Answer:
187,500N/mExplanation:
From the question, the kinectic energy of the train will be equal to the energy stored in the spring.
Kinetic energy = 1/2 mv² and energy stored in a spring E = 1/2 ke².
Equating both we will have;
1/2 mv² = 1/2ke²
mv² = ke²
m is the mass of the train
v is the velocity of then train
k is the spring constant
e is the extension caused by the spring.
Given m = 30000kg, v = 4 m/s, e = 4 - 2.4 = 1.6m
Substituting this values into the formula will give;
30000*4² = k*1.6²
[tex]k = \frac{30,000*16}{1.6^2}\\ \\k = \frac{480,000}{2.56}\\ \\k = 187,500Nm^{-1}[/tex]
The value of the spring constant is 187,500N/m
At a playground, two young children are on identical swings. One child appears to be about twice as heavy as the other. Part A If you pull them back together the same distance and release them to start them swinging, what will you notice about the oscillations of the two children
Answer:
The motion of the lighter child would look faster than that of the heavier child, but both have the same period of oscillation.
Explanation:
Oscillation is a type of simple harmonic motion which involves the to and fro movement of an object. The oscillation takes place at a required time called the period of oscillation.
Since the swings are similar, the period of oscillation of the two children are the same and they would complete one oscillation in the same time. Though the oscillation of the lighter child seems faster than that of the heavy child, their masses does not affect the period of oscillation.
When a heavy object oscillates, its mass increases the drag or damping force, but not the period of oscillation. Thus, it oscillate slowly.
An ac series circuit contains a resistor of 20 ohms, a capacitor of 0.75 microfarads of 120 x 10-3 H. If an effective (rms) voltage of 120 V is applied, what is the effective (rms) current when the circuit is in resonance
Answer:
The effective (rms) current when the circuit is in resonance is 6 A
Explanation:
Given;
resistance of the resistor, R = 20 ohms
capacitance of the capacitor, C = 0.75 microfarads
inductance of the inductor, L = 0.12 H
effective rms voltage, [tex]V_{rms}[/tex] = 120
At resonance, the impedance Z = R, Since the capacitive reactance (Xc) is equal to inductive reactance (XL).
The effective (rms) current, = [tex]V_{rms}[/tex] / R
= 120 / 20
= 6 A
Therefore, the effective (rms) current when the circuit is in resonance is 6 A
The vector indicates the instantaneous displacement of a projectile from the origin. At the instant when the projectile is at , its velocity and acceleration vectors are and . Which statement is correct?
Answer:
The only force acts on a projectile is gravitational force {Fg}, therefore its acceleration a=Fg/m will always directed towards the direction of force i.e. vertically downwards. Therefore it will always be perpendicular to the x direction or here we can say that a is always perpendicular to Vx}.
Explanation:
The vector r indicates the instantaneous displacement of a projectile from the origin. At the instant when the projectile is at r , its velocity and acceleration vectors are v and a . Which statement is correct?
in a certain region of space, the gravitational field is given by -k/r,where r=distance,k=const.if gravitational potential at r=r0 be v0,then what is the expression for the gravitational potential v?
options
1)k log(r/ro)
2)k log(ro/r)
3)vo+k log(r/ro)
4)vo+k log(ro/r)
plz help me out
I will mark u as brainliest if u answer correct
Answer:
The correct answer is option 3 .
Please check the answer once :)
A uniform electric field stack \rightarrow Ei subscript I with rightwards arrow on top is present in the region between infinite parallel plane plates A and B and a uniform electric field stack\rightarrow Eii subscript I I end subscript with rightwards arrow on top is present in the region between infinite parallel plane plates B and C. When the plates are vertical, stack \rightarrow Ei subscript I with rightwards arrow on top is directed to the right and stack \rightarrow Eii subscript I I end subscript with rightwards arrow on top to the left. The signs of the charges on plates A, B and C may be
a. ?, ?, ?.
b. +, ?, ?.
c. +, ?, +.
d. +, +, +.
e. any one of the above
Answer:
e. any one of the above
Explanation:
7. Which statement is true about teens that are in Marcia’s final state of identity formation?
Answer:
D. All of the above
Explanation:
The last stage in the Marcia's identity formation theory is Identity achievement. In this last stage, teens have made a thorough search or exploration about their identity and have made a commitment to that identity. This identity represents their values, beliefs, and desired goals. At this point, they know want they want in life, and can now make informed decisions based on their belief and ideology.
