is reflected and moves off to the right with its original amplitude
- the two pulses combine into a single pulse moving to the right
Answer:
the large amplitude wave keeps moving to the right
the small amplitude wave continues to move to the left.
When checking the answers, the correct ones are 1, 2
Explanation:
The waves fulfill the principle of superposition, which states that the value of the function at a point is the algebraic sum of the waves at a given instant.
The two waves in this exercise travel in the opposite direction, so when they are close, the resulting wave is the sum of the two waves, having a complicated shape. But when the waves follow their movement, they give in the same way as the initial a,
the large amplitude wave keeps moving to the right
the small amplitude wave continues to move to the left.
When checking the answers, the correct ones are 1, 2
If a sample emits 2000 counts per second when the detector is 1 meter from the sample, how many counts per second would be observed when the detector is 3 meters from the sample?
Using the sample in above question how many counts per second would be observed when the detector is 10 meters away from the sample?
Answer:
At 3 meter distance, the per-second count is 222.22 and at a 10 meter distance, the per-second count is 20.
Explanation:
The number of particles (N) counts are inversely proportional to the distance between the source and the detector.
By using the below formula we can find the number of counts.
[tex]N2 = \frac{(D1)^2}{(D2)^2} \times N1 \\N1 = 2000 \\D 1 = 1 \ meter \\D2 = 3 \\[/tex]
The number of count per second, when the distance is 3 meters.
[tex]= \frac{1}{3^2} \times 2000 \\= 222.22[/tex]
Number of count per second when the distance is 10 meters.
[tex]= \frac{1}{10^2} \times 2000 \\= 20[/tex]
A pool ball moving 1.83 m/s strikes an identical ball at rest. Afterward, the first ball moves 1.15 m/s at a 23.3 degrees angle. What is the y-component of the velocity of the second ball?
Answer:
v_{1fy} = - 0.4549 m / s
Explanation:
This is an exercise of conservation of the momentum, for this we must define a system formed by the two balls, so that the forces during the collision have internal and the momentum is conserved
initial. Before the crash
p₀ = m v₁₀
final. After the crash
[tex]p_{f}[/tex] = m [tex]v_{1f}[/tex] + m v_{2f}
Recall that velocities are a vector so it has x and y components
p₀ = p_{f}
we write this equation for each axis
X axis
m v₁₀ = m v_{1fx} + m v_{2fx}
Y Axis
0 = -m v_{1fy} + m v_{2fy}
the exercise tells us the initial velocity v₁₀ = 1.83 m / s, the final velocity v_{2f} = 1.15, let's use trigonometry to find its components
sin 23.3 = v_{2fy} / v_{2f}
cos 23.3 = v_{2fx} / v_{2f}
v_{2fy} = v_{2f} sin 23.3
v_{2fx} = v_{2f} cos 23.3
we substitute in the momentum conservation equation
m v₁₀ = m v_{1f} cos θ + m v_{2f} cos 23.3
0 = - m v_{1f} sin θ + m v_{2f} sin 23.3
1.83 = v_{1f} cos θ + 1.15 cos 23.3
0 = - v_{1f} sin θ + 1.15 sin 23.3
1.83 = v_{1f} cos θ + 1.0562
0 = - v_{1f} sin θ + 0.4549
v_{1f} sin θ = 0.4549
v_{1f} cos θ = -0.7738
we divide these two equations
tan θ = - 0.5878
θ = tan-1 (-0.5878)
θ = -30.45º
we substitute in one of the two and find the final velocity of the incident ball
v_{1f} cos (-30.45) = - 0.7738
v_{1f} = -0.7738 / cos 30.45
v_{1f} = -0.8976 m / s
the component and this speed is
v_{1fy} = v1f sin θ
v_{1fy} = 0.8976 sin (30.45)
v_{1fy} = - 0.4549 m / s
At the pizza party you and two friends decide to go to Mexico City from El Paso, TX where y'all live. You volunteer your car if everyone chips in for gas. Someone asks how much the gas will cost per person on a round trip. Your first step is to call your smarter brother to see if he'll figure it out for you. Naturally he's too busy to bother, but he does tell you that it is 2015 km to Mexico City, there's 11 cents to the peso, and gas costs 5.8 pesos per liter in Mexico. You know your car gets 21 miles to the gallon, but we still don't have a clue as to how much the trip is going to cost (in dollars) each person in gas ($/person).
