Answer:
The tub turns 37.520 revolutions during the 25-second interval.
Explanation:
The total number of revolutions done by the tub of the washer ([tex]\Delta n[/tex]), in revolutions, is the sum of the number of revolutions done in the acceleration ([tex]\Delta n_{1}[/tex]), in revolutions, and deceleration stages ([tex]\Delta n_{2}[/tex]), in revolutions:
[tex]\Delta n = \Delta n_{1} + \Delta n_{2}[/tex] (1)
Then, we expand the previous expression by kinematic equations for uniform accelerated motion:
[tex]\Delta n = \frac{1}{2}\cdot ( \ddot n_{1}\cdot t_{1}^{2} - \ddot n_{2} \cdot t_{2}^{2})[/tex] (1b)
Where:
[tex]\ddot n_{1}, \ddot n_{2}[/tex] - Angular accelerations for acceleration and deceleration stages, in revolutions per square second.
[tex]t_{1}, t_{2}[/tex] - Acceleration and deceleration times, in seconds.
And each acceleration is determined by the following formulas:
Acceleration
[tex]\ddot n_{1} = \frac{\dot n}{t_{1}}[/tex] (2)
Deceleration
[tex]\ddot n_{2} = -\frac{\dot n}{t_{2} }[/tex] (3)
Where [tex]\dot n[/tex] is the maximum angular velocity of the tub of the washer, in revolutions per second.
If we know that [tex]\dot n = 3\,\frac{rev}{s}[/tex], [tex]t_{1} = 13\,s[/tex] and [tex]t_{2} = 12\,s[/tex], then the quantity of revolutions done by the tub is:
[tex]\ddot n_{1} = \frac{3\,\frac{rev}{s} }{13\,s}[/tex]
[tex]\ddot n_{1} = 0.231\,\frac{rev}{s^{2}}[/tex]
[tex]\ddot n_{2} = -\frac{3\,\frac{rev}{s} }{12\,s}[/tex]
[tex]\ddot n_{2} = -0.25\,\frac{rev}{s^{2}}[/tex]
[tex]\Delta n = \frac{1}{2}\cdot ( \ddot n_{1}\cdot t_{1}^{2} + \ddot n_{2} \cdot t_{2}^{2})[/tex]
[tex]\Delta n = \frac{1}{2}\cdot \left[\left(0.231\,\frac{rev}{s^{2}} \right)\cdot (13\,s)^{2}-\left(-0.25\,\frac{rev}{s^{2}} \right)\cdot (12\,s)^{2}\right][/tex]
[tex]\Delta n = 37.520\,rev[/tex]
The tub turns 37.520 revolutions during the 25-second interval.
6. (6P) Formula Based Problem: A car is speeding down a highway at 30 m/s. A stray dog
runs into the road ahead of the driver. The driver hits the brake decelerating at a rate of
10m/s. The driver managed to stop the car right as it was about to hit the car. How far
away was the dog when the driver first hit the brake?
Answer:
45 m
Explanation:
From the question given above, the following data were obtained:
Initial velocity (u) = 30 m/s
Deceleration (a) = –10 m/s²
Final velocity (v) = 0 m/s
Distance (s) =?
The distance of the dog from the car before the driver hits the brake can be obtained as follow:
v² = u² + 2as
0² = 30² + (2 × –10 × s)
0 = 900 + (–20s)
0 = 900 – 20s
Collect like terms
0 – 900 = –20s
–900 = –20s
Divide both side by –20
s = –900 / –20
s = 45 m
Therefore, the dog was 50 m away when the driver first hit the brake
An insulated tank contains 50 kg of water, which is stirred by a paddle wheel at 300 rpm while transmitting a torque of 0.1 kN-m. At the same time, an electric resistance heater inside the tank operates at 110 V, drawing a current of 2 A. Determine the rate of heat transfer after the system achieves steady state.
Answer:
the rate of heat transfer after the system achieves steady state is -3.36 kW
Explanation:
Given the data in the question;
mass of water m = 50 kg
N = 300 rpm
Torque T = 0.1 kNm
V = 110 V
I = 2 A
Electric work supplied W₁ = PV = 2 × 110 = 220 W = 0.22 kW
Now, work supplied by paddle wheel W₂ is;
W₂ = 2πNT/60
W₂ = (2π × 0.1 × 300) / 60
W₂ = 188.495559 / 60
W₂ = 3.14 kW
So the total work will be;
W = 0.22 + 3.14
W = 3.36 kW
Hence total work done on the system is 3.36 kW.
