Answer:
The system will take approximately 0.255 seconds to reach the (new) equilibrium position.
Explanation:
We notice that block-spring system depicts a Simple Harmonic Motion, whose equation of motion is:
[tex]y(t) = A\cdot \cos \left(\sqrt{\frac{k}{m} }\cdot t +\phi\right)[/tex] (1)
Where:
[tex]y(t)[/tex] - Position of the mass as a function of time, measured in meters.
[tex]A[/tex] - Amplitude, measured in meters.
[tex]k[/tex] - Spring constant, measured in newtons per meter.
[tex]m[/tex] - Mass of the block, measured in kilograms.
[tex]t[/tex] - Time, measured in seconds.
[tex]\phi[/tex] - Phase, measured in radians.
The spring is now calculated by Hooke's Law, that is:
[tex]k = \frac{m\cdot g}{\Delta y}[/tex] (2)
Where:
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]\Delta y[/tex] - Deformation of the spring due to gravity, measured in meters.
If we know that [tex]m=1.65\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]\Delta y = 0.260\,m[/tex], then the spring constant is:
[tex]k = \frac{(1.65\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}{0.260\,m}[/tex]
[tex]k = 62.237\,\frac{N}{m}[/tex]
If we know that [tex]A = 0.130\,m[/tex], [tex]k = 62.237\,\frac{N}{m}[/tex], [tex]m=1.65\,kg[/tex], [tex]x(t) = 0\,m[/tex] and [tex]\phi = 0\,rad[/tex], then (1) is reduced into this form:
[tex]0.130\cdot \cos (6.142\cdot t)=0[/tex] (1)
And now we solve for [tex]t[/tex]. Given that cosine is a periodic function, we are only interested in the least value of [tex]t[/tex] such that mass reaches equilibrium position. Then:
[tex]\cos (6.142\cdot t) = 0[/tex]
[tex]6.142\cdot t = \cos^{-1} 0[/tex]
[tex]t = \frac{1}{6.142}\cdot \left(\frac{\pi}{2} \right)\,s[/tex]
[tex]t \approx 0.255\,s[/tex]
The system will take approximately 0.255 seconds to reach the (new) equilibrium position.
The time taken for the spring to reach new equilibrium position is 0.26 s.
The given parameters;
mass, m = 1.65 kgextension of the string, x = 0.26 mdisplacement with time x(t) = 0.13The general wave equation is given as;
[tex]y(t) = A\ cos(\omega t \ + \phi)[/tex]
The angular frequency is given as follows;
[tex]\omega = \sqrt{\frac{k}{m} } \\\\[/tex]
The spring constant is calculated as;
[tex]F = mg\\\\kx = mg\\\\k = \frac{mg}{x} \\\\k = \frac{1.65 \times 9.8}{0.26} \\\\k = 62.2 \ N/m[/tex]
The angular frequency is calculated as follows;
[tex]\omega = \sqrt{\frac{k}{m} } \\\\\omega = \sqrt{\frac{62.2}{1.65} }\\\\\omega = 6.14 \ rad/s[/tex]
The time taken for the spring to reach new equilibrium position is calculated as follows;
[tex]y(t) = A \ cos(\omega t)\\\\0 = 0.13 \times cos(6.14t)\\\\6.14t = cos^{-1}(0)\\\\6.14t = 1.57 \ rad\\\\t = \frac{1.57 \ rad}{6.14 \ rad/s} \\\\t = 0.26 \ s[/tex]
Thus, the time taken for the spring to reach new equilibrium position is 0.26 s.
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Calculate the wavelengths of a 1530 kHz AM radio signal, a 105.1 MHz FM radio signal, and a 1.90 GHz cell phone signal.
Answer:
196.07 m, 2.85 m and 0.15 m
Explanation:
Frequency of AM radio signal, f₁ = 1530 kHz
Its wavelength can be given by :
[tex]\lambda_1=\dfrac{c}{f_1}\\\\=\dfrac{3\times 10^8}{1530\times 10^3}\\\\=196.07\ m[/tex]
Frequency of FM radio signal, f₂ = 105.1 MHz
Its wavelength can be given by :
[tex]\lambda_2=\dfrac{c}{f_2}\\\\=\dfrac{3\times 10^8}{105.1\times 10^6}\\\\=2.85\ m[/tex]
Frequency of cell phone signal, f₃ = 1.9 GHz
Its wavelength can be given by :
[tex]\lambda_2=\dfrac{c}{f_2}\\\\=\dfrac{3\times 10^8}{1.9\times 10^9}\\\\=0.15\ m[/tex]
Hence, the required wavelengths are 196.07 m, 2.85 m and 0.15 m respectively.
