Answer:
Tenemos dos problemas a resolver acá:
Primero, debemos encontrar la velocidad con la que la pelota impacta el suelo.
Acá podemos usar la conservación de la energía.
E = U + K
U = energía potencial = m*g*H
m = masa
g = aceleración gravitatoria = 9.8m/s^2
H = altura
K = energía cinética = (m/2)*V^2
donde V es la velocidad.
Inicialmente, cuando la pelota es soltada, su velocidad es cero, entonces solo tenemos energía potencial:
Ei = U = m*(9.8m/s^2)*2.5m
Al final, cuando la pelota esta por impactar el suelo, la altura tiende a cero, entonces ya no hay energía potencial, solo hay energía cinética:
Ef = (m/2)*V^2
Y como la energía se conserva, la energía final es igual a la inicial, entonces:
m*(9.8m/s^2)*2.5m = (m/2)*V^2
Podemos resolver esto para V, y asi obtener la velocidad con la que la pelota impacta el suelo.
V = √(2*(9.8m/s^2)*2.5m) = 7m/s
Ahora respondamos la segunda parte.
Una vez la pelota rebota, su aceleración va a estar dada solamente por la aceleración gravitatoria, entonces tenemos:
A(t) = -9.8m/s^2
Para obtener su velocidad integramos:
V(t) = (-9.8m/s^2)*t + V0
donde V0 es la velocidad con la que la pelota reboto, que sabemos que es 3/4 de 7m/s
V0 = (3/4)*7m/s = (21/4) m/s
Así, la ecuación de la velocidad es:
V(t) = (-9.8m/s^2)*t + (21/4) m/s
Sabemos que la altura máxima se da cuando la velocidad es igual a cero, entonces primero calculemos el valor de t tal que esto ocurra:
V(t) = 0 = (-9.8m/s^2)*t + (21/4) m/s
t = (21/4) m/s/9.8m/s^2 = 0.54 s
Ahora debemos encontrar la ecuación de la posición y evaluarlo en este tiempo.
Para ello integramos de vuelta:
P(t) = (1/2)(-9.8m/s^2)*t^2 + (21/4 m/s)*t + P0
donde P0 es la posición inicial, como la pelota rebota en el suelo, la posición inicial es el suelo, el cual representamos con 0, entonces la ecuación de la posición es:
P(t) = (1/2)(-9.8m/s^2)*t^2 + (21/4 m/s)*t
La altura máxima estará dada por esta ecuación evaluada en t = 0.54 s
P(0.54s) = (1/2)(-9.8m/s^2)*(0.54s)^2 + (21/4 m/s)*0.54s = 1.81 m
La altura máxima es 1.81 metros.
Y entre que rebota y llega a esta altura máxima, transcurren 0.54 segundos.
what change occurs to the mass of an object when a unbalanced
Answer:
The mass decreases
Explanation:
Just smart
Kulsum’s TV uses 45 W. How much does it cost her to watch TV for one month (30 days). She watches TV for 4 hours/day during mid-peak time (10.4 cents/kWh).
Answer:
Total cost = 56.16 cents
Explanation:
Given the following data;
Power = 45 Watts
Time = 4 hours
Number of days = 30 days
Cost = 10.4 cents
To find how much does it cost her to watch TV for one month;
First of all, we would determine the energy consumption of the TV;
Energy = power * time
Energy = 45 * 4
Energy = 180 Watt-hour = 180/1000 = 0.18 Kwh (1 Kilowatts is equal to 1000 watts).
Energy consumption = 0.18 Kwh
Next, we find the total cost;
Total cost = energy * number of days * cost
Total cost = 0.18 * 30 * 10.4
Total cost = 56.16 cents
A pilot drops a package from a plane flying horizontally at a constant speed. Neglecting air resistance, when the package hits the ground the horizontal location of the plane will Group of answer choices be behind the package. be over the package. be in front of the package depend of the speed of the plane when the package was released.
Answer:
The location of helicopter is behind the packet.
Explanation:
As the packet also have same horizontal velocity as same as the helicopter, and also it has some vertical velocity as it hits the ground.
The horizontal velocity remains same as there is no force in the horizontal direction. The vertical velocity goes on increasing as acceleration due to gravity acts.