At steady state, the properties of the system does not change so the heat transfer will be 3.36 KW.
The heat will be rejected by the system so the sign of heat will be negative.
i.e Q = -3.36 kW
Therefore, the rate of heat transfer after the system achieves steady state is -3.36 kW
A magnetic field of 0.276 T exists in the region enclosed by a solenoid that has 517 turns and a diameter of 10.5 cm. Within what period of time must the field be reduced to zero if the average magnitude of the induced emf within the coil during this time interval is to be 12.6 kV
Answer:
The period the field must be reduced to zero is 9.81 x 10⁻⁵ s
Explanation:
Given;
initial value of the magnetic field, B₁ = 0.276 T
number of turns of the solenoid, N = 517 turns
diameter of the solenoid, d = 10.5 cm = 0.105 m
induced emf, = 12.6 kV = 12,600 V
when the field becomes zero, then the final magnetic field value, B₂ = 0
The induced emf is given by Faraday's law;
[tex]emf = -\frac{NA\Delta B}{t} \\\\emf = -\frac{NA (B_2 -B_1)}{t} \\\\t = -\frac{NA (B_2 -B_1)}{emf}\\\\t = \frac{NA (B_1 -B_2)}{emf}\\\\where;\\\\t \ is \ the \ time \ when \ B = 0 \ \ (i.e\ B_2 = 0)\\\\A \ is \ the \ area \ of \ the \ coil\\\\A = \frac{\pi d^2}{4} = \frac{\pi (0.105)^2}{4} = 0.00866 \ m^2\\\\t= \frac{(517) \times (0.00866)\times (0.276 -0)}{12,600}\\\\t = 9.81 \times 10^{-5} \ s[/tex]
Therefore, the period the field must be reduced to zero is 9.81 x 10⁻⁵ s
In another version of the "Giant Swing", the seat is connected to two cables as shown in the figure (Figure ), one of which is horizontal. The seat swings in a horizontal circle at a rate of 39.3 rev / m * i * n
If the seat weighs 268 N and a 896-N person is sitting in it, find the tension in the horizontal cable
If the seat weighs 268 N and a 896-N person is sitting in it, find the tension in the inclined cable
Answer:Solution to 47E Step 1 Angular velocity of the swing=32rpm Weight of the seat =255N Weight of the person =825N Total weight =255+825=1080N Radius =7.5m
Explanation:
A typical ceiling fan running at high speed has an airflow of about 2.00 ✕ 103 ft3/min, meaning that about 2.00 ✕ 103 cubic feet of air move over the fan blades each minute.
Determine the fan's airflow in m3/s.
Answer:
0.94 m³/s
Explanation:
From the question given above, the following data were obtained:
Air flow (in ft³/min) = 2×10³ ft³/min
Air flow (in m³/s) =.?
Next, we shall convert 2×10³ ft³/min to m³/min. This can be obtained as follow:
35.315 ft³/min = 1 m³/min
Therefore,
2×10³ ft³/min = 2×10³ ft³/min × 1 m³/min / 35.315 ft³/min
2×10³ ft³/min = 56.63 m³/min
Finally, we shall convert 56.63 m³/min to m³/s. This can be obtained as follow:
1 m³/min = 1/60 m³/s
Therefore,
56.63 m³/min = 56.63 m³/min × 1/60 m³/s ÷ 1 m³/min
56.63 m³/min = 0.94 m³/s
Thus, 2×10³ ft³/minis equivalent to 0.94 m³/s.
The decibel level of a jackhammer is 125 dB relative to the threshold of hearing. Determine the decibel level if three jackhammers operate side by side.
