Answer:
Período del tambor: [tex]T = 0.25\,s[/tex], fuerza sobre la prenda: [tex]F \approx 80.852\,N[/tex], velocidad lineal del tambor: [tex]v \approx 10.053\,\frac{m}{s}[/tex], velocidad angular del tambor: [tex]\omega \approx 25.133\,\frac{rad}{s}[/tex].
Explanation:
La expresión tiene un error por omisión, su forma correcta queda descrita a continuación:
"Una prenda de 320 gramos de ropa gira en el interior de una lavadora si dicha lavadora tiene un radio de 40 centímetros y gira con una frecuencia de 4 hertz. Halle a) el período, b) la velocidad angular, c) la fuerza con la que gira la prenda y d) la velocidad lineal de la lavadora."
El tambor gira a velocidad angular constante ([tex]\omega[/tex]), en radianes por segundo, lo cual significa que la prenda experimenta una aceleración centrífuga ([tex]a[/tex]), en metros por segundo al cuadrado. En primer lugar, calculamos el período de rotación del tambor ([tex]T[/tex]), en segundos:
[tex]T = \frac{1}{f}[/tex] (1)
Donde [tex]f[/tex] es la frecuencia, en hertz.
([tex]f = 4\,hz[/tex])
[tex]T = \frac{1}{4\,hz}[/tex]
[tex]T = 0.25\,s[/tex]
Ahora determinamos la fuerza aplicada sobre la prenda ([tex]F[/tex]), en newtons:
[tex]F = m\cdot a[/tex] (2)
[tex]F = \frac{4\pi^{2}\cdot m \cdot r}{T^{2}}[/tex] (2b)
Donde:
[tex]m[/tex] - Masa de la prenda, en kilogramos.
[tex]r[/tex] - Radio interior del tambor, en metros.
([tex]m = 0.32\,kg[/tex], [tex]r = 0.4\,m[/tex], [tex]T = 0.25\,s[/tex])
[tex]F = \frac{4\pi^{2}\cdot (0.32\,kg)\cdot (0.4\,m)}{(0.25\,s)^{2}}[/tex]
[tex]F \approx 80.852\,N[/tex]
La velocidad lineal de la lavadora es:
[tex]v = \frac{2\pi\cdot r}{T}[/tex] (3)
([tex]r = 0.4\,m[/tex], [tex]T = 0.25\,s[/tex])
[tex]v = \frac{2\pi\cdot (0.4\,m)}{0.25\,s}[/tex]
[tex]v \approx 10.053\,\frac{m}{s}[/tex]
Y la velocidad angular del tambor de la lavadora:
[tex]\omega = \frac{2\pi}{T}[/tex]
([tex]T = 0.25\,s[/tex])
[tex]\omega = \frac{2\pi}{0.25\,s}[/tex]
[tex]\omega \approx 25.133\,\frac{rad}{s}[/tex]
A 500-kg crate sits on a 10-degree ramp. If friction between the ramp and the crate is 800 N, what is the acceleration of the crate?
By Newton's second law, the net force acting on the crate parallel to the surface is
∑ F = mg sin(10°) - 800 N = ma
where m = 500 kg is the mass of the crate and a is the acceleration.
Solve for a :
a = ((500 kg) (9.80 m/s^2) sin(10°) - 800 N) / (500 kg)
a ≈ 0.102 m/s^2
A machine has mechanical advantage 2.What does that mean
Answer:
Mechanical advantage is the measure of the force amplification achieved by using a tool , mechanical device or machine . The machine preserve the input power and supply trade off force. against movement to obtain a desired amplification in the output force .
Answer:
The ratio of load overcome by the machine to the effort applied is called mechanical advantage.
The unit of area is a derived unit. Why?
Explanation:
area=length(m) ×breadth(m) . The unit of area is expressed in terms of fundamental units m^2.thus it is derived unit
Why is a person not a good blackbody radiator?
O A. A person emits only visible light.
OB. A person emits only infrared radiation.
O C. A person absorbs most of the light that hits him or her.
O D. A person reflects little of the light that hits him or her.
Answer:
O C. A person absorbs most of the light that hits him or her.
Explanation:
Answer:
Option D hope it's helpful mark me as brainlistEn una sala de juntas hay mesas, sillas y otras personas. ¿Cuál de ellas tienen temperaturas
a) menores, b) mayores y d) iguales que la del aire?
Answer:
table and chair
Explanation:
In a meeting room there are tables, chairs, and other people. Which of them have temperatures
a) smaller, b) bigger and d) equal to that of air?
the temperature of tables and chairs is same as air.