Answer:
cost_cost = $ 96
Explanation:
In this exercise we have units in the groin system and the SI system, to avoid problems let's reduce everything to the SI system
performance = 21 miles / gallon (1,609 km / 1 mile) (1 gallon / 3,785 l)
perfomance= 8,927 km / l
now let's use a direct rule of proportions (rule of three). If a liter travels 8,927 km, how many liters are needed to travel the 2015 km
#_gasoline = 2015 km (1l / 8.927 km) = 225.72 liters
Now let's find the total cost of fuel. Ns indicates that $ 0.11 = 1 peso and the liter of fuel costs 5.8 pesos
cost_litre = 5.8 peso ($ 0.11 / 1 peso) = $ 0.638
cost_gasoline = #_gasoline cost_litro
cost_gasoline = 225.72 0.638
cost_gasoline = $ 144
This cost is for the one way trip, the total round trip cost is
cost_total = 2 cost_gasoline
cost_total = $ 288
Now let's look for the cost in the vehicle, you and two people will go, for which a total of 3 people will go, so the cost per person is
cost_person = total_cost / #_people
cost_person = 288/3
cost_cost = $ 96
A transformer consists of a 500-turn primary coil and a 2000-turn secondary coil. If the current in the secondary is 3.0 A, what is the current in the primary
Answer:
12AExplanation:
Formula for calculating the relationship between the electromotive force (emf), current and number of turns of a coil in a transformer is expressed as shown:
[tex]\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p} = \dfrac{I_p}{I_s}[/tex] where;
Vs and Vp are the emf in the secondary and primary coil respectively
Ns and Np are the number if turns in the secondary and primary coil respectively
Ip and Is are the currents in the secondary and primary coil respectively
Since the are all equal to each other, then we can equate any teo of the expression as shown;
[tex]\dfrac{N_s}{N_p} = \dfrac{I_p}{I_s}[/tex]
Given parameters
Np = 500-turns
Ns = 2000-turns
Is = 3.0Amp
Required
Current in the primary coil (Ip)
Using the relationship [tex]\dfrac{N_s}{N_p} = \dfrac{I_p}{I_s}[/tex]
[tex]I_p = \dfrac{N_sI_s}{N_p}[/tex]
[tex]I_p = \dfrac{2000*3}{500} \\\\I_p = \frac{6000}{500}\\ \\I_p = 12A\\[/tex]
Hence the current in the primary coil is 12Amp
The maximum gauge pressure in a hydraulic system is 15 atm. What is the largest mass that could be lifted by this system if the diameter of the piston is 65 cm
Answer:
The maximum force that can be lifted by this system is 51,478.4 kg
Explanation:
Given;
maximum gauge pressure of the hydraulic system, Hp = 15 atm = 1.52 x 10⁶ N/m²
diameter of the piston, d = 65 cm = 0.65 m
The maximum gauge pressure of the piston is given as;
[tex]Hp = \frac{F}{A}[/tex]
Where;
F is the maximum force of the piston
A is the area of the piston
[tex]A = \pi (\frac{0.65}{2} )^2\\\\A = 0.3319 \ m^2[/tex]
F = Hp x A
F = 1.52 x 10⁶N/m² x 0.3319m²
F = 504488 N
Force is given as;
F = mg
m = F/g
m = 504488/9.8
m = 51,478.4 kg
Therefore, the maximum force that can be lifted by this system is 51,478.4 kg
A student uses a spring scale attached to a textbook to compare the static and kinetic friction between the textbook and the top of a lab
table. If the scale measures 1,580 g while the student is pulling the sliding book along the table, which reading on the scale could have been
possible at the moment the student overcame the static friction? (1 point)
1,860 g
820 g
1,580 g
1,140 g
Answer:
1,860 g
Explanation:
In a system, the coefficient of static friction is usually higher than the coefficient of kinetic friction. This means that the kinetic friction is usually less than the static friction. From the question, since the book is already sliding, it means that kinetic friction is the friction in play. This means that before the reading on the scale that could have been possible at the moment the student overcame the static friction must be greater than the reading on the scale during sliding. The only option above 1580 g is 1860 g
A bus carrying 10 people has over turned on a remote hillside during an intense thunderstorm. What three factors could contribute to creating a delay in advanced care
Answer:
The three factors that can contribute to creating a delay in advanced care for the passengers in the overturned bus include:
1. Lack of communication: Since the accident happened on the remote hillside, there is a possibility that, there would be no communication network which could have afforded them the opportunity to call medical or technical team.