A rock is thrown with an initial vertical velocity 50 m/s at an angle of 40 degrees.
a. What is the horizontal component of the velocity?
b. What is the vertical component of the velocity?
c. What is the hang timel?
d. What is the peak height?
e. What is the range?
Answer:
[tex]38.3\ \text{m/s}[/tex]
[tex]32.14\ \text{m/s}[/tex]
6.55 seconds
[tex]52.65\ \text{m}[/tex]
[tex]254.84\ \text{m}[/tex]
Explanation:
u = Initial velocity of rock = 50 m/s
[tex]\theta[/tex] = Angle of throw = [tex]40^{\circ}[/tex]
g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]
Horizontal component is given by
[tex]u_x=u\cos\theta\\\Rightarrow u_x=50\times \cos40^{\circ}\\\Rightarrow u_x=38.3\ \text{m/s}[/tex]
The horizontal component of the velocity is [tex]38.3\ \text{m/s}[/tex]
Vertical component is given by
[tex]u_y=u\sin\theta\\\Rightarrow u_y=50\times \sin40^{\circ}\\\Rightarrow u_y=32.14\ \text{m/s}[/tex]
The horizontal component of the velocity is [tex]32.14\ \text{m/s}[/tex]
Time of flight is given by
[tex]t=\dfrac{2u\sin\theta}{g}\\\Rightarrow t=\dfrac{2\times 50\sin40^{\circ}}{9.81}\\\Rightarrow t=6.55\ \text{s}[/tex]
The hang time of the rock is 6.55 seconds
Maximum height is given by
[tex]h=\dfrac{u^2\sin^2\theta}{2g}\\\Rightarrow h=\dfrac{50^2\sin^240^{\circ}}{2\times 9.81}\\\Rightarrow h=52.65\ \text{m}[/tex]
Maximum height is [tex]52.65\ \text{m}[/tex]
Range is given by
[tex]d=\dfrac{u^2\sin2\theta}{g}\\\Rightarrow d=\dfrac{50^2\sin(2\times40)^{\circ}}{9.81}\\\Rightarrow d=254.84\ \text{m}[/tex]
The range is [tex]254.84\ \text{m}[/tex]
help asap
How does the scientist know that the substance goes through a change of state during the time period labeled B on the graph?
a.the temperature remains the same
b.the temperature increases gradually
c.the particles start slowing down
d.the particles start moving faster
Answer:
The answer is A. The temperature remains the same.
A fish is able to jump vertically out of the water with a speed of 4.45 m/s. What is the speed of the fish as it passes a point 0.6 m above the water?
How much time, after leaving the water does it take for the fish to pass a point 0.6 m above the water while it is on its way down?
Answer:
Explanation:
given
initial velocity u = 4.45m/s
Height = 0.6m
g = 9.8m/s²
Required
final velocity v
Using the equation of motion;
v² = u²-2gH (upward motion of the fish makes g to be negative)
v² = 4.45²-2(9.8)(0.6)
v² = 19.8025-11.76
v² = 8.0425
v = 2.84 m/s
Hence the speed of the fish as it passes a point 0.6 m above the water is 2.84m/s
To get the time, we will use the formula
v = u - gt
2.84 = 4.45 - 9.8t
2.84-4.45 = -9.8t
-1.61 = -9.8t
t = 1.61/9.8
t = 0.164secs
Hence the time taken is 0.164secs
At what final position would you be if you began at initial position of 4m and drove at 30m/s for 20 seconds?
Answer:
604m
Explanation:
30 * 20=600
600 + 4 = 604m
if you double the magnitude of a vector, does it follow that the magnitude of the components double? Give example.
Answer:
The magnitude of the vector clearly doubles if each of its components is doubled.