James Marcia is a psychologist known mainly for his research and theories in human identity. Identity according to him is the sum total of a person's beliefs, values, and ideologies that shape what a person actually becomes and is known for. Occupation and Ideologies primarily determine identity. The four stages of Identity status include, Identity diffusion, foreclosure, moratorium, and achievement.
A loaded ore car has a mass of 950 kg and rolls on rails with negligible friction. It starts from rest and is pulled up a mine shaft by a cable connected to a winch. The shaft is inclined at 28.0° above the horizontal. The car accelerates uniformly to a speed of 2.35 m/s in 14.0 s and then continues at constant speed.(A) What power must the winch motor provide when the car is moving at constant speed? kW(B) What maximum power must the motor provide? kW(C) What total energy transfers out of the motor by work by the time the car moves off the end of the track, which is of length 1,250 m?
Answer:
a) P = 10.27 kW
b) Pmax = 10.65 kW
c) E = 5.47 MJ
Explanation:
Mass of the loaded car, m = 950 kg
Angle of inclination of the shaft, θ = 28°
Acceleration due to gravity, g = 9.8 m/s²
The speed of the car, v = 2.35 m/s
Change in time, t = 14.0 s
a) The power that must be provided by the winch motor when the car is moving at constant speed.
P = Fv
The force exerted by the motor, F = mg sinθ
P = mgv sinθ
P = 950 * 9.8 *2.35* sin28°
P = 10,271.3 W
P = 10.27 kW
b) Maximum power that the motor must provide:
[tex]P = mv\frac{dv}{dt} + mgvsin \theta\\dv/dt = \frac{2.35 - 0}{14} \\dv/dt = 0.168 m/s^2\\P = (950*2.35*0.168) + (950*9.8*2.35* sin28)\\P = 374.74 + 10271.3\\P = 10646.04 W\\10.65 kW[/tex]
c) Total energy transferred:
Length of the track, d = 1250 m
[tex]E = 0.5 mv^2 + mgd sin \theta\\E = (0.5 * 950 * 2.35^2) + (950 * 9.8 * 1250 * sin 28)\\E = 2623.19 + 5463475.31\\E = 5466098.50 J\\E = 5.47 MJ[/tex]
How would a spinning disk's kinetic energy change if its moment of inertia was five times larger but its angular speed was five times smaller
Answer:
The kinetic energy of a spinning disk will be reduced to a tenth of its initial kinetic energy if its moment of inertia is made five times larger, but its angular speed is made five times smaller.
Explanation:
Let us first consider the initial characteristics of the angular motion of the disk
moment of inertia = [tex]I[/tex]
angular speed = ω
For the second case, we consider the characteristics to now be
moment of inertia = [tex]5I[/tex] (five times larger)
angular speed = ω/5 (five times smaller)
Recall that the kinetic energy of a spinning body is given as
[tex]KE = \frac{1}{2}Iw^{2}[/tex]
therefore,
for the first case, the K.E. is given as
[tex]KE = \frac{1}{2}Iw^{2}[/tex]
and for the second case, the K.E. is given as
[tex]KE = \frac{1}{2}(5I)(\frac{w}{5} )^{2} = \frac{5}{50}Iw^{2}[/tex]
[tex]KE = \frac{1}{10}Iw^{2}[/tex]
this is one-tenth the kinetic energy before its spinning characteristics were changed.
This implies that the kinetic energy of the spinning disk will be reduced to a tenth of its initial kinetic energy if its moment of inertia is made five times larger, but its angular speed is made five times smaller.
A spinning disk's kinetic energy will change to one-tenth if its moment of inertia was five times larger but its angular speed was five times smaller.
Relation between Kinetic energy and Moment of Inertia:
Rotational kinetic energy is directly proportional to the rotational inertia and the square of the magnitude of the angular velocity.Now, let's consider moment of inertia = I and angular speed = ω
It is asked that what would be change in Kinetic energy if
moment of inertia = (five times larger)
angular speed = ω/5 (five times smaller)
The kinetic energy of a spinning body is given as:
[tex]K.E.=\frac{1}{2} I. w^2[/tex]
On substituting the values, we will get:
[tex]K.E.= \frac{1}{2} (5I) (\frac{w}{5} )^2 \\\\K.E. =\frac{1}{10} I. w^2[/tex]
Kinetic energy will be one-tenth to the kinetic energy before its spinning characteristics were changed.