3- Define light year. What quantity does it measure? what is one light year equal to in sl unit?
Three voltmeters V, V₁ and V₂ are connected as in
Figure 37.9. a If V reads 18V and V, reads 12V, what does V₂ read?
b If the ammeter A reads 0.5A, how much electrical energy is changed to heat and light in lamp L₁ in one minute? c Copy Figure 37.9 and mark with a + the positive terminals of the ammeter and voltmeters for correct
connection.
Answer:
a. V₂ = 6 V
b. 360 joules
c. The positive terminals of both the voltmeter and ammeter are connected to the positive terminal of the power source
Please see the attached drawing created with MS Visio
Explanation:
a. The voltmeter readings are;
V₁ = 12 V, V = 18 V
Given that the voltage reading, 'V', is the voltage reading across two loads with voltages, V₁ and V₂ connected in series, we have;
V = V₁ + V₂
V₂ = V - V₁
V₂ = 18 V - 12 V = 6 V
b. The reading of the ammeter, A, I = 0.5 A
The heat energy, Q = I·V·t
Where;
t = The time = 1 minute (60 seconds)
Therefore, for the lamp L₁, where V = 12 V, we have;
Q₁ = 0.5 A × 12 V × 60 s = 360 Joules
The amount of electrical energy changed into heat and light in lamp L₁, Q₁ = 360 joules
c. The positive terminals of the voltmeter and ammeter are connected to the positive terminal of the power source
Please see attached drawing created with MS Visio
if a train starts from rest and attains a velocity of 100m/s in 25 seconds. calculate the acceleration produced by the train.
ball of mass M kg is dropped from rest off a high building and strikes the ground 10 seconds later. Calculate the height of the building assuming that upwards is positive in this coordinate system
Answer:
h = 1/2gt^2
Is the formula for the height of a free falling object.
Put in the numbers:
h = 1/2*9.8 (gravitational constant) * 10^2
h=490 meters.
Please note that the gravitational constant is approximate, you could also use a more specific value or 10
A HIGH SPEED TRAIN IS 180M LONG AND IT IS TRAVELLING AT 50M/S.HOW LONG WILL IT TAKE TO PASS A PERSON STANDING AT A LEVEL CROSSING?
B-HOW LONG WILL IT TAKE TO PASS COMPLETELY THROUGH A STATION WHOSE PLATFORMS ARE 220M IN LENGTH?
Answer:
a. Time = 3.6 seconds
b. Time = 4.4 seconds
Explanation:
Given the following data;
Distance = 180 m
Speed = 50 m/s
a. To find the time;
Speed can be defined as distance covered per unit time. Speed is a scalar quantity and as such it has magnitude but no direction.
Mathematically, speed is given by the formula;
[tex]Speed = \frac{distance}{time}[/tex]
Making time the subject of formula, we have;
[tex]Time = \frac{distance}{speed}[/tex]
Substituting into the equation, we have;
[tex]Time = \frac{180}{50}[/tex]
Time = 3.6 seconds
b. Distance = 220 meters
Speed = 50 m/s
To find the time;
[tex]Time = \frac{distance}{speed}[/tex]
Substituting into the equation, we have;
[tex]Time = \frac{220}{50}[/tex]
Time = 4.4 seconds
Select the correct answer.
Which quantity is a vector quantity?
ОА. .
acceleration
OB.
mass
OC.
speed
O D.
volume
Reset
Next
Answer:
acceleration is the vector quantity because it depends on particular direction and has magnitude
Find the emitted power per square meter and wavelength of peak intensity for a 3000 K object that emits thermal radiation.
Answer:
power per square meter = 4.593 × 10^(6) W/m²
Wavelength of peak intensity = 9.67 × 10^(-7) m
Explanation:
From Stefan-Boltzmann law, total emitted power per square meter is given as;
P/A = eσT⁴
where;
P is power
A is surface area
σ = Stefan-Boltzmann constant = 5.67 × 10^(-8) W/m².k⁴
T = temperature of the body = 3000 K
e = emissivity of the substance (for ideal radiation, it has a value = 1)
Thus, Plugging in the relevant values we have;
P/A = 1 × 5.67 × 10^(-8) × (3000)^(4)
P/A = 4.593 × 10^(6) W/m²
Let's find the wavelength of peak intensity.
From wiens displacement law, we know that;
λ_m × T = b
where;
λ_m = maximum wavelength
T = Temperature
b is Wien's displacement constant = 2.9 × 10^(−3) m/K
thus;
λ_m = b/T = (2.9 × 10^(−3))/3000 = 9.67 × 10^(-7) m
A car takes a full round of journey in a roundabout with constant speed, as the driver got confused with the route. Can we consider it as a uniform motion? Why?