2. Steep Nature of the Hill: This is another factor which will affect the care which they could have received. Steeply area tends to be difficult for climbing in or out from.
3. Thunderstorm: This factor is another reason which could contribute to delay in receiving advance care. Thunderstorm create barriers for location f the area where the bus overturned or in other situation complicate the rescue efforts of the team sent out to rescue.
Explanation:
As you finish listening to your favorite compact disc (CD), the CD in the player slows down to a stop. Assume that the CD spins down with a constant angular acceleration. If the CD rotates clockwise (let's take clockwise rotation as positive) at 500 rpm (revolutions per minute) while the last song is playing, and then spins down to zero angular speed in 2.60 s with constant angular acceleration, the angular acceleration of the CD, as it spins to a stop at -20.1 rad/s 2. How many revolutions does the CD make as it spins to a stop?
Answer:
10.8rev
Explanation:
Using
Wf²-wf = 2 alpha x theta
0²- 56.36x56.36/ 2(-20.13) x theta
Theta = 68.09 rad
But 68.09/2π
>= 10.8 revolutions
Explanation:
A beam of light from a laser illuminates a glass how long will a short pulse of light beam take to travel the length of the glass.
Answer:
The time of short pulse of light beam is [tex]2.37\times10^{-9}\ sec[/tex]
Explanation:
Given that,
A beam of light from a laser illuminates a glass.
Suppose, the length of piece is [tex]L=25.21\times10^{-2}\ m[/tex]
Index of refraction is 2.83.
We need to calculate the speed of light pulse in glass
Using formula of speed
[tex]v=\dfrac{c}{\mu}[/tex]
Put the value into the formula
[tex]v=\dfrac{3\times10^{8}}{2.83}[/tex]
[tex]v=1.06\times10^{8}\ m/s[/tex]
We need to calculate the time of short pulse of light beam
Using formula of velocity
[tex]v=\dfrac{d}{t}[/tex]
[tex]t=\dfrac{d}{v}[/tex]
Put the value into the formula
[tex]t=\dfrac{25.21\times10^{-2}}{1.06\times10^{8}}[/tex]
[tex]t=2.37\times10^{-9}\ sec[/tex]
Hence, The time of short pulse of light beam is [tex]2.37\times10^{-9}\ sec[/tex]
The actual depth of a shallow pool 1.00 m deep is not the same as the apparent depth seen when you look straight down at the pool from above. How deep (in cm) will it appear to be
Answer:
d' = 75.1 cm
Explanation:
It is given that,
The actual depth of a shallow pool is, d = 1 m
We need to find the apparent depth of the water in the pool. Let it is equal to d'.
We know that the refractive index is also defined as the ratio of real depth to the apparent depth. Let the refractive index of water is 1.33. So,
[tex]n=\dfrac{d}{d'}\\\\d'=\dfrac{d}{n}\\\\d'=\dfrac{1\ m}{1.33}\\\\d'=0.751\ m[/tex]
or
d' = 75.1 cm
So, the apparent depth is 75.1 cm.
A steel ball attached to a spring moves in simple harmonic motion. The amplitude of the ball's motion is 11.0 cm, and the spring constant is 6.00 N/m. When the ball is halfway between its equilibrium position and its maximum displacement from equilibrium, its speed is 26.1 cm/s. (a) What is the mass of the ball (in kg)? kg (b) What is the period of oscillation (in s)? s (c) What is the maximum acceleration of the ball? (Enter the magnitude in m/s2.) m/s2
Answer:
a) m = 0.626 kg , b) T = 2.09 s , c) a = 1.0544 m / s²
Explanation:
In a spring mass system the equation of motion is
x = A cos (wt + Ф)
with w = √(k / m)
a) velocity is defined by
v = dx / dt
v = - A w sin (wt + Ф) (1)