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Although electromagnetic waves can always be represented as either photons or waves, in the radio part of the spectrum we typically do not discuss photons (like we do in the visible) because they are at such a low energy. Nevertheless. they exist. Consider such a photon in a radio wave from an AM station has a 1545 kHz broadcast frequency.
Required:
a. What is the energy, in joules, of the photon?
b. What is the energy, in electron volts. of the photon?
Answer:
a. E = 1.02*10^-27 J
b. E = 6.39*10^-9eV
Explanation:
a. In order to calculate the energy of the radio photon, you use the following formula:
[tex]E=hf[/tex] (1)
h: Planck's constant = 6.626*10^-34 Js
f: frequency of the photon = 1545kHz = 1.545*10^6 Hz
Then, by replacing you obtain the energy of the photon:
[tex]E=(6.626*10^{-34}Js)(1.545*10^6s^{-1})=1.02*10^{-27}J[/tex]
b. In electron volts, the energy of the photon is:
[tex]E=1.02*10^{-27}J*\frac{6.242*10^{18}eV}{1J}=6.39*10^{-9}eV[/tex]
In an RC circuit, how many time constants must elapse if an initially uncharged capacitor is to reach 80% of its final potential difference
Answer:
1.6 time constants must elapse
Explanation:
voltage on a cap, charging is given as
v = v₀[1–e^(–t/τ)]
Where R is resistance in ohms,
C is capacitance in farads
t is time in seconds
RC = τ = time constant
v = v₀[1–e^(–t/τ)]
1–e^(–t/τ) = 0.8
e^(–t/τ) = 0.2
–t/τ = –1.609
t = 1.609τ
Find the period of revolution for the planet Mercury, whose average distance from the Sun is 5.79 x 1010 m.
Answer:
T = 7.61*10^6 s
Explanation:
In order to calculate the Mercury's period. in its orbit around the sun, you take into account on the Kepler's law. You use the following formula:
[tex]T=\sqrt{\frac{4\pi^2r^3}{GM_s}}[/tex] (1)
T: period of Mercury
r: distance between Mercury and Sun
Ms: mass of the sun = 1.98*10^30 kg
G: Cavendish's constant = 6.674*10^-11 m^3 kg^-1 s^-2
You replace the values of all parameters in the equation (1):
[tex]T=\sqrt{\frac{4\pi^2(5.79*10^{10}m)^3}{(6.674*10^{-11}m^3kg^{-1}s^{-2})(1.98*10^{30}kg)}}\\\\T=7.61*10^6s*\frac{1h}{3600s}*\frac{1d}{24h}=88.13\ days[/tex]
The period of Mercury is 7.61*10^6 s, which is approximately 88.13 Earth's days
I WILL MARK YOU AS BRAINLIEST!!! An object is launched straight up into the air with an initial velocity of 40 meters per second, from a height 30 m above the ground. Assuming that gravity pulls it down, changing its position by about 4.9 /2, after how many seconds will the object hit the ground? Enter your answer as a number rounded to the nearest tenth, such as: 42.5
Answer:
8.9 seconds
Explanation:
The height of the object at time t is:
y = h + vt − 4.9t²
where h is the initial height, and v is the initial velocity.
Given h = 30 and v = 40:
y = 30 + 40t − 4.9t²
When y = 0:
0 = 30 + 40t − 4.9t²
4.9t² − 40t − 30 = 0
Solving with quadratic formula:
t = [ -(-40) ± √((-40)² − 4(4.9)(-30)) ] / 2(4.9)
t = [ 40 ± √(1600 + 588) ] / 9.8
t = 8.9
It takes 8.9 seconds for the object to land.
Jerome solves a problem using the law of conservation of momentum. What should Jerome always keep constant for each object after the objects collide and bounce apart?
Jerome solves a problem using the law of conservation of momentum. What should Jerome always keep constant for each object after the objects collide and bounce apart?
a-velocity
b-mass
c-momentum
d-direction
Answer:
b. Mass
Explanation:
This question has to do with the principle of the law of conservation of momentum which states that the momentum of a system remains constant if no external force is acting on it.
As the question states, two objects collide with each other and eventually bounce apart, so their momentum may not be conserved but the mass of the objects is constant for each non-relativistic motion. Because of this, the mass of each object prior to the collision would be the same as the mass after the collision.