So, the helicopter is behind the packet.
help me with this question
Explanation:
Let's set the x-axis to be parallel to the and positive up the plane. Likewise, the y-axis will be positive upwards and perpendicular to the plane. As the problem stated, we are going to assume that m1 will move downwards so its acceleration is negative while m2 moves up so its acceleration is positive. There are two weight components pointing down the plane, [tex]m_1g \sin \theta[/tex] and [tex]m_2g \sin \theta[/tex] and two others pointing up the plane, the two tensions T along the strings. There is a normal force N pointing up from the plane and two pointing down, [tex]m_1g \sin \theta[/tex] and [tex]m_2g \sin \theta[/tex]. Now let's apply Newton's 2nd law to this problem:
x-axis:
[tex]m1:\:\:\:\displaystyle \sum_i F_i = T - m_1g \sin \theta = - m_1a\:\:\:\:(1)[/tex]
[tex]m2:\:\:\:\displaystyle \sum_i F_i = T - m_2g \sin \theta = m_2a\:\:\:\:(2)[/tex]
y-axis:
[tex]\:\:\:\displaystyle \sum_i F_i = N - m_1g \cos \theta - m_2g \cos \theta = 0[/tex]
Use Eqn 1 to solve for T,
[tex]T = m_1(g \sin \theta - a)[/tex]
Substitute this expression for T into Eqn 2,
[tex]m_1g \sin \theta - m_1a - m_2g \sin \theta = m_2a[/tex]
Collecting all similar terms, we get
[tex](m_1 + m_2)a = (m_1 - m_2)g \sin \theta[/tex]
or
[tex]a = \left(\dfrac{m_1 - m_2}{m_1 + m_2} \right)g \sin \theta[/tex]
Your car breaks down in the middle of nowhere. A tow truck weighing 4000 lbs. comes along and agrees to tow your car, which weighs 2000 lbs., to the nearest town. The driver of the truck attaches his cable to your car at an angle of 20 degrees to horizontal. He tells you that his cable has a strength of 500 lbs. He plans to take 10 secs to tow your car at a constant acceleration from rest in a straight line along a flat road until he reaches the maximum speed of 45 m.p.h. Can the driver carry out the plan
Answer:
F = 1010 Lb
the tension on the cable is greater than its resistance, which is why the plan is not viable
Explanation:
For this exercise we can use the kinematic relations to find the acceleration and with Newton's second law find the force to which the cable is subjected.
v = v₀ + a t
how the car comes out of rest v₀ = 0
a = v / t
let's reduce to the english system
v = 45 mph (5280 ft / 1 mile) (1h / 3600) = 66 ft / s
let's calculate
a = 66/10
a = 6.6 ft / s²
now let's write Newton's second law
X axis
Fₓ = ma
with trigonometry
cos 20 = Fₓ / F
Fₓ = F cos 20
we substitute
F cos 20 = m a
F = m a / cos20
W = mg
F = [tex]\frac{W}{g} \ \frac{a}{cos 20}[/tex]
let's calculate
F = [tex]\frac{2000}{32} \ \frac{6.6 }{cos20}[/tex](2000/32) 6.6 / cos 20
F = 1010 Lb
Under these conditions, the tension on the cable is greater than its resistance, which is why the plan is not viable.
why are you teachers regarded as professionals
Answer:
coz teaching is their profession.
g you hang an object of mass m on a spring with spring constant k and find that it has a period of T. If you change the spring to one that has a spring constant of 2 k, the new period is
Answer:
a) T = 2π [tex]\sqrt{\frac{m}{k} }[/tex], b) T ’= [tex]\frac{1}{\sqrt{2} } T[/tex]
Explanation:
a) A system formed by a mass and a spring has a simple harmonic motion with angular velocity
w² = k / m
angular velocity and period are related
w = 2π /T
we substitute
4π²/ T² = k / m
T = 2π [tex]\sqrt{\frac{m}{k} }[/tex]
b) We change the spring for another with k ’= 2 k, let's find the period
T ’= 2π [tex]\sqrt{\frac{m}{k'} }[/tex]
T ’= 2π [tex]\sqrt{ \frac{m}{2k} }[/tex]
T ’= [tex]\frac{1}{\sqrt{2} } T[/tex]
why material selection is important to design and manufacturing?