Answer: The given statement is True
Explanation:
A uniform motion is defined as the motion where an object is moving at a constant speed.
A non-uniform motion is defined as the motion where an object keeps changing its position and does not move at a constant speed.
We are given:
A car takes a full round of journey in a roundabout with constant speed
As the speed remains constant in a circular path, it is considered a uniform motion.
Hence, the given statement is True
what is potential difference??. Please give a detailed explanation.
Explanation:
The difference of electrical potential between two points is called the potential difference.
It is also defined as the work done in the transfer of a unit quantity of electricity from one point to the other.
The SI unit of electric potential is volt. It is denoted by V.
Sean and Tommy are debating the positive and negative impacts of technology on environmental quality. Tommy is listing the negative impacts. Which of the following would help his argument?
A. Solar panels have become more effective due to technological improvements.
B. Technology has helped make recycling to be more efficient.
C. It has created new ways to clean up oil spills.
D. Factories contribute to air and water pollution
6. A warehouse employee is pushing a 15.0 kg desk across a floor at a constant speed of 0.50 m/s. How much work must the employee do on the desk to change the speed to 1.00 m/s?
Answer:
7.5 J
Explanation:
To answer the question given above, we need to determine the energy that will bring about the speed of 1 m/s. This can be obtained as follow:
Mass (m) = 15 Kg
Velocity (v) = 1 m/s
Energy (E) =?
E = ½mv²
E = ½ × 15 × 1²
E = ½ × 15 × 1
E = ½ × 15
E = 7.5 J
Therefore, to change the speed to 1 m/s, the employee must do a work of 7.5 J.
A bullet of mass 0.5 kg is moving horizontally with a speed of 50 m/s when it hits a block
of mass 3 kg that is at rest on a horizontal surface with a coefficient of friction of 0.2.
After the collision the bullet becomes embedded in the block.
A) What is the net momentum of the bullet-block system before the collision?
B) Find the total energy of the bullet-block system before the collision?
C) What is the speed of the bullet-block system after the collision?
D) *Find the total energy of the bullet-block system after the collision?
E) *How much work must be done to stop the bullet-block system?
F) *Find the maximum traveled distance of the bullet-block after the collision?
Answer:
a) The net momentum of the bullet-block system before the collision is 25 kilogram-meters per second.
b) The initial translational kinetic energy of the bullet before the collision is 625 joules.
c) The final speed of the bullet-block system after the collision is 7.143 meters per second.
d) The total energy of the bullet-block system after the collision is 89.289 joules.
e) 89.289 joules must be done to stop the bullet-block system.
f) The bullet-block system will travel 13.007 meters before stopping.
Explanation:
a) Since no external forces are applied on the system defined by the bullet and the block, then the net momentum is conserved and can be calculated by the initial momentum of the bullet:
[tex]p = m\cdot v_{o}[/tex] (1)
Where:
[tex]p[/tex] - Net momentum, in kilogram-meters per second.
[tex]m[/tex] - Mass of the bullet, in kilograms.
[tex]v_{o}[/tex] - Initial speed of the bullet, in meters per second.
If we know that [tex]m = 0.5\,kg[/tex] and [tex]v_{o} = 50\,\frac{m}{s}[/tex], then the net momentum of the bullet-block system before the collision is:
[tex]p = (0.5\,kg)\cdot \left(50\,\frac{m}{s} \right)[/tex]
[tex]p = 25\,\frac{kg\cdot m}{s}[/tex]
The net momentum of the bullet-block system before the collision is 25 kilogram-meters per second.
b) The total energy of the bullet before the collision is its initial translational kinetic energy ([tex]K[/tex]), in joules:
[tex]K = \frac{1}{2}\cdot m \cdot v_{o}^{2}[/tex] (2)
[tex]K = \frac{1}{2}\cdot (0.5\,kg)\cdot \left(50\,\frac{m}{s} \right)^{2}[/tex]
[tex]K = 625\,J[/tex]
The initial translational kinetic energy of the bullet before the collision is 625 joules.
c) Both the bullet and the block experiments a complete inelastic collision, then the final speed of the bullet-block system is calculated solely by the Principle of Momentum Conservation:
[tex]v_{f} = \frac{m\cdot v_{o}}{m+M}[/tex] (3)
Where:
[tex]v_{f}[/tex] - Final speed, in meters per second.