Therefore, the correct answer is B. Mass.
This question involves the concept of the law of conservation of momentum.
Jerome should always keep the "mass" of each object constant after the objects collide and bounce apart.
The law of conservation of momentum states that the momentum of a system of objects must remain constant before and after the collision has taken place.
Mathematically,
[tex]m_1u_1+m_2u_2=m_1v_1+m_v_2[/tex]
where,
m₁ = mass of the first object
m₂ = mass of the second object
u₁ = velocity of the first object before the collision
u₂ = velocity of the second object before the collision
v₁ = velocity of the first object after the collision
v₂ = velocity of the second object after the collision
Hence, it is clear from the formula that the only thing unchanged before and after the collision is the mass of each object.
Learn more about the law of conservation of momentum here:
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The attached picture illustrates the law of conservation of momentum.
Help asap thank you!!
Consider a loop of wire placed in a uniform magnetic field. Which factors affect the magnetic flux Φm through the loop?
Answer:
* The value of the magnetic field changes either in time or space
* The waxed area changes, the bow is fitting in size
* The angle between the field and the area changes
Explanation:
Magnetic flux is the scalar product of the magnetic field over the area
Ф = ∫ B. dA
where B is the magnetic field and A is the area
Let's look at stationary, for which factors affect flow
* The value of the magnetic field changes either in time or space
* The waxed area changes, the bow is fitting in size
* The angle between the field and the area changes
two 200 pound lead balls are separated by a distance 1m. both balls have the same positive charge q. what charge will produce an electrostatic force.between the balls that is of the same order of magnitude as the weight of one ball?
Answer:
The charge is [tex]q = 3.14 *10^{-4} \ C[/tex]
Explanation:
From the question we are told that
The mass of each ball is [tex]m = 200 \ lb = \frac{200}{2.205} = 90.70 \ kg[/tex]
The distance of separation is [tex]d = 1 \ m[/tex]
Generally the weight of the each ball is mathematically represented as
[tex]W = m * g[/tex]
where g is the acceleration due to gravity with a value [tex]g = 9.8 m/s^2[/tex]
substituting values
[tex]W = 90.70 * 9.8[/tex]
[tex]W = 889 \ N[/tex]
Generally the electrostatic force between this balls is mathematically represented as
[tex]F_e = \frac{k * q_1* q_2 }{d^2}[/tex]
given that the the charges are equal we have
[tex]q_1= q_2 = q[/tex]
So
[tex]F_e = \frac{k * q^2 }{d^2}[/tex]
Now from the question we are told to find the charge when the weight of one ball is equal to the electrostatic force
So we have
[tex]889 = \frac{9*10^9 * q^2}{1^2}[/tex]
=> [tex]q = 3.14 *10^{-4} \ C[/tex]
The magnitude of charge on the balls is [tex]3.14 \times 10^{-4} \;\rm C[/tex].
Given data:
The masses of two lead balls are, m = 200 lb = 200/2.205 = 90.70 kg.
The distance of separation of two balls is, d = 1 m.
First of all we need to obtain the weight of ball. The weight of the ball is expressed as,
W = mg
Here,
g is the gravitational acceleration.
Solving as,
W = 90.70 × 9.8
W = 888.86 N
The expression for the electrostatic force between this balls is mathematically represented as,
[tex]F = \dfrac{k \times q_{1} \times q_{2}}{d^{2}}[/tex]
Since, the charges are equal then,
[tex]q_{1} =q_{2}=q[/tex]
Also, the magnitude of force between the balls is same as the weight of one ball. Then,
F = W
Solving as,
[tex]F =W= \dfrac{(9 \times 10^{9}) \times q^{2}}{1^{2}}\\\\889= \dfrac{(9 \times 10^{9}) \times q^{2}}{1^{2}}\\\\q = 3.14 \times 10^{-4} \;\rm C[/tex]
Thus, we can conclude that the magnitude of charge on the balls is [tex]3.14 \times 10^{-4} \;\rm C[/tex].
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Determine the magnitude of the force between two 11 m-long parallel wires separated by 0.033 m, both carrying 5.2 A in the same direction.