Explanation:
When the rate of the forward reaction equals the rate of the backward reaction, the system is said to be in
Answer:
equilibrium
Explanation:
bcoz forward reaction equals backward reaction
Consider an underwater spherical air bubble (you can assume the index of refraction of air is 1 and water is 1.33) with radius R. This bubble can act as a lens. What is its focal length?
Answer:
-2R
Explanation:
The focal length can be calculated using this formula
1/f = (n1/n2-1)(1/r1-1/r2)
1/f = (1/1.33-1)(1/r + 1/r)
= -0.133/1.33(2/R)
F = -2R
The focal length is this
What is energy?
A. A change that appears in an object when forced is applied.
B. The property of a body that gives it mess.
C. The amount of heat produced by the body.
D. The ability of an object to undergo change.
E. The ability of a body to move.
Answer:
A
Explanation:
Answer: a or e
Explanation:
Most generation of electricity involves turning turbines to spin coils of metal. However some do not. Give a type of energy that does not involve spinning coils of metal. How does it generate electricity?
Answer:
Photovoltaic Energy
Explanation:
There are other forms of energy that do not involve spinning of metals to generate electricity, one of many such is the Solar Photovoltaic Energy.
Photovoltaic is a form of energy in which sunlight Is directly into electricity. This type of energy is often used to power devices such as small wrist watches, rechargeable lamps, and others. They are connected together to panels that are joined together in arrays to power people or heavy power plants. Photovoltaic power plants are happening to be one of the fastest growing sources of electricity generation around the world.
7.
As the distance from the Sun increases the time to orbit the Sun
a. Increases
b. Decreases
c. Stays the same
As the distance from the Sun increases the time to orbit the Sun increases. (a)
Examples:
-- Nearest planet to the sun: Mercury. Time to orbit the sun: 88 days
-- 3rd planet from the sun: Earth. Time to orbit the sun: 1 year
-- 5th planet from the sun: Jupiter. Time to orbit the sun: 12 years
-- 9th closest object to the sun: Pluto. Time to orbit the sun: 248 years
Another pair of examples:
-- Object in a near Earth orbit: International Space Station
Time to orbit the Earth: 90 minutes
-- Object in a far Earth orbit: the Moon
Time to orbit the Earth: 27.3 days
The reason for all of this is: Two things about orbits.
1). The larger the orbit is, the farther the object has to travel around it.
2). The farther out the object is, the slower it travels in its orbit.
This is simply the way gravity works.
A tennis rackets delivers 500 N on impact with a tennis ball. What is the force of the tennis ball on the racket?
A: this cannot be determined without knowing the mass of the tennis ball
B: the force on the tennis racket is more than 500 N
C: the force on the tennis racket is 500 N
D: the force on the tennis racket is less than 500 N
Answer:
the a.
Explanation:
because if you compute you can't know the distance
Emmett is lifting a box vertically. Which forces are necessary for calculating the total force?
Fp, Ff, and Fg
Fp and FN
Fp, Fg, and Ff
Fp and Fg
Answer:
Answer is D: D.Fp and Fg
Explanation:
I did the quiz some time ago and rember it hope it helps.
Answer:
i just finish its d
Explanation:
A basketball is shot with an initial velocity of 16 m/s at an angle of 55° to the
horizontal. What is the approximate horizontal distance that the ball travels in
1.5 s?
Answer:
approximately 13.77 m
Explanation:
We use the kinematic equations for the horizontal motion which is a uniform non-accelerated motion, and with initial velocity equal to 16 * cos(55). This is described by the equation:
[tex]x=16 * cos(55^o) * t = 9.177 * t[/tex]
then at t = 1.5 seconds, the covered distance becomes:
x = 9.177 * (1.5) = 13.7655 m
Round the answer to the number of decimals that the problem asks.
HELP!!
You are traveling at a speed of 70 miles per hour. How long did it take you to travel 2400 miles?
Answer:
So do 2400 divided by 70. I got 34.285714 and the numbers behind the decimal are repeating. If you round it you get 34.3
To exercise, a man attaches a 4.0 kg weight to the heel of his foot. When his leg is stretched out before him, what is the torque exerted by the weight about his knee, 40 cm away from the weight? Use g = 10 m/s2. A. 160 Nâm B. 16 Nâm C. 1600 Nâm D. 1.6 Nâm
Answer:
B. τ = 16 Nm
Explanation:
In order to find the torque exerted by the weight attached to the heel of man's foot, when his leg is stretched out. We use following formula:
τ = Fd
here,
τ = Torque = ?