Answer:
. You want your product to be as strong and as long lasting as possible. There are also the safety implications to consider. You see, dangerous failures arising from poor material selection are still an all too common occurrence in many industries. yep that the answer have a Great day
Explanation:
(◕ᴗ◕✿)
Give the missing ammeter reading a and b. suggest why more current flow through some bulbs than through others Grade 10 question and Answer
Answer:
becaude of electricity
help asap PLEASE I will give u max everything all that
steps if possible
Explanation:
2. [tex]R_T = R_1 + R_2 + R_3 = 625\:Ω + 330\:Ω + 1500\:Ω[/tex]
[tex]\:\:\:\:\:\:\:= 2455\:Ω = 2.455\:kΩ[/tex]
3. Resistors in series only need to be added together so
[tex]R_T = 8(140\:Ω) = 1120\:Ω = 1.12\:kΩ[/tex]
A 5.0-kg mass is placed at (3.0, 4.0) m, and a 6.0-kg mass is placed at (3.0, -4.0) m. What is the moment of inertia of this system of masses about the y-axis?
Answer:
the moment of inertia of this system of masses about the y-axis is 99 kgm²
Explanation:
Given the data in the question;
mass m₁ = 5.0 kg at point ( 3.0, 4.0 )
mass m₂ = 6.0 kg at point ( 3.0, -4.0 )
Now, Moment of inertia [tex]I[/tex] of this system of masses about the y-axis will be;
Moment of inertia [tex]I[/tex]ₓ = mixi²
Moment of inertia [tex]I[/tex] = m₁x₁² + m₂x₂²
we substitute
Moment of inertia [tex]I[/tex] = [ 5.0 × ( 3 )² ] + [ 6.0 × ( 3 )² ]
Moment of inertia [tex]I[/tex] = [ 5.0 × 9 ] + [ 6.0 × 9 ]
Moment of inertia [tex]I[/tex] = 45 + 54
Moment of inertia [tex]I[/tex] = 99 kgm²
Therefore, the moment of inertia of this system of masses about the y-axis is 99 kgm²
An object with mass m = 0.56 kg is attached to a string of length r = 0.72 m and is rotating with an angular velocity ω = 1.155 rad/s. What is the centripetal force acting in the object?
Answer:
The centripetal force is 0.54 N.
Explanation:
mass, m = 0.56 kg
radius, r = 0.72 m
angular speed, w = 1.155 rad/s
The centripetal force is given by
[tex]F = m r w^2\\\\F =0.56\times 0.72\times 1.155\times 1.155\\\\F = 0.54 N[/tex]
A 70-kg boy is surfing and catches a wave which gives him an initial speed of 1.6 m/s. He then drops through a height of 1.60 m, and ends with a speed of 8.5 m/s. How much nonconservative work (in kJ) was done on the boy
Answer:
3.6 KJ
Explanation: Given that a 70-kg boy is surfing and catches a wave which gives him an initial speed of 1.6 m/s. He then drops through a height of 1.60 m, and ends with a speed of 8.5 m/s. How much nonconservative work (in kJ) was done on the boy
The workdone = the energy.
There are two different energies in the scenario - the potential energy (P.E ) and the kinetic energy ( K.E )
P.E = mgh
P.E = 70 × 9.8 × 1.6
P.E = 1097.6 J
P.E = 1.098 KJ
K.E = 1/2mv^2
K.E = 1/2 × 70 × 8.5^2
K.E = 2528.75 J
K.E = 2.529 KJ
The non conservative workdone = K.E + P.E
Work done = 1.098 + 2.529
Work done = 3.63 KJ
Therefore, the non conservative workdone is 3.6 KJ approximately
One way families influence healthy technology use is when siblings explain the use of media to each other. Which of these outfits would you expect if this guideline was followed?
Answer:
The answer would be C.
Explanation:
This is what I would expect when you show someone else how to do something then is also known as teaching.
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Hope this Helps
Increasing the surfactant concentration above the critical micellar concentration
will result in: Select one:
1.An increase in surface tension
2. A decrease in surface tension
3. No change in surface tension
4.None of the above
Answer:
Explanation:no change in surface tension
An increase in the surfactant concentration above the critical micellar concentration will result in no change in surface tension.