Answer:
[tex]F=1.8\times 10^{-3}\ N[/tex]
Explanation:
We have,
Length of wires is 11 m
Separation between wires is 0.033 m
Current in both the wires is 5.2 A
It is required to find the magnitude of force between two wires. The force between wires is given by :
[tex]F=\dfrac{\mu_o I_1I_2l}{2\pi r}\\\\F=\dfrac{4\pi \times 10^{-7}\times 5.2\times 5.2\times 11}{2\pi \times 0.033}\\\\F=1.8\times 10^{-3}\ N[/tex]
So, the magnitude of force between wires is [tex]1.8\times 10^{-3}\ N[/tex]
If a diver below the water's surface shines a light up at the bottom of the oil film, at what wavelength (as measured in water) would there be constructive interference in the light that reflects back downward
Answer:
see that the entire spectrum of the visible is between the integers from 3 to 5 so only three wavelengths are reflected with constructive interference
Explanation:
This is an interference problem in thin films, the refractive index of water is 1.33 and the refractive index of oil is 1.5
Let's analyze the light beam path emitted by the diver.
* when the beam passes from the water to the oil with the highest refractive index, it has a phase change of 180º
* also the wavelength of light in a material medium changes
λ_n = λ / n
where λ_n is the wavelength in the material and λ the wavelength in the vacuum air and n the refractive index.
If we include these aspects, the constructive interference equation is
2t = (m + ½) λ_n
2nt = (m + ½) λ
let's apply this equation to our case
λ = 2nt / (m + ½)
The incidence of replacement of the oil with respect to water is
n = n_oil / n_water = 1.5 / 1.33
n = 1,128
let's calculate
λ = 2 1,128 t / (m + ½)
λ = 2,256 t / (m + ½)
In your statement you do not include the value of the oil layer that is the thin film, suppose a value to finish the calculation
t = 0.001 mm = 1 10⁻⁶ m
the formula remains
λ = 2,256 10⁻⁶ / (m + ½)
Let's find what values of m we have to cut light in the visible range (400 to 700) 10⁻⁹ m
m + ½ = 2,256 10⁻⁶ / λ
m = 2,256 10⁻⁶ / λ - ½
light purple lan = 400 10⁻⁹m
m = 2,256 10-6 / 400 10⁻⁹ - ½
m = 5.64 - 0.5
m = 5.14
m = 5
red light λ = 700 10⁻⁹m
m = 2,256 1-6 / 700 10⁻⁹ - ½
m = 3.22 - 0.5
m = 2.72
m = 3
we see that the entire spectrum of the visible is between the integers from 3 to 5 so only three wavelengths are reflected with constructive interference
lens is to be used to allow this eye to clearly focus on objects 25 cm in front of it. What should be the focal length of this lens
The nearpoint of an eye is 151 cm. A corrective lens is to be used to allow this eye to clearly focus on objects 25 cm in front of it. What should be the focal length of this lens?
Answer:
29.96cm
Explanation:
Using the corrective lens, the image should be formed at the front of the eye and be upright and virtual.
Now using the lens equation as follows;
[tex]\frac{1}{f} = \frac{1}{v} + \frac{1}{u}[/tex] -------------(i)
Where;
f = focal length of the lens
v = image distance as seen by the lens
u = object distance from the lens
From the question;
v = -151cm [-ve since the image formed is virtual]
u = 25cm
Rewrite equation (i) to have;
[tex]f = \frac{uv}{u+v}[/tex]
Substitute the values of v and u into the equation;
[tex]f = \frac{25*(-151)}{25-151}[/tex]
[tex]f = \frac{-3775}{-126}[/tex]
f = 29.96cm
The focal length should be 29.96cm
A student stretches an elastic band by 0.8 m in 0.5 seconds. The spring constant of the elastic band is 40 N/m. What was the power exerted by the student
Answer:
The power exerted by the student is 51.2 W
Explanation:
Given;
extension of the elastic band, x = 0.8 m
time taken to stretch this distance, t = 0.5 seconds
the spring constant, k = 40 N/m
Apply Hook's law;
F = kx
where;
F is the force applied to the elastic band
k is the spring constant
x is the extension of the elastic band
F = 40 x 0.8
F = 32 N
The power exerted by the student is calculated as;
P = Fv
where;
F is the applied force
v is velocity = d/t
P = F x (d/t)
P = 32 x (0.8 /0.5)
P = 32 x 1.6
P = 51.2 W
Therefore, the power exerted by the student is 51.2 W
The mass of M1 = 12 Daltons and it has a speed of v1 = 200 m/s. The mass of M2 = 4 Daltons. What was the total momentum of the system consisting of both masses before the collision (in Dalton meters per second, assume positive to the right and negative to the left)?