give us that the speed is
v = 26.1 m / s
for the point
x = a / 2
the range of motion is a = 11.0 cm
x = 11.0 / 2
x = 5.5 cm
Let's find the time it takes to get to this distance
wt + Ф = cos⁻¹ (x / A)
wt + Ф = cos 0.5
wt + Ф = 0.877
In the exercise they do not indicate that the body started its movement with any speed, therefore we assume that for the maximum elongation the body was released, therefore the phase is zero f
Ф = 0
wt = 0.877
t = 0.877 / w
we substitute in equation 1
26.1 = -11.0 w sin (w 0.877 / w)
w = 26.1 / (11 sin 0.877))
w = 3.096 rad / s
from the angular velocity equation
w² = k / m
m = k / w²
m = 6 / 3,096²
m = 0.626 kg
b) angular velocity and frequency are related
w = 2π f
frequency and period are related
f = 1 / T
we substitute
w = 2π / T
T = 2π / w
T = 2π / 3,096
T = 2.09 s
c) maximum acceleration
the acceleration of defined by
a = dv / dt
a = - Aw² cos (wt)
the acceleration is maximum when the cosine is ±1
a = A w²
a = 11 3,096²
a = 105.44 cm / s²
we reduce to m / s
a = 1.0544 m / s²
Design a voltage divider to provide the following approximate voltages with respect to ground using a 30 V source: 8.18 V, 14.7 V, and 24.6 V. The current drain on the source must be limited to no more than 1 mA. The number of resistors, their values, and their wattage ratings must be specified. A schematic showing the circuit arrangement and resistor placement must be provided
Answer:
R₁ = 14.7 10³ Ω , R₂ = 8.18 10³ Ω , R₃ = 1.72 10³ Ω , R₄ = 5.4 10³ Ω 1/8 W resistor
Explanation:
For this exercise we must use a series circuit since the sum of the voltage on each resin is equal to the source voltage (V = 30 V)
Therefore we build a circuit with 4 resistors in series, in such a way that
V = i R
let the voltage
1st resistance
V = i R
R₁ = V / i
R₁ = 14.7 / 1 10⁻³
R₁ = 14.7 10³ Ω
power is
P = V i
P = 14.7 1 10⁻³
P = 14.7 10⁻³ W = 0.0147 W
a resistance of ⅛ W is indicated
2nd resistance
R₂ = 8.18 / 1 10⁻³
R₂ = 8.18 10³ Ω
Power
P = 8.18 1 10⁻³
P = 0.00818W
a 1/8 W resistor
3rd resistance
this resistance is calculated in such a way that
V₁ + V₂ + V₃ = 24.6
V₃ = 24.6 - V₁ -V₂
V₃ = 24.6 - 14.7 - 8.18
V₃ = 1.72 V
R₃ = 1.72 / 1 10⁻³
R₃ = 1.72 10³ Ω
power
P = Vi
P = 1.72 10⁻³
P = 0.00172 W
a resistance of ⅛ W
To obtain the voltage of 24.6 we use this three resistors together
4th resistance
The value of this resistance is calculated so that the sum of all the voltages reaches the source voltage
30 = V₁ + V₂ + V₃ + V₄
V₄ = 30 - V₁ -V₂ -V₃
V₄ = 30 -14.7 - 8.18 - 1.72
V₄ = 5.4 V
R₄ = 5.4 / 1 10⁻³
R₄ = 5.4 10³ Ω
Power
P = V i
P = 5.4 10⁻³
P = 0.0054 W
⅛ W resistance
The values of these resistance are commercially
Let's check the consumption of the circuit
R_total = R₁ + R₂ + R₃ + R₄
R_total = (14.7 + 8.18 + 1.72 + 5.4) 10³
R_total = 30 10³
the current circulating in the circuit is
i = V / R_total
i = 30/30 10³
i = 1 10⁻³ A
therefore it is within the order requirement.
for connections see attached diagram
A city of Punjab has a 15 percent chance of wet weather on any given day. What is the probability that it will take a week for it three wet weather on 3 separate days?
Answer: 0.0617
Explanation:
Given: The probability of wet weather on any given day in a city of Punjab : p=15%=0.15
Let X be a binomial variable that represents the number of days having wet weather.
Binomial probability formula : [tex]P(X=x)=^nC_xp^x(1-p)^x[/tex], where n= total outcomes, p = probability of success in each outcomes.