Answer:
130 dB
Explanation:
The equation for decibel level is given by:
[tex]D=10log(\frac{I}{I_n} )\\\\Where\ D\ is\ the \ decibel\ level\ in\ dB, I\ is\ the\ intensity\ in \ W/m^2, \\I_n\ is\ threshold\ intensity\ to\ the\ human\ ear=1*10^{-12}W/m^2\\\\Given\ that\ D=125dB, hence:\\\\125=10log(\frac{I}{1*10^{-12}} )\\\\12.5=log(\frac{I}{1*10^{-12}} )\\\\I=3.2\ W/m^2[/tex]
The intensity for 1 jack hammer is 3.2 W/m², therefore for 3 jack hammers, the intensity = 3 * 3.2 = 9.6 W/m²
[tex]D=10\ log(\frac{I}{I_n} )\\\\D=10*log(\frac{9.6}{1*10^{-12}} )\\\\D=130\ dB[/tex]
A Michelson interferometer operating at a 400 nm wavelength has a 3.70-cm-long glass cell in one arm. To begin, the air is pumped out of the cell and mirror M2 is adjusted to produce a bright spot at the center of the interference pattern. Then a valve is opened and air is slowly admitted into the cell. The index of refraction of air at 1.00 atm pressure is 1.00028.
How many bright-dark-bright fringe shifts are observed as the cell fills with air?
Answer:
[tex]m=42\ fringes[/tex]
Explanation:
From the question we are told that:
Wavelength [tex]\lambda=400nm[/tex]
Length of cell arm [tex]h=3.70cm[/tex]
Refraction of air at at 1.00 atm pressure [tex]n=1.00028.[/tex]
Generally the equation for Number of shifts is mathematically given by
[tex]m=N-N_o[/tex]
Since
[tex]N_0=\frac{2t}{\lambda_0}[/tex]
Therefore
[tex]m=\frac{2t}{\lambda_0/n}-\frac{2t}{\lambda_0}[/tex]
[tex]m=\frac{2t}{\lambda_0} n-1[/tex]
[tex]m=\frac{2(3.7*10^{-2})}{400*10^{-9}}*(1.00028-1)[/tex]
[tex]m=51.8[/tex]
[tex]m=42\ fringes[/tex]
Which of the following would experience induced magnestism mostly easily
Answer:
Copper would experience induced magnestism mostly easily.
Explanation:
Permalloy would experience induced magnetism most easily.
I don't know the options but that is correct too!!
A rod of 3.0-m length and a square (2.0 mm X 2.0 mm) cross section is made of a material with a resistivitYof 6.0 X 10-8 Ω m. If a potential difference of 0.60 V is placed across the ends of the rod, at what rate is heat generated in the rod in watt.
Select one:
a. 24
b. 12
c. 4
d. 8
On a particle level, what happens when thermal conduction occurs within a
solid?
How would an observer on train A, which is moving at nearly the speed of light, view a clock on train B, which is moving at the same speed and in the same direction?
A. The clock on train B would appear narrower and run more slowly.
B. The clock on train B would appear to be the same width and to run at the same rate. C. The clock on train B would appear narrower and run faster.
D. The clock on train B would appear wider and run more slowly.
Answer:
B.
Explanation:
If both train A and train B are moving in the same direction and the exact same speed then from an observer's viewpoint within either train everything would seem as though it is not moving. Therefore, the clock on train B would appear to be the same width and to run at the same rate. In order for this to be the case the speed of both trains would need to be exactly the same, any difference in speed will cause the clock on the opposite train to appear distorted and run either faster or slower depending on the speed of the train you are on.
Answer:
The trains are moving at the same speed, so the answer is A.
An air-track glider attached to a spring oscillates between the 14.0 cm mark and the 71.0 cm mark on the track. The glider completes 12.0 oscillations in 34.0 s . You may want to review (Pages 391 - 393) . Part A What is the period of the oscillations
Answer:
A = 2,8333 s
Explanation:
El periodo es definido como el tiene que toma de dar una oscilación.
En este caso realiza varias osicilacion por lo cual debemos encontrar el promedio del perdono.
T = t/n
calculemos
A = 34,0/ 12,0
A = 2,8333 s
Two water-slide riders, A and B, start from rest at the same time and same height, h but on differently shaped slides.
Required:
a. Which rider is traveling faster at the bottom?
b. Which rider makes it to the bottom first? Ignore friction and assume both slides have the same path length.
Answer:
a. None
b. Both
Explanation:
a. Which rider is traveling faster at the bottom?
Since both riders fall from the same height, h, their potential energy, U at the top equals their kinetic energy, K at the bottom.