Answer:
The total momentum is [tex]p__{T }} =(2400 -4 v_2) \ Dalton \cdot m/s[/tex]
Explanation:
The diagram illustration this system is shown on the first uploaded image (From physics animation)
From the question we are told that
The mass of the first object is [tex]M_1 = 12 \ Dalton[/tex]
The speed of the first mass is [tex]v_1 = 200 \ m/s[/tex]
The mass of the second object is [tex]M_2 = 4 \ Dalton[/tex]
The speed of the second object is assumed to be [tex]- v_2[/tex]
The total momentum of the system is the combined momentum of both object which is mathematically represented as
[tex]p__{T }} = M_1 v_1 + M_2 v_2[/tex]
substituting values
[tex]p__{T }} = 12 * 200 + 4 * (-v_2)[/tex]
[tex]p__{T }} =(2400 -4 v_2) \ Dalton \cdot m/s[/tex]
A 100 cm length of nichrome wire has a radius of 0.50 mm, a resistivity LaTeX: \rho_0ρ 0= 1.0 × 10-6 Ω ∙ m , and a temperature coefficient LaTeX: \alphaα = 0.4 × 10-3 (oC)-1. At T0 = 20 oC the wire carries current of 0.50 A. How much power does the wire dissipate at a temperature T = 350 oC? Assume the potential difference across the ends of the wire remains constant. Group of answer choices
Answer:
P₃₅₀ = 0.28 watt
Explanation:
First we find the resistance of the wire at 20°C:
R₀ = ρL/A
where,
ρ = resistivity = 1 x 10⁻⁶ Ωm
L = Length of wire = 100 cm = 1 m
A = cross-sectional area of wire = πr² = π(0.5 x 10⁻³ m)² = 0.785 x 10⁻⁶ m²
Therefore,
R₀ = (1 x 10⁻⁶ Ωm)(1 m)/(0.785 x 10⁻⁶ m²)
R₀ = 1.27 Ω
Now, from Ohm's Law:
V = I₀R₀
where,
V = Potential Difference = ?
I₀ = Current Passing at 20°C = 0.5 A
Therefore,
V = (0.5 A)(1.27 Ω)
V = 0.64 volts
Now, we need to find the resistance at 350°C:
R₃₅₀ = R₀(1 + αΔT)
where,
R₃₅₀ = Resistance at 350°C = ?
α = temperature coefficient of resistance = 0.4 x 10⁻³ °C⁻¹
ΔT = Difference in Temperature = 350°C - 20°C = 330°C
Therefore,
R₃₅₀ = (1.27 Ω)[1 + (0.4 x 10⁻³ °C⁻¹)(330°C)]
R₃₅₀ = 1.44 Ω
Now, for power at 350°C:
P₃₅₀ = VI₃₅₀
where,
P₃₅₀ = Power dissipation at 350°C = ?
V = constant potential difference = 0.64 volts
I₃₅₀ = Current at 350°C = V/R₃₅₀ (From Ohm's Law)
Therefore,
P₃₅₀ = V²/R₃₅₉
P₃₅₀ = (0.64 volts)²/(1.44 Ω)
P₃₅₀ = 0.28 watt
A force of only 150 N can lift a 600 N sack of flour to a height of 0.50 m when using a lever as shown in the diagram below. a. Find the work done on the sack of flour (in J). b. Find the distance you must push with the 150 N force-on the left side (in m). c. Briefly explain the benefit of using a lever to lift a heavy object.
What fundamental frequency would you expect from blowing across the top of an empty soda bottle that is 24 cm deep, if you assumed it was a closed tube
Answer:
f = 357.29Hz
Explanation:
In order to calculate the fundamental frequency in the closed tube, you use the following formula:
[tex]f_n=\frac{nv}{4L}[/tex] (1)
n: order of the mode = 1
v: speed of sound = 343m/s
L: length of the tube = 24cm = 0.24m
You replace the values of the parameters in the equation (1):
[tex]f_1=\frac{(1)(343m/s)}{4(0.24m)}=357.29Hz[/tex]
The fundamental frequency of in the tube is 357.29Hz
The disks you will be using in the lab have the following parameters. All disks are solid with outer radius R1 = 0 0632 m and inner radius R2 = 0 0079 m The masses of the disks are: MT starless steel = 1357kg, MBottom starless stee,ã¼1.344kg. Calculate the moments of inertia of the disks.