F = Force exerted by the weight = Weight = mg
F = mg = (4 kg)(10 m/s²) = 40 N
d = distance from knee to weight = 40 cm = 0.4 m
Therefore,
τ = (40 N)(0.4 m)
B. τ = 16 Nm
The torque exerted by the weight about his knee, 40 cm away from the weight is: D. 1.6 Nm
Given the following data:
Mass = 4.0 kgDistance = 40 cm to m = 0.04 metersAcceleration due to gravity (g) = 10 [tex]m/s^2[/tex].To determine the torque exerted by the weight about his knee, 40 cm away from the weight:
First of all, we would calculate the force being exerted by the weight on his heel.
[tex]Force = mg\\\\Force = 4 \times 10[/tex]
Force = 40 Newton
Mathematically, torque is given by the formula:
[tex]Torque = force \times distance[/tex]
Substituting the given parameters into the formula, we have;
[tex]Torque = 40 \times 0.04[/tex]
Torque = 1.6 Nm
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if it took 3.7 seconds for the sound to reach john how far away was the firework shell when it exploded in kilometers assume that the speed of sound in air is 1,236 km/h
Answer:
1.2703 km
Explanation:
The speed can be calculated using the formula;
Speed (m/s) = distance (m) ÷ time (s)
Based on the information in this question, it took 3.7 seconds for a sound with speed of 1236 km/h to reach John. The distance will be:
Distance = speed × time
However, we need to convert the time in seconds (s) to hour (hr).
1 second = 0.000277778 hour
3.7 seconds = 0.00102778 hours.
Hence,
distance = 1236 × 0.00102778
Distance = 1.2703 km
A flat loop of wire consisting of a single turn of cross-sectional area 8.00 cm2 is perpendicular to a magnetic field that increases uniformly in magnitude from 0.500 G to 1.60 M in 0.99 s. What is the resulting induced current if the loop has a resistance of 1.20 Ω?
Complete Question
A flat loop of wire consisting of a single turn of cross-sectional area 8.00 cm2 is perpendicular to a magnetic field that increases uniformly in magnitude from 0.500 T to 1.60 T in 0.99 s. What is the resulting induced current if the loop has a resistance of [tex]1.20 \ \Omega[/tex]
Answer:
The current is [tex]I = 0.0007 41 \ A[/tex]
Explanation:
From the question we are told that
The area is [tex]A = 8.00 \ cm^2 = 8.0 *10^{-4} \ m^2[/tex]
The initial magnetic field at [tex]t_o = 0 \ seconds[/tex] is [tex]B_i = 0.500 \ T[/tex]
The magnetic field at [tex]t_1 = 0.99 \ seconds[/tex] is [tex]B_f = 1.60 \ T[/tex]
The resistance is [tex]R = 1.20 \ \Omega[/tex]
Generally the induced emf is mathematically represented as
[tex]\epsilon = A * \frac{B_f - B_i }{ t_f - t_o }[/tex]
=> [tex]\epsilon = 8.0 *10^{-4} * \frac{1.60 - 0.500 }{ 0.99- 0 }[/tex]
=> [tex]\epsilon = 0.000889 \ V[/tex]
Generally the current induced is mathematically represented as
[tex]I = \frac{\epsilon}{R }[/tex]
=> [tex]I = \frac{0.000889}{ 1.20 }[/tex]
=> [tex]I = 0.0007 41 \ A[/tex]
A volume of air increases 0.227 m^3 at a net pressure of 2.07 x 10^7 Pa. How much work is done on the air?
Answer:
The work done on the air is 4.699 x 10⁶ Joules
Explanation:
Given;
increase in air volume, ΔV = 0.227 m³
net pressure of the air, P = 2.07 x 10⁷ Pa
The work done on the air is given by;
W = PΔV
Where;
W is the work done on the air
P is the net pressure
ΔV is the increase in air volume
Substitute the given values and solve for work done;
W = (2.07 x 10⁷ Pa) (0.227 m³)
W = 4.699 x 10⁶ Joules
Therefore, the work done on the air is 4.699 x 10⁶ Joules
How are all paths that have a displacement of zero similar?