In water-gas interface, surfactant reduces the surface tension of water by adsorbing at the liquid–gas interface.
Also, in oil-water interface, surfactant reduces the interfacial tension between oil and water by adsorbing at the oil-water interface.
The concentration of the surfactant can increase to a level called critical micellar concentration, which is an important characteristic of a surfactant.
As the concentration of the surfactant increases before critical micellar concentration, the surface tension changes strongly with an increase in the concentration of the surfactant. After reaching the critical micellar concentration, any further increase in the concentration will result in no change of the surface tension, that is the surface tension will be constant.Thus, increasing the surfactant concentration above the critical micellar concentration will result in no change in surface tension.
Learn more here: https://brainly.com/question/15785205
plz answer the question
Answer:
Ray A = Incidence ray
Ray B = Reflected ray
Explanation:
From the law of reflection,
Normal: This is the line that makes an angle of 90° with the reflecting surface.
Ray A is the incidence ray: This is the ray that srikes the surface of a reflecting surface. The angle formed between the normal and the incidence ray is called the incidence angle
Ray B is the reflected ray: This is the ray leaves the surface of a reflecting surface. The angle formed between the reflected ray and the normal is called reflected angle
If you are driving a car with a velocity of -25 m/s and you have an acceleration of -2 m/s^2, are you speeding up or slowing down? Why?
Answer:
Hmmm...
This is a bit tricky
Ok...
Negative Velocity means you're Moving in the Opposite direction....
Negative Acceleration (deceleration) means you're slowing down.
Deceleration would mean slowing down if you were Moving with a Positive velocity.
But In this case...
You're Moving with negative velocity and Negative acceleration...
This simply means that the acceleration and velocity vector are in the same direction....
Its means that...
"YOU'RE SPEEDING UP"
Just that you're doing it in the opposite direction.
Hope this helps.
LC-circuit of the radio receiver consists of variable capacitor (Cmin= 1 pF, Cmax=10 pF) and inductor
with inductance 1 µH. Determine the wavelength range of this radio receiver.
Answer:
the radio can tune wavelengths between 1.88 and 5.97 m
Explanation:
The signal that can be received is the one that is in resonance as the impedance of the LC circuit.
X = X_c - X_L
X = 1 / wC - w L
at the point of resonance the two impedance are equal so their sum is zero
X_c = X_L
1 / wC = w L
w² = 1 / CL
w = [tex]\sqrt{\frac{1}{CL} }[/tex]
let's look for the extreme values
C = 1 10⁻¹² F
w = [tex]\sqrt{\frac{1}{ 1 \ 10^{-12} \ 1 \ 10^{-6}} }[/tex]
w = [tex]\sqrt{1 \ 10^{18}}[/tex]
w = 10⁹ rad / s
C = 10 10⁻¹² F
w = [tex]\sqrt{\frac{1}{10 \ 10^{-12} \ 1 \ 10^{-6}} }[/tex]Ra 1/10 10-12 1 10-6
w = [tex]\sqrt{0.1 \ 10^{18}}[/tex]Ra 0.1 1018
w = 0.316 10⁹ rad / s
Now the angular velocity and the frequency are related
w = 2π f
f = w / 2π
the light velocity is
c = λ f
λ = c / f
we substitute
λ = c 2π/w
we calculate the two values
C = 1 pF
λ₁ = 3 10⁸ 2π / 10⁹
λ₁= 18.849 10⁻¹ m
λ₁ = 1.88 m
C = 10 pF
λ₂ = 3 10⁸ 2π / 0.316 10⁹
λ₂ = 59.65 10⁻¹ m
λ₂ = 5.97 m
so the radio can tune wavelengths between 1.88 and 5.97 m
A police car in hot pursuit goes speeding past you. While the siren is approaching, the frequency of the sound you hear is 5500 Hz. When the siren is receding away from you, the frequency of the sound is 4500 Hz. Use the Doppler formula to determine the velocity of the police car. Use vsound=330 m/s.
What is the velocity v of the police car ?
When a police car in hot pursuit goes speeding past you, the velocity v of the police car is 33 m/s.