Here, n= 7 ( 1 week = 7 days)
The probability that it will take a week for it three wet weather on 3 separate days:
[tex]P(X=3)^=\ ^7C_3(0.15)^3(1-0.15)^{7-3}\\\\=\dfrac{7!}{3!(7-3)!}(0.15)^3(0.85)^4\\\\=\dfrac{7\times6\times5}{3\times2}\times 0.003375\times0.52200625\approx0.0617[/tex]
Hence, the required probability =0.0617
a toy propeller fan with a moment of inertia of .034 kg x m^2 has a net torque of .11Nxm applied to it. what angular acceleration does it experience
Answer:
The angular acceleration is [tex]\alpha = 3.235 \ rad/s ^2[/tex]
Explanation:
From the question we are told that
The moment of inertia is [tex]I = 0.034\ kg \cdot m^2[/tex]
The net torque is [tex]\tau = 0.11\ N \cdot m[/tex]
Generally the net torque is mathematically represented as
[tex]\tau = I * \alpha[/tex]
Where [tex]\alpha[/tex] is the angular acceleration so
[tex]\alpha = \frac{\tau }{I}[/tex]
substituting values
[tex]\alpha = \frac{0.1 1}{ 0.034}[/tex]
[tex]\alpha = 3.235 \ rad/s ^2[/tex]
How many turns of wire are needed in a circular coil 13 cmcm in diameter to produce an induced emf of 5.6 VV
Answer:
Number of turns of wire(N) = 3,036 turns (Approx)
Explanation:
Given:
Diameter = 13 Cm
emf = 5.6 v
Note:
The given question is incomplete, unknown information is as follow.
Magnetic field increases = 0.25 T in 1.8 (Second)
Find:
Number of turns of wire(N)
Computation:
radius (r) = 13 / 2 = 6.5 cm = 0.065 m
Area = πr²
Area = (22/7)(0.065)(0.065)
Area = 0.013278 m²
So,
emf = (N)(A)(dB / dt)
5.6 = (N)(0.013278)(0.25 / 1.8)
5.6 = (N)(0.013278)(0.1389)
N = 3,036.35899
Number of turns of wire(N) = 3,036 turns (Approx)
Air bags greatly reduces the chance og injury in a car accident.explain how they do si in terms of energy transfer
Answer:
in an accident, when the body collides with the air bags, the collision time of impact between the two bodies will increase due to the presence of air bags in the car. Larger is the impact time smaller is the transformation of energy between the body and air bag. That is why air bags greatly reduce the chance of injury in a car accident.
A ball is thrown from the ground so that it’s initial vertical and horizontal components of velocity are 40m/s and 20m/s respectively. Find the ball’s total time of flight and distance it traverses before hitting the ground.
Answer:
8 seconds
160 meters
Explanation:
Given in the y direction:
Δy = 0 m
v₀ = 40 m/s
a = -10 m/s²
Find: t
Δy = v₀ t + ½ at²
0 m = (40 m/s) t + ½ (-10 m/s²) t²
0 = 40t − 5t²
0 = 5t (8 − t)
t = 0 or 8
Given in the x direction:
v₀ = 20 m/s
a = 0 m/s²
t = 8 s
Find: Δx
Δx = v₀ t + ½ at²
Δx = (20 m/s) (8 s) + ½ (0 m/s²) (8 s)²
Δx = 160 m
A velocity selector in a mass spectrometer uses a 0.100-T magnetic field. (a) What electric field strength is needed to select a speed of 4.00 . 106 m/s
Answer:
The electric field strength needed is 4 x 10⁵ N/C
Explanation:
Given;
magnitude of magnetic field, B = 0.1 T
velocity of the charge, v = 4 x 10⁶ m/s
The velocity of the charge when there is a balance in the magnetic and electric force is given by;
[tex]v = \frac{E}{B}[/tex]
where;
v is the velocity of the charge
E is the electric field strength
B is the magnetic field strength
The electric field strength needed is calculated as;
E = vB
E = 4 x 10⁶ x 0.1
E = 4 x 10⁵ N/C
Therefore, the electric field strength needed is 4 x 10⁵ N/C
Based on the passage, why is it important that different ethnic groups worked together on the strike? The groups needed to avoid speaking to one another because they wouldn’t understand. The different ethnic groups believed in being separate. The groups needed to trick the owners. They needed to be able to unite even though they spoke different languages.
Answer:D
Explanation:I got it right
Answer:
They needed to be able to unite even though they spoke different languages.
Explanation:
A rigid container holds 4.00 mol of a monatomic ideal gas that has temperature 300 K. The initial pressure of the gas is 6.00 * 104 Pa. What is the pressure after 6000 J of heat energy is added to the gas?
Answer:
The final pressure of the monoatomic ideal gas is 8.406 × 10⁶ pascals.
Explanation:
When a container is rigid, the process is supposed to be isochoric, that is, at constant volume. Then, the equation of state for ideal gases can be simplified into the following expression:
[tex]\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}[/tex]
Where:
[tex]P_{1}[/tex], [tex]P_{2}[/tex] - Initial and final pressures, measured in pascals.