Answer:
MT disc I = 2,752 10-3 kg m²
MB disc I = 2,726 10⁻³ kg m²
Explanation:
The moment of inertia given by the expression
I = ∫ r² dm
for bodies with high symmetry it is tabulated
for a hollow disk it is
I = ½ M (R₁² + R₂²)
let's apply this equation to our case
disc MT = 1,357 kg
I = ½ 1,357 (0.0079² + 0.0632²)
I = 2,752 10-3 kg m²
disk MB = 1,344 kg
I = ½ 1,344 (0.0079² + 0.0632²)
I = 2,726 10⁻³ kg m²
What are the two types of long-term memory?
A. sensory and short-term
B. iconic and echoic
C. explicit and implicit
D. recency and primacy
Answer:
C. explicit and implicit
Explanation:
E20
If you were to experimentally determine the length of the pendulum, why would you not get the same length in Iowa?
Answer:
The length of the pendulum depends on acceleration due to gravity (g) which varies in different Earth's location beacuse Earth is not perfectly spherical.
Explanation:
The period of oscillation is calculated as;
[tex]T = 2\pi\sqrt{\frac{l}{g} }[/tex]
where;
L is the length of the pendulum bob
g is acceleration due to gravity
If we make L the subject of the formula in the equation above, we will have;
[tex]T = 2\pi\sqrt{\frac{l}{g}}\\\\\sqrt{\frac{l}{g} } = \frac{T}{2\pi} \\\\\frac{l}{g} = (\frac{T}{2\pi} \)^2\\\\\frac{l}{g} =\frac{T^2}{4\pi^2}\\\\L = \frac{gT^2}{4\pi^2}[/tex]
The length of the pendulum depends on acceleration due to gravity (g).
Acceleration due to gravity is often assumed to be the same everywhere on Earth, but it varies because Earth is not perfectly spherical. The variation of acceleration due to gravity (g) as a result of Earth's geometry, will also cause the length of the pendulum to vary.
A ball is shot at an angle of 45 degrees into the air with initial velocity of 46 ft/sec. Assuming no air resistance, how high does it go
Answer:
5.02 m
Explanation:
Applying the formula of maximum height of a projectile,
H = U²sin²Ф/2g...................... Equation 1
Where H = maximum height, U = initial velocity, Ф = angle, g = acceleration due to gravity.
Given: U = 46 ft/sec = 14.021 m/s, Ф = 45°
Constant: g = 9.8 m/s²
Substitute these values into equation 1
H = (14.021)²sin²45/(2×9.8)
H = 196.5884×0.5/19.6
H = 5.02 m.
Hence the ball goes 5.02 m high
The ball reaches the maximum height of 54 feet
The question is about projectile motion,
the ball is shot at an angle α = 45°, and
the initial velocity u = 46 ft/s.
Under the projectile motion, the maximum height H is given by:
[tex]H=\frac{u^2sin^2\alpha }{2g} [/tex]
where, g = 9.8 m/s²
substituting the given values we get:
[tex]H=\frac{46^2sin^{2}(45)}{2*9.8}\\ \\ H=\frac{46*46*(1/2)}{2*9.8}\\ \\ H=54 feet[/tex]
Hence, the maximum height is 54 feet.
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Point charges q1=50μCq1=50μC and q2=−25μCq2=−25μC are placed 1.0 m apart. (a) What is the electric field at a point midway between them? (b) What is the force on a charge q3=20μCq3=20μC situated there?
Answer:
a) E = 2.7x10⁶ N/C
b) F = 54 N
Explanation:
a) The electric field can be calculated as follows:
[tex] E = \frac{Kq}{d^{2}} [/tex]
Where:
K: is the Coulomb's constant = 9x10⁹ N*m²/C²
q: is the charge
d: is the distance
Now, we need to find the electric field due to charge 1:
[tex] E_{1} = \frac{9 \cdot 10^{9} N*m^{2}/C^{2}*50 \cdot 10^{-6} C}{(0.5 m)^{2}} = 1.8 \cdot 10^{6} N/C [/tex]
The electric field due to charge 2 is:
[tex]E_{2} = \frac{9 \cdot 10^{9} N*m^{2}/C^{2}*(-25) \cdot 10^{-6} C}{(0.5 m)^{2}} = -9.0 \cdot 10^{5} N/C[/tex]
The electric field at a point midway between them is given by the sum of E₁ and E₂ (they are in the same direction, that is to say, to the right side):
[tex]E_{T} = E_{1} + E_{2} = 1.8 \cdot 10^{6} N/C + 9.0 \cdot 10^{5} N/C = 2.7 \cdot 10^{6} N/C to the right side[/tex]
Hence, the electric field at a point midway between them is 2.7x10⁶ N/C to the right side.