U = mgh and K = 1/2mv²
Since U is he same for both water-slide riders, then K will be the same and thus their speed at the bottom will be the same. This is shown below.
K = U
1/2mv² = mgh
v² = 2gh
v =√(2gh) where v = speed of rider at the bottom, g = acceleration due to gravity and h = height of slide.
Since the height is the same, so their speed at the bottom is the same. So, none of the riders travels faster than the other since they have the same speed at the bottom.
b. Which rider makes it to the bottom first? Ignore friction and assume both slides have the same path length.
Since the path length of the water slides are the same and friction is neglected, both water-slide rider get to the bottom at the same time since the distance moved is the same and they both start from rest.
So, both riders make it to the bottom at the same time.
The acceleration due to gravity acts vertically downwards, and the component of gravity acceleration is larger when the slope is steeper.
a. Rider Bb. Rider B
Reasons:
The acceleration of the riders are due to gravity
The component acceleration due to gravity acting on a slope is a = g·sin(θ)
As the steepness of the slope increase, the angle, θ, and sin(θ) increases, therefore, the acceleration increases.
Rider A is on a slide with gentle slope, such that if the slide is flat, rider A will be stationary.
The shape of the water slide rider B is on is steeper, and therefore, rider B is accelerating more than rider A. The higher acceleration of rider B, gives rider B a higher speed than rider, such that rider B, is riding faster than rider ATherefore;
a. The rider that is travelling faster at the bottom is rider B
b. Given that friction is ignored, and the path have the same length to the
bottom, the rider that makes to the bottom first is the rider that is moving
faster, which is rider B
Learn more here:
https://brainly.com/question/13218675
You throw a stone straight down from the top of a tall tower. It leaves your hand moving at 8.00 m/s, Air resistance can be neglected. Take the positive -direction to be upward, and choose y 0 to be the point where the stone leaves your hand. Find the stone's position 1.50s after it leaves your hand.
Express your answer with the appropriate units.
Find the y-component of the stone's velocity 1.50 s after it leaves your hand. Express your answer with t0he appropriate units.
Answer:
The velocity after 1.5 s is 22.7 m/s downwards.
Explanation:
Initial velocity = - 8 m/s
acceleration, a = - 9.8 m/s2
time, t = 1.5 s
Use first equation of motion
v = u + at
v = - 8 - 9.8 x 1.5
v = - 8 - 14.7
v = - 22.7 m/s
Thus, the velocity after 1.5 s is 22.7 m/s downwards.
In higher mass stars, repeating cycles of fusion will create heavier elements in layers
until which element is created at the center of the core?
hydrogen
iron
uranium
helium
g Is a nucleus that absorbs at 4.13 δ more shielded or less shielded than a nucleus that absorbs at 11.45 δ? _________ Does the nucleus that absorbs at 4.13 δ require a stronger applied field or a weaker applied field to come into resonance than the nucleus that absorbs at 11.45 δ?
Answer: A nucleus that absorbs at [tex]11.45\delta[/tex] is less shielded and a nucleus that absorbs at [tex]4.13\delta[/tex] will require a stronger applied field
Explanation:
While interpreting the data in NMR, the positions of signals are studied.
The nucleus/ protons having a higher value of [tex]\delta[/tex] are said to be less shielded. They are said to be upfield.
The nucleus/protons having a lower value of [tex]\delta[/tex] are said to be more shielded. They are said to be downfield.
So, a nucleus that absorbs at [tex]11.45\delta[/tex] is less shielded by the nucleus that absorbs at [tex]4.13\delta[/tex]
Also, the less shielded nucleus/protons will require a weak applied field to come into resonance than the more shielded nucleus/protons
So, a nucleus that absorbs at [tex]4.13\delta[/tex] will require a stronger applied field to come into resonance than the nucleus that absorbs at [tex]11.45\delta[/tex]
what is simple definition of democracy
it's a form of government where people elect their representatives
Answer:
The word democracy itself means rule by the people.
Q}Write any two conversation which are followed while writing the units and symbols?Please tell me the answer of this question
Answer:
Units named after scientists are written in lowercase but their symbols are written as capital
Unit of power is watt, since it is named after a scientist, then symbol will be written as W
Farad, symbol = F
Henry, symbol = H
Units whose names aren't culled from that of scientists are written in lower case and their symbols are also in lower case.