[tex]M[/tex] - Mass of the block, in kilograms.
If we know that [tex]m = 0.5\,kg[/tex], [tex]v_{o} = 50\,\frac{m}{s}[/tex] and [tex]M = 3\,kg[/tex], then the final speed of the bullet-block system is:
[tex]v_{f} = \left(\frac{0.5\,kg}{0.5\,kg + 3\,kg} \right)\cdot \left(50\,\frac{m}{s} \right)[/tex]
[tex]v_{f} = 7.143\,\frac{m}{s}[/tex]
The final speed of the bullet-block system after the collision is 7.143 meters per second.
d) The total energy of the bullet-block system after the collision is the translational kinetic energy of the system ([tex]K[/tex]), in joules, is:
[tex]K = \frac{1}{2}\cdot (m + M)\cdot v_{f}^{2}[/tex] (4)
[tex]K = \frac{1}{2}\cdot (0.5\,kg + 3\,kg)\cdot \left(7.143\,\frac{m}{s} \right)^{2}[/tex]
[tex]K = 89.289\,J[/tex]
The total energy of the bullet-block system after the collision is 89.289 joules.
e) By Work-Energy Theorem, magnitude of the work done by friction must be equal to the magnitude of the translational kinetic energy of the system. Hence, 89.289 joules must be done to stop the bullet-block system.
f) The maximum travelled distance of the bullet-block after the collision can be determined by means of Work-Energy Theorem and definition of work:
[tex]W = \mu_{k}\cdot (m+M)\cdot g\cdot s[/tex] (5)
Where:
[tex]W[/tex] - Work done by friction, in joules.
[tex]g[/tex] - Gravitational acceleration, in meters per square second.
[tex]s[/tex] - Travelled distance, in meters.
[tex]\mu_{k}[/tex] - Kinetic coefficient of friction, no unit.
If we know that [tex]m = 0.5\,kg[/tex], [tex]M = 3\,kg[/tex], [tex]\mu_{k} = 0.2[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]W = 89.289\,J[/tex], then the travelled distance of the bullet-block system is:
[tex]s = \frac{W}{\mu_{k}\cdot (m+M)\cdot g}[/tex]
[tex]s = \frac{89.289\,J}{0.2\cdot (0.5\,kg + 3\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}[/tex]
[tex]s = 13.007\,m[/tex]
The bullet-block system will travel 13.007 meters before stopping.
Explain: What happens to the velocity of a stream as the size of the sediment increases?
Answer:
Also, as stream depth increases, the hydraulic radius increases thereby making the stream more free flowing. Both of these factors lead to an increase in stream velocity. The increased velocity and the increased cross-sectional area mean that discharge increases.
13) Un móvil A parte de una ciudad a las 12 horas, con una velocidad de 40 Km/h. 2 horas después parte otro con una velocidad de 60 Km/h. Averiguar a qué hora se encuentran y a que distancia de la ciudad
Answer:
¿Podrías poner la pregunta en inglés por favor?
Explanation:
A bicyclist rides 2.93 km due east, while the resistive force from the air has a magnitude of 8.65 N and points due west. The rider then turns around and rides 2.93 km due west, back to her starting point. The resistive force from the air on the return trip has a magnitude of 8.65 N and points due east. Find the work done by the resistive force during the round trip. Number Type your answer here Units Choose your answer here
Answer:
-50.6 kJ
Explanation:
The work done (W) on an object is given by:
W = (Fcosθ) * S
where F is the force, S is the displacement and θ is the angle between the force and displacement.
i) During the first trip riding east, S₁ = 2.93 km = 2930 m, F₁ = 8.65 N.
The displacement is due east and the force is due west, hence θ₁ = 180°. Therefore:
W₁ = (F₁ * cosθ₁)S₁ = (8,65 * cos(180))2930 = -25.3 kJ
ii) i) During the second trip riding west, S₂ = 2.93 km = 2930 m, F₂ = 8.65 N.
The displacement is due west and the force is due east, hence θ₂ = 180°. Therefore:
W₂ = (F₂ * cosθ₂)S₂ = (8,65 * cos(180))2930 = -25.3 kJ
work done by the resistive force during the round trip is:
W = W₁ + W₂ = -25.3 kJ + (-25.3 kJ) = -50.6 kJ
How many players are allowed to play for one team during a game of
5 points
Netball?
Answer:
Netball is a ball sport for two teams of seven players it's rules are published in print and online by the international netball federation Games are played on a rectangular court divided into thirds with a raised goal at coach short end
Explanation:
It will help you
Melanie gets into an accident on the highway that sends her to the hospital for three weeks with multiple broken bones. Her hospital bill totals over $32,000, but she discovers that the woman who hit her only has $25,000 worth of liability insurance.