What is the Doppler formula?The formula is used when there exists a Doppler shift. The Doppler shift is due to the relative motion of sound waves between the source and observer.
The frequency increase by the Doppler effect is represented by the formula
f' = [tex]\dfrac{v-v_{o} }{v-v_{s} }[/tex]× f
Given the frequency of source f' is 5500 Hz . Velocity of the observer v₀ is 0.
Substituting the value into the equation will give us the velocity of the police car.
[tex]5500 = \dfrac{330}{330-v} \times f[/tex]...........(1)
When the car is receding, the frequency of the receiving signal f = 4500 Hz.
[tex]4500 = \dfrac{330}{330+v} \times f[/tex]..........(2)
Solving both equation, we get the velocity of a police car.
v = 33 m/s
Therefore, the velocity v of the police car is 33 m/s.
Learn more about Doppler equation.
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Two children stretch a jump rope between them and send wave pulses back and forth on it. The rope is 3.3 m long, its mass is 0.52 kg, and the force exerted on it by the children is 47 N. (a) What is the linear mass density of the rope (in kg/m)
Answer:
The linear mass density of rope is 0.16 kg/m.
Explanation:
mass, m = 0.52 kg
force, F = 47 N
length, L = 3.3 m
(a) The linear mass density of the rope is defined as the mass of the rope per unit length.
Linear mass density = m/L = 0.52/3.3 = 0.16 kg/m
nariz (am
miria amy
0 = 0 +260 + (0)
U= 29 mb
6= ut +1 (04)
Car I was sitting at rest when it nous hit from
the rear by car 2 of identical mass. Both cant had
their heaks on and they stidled together Guy
in the original directioned of motion. If the stopping
force is notx (Combined weight of the cars), die
u=0 to find the approximate speed of car a just
before the collision took place on
Answer:
33 mph
Explanation:
My best guess
A camera lens with focal length f = 50 mm and maximum aperture f>2
forms an image of an object 9.0 m away. (a) If the resolution is limited
by diffraction, what is the minimum distance between two points on the
object that are barely resolved? What is the corresponding distance
between image points? (b) How does the situation change if the lens is
“stopped down” to f>16? Use λ= 500 nm in both cases
Answer:
The minimum distance between two points on the object that are barely resolved is 0.26 mm
The corresponding distance between the image points = 0.0015 m
Explanation:
Given
focal length f = 50 mm and maximum aperture f>2
s = 9.0 m
aperture = 25 mm = 25 *10^-3 m
Sin a = 1.22 *wavelength /D
Substituting the given values, we get –
Sin a = 1.22 *600 *10^-9 m /25 *10^-3 m
Sin a = 2.93 * 10 ^-5 rad
Now
Y/9.0 m = 2.93 * 10 ^-5
Y = 2.64 *10^-4 m = 0.26 mm
Y’/50 *10^-3 = 2.93 * 10 ^-5
Y’ = 0.0015 m
Determine the tension in the string that connects M2 and M3.
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Who stated that man is an animal
plz answer the question
Answer:
Ray A - incident ray
Ray B - reflected ray
A helicopter is ascending vertically witha speed of 5.40 m/s. At a height of 105 m above the earth a package is dropped from the helicopter. How much time does is take for the package to reach the ground
Answer: 5.21 s
Explanation:
Given
Helicopter ascends vertically with [tex]u=5.4\ m/s[/tex]
Height of helicopter [tex]h=105\ m[/tex]
When the package leaves the helicopter, it will have the same vertical velocity
Using equation of motion
[tex]\Rightarrow h=ut+\dfrac{1}{2}at^2\\\\\Rightarrow 105=-5.4t+0.5\times 9.8t^2\\\Rightarrow 4.9t^2-5.4t-105=0\\\\\Rightarrow t=\dfrac{5.4\pm \sqrt{5.4^2+4\times 4.9\times 105}}{2\times 4.9}\\\\\Rightarrow t=\dfrac{5.4\pm 45.68}{9.8}\\\\\Rightarrow t=5.21\ s\quad \text{Neglect negative value}[/tex]
So, package will take 5.21 s to reach the ground
which of the following is a correct statement. a. In dc steady state conditions, the voltages across the capacitors are constant and the currents through the capacitance are zero. The current through the inductors are constant and the voltage across the inductances are constant. b. In dc steady state conditions, the voltages across the capacitors are zero and the currents through the capacitance are constant. The current through the inductors are constant and the voltage across the inductances are zero. c. In dc steady state conditions, the voltages across the capacitors are constant and the currents through the capacitance are zero. The current through the inductors are zero and the voltage across the inductances are constant. d. WIn dc steady state conditions, the voltages across the capacitors are constant and the currents through the capacitance are zero. The current through the inductors are constant and the voltage across the inductances are zero.