Units such as meter, kilogram should be represented with symbols in small letters as: m and kg respectively.
Explanation:
Kindly check answer
If the open body is postively charged and another body is negatively charged ,free electrons tend to
Answer:
The free electrons tend to Negatively Charged body to Positively Charged body
Explanation:
As they have different charges, Electrons are more attracted to the Positive or it's opposite charge.
What is the minimum value of the friction coefficient between the boxes that will keep them from slipping when the 100 N force is applied
Answer:
The friction coefficient's minimum value will be "0.173".
Explanation:
The given query seems to be incomplete. Below is the attached file of the complete question.
According to the question,
(a)
The net friction force's magnitude will be:
⇒ [tex]F_{net}=ma[/tex]
[tex]=5\times 1.7[/tex]
[tex]=8.5 \ N[/tex]
(b)
For m₃,
⇒ [tex]ma=\mu m_3 g[/tex]
Or,
⇒ [tex]\mu=\frac{a}{g}[/tex]
[tex]=\frac{1.7}{9.8}[/tex]
[tex]=0.173[/tex]
galileo was a contemporary of
Suppose 4 J of work are needed to stretch a spring 14 cm from its natural position. How much work is needed to stretch it 20 cm from its natural position
Answer:
8.16 J
Explanation:
Applying,
W = ke²/2................. Equation 1
Where W = work done in stretching the spring, k = spring constant, e = extension
make k the subject of the equation
k = 2W/e²................ Equation 2
From the question,
Given: W = 4 J, e = 14 cm = 0.14 m
Substitute these values into equation 2
k = 2(4)/0.14²
k = 8/0.0196
k = 408.16 N/m
If stretchd 20 cm from its natural length,
Therefore,
e = 20 cm = 0.2 m
W = 408.16(0.2²)/2
W = 8.16 J
The v-t graph of a moving body is given below. The distance covered by the body in
the first 40s is _________and its acceleration during the last 40s is _____________.
a) 3200m; 2ms-1
b) 700m; -2ms-2
c) 900m; - 2ms-2
d) 900m; 2ms-2
Answer: The correct answer is a) 3200 m, [tex]-2m/s^2[/tex]
Explanation:
Speed is defined as the ratio of distance travelled to the time taken. The equation follows:
[tex]\text{Speed}=\frac{\text{Distance travalled}}{\text{Time taken}}[/tex]
From the graph:
Speed for the first 40 s, v = 80 m/s
Time taken, t = 40 s
Putting values in above equation, we get:
[tex]\text{Distance travalled}=(80 m/s\times 40s)=3200 m[/tex]
Acceleration is defined as the ratio of change of velocity to the change of time. The equation follows:
[tex]\text{Acceleration}=\frac{\Delta v}{\Delta t}=\frac{v_2-v_1}{t_2-t_1}[/tex]
From the graph, for the last 40 sec:
Initial velocity, [tex]v_1[/tex] = 80 m/s
Final velocity, [tex]v_2[/tex] = 0 m/s
Initial time, [tex]t_1[/tex] = 40 s
Final time, [tex]t_2[/tex] = 80 s
Putting values in above equation, we get:
[tex]\text{Acceleration}=\frac{(0-80)m/s}{(80-40)s}\\\\\text{Acceleration}=\frac{-80m/s}{40s}=-2m/s^2[/tex]
Hence, the correct answer is a) 3200 m, [tex]-2m/s^2[/tex]
An eagle is flying horizontally at a speed of 3.40 m/s when the fish in her talons wiggles loose and falls into the lake 5.50 m below. Calculate the velocity (in m/s) of the fish relative to the water when it hits the water. (Assume that the eagle is flying in the x-direction and that the y-direction is up.)
Answer:
The velocity of the fish when it hits the water is:
v = 10.93 m/s and 71.88 ° below the x-direction.
Explanation:
Let's find the velocity of the fish in the y-direction.
[tex]v_{fy}^{2}=v_{iy}^{2}-2gh[/tex]
Here, v(iy) of the fish is zero, and the heigh h = 5.50 m, then the velocity will be:
[tex]v_{fy}^{2}=0-2(9.81)(5.50)[/tex]
[tex]v_{fy}^{2}=-2(9.81)(5.50)[/tex]
[tex]v_{fy}=-10.39 \: m/s[/tex]
Now, we know that the velocity in the x-direction is constant, so we can calculate the velocity of the fish when it hits the water.