[tex]T_{1}[/tex], [tex]T_{2}[/tex] - Initial and final temperatures, measured in Kelvins.
In addtion, the specific heat at constant volume for monoatomic ideal gases, measured in joules per mole-Kelvin is given by:
[tex]\bar c_{v} = \frac{3}{2}\cdot R_{u}[/tex]
Where:
[tex]R_{u}[/tex] - Ideal gas constant, measured by pascal-cubic meters per mole-Kelvin.
If [tex]R_{u} = 8.314\,\frac{Pa\cdot m^{3}}{mol\cdot K}[/tex], then:
[tex]\bar c_{v} = \frac{3}{2}\cdot \left(8.314\,\frac{Pa\cdot m^{2}}{mol\cdot K} \right)[/tex]
[tex]\bar c_{v} = 12.471\,\frac{J}{mol\cdot K}[/tex]
And change in heat energy ([tex]Q[/tex]), measured by joules, by:
[tex]Q = n\cdot \bar c_{v}\cdot (T_{2}-T_{1})[/tex]
Where:
[tex]n[/tex] - Molar quantity, measured in moles.
The final temperature of the monoatomic ideal gas is now cleared:
[tex]T_{2} = T_{1} + \frac{Q}{n\cdot \bar c_{v}}[/tex]
Given that [tex]T_{1} = 300\,K[/tex], [tex]Q = 6000\,J[/tex], [tex]n = 4\,mol[/tex] and [tex]\bar c_{v} = 12.471\,\frac{J}{mol\cdot K}[/tex], the final temperature is:
[tex]T_{2} = 300\,K + \frac{6000\,J}{(4\,mol)\cdot \left(12.471\,\frac{J}{mol\cdot K} \right)}[/tex]
[tex]T_{2} = 420.279\,K[/tex]
The final pressure of the system is calculated by the following relationship:
[tex]P_{2} = \left(\frac{T_{2}}{T_{1}}\right) \cdot P_{1}[/tex]
If [tex]T_{1} = 300\,K[/tex], [tex]T_{2} = 420.279\,K[/tex] and [tex]P_{1} = 6.00\times 10^{4}\,Pa[/tex], the final pressure is:
[tex]P_{2} = \left(\frac{420.279\,K}{300\,K} \right)\cdot (6.00\times 10^{4}\,Pa)[/tex]
[tex]P_{2} = 8.406\times 10^{4}\,Pa[/tex]
The final pressure of the monoatomic ideal gas is 8.406 × 10⁶ pascals.
Q9 A physics book slides off a horizontal tabletop with a speed of 1.10 m/s. It strikes the floor in 0.350s. ignore air resistance. Find (a) the height of the tabletop above the floor; (b) the horizontal distance from the edge of the table to the point where the book strikes the floor; (c) the horizontal and vertical components of the book's velocity, and the magnitude and direction of its velocity, just before the book reaches the floor.
Answer:
(a) 0.613 m
(b) 0.385 m
(c) vₓ = 1.10 m/s, vᵧ = 3.50 m/s
v = 3.68 m/s², θ = 72.6° below the horizontal
Explanation:
(a) Take down to be positive.
Given in the y direction:
v₀ = 0 m/s
a = 10 m/s²
t = 0.350 s
Find: Δy
Δy = v₀ t + ½ at²
Δy = (0 m/s) (0.350 s) + ½ (10 m/s²) (0.350 s)²
Δy = 0.613 m
(b) Given in the x direction:
v₀ = 1.10 m/s
a = 0 m/s²
t = 0.350 s
Find: Δx
Δx = v₀ t + ½ at²
Δx = (1.10 m/s) (0.350 s) + ½ (0 m/s²) (0.350 s)²
Δx = 0.385 m
(c) Find: vₓ and vᵧ
vₓ = aₓt + v₀ₓ
vₓ = (0 m/s²) (0.350 s) + 1.10 m/s
vₓ = 1.10 m/s
vᵧ = aᵧt + v₀ᵧ
vᵧ = (10 m/s²) (0.350 s) + 0 m/s
vᵧ = 3.50 m/s
The magnitude is:
v² = vₓ² + vᵧ²
v = 3.68 m/s²
The direction is:
θ = atan(vᵧ / vₓ)
θ = 72.6° below the horizontal
An electromagnetic flowmeter is useful when it is desirable not to interrupt the system in which the fluid is flowing (e.g. for the blood in an artery during heart surgery). Such a device is illustrated. The conducting fluid moves with velocity v in a tube of diameter d perpendicular to which is a magnetic field B. A voltage V is induced between opposite sides of the tube. Given B = 0.120 T, d = 1.2 cm., and a measured voltage of 2.88 mV, determine the speed of the blood.