Answer:
d. In dc steady state conditions, the voltages across the capacitors are constant and the currents through the capacitance are zero. The current through the inductors are constant and the voltage across the inductances are zero.
Explanation:
The current through a capacitor is given by i = CdV/dt where C = capacitance of capacitor and V = voltage across capacitor. At steady state dV/dt = 0 and V = constant. So, i = CdV/dt = C × 0 = 0.
So, in dc steady state, the voltage across a capacitor is constant and the current zero.
The voltage across an inductor is given by V = Ldi/dt where L = inductance of inductor and i = current through inductor. At steady state di/dt = 0 and V = constant. So, V = Ldi/dt = L × 0 = 0.
So, in dc steady state, the voltage across an inductor is zero and the current constant.
So, In dc steady state conditions, the voltages across the capacitors are constant and the currents through the capacitance are zero. The current through the inductors are constant and the voltage across the inductances are zero.
The answer is d.
Copy the diagram. add a voltmeter to show how you would measure the voltage of the cell
Answer: the answer is 23voltage
Explanation: because the voltage and time put together is 23
An investigator collects a sample of a radioactive isotope with an activity of 490,000 Bq.48 hours later, the activity is 110,000 Bq. Part A For the steps and strategies involved in solving a similar problem, you may view a Video Tutor Solution What is the half-life of the sample?
Answer:
The correct answer is "22.27 hours".
Explanation:
Given that:
Radioactive isotope activity,
= 490,000 Bq
Activity,
= 110,000 Bq
Time,
= 48 hours
As we know,
⇒ [tex]A = A_0 e^{- \lambda t}[/tex]
or,
⇒ [tex]\frac{A}{A_0}=e^{-\lambda t}[/tex]
By taking "ln", we get
⇒ [tex]ln \frac{A}{A_0}=- \lambda t[/tex]
By substituting the values, we get
⇒ [tex]-ln \frac{110000}{490000} = -48 \lambda[/tex]
⇒ [tex]-1.4939=-48 \lambda[/tex]
[tex]\lambda = 0.031122[/tex]
As,
⇒ [tex]\lambda = \frac{ln_2}{\frac{T}{2} }[/tex]
then,
⇒ [tex]\frac{ln_2}{T_ \frac{1}{2} } =0.031122[/tex]
⇒ [tex]T_\frac{1}{2}=\frac{ln_2}{0.031122}[/tex]
[tex]=22.27 \ hours[/tex]
A child throws a ball vertically upward to a friend on a balcony 28 m above him. The friend misses the ball on its upward flight but catches it as it is falling back to earth. If the friend catches the ball 3.0 s after it is thrown, at what time did it pass him on its upward flight
Answer:
[tex]t=1.9 sec[/tex]
Explanation:
From the question we are told that:
Height [tex]h=28m[/tex]
Time [tex]t=3s[/tex]
Generally the Newton's equation for Initial velocity upward is mathematically given by
[tex]s=ut+\frtac{1}{2}at^2[/tex]
[tex]28=3u-\frac{1}{2}*9.8*3^2[/tex]
[tex]u=24.03m/s[/tex]
Generally the velocity at elevation and depression occurs as ball arrives and passes through S=28
[tex]v=\sqrt{24.03-2*9.8*28}[/tex]
[tex]v=5.35m/s and -5.35m/s[/tex]
Generally the Newton's equation for time to reach initial velocity is mathematically given by
[tex]v=u+at[/tex]
[tex]5.35=24.03-9.8t[/tex]
[tex]t=\frac{28.03-5.35}{9.8}[/tex]
[tex]t=1.9 sec[/tex]