b) The force on a charge q₃ situated there is given by:
[tex]E_{T} = \frac{F_{T}}{q_{3}} \rightarrow F_{T} = E_{T}*q_{3}[/tex]
[tex] F = 2.7 \cdot 10^{6} N/C*20 \cdot 10^{-6} C = 54 N [/tex]
Therefore, the force on a charge q₃ situated there is 54 N.
I hope it helps you!
(a) The electric field at a point midway between [tex]q_1[/tex] and [tex]q_2[/tex] is obtained to be [tex]2.7\times 10^6 \,N/C[/tex].
(b) The electrostatic force on the third charge [tex]q_3[/tex] situated between [tex]q_1[/tex] and [tex]q_2[/tex] is obtained as 54 N.
The answer can be explained as follows.
Electric FieldGiven that the two charges are;
[tex]q_1 = 50\times 10^{-6}\,C[/tex] and [tex]q_2 = -25\times 10^{-6}\,C[/tex](a) At the midpoint; [tex]r = 0.5\,m[/tex].
We know that the electric field due to charge [tex]q_1[/tex].
[tex]E_1 = k\,\frac{q_1}{r^2}[/tex]Where, [tex]k=9\times 10^9\,Nm^2/C[/tex]
[tex]E_1 = (9\times 10^9) \times\frac{(50 \times 10^{-6})}{(0.5)^2}=1.8\times 10^6N/C[/tex]The electric field due to charge [tex]q_2[/tex] is given by;
[tex]E_2 = (9\times 10^9) \times\frac{(-25 \times 10^{-6})}{(0.5)^2}=-9\times 10^5\,N/C[/tex]Therefore, the net electric field in the midpoint is given by;
[tex]E_{net} =E_2+E_1[/tex][tex]\implies E_{net}=1.8 \times 10^6 N/C + 9 \times 10^5\,N/C=2.7\times 10^6\,N/C[/tex]The direction is towards the right side.
Electrostatic Force(b) Now, there is another charge [tex]q_3=20\times 10^{-6}[/tex] in the midpoint.
So the force on the charge is ;
[tex]F=E_{net} \times q_3=(2.7 \times 10^6\,N/C) \times (20\times 10^{-6}\,C)=54\,N[/tex]Find out more about electrostatic force and fields here:
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"On earth, you have a pendulum of length L that oscillates with period T. Your friend lives on a planet where the acceleration of gravity is four times as big as it is on the earth. What should be the length of a pendulum on your friend s planet so that it also oscillates with the same period T
Answer:
4L
Explanation:
Data provided in the question according to the question is as follows
Length = L
Gravity = G
For friend
Length = ?
Growth = 4G
Moreover,
[tex]T_1 = T_2[/tex]
Based on the above information ,
Now we have to apply the simple pendulum formula which is shown below:
[tex]T = 2\pi \frac{L}{G}[/tex]
Now equates these equations in both sides
[tex]2\pi \frac{L}{G} = 2\pi \frac{L}{4G}[/tex]
So after solving this, the length of the pendulum is 4L
Answer:
the length of a pendulum on your friend s planet should be 4 times than that on earth
Explanation:
We know that time period of simple pendulum is given by
[tex]T= 2\pi\sqrt{\frac{L}{g} }[/tex]
L= length of pendulum
g= acceleration due to gravity
therefore, Let T_1 and T_2 be the time period of the earth and other planet respectively.
[tex]\frac{T_1}{T_2} =\sqrt(\frac{L_1}{L_2}\times\frac{g_2}{g_1})[/tex]
ATQ
T_1=T_2=T, g_2=4g_1
Putting the values in above equation and solving we get
[tex]\frac{L_1}{L_2} =\frac{1}{4}[/tex]