Answer:
2 m/s
Explanation:
The electromagnetic flow-metre work on the principle of electromagnetic induction. The induced voltage is given as
[tex]E = Blv[/tex]
where [tex]E[/tex] is the induced voltage = 2.88 mV = 2.88 x 10^-3 V
[tex]l[/tex] is the distance between the electrodes in this field which is equivalent to the diameter of the tube = 1.2 cm = 1.2 x 10^-2 m
[tex]v[/tex] is the velocity of the fluid through the field = ?
[tex]B[/tex] is the magnetic field = 0.120 T
substituting, we have
2.88 x 10^-3 = 0.120 x 1.2 x 10^-2 x [tex]v[/tex]
2.88 x 10^-3 = 1.44 x 10^-3 x [tex]v[/tex]
[tex]v[/tex] = 2.88/1.44 = 2 m/s
Heat and thermodynamics Numerical
Answer:
K = 227.04 W/m.°C
Explanation:
First we need to find the heat required to melt the ice:
q = m H
where,
q = heat required = ?
m = mass of the ice = 8.5 g = 8.5 x 10⁻³ kg
H = Latent heat of fusion of ice = 3.34 x 10⁵ J/kg
Therefore,
q = (8.5 x 10⁻³ kg)(3.34 x 10⁵ J/kg)
q = 2839 J
Now, we find the heat transfer rate through rod:
Q = q/t
where,
t = time = (10 min)(60 s/1 min) = 600 s
Q = Heat Transfer Rate = ?
Therefore,
Q = 2839 J/600 s
Q = 4.73 W
From Fourier's Law of Heat Conduction:
Q = KA ΔT/L
where,
K = Thermal Conductivity = ?
A = cross sectional area = 1.25 cm² = 1.25 x 10⁻⁴ m²
L = Length of rod = 60 cm = 0.6 m
ΔT = Difference in temperature = 100°C - 0°C = 100°C
Therefore,
4.73 W = K(1.25 X 10⁻⁴ m²)(100°C)/0.6 m
K = (4.73 W)/(0.0208 m.°C)
K = 227.04 W/m.°C
15.Restore the battery setting to 10 V. Now change the number of loops from 4 to 3. Explain what happens to the magnitude and direction of the magnetic field. Now change to 2 loops, then to 1 loop. What do you observe the relationship to be between the magnitude of the magnetic field and the number of loops for the same current
Answer:
we see it is a linear relationship.
Explanation:
The magnetic flux is u solenoid is
B = μ₀ N/L I
where N is the number of loops, L the length and I the current
By applying this expression to our case we have that the current is the same in all cases and we can assume the constant length. Consequently we see that the magnitude of the magnetic field decreases with the number of loops
B = (μ₀ I / L) N
the amount between paracentesis constant, in the case of 4 loop the field is worth
B = cte 4
N B
4 4 cte
3 3 cte
2 2 cte
1 1 cte
as we see it is a linear relationship.
In addition, this effect for such a small number of turns the direction of the field that is parallel to the normal of the lines will oscillate,
If this is the only water being used in your house, how fast is the water moving through your house's water supply line, which has a diameter of 0.021 m (about 3/4 of an inch)?
Answer:
0.273m/s
Explanation:
first find out the meaning of 0.90×10−4m3/s
literally, that is 0.9x6 = 5.4m3/s = 3•5.4m/s or 16.2 m/s
1.5 gal/min = 0.00009464 m³/s, perhaps that is what you mean?
cross-sectional area of pipe is πr² = 0.0105²π = 0.0003464 m²
so you have a a flow of 0.00009464 m³/s flowing through an area of 0.0003464 m²
they divide to 0.00009464 m³/s / 0.0003464 m² = 0.273 m/s
g A projectile is fired from the ground at an angle of θ = π 4 toward a tower located 600 m away. If the projectile has an initial speed of 120 m/s, find the height at which it strikes the tower
Answer:
The projectile strikes the tower at a height of 354.824 meters.