Em um fio condutor uma carga de 6.000 C atravessa uma secção transversal em 5 minutos. Determinando-se a corrente no fio, encontraremos o valor de?
Answer:
I = 20 A
Explanation:
The question says that, "A load of 6,000 C is conducted through a cross section in 5 minutes. Determining-if a current is not correct, we will find the value of?"
We have,
Charge, q = 6,000 C
Time, t = 5 minutes = 300 s
We need to find the current. We know that, the charge flowing per unit time is equal to current. So,
[tex]I=\dfrac{q}{t}\\\\I=\dfrac{6000}{300}\\\\I=20\ A[/tex]
So, the current flowing through the circuit is 20 A.
On which planet would your weight be the most and the least?
a) Jupiter and Mercury
b. Jupiter and neptune
c. Saturn and Neptune
d. Saturn and uranus
Answer:
Jupiter and neptune
Explanation:
Neha and Reha are playing see-saw.Neha is sitting 60cm away from the fulcrum and Reha is sitting 40cm away from the fulcrum.Calculate the effort that Reha should apply to lift Neha.The weight of Neha is 360N.
Answer:
Effort = 540 Newton
Explanation:
Given the following data;
Load arm = 60 cm
Effort arm = 40 cm
Load = 360 N
Conversion:
100 cm = 1 meters
40 cm = 40/100 = 0.4 meters
60 cm = 60/100 = 0.6 meters
To calculate the effort that Reha should apply to lift Neha, we would use the expression;
Effort * effort arm = load * load arm
Substituting into the expression, we have;
Effort * 0.4 = 360 * 0.6
Effort * 0.4 = 216
Effort = 216/0.4
Effort = 540 Newton
A runner has a speed of 5 m/s and a mass of 130 kg. What is his kinetic
energy?
O A. 1625 J
B. 3250 J
C. 875 J
D. 325 J
The auroras occur in the
C.
a. troposphere
b. stratosphere
mesophere
d. ionosphere
Answer:
Ionosphere
Explanation:
The thermosphere reaches 600 kilometres just above mesosphere and begins immediately above the mesosphere. This layer is where the aurora and satellites appear.
The ionosphere is the comprehensive career of the mesosphere because most of the thermosphere, located 80–400 kilometres just above ground atmosphere.
Auroras — magnificent flowing streaks of light seen in the night sky – appear in this location.
Một vật có khối lượng 2 kg rơi tự do xuống đất trong khoảng thời gian 0,5 s. Độ biến thiên động lượng của vật trong khoảng thời gian đó là bao nhiêu ? Cho g = 10 m/s2.
Answer: The change in momentum is +20 kg.m/s
Explanation:
To calculate the final velocity of object, we use the first equation of motion:
[tex]v=u+at[/tex]
where,
v = final velocity
u = initial velocity = 0 m/s
a = acceleration = [tex]10m/s^2[/tex]
t = time = 0.5 s
Putting values in above equation, we get:
[tex]v=0+(10\times 0.5)\\\\v=5m/s[/tex]
Momentum is defined as the product of the mass and velocity of an object. It is given by the equation:
[tex]p=mv[/tex]
where,
p = momentum
m = mass of object = 2 kg
Let the upward velocity be positive and the downward velocity be negative
When the object is dropped, the velocity is downward
v = -5m/s
Initial momentum = [tex]2kg\times (-5m/s)=-10kg.m/s[/tex]
When the object is bounced back, the velocity is upward
v = +5m/s
Final momentum = [tex]2kg\times (+5m/s)=10kg.m/s[/tex]
Change in momentum = Final - Inital
Change in momentum = [10 - (-10)] = +20 kg.m/s
Hence, the change in momentum is +20 kg.m/s
The earth's orbital is oval in shape. Explain how the magnitude of the gravitational force between the earth and the sun changes as the earth moves from position A to B as shown in the figure.
The gravitational force does change believe it or not, but the explaination for this is because the earths orbit is an oval (or a not circle) the closer it nears its self to the sun.
In what way is Height related to Potential Energy?
Explanation:
Lets say you have a ball in your hand, you raise your hand to just above your head. Now, when you did that you created potential energy that is ready to be released. you drop the ball and the ball bounces a few times off the ground. Now lets say you got a ladder and doubled the ball's height doubling the energy now stored in the ball, when you drop it the ball should bounce much higher after hitting the ground as a result of more energy being released.
Hope this helped.