[tex]v=\sqrt{v_{x}^{2}+v_{y}^{2}}[/tex]
[tex]v=\sqrt{3.40^{2}+(-10.39)^{2}}[/tex]
[tex]v=10.93 \: m/s[/tex]
And the direction will be:
[tex]\theta=tan^{-1}(\frac{10.39}{3.40})[/tex]
[tex]\theta=71.88^{\circ}[/tex]
The angle is 71.88 ° belox the x-direction.
I hope it helps you!
From 2 King 6:1-6, one of the disciples of Elisha was cutting a tree and the ax head fell into the water. While we do not know how high the ax head was when it fell into the water, we will work through a physics example of the ax head's vertical motion as if it were dropped into the water. ( Due date 09/07)
Write your name and date. The due date of this assignment is the height the ax head falls from in meters into the water. For example, if the due date is July 15, then the ax head fell 15 meters to the water.
Write Newton’s 2nd Law in Equation Form.
Write the quantity and units of average gravitational acceleration on the surface of Earth.
Given the ax head mentioned in the opening portion with the height being equal in numerical value of the due day of this assignment. How long does it take for the ax to fall to the river surface?
Compute the final speed of the ax when it hits the water.
Answer:
time of fall is 1.75 s and the velocity with which it strikes the water is 17.15 m/s.
Explanation:
Height, h = 15 m
Newton's second law
Force = mass x acceleration
The unit of gravitational force is Newton and the value is m x g.
where, m is the mas and g is the acceleration due to gravity.
Let the time of fall is t.
Use second equation of motion
[tex]s= u t +0.5 at^2\\\\15 = 0 +0.5\times 9.8\times t^{2}\\\\t = 1.75 s[/tex]
Let the final speed is v.
Use third equation of motion
[tex]v^2 = u^2 + 2 a s\\\\v^2 = 0 + 2 \times 9.8\times 15\\\\v =17.15 m/s[/tex]
A balloon pops, making a loud noise that startles you. What kind of energy best describes this experience?
A. Thermal Energy
B. Sound Energy
C. Gravitational Energy
D. Radiant Energy
factors that favour mining in South Africa
Answer:
According to the data, factors influencing mining investment in South Africa's favour are the availability of labour and skills, the quality of the country's infrastructure, the quality of its geological database, and the State's environmental regulations.
Our solar system is made up of the Sun, 8 planets, and other bodies such as
asteroids orbiting the Sun. The solar system is very large compared to anything we see on
Earth. The distance between planets is measured in astronomical units (AU). One AU is
equal to 149.6 million kilometers, the average distance between the Sun and Earth. Scale
models are useful for helping us understand the size of the solar system.
Mr. Wilson’s science class made a scale model of the solar system. They went out to the
school’s football field, and they used the chart shown below to mark out the scale distance
from the Sun to each planet
The average orbital radius for a planet is its distance from the Sun. Which statement BEST
describes the relationship between the planets and their average orbital radii?
A. The planets are evenly spaced in the solar system.
B. Closer to the Sun, the planets are regularly spaced apart.
C. The planets closest to each other are the ones farthest from the sun.
D. Farther from the Sun, the planets are spaced farther apart from each other.
the answer to the question is D
A cheetah is running in a straight line in pursuit of prey. The cheetah's mass is 63.9 kg. Through its running motion, assume the cheetah experiences a constant forward force of 609.1 N. Also, assume the cheetah experiences a constant 107.9 N air resistance force that is opposite its motion. What is the magnitude of the cheetah's acceleration, in units of m/s2?
Answer:
a = 7.84 m / s²
Explanation:
For this exercise let's use Newton's second law
F - fr = m a
indicate that the force is F = 609.1 N and the friction force is fr = 107.9 N and is constant
a = [tex]\frac{F - fr}{m}[/tex]
let's calculate
a = [tex]\frac{ 609.1 - 107.9 }{63.9}[/tex]
a = 7.84 m / s²
resolve the vector shown below into its components
Answer:
ans: option A
Explanation:
components along x- axis is -3x and along x- axes is -2y
now use triangle law of vector addition