Explanation:
The projectile experiments a parabolic motion, which consist of a horizontal motion at constant speed and a vertical uniformly accelerated motion due to gravity. The equations of motion are, respectively:
Horizontal motion
[tex]x = x_{o}+v_{o}\cdot t \cdot \cos \theta[/tex]
Vertical motion
[tex]y = y_{o} + v_{o}\cdot t \cdot \sin \theta +\frac{1}{2} \cdot g \cdot t^{2}[/tex]
Where:
[tex]x_{o}[/tex], [tex]x[/tex] - Initial and current horizontal position, measured in meters.
[tex]y_{o}[/tex], [tex]y[/tex] - Initial and current vertical position, measured in meters.
[tex]v_{o}[/tex] - Initial speed, measured in meters per second.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]t[/tex] - Time, measured in seconds.
The time spent for the projectile to strike the tower is obtained from first equation:
[tex]t = \frac{x-x_{o}}{v_{o}\cdot \cos \theta}[/tex]
If [tex]x = 600\,m[/tex], [tex]x_{o} = 0\,m[/tex], [tex]v_{o} = 120\,\frac{m}{s}[/tex] and [tex]\theta = \frac{\pi}{4}[/tex], then:
[tex]t = \frac{600\,m-0\,m}{\left(120\,\frac{m}{s} \right)\cdot \cos \frac{\pi}{4} }[/tex]
[tex]t \approx 7.071\,s[/tex]
Now, the height at which the projectile strikes the tower is: ([tex]y_{o} = 0\,m[/tex], [tex]t \approx 7.071\,s[/tex], [tex]v_{o} = 120\,\frac{m}{s}[/tex] and [tex]g = -9.807\,\frac{m}{s^{2}}[/tex])
[tex]y = 0\,m + \left(120\,\frac{m}{s} \right)\cdot (7.071\,s)\cdot \sin \frac{\pi}{4}+\frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right) \cdot (7.071\,s)^{2}[/tex]
[tex]y \approx 354.824\,m[/tex]
The projectile strikes the tower at a height of 354.824 meters.
In a physics lab, Asha is given a 11.5 kg uniform rectangular plate with edge lengths 62.9 cm by 46.9 cm . Her lab instructor requires her to rotate the plate about an axis perpendicular to its plane and passing through one of its corners, and then prepare a report on the project. For her report, Asha needs the plate's moment of inertia ???? with respect to given rotation axis. Calculate ???? .
Answer:
6.9kgm²
Explanation:
For an axis through the center of the rectangle, I = m[(w²+L²)/12
Using the parallel axis theorem, the added value of I = mR² = m[(w²/4 + L²/4]
Adding the 2 expressions,
I = (m/3)*(w²+L²)
I =6.95 kg∙m²
A 5.0-µC point charge is placed at the 0.00 cm mark of a meter stick and a -4.0-µC point charge is placed at the 50 cm mark. At what point on a line joining the two charges is the electric field due to these charges equal to zero?
Answer:
Electric field is zero at point 4.73 m
Explanation:
Given:
Charge place = 50 cm = 0.50 m
change q1 = 5 µC
change q2 = 4 µC
Computation:
electric field zero calculated by:
[tex]E1 =k\frac{q1}{r^2} \\\\E2 =k\frac{q2}{R^2} \\\\[/tex]
Where electric field is zero,
First distance = x
Second distance = (x-0.50)
So,
E1 = E2
[tex]k\frac{q1}{r^2}=k\frac{q2}{R^2} \\\\[/tex]
[tex]\frac{5}{x^2}=\frac{4}{(x-50)^2} \\\\[/tex]
x = 0.263 or x = 4.73
So,
Electric field is zero at point 4.73 m
A certain resistor dissipates 0.5 W when connected to a 3 V potential difference. When connected to a 1 V potential difference, this resistor will dissipate:
Answer:
0.056 WExplanation:
[tex]Power = IV[/tex]
From ohms law we know that
[tex]V= IR\\\\I= \frac{V}{R} \\\\Power= \frac{V}{R}*V\\\\Power= \frac{V^2}{R}[/tex]
Given data
P1 = 0.5 Watt
P2 = ?
V1= 3 Volts
V2= 1 Volt
Thus we can solve for the power dissipated as follows
[tex]P1= \frac{V1^2}{R1}\\\\P2= \frac{V2^2}{R2}[/tex]
[tex]\frac{P1}{P2} = \frac{V1^2}{V2^2}\\\\ P2=\frac{ V2^2}{ V1^2} *P1\\\\ P2=\frac{ 1^2}{ 3^2} *0.5= 0.055= 0.056 W[/tex]
The resistor will dissipate 0.056 Watt