Answer:
Any situation that has a path that stops at the same position that it started from has a displacement of zero
Explanation:
hope it helps
Rain drops are formed when water vapor _____.
condenses on grass
evaporates
condenses on dust
freezes
Answer:
Condenses on dust
Explanation:
Pls brainiest
Answer:
condenses on dust
Explanation:
Rebecca heated 50mL of water from 0 degrees Celsius to 60 degrees Celsius. How much energy did she use to heat the water? Remember: cal=m*deltaT
(WRITE THE WORK!/Give explanation.)
Answer:
Q = 12540 J
Explanation:
It is given that,
Mass of water, m = 50 mL = 50 g
It is heated from 0 degrees Celsius to 60 degrees Celsius.
We need to find the energy required to heat the water. The formula use to find it as follows :
[tex]Q=mc\Delta T[/tex]
Where c is the specific heat of water, c = 4.18 J/g°C
Put all the values,
[tex]Q=50\times 4.18\times (60-0)\\Q=12540\ J[/tex]
So, 12540 J of energy is used to heat the water.
A 4 kilogram box is at rest on a frictionless floor. A net force of 12 newtons acts on it. What is the acceleration of the box?
Answer:
3 m/s^2Explanation:
Step one:
given
Mass m= 4kg
Force F= 12N
Required
Acceleration the relation between force, acceleration, and mass is Newton's first equation of motion, which says a body will continue to be at rest or uniform motion unless acted upon by an external force
F=ma
a=F/m
a=12/4
a=3 m/s^2
All of the following are categories used to classify problems except problems of
A. arrangement
B. inducing structure
C. transformation
D. schematic completion
Answer:D. schematic completion
Explanation:The rest are actual categoroes
Answer:
schematic completion
Explanation:
got it right on edge
Integrated Concepts A lightning bolt strikes a tree, moving 20.0 C of charge through a potential difference of 1.00×102 MV . (a) What energy was dissipated? (b) What mass of water could be raised from 15ºC to the boiling point and then boiled by this energy? (c) Discuss the damage that could be caused to the tree by the expansion of the boiling steam.
Answer:
a) 2*10^9 J
b) 764.8 kg
Explanation:
Given that
Energy of charge, q = 20 C
Potential difference, ΔV = 1*10^2 MV = 1*10^2 * 10^6 V = 1*10^8 V
a)
To find the energy dissipated, we use the formula
ΔU = qΔV
ΔU = 20 * 1*10^8
ΔU = 2*10^9 J
b)
Change in temperature, ΔT = 100 - 15°
ΔT = 85° C
Change in energy, ΔU = 2*10^9 J
Specific heat of water, C = 4180 j./Kg.K
Latent heat of vaporization, L(v) = 2.26*10^6 J/Kg
Q1 = mcΔT
Q2 = mL(v)
Net energy needed, U = Q1 + Q2
U = mcΔT + mL(v)
U = m (cΔT + L(v))
m = U /[cΔT + L(v)]
Being that we have all the values, we then substitute
m = 2*10^9 / [(4180 * 85) + 2.26*10^6]
m = 2*10^9 / (3.553*10^5 + 2.26*10^6]
m = 2*10^9 / 2.615*10^6
m = 764.8 kg
c)
Having 765 kg of steam at the temperature would have extreme effect on the tree, damaging it permanently. Possibly even blowing it to pieces
Rebecca heated 50mL of water from 0 degrees Celsius to 60 degrees Celsius. How much energy did she use to heat the water? Remember: cal=m*deltaT
(WRITE THE WORK!/Give explanation.)
I’ll give BRAINLIEST.
Answer:
12552 J or 3000 calories
Explanation:
Q = m × c × ∆T
Where;
Q = amount of heat energy (J)
m = mass of water (g)
c = specific heat capacity (4.184 J/g°C)
∆T = change in temperature
For 50mL of water, there are 50g, hence, m = 50g, c = 4.184 J/g°C, initial temperature = 0°C, final temperature = 60°C.
Q = m × c × ∆T
Q = 50 × 4.184 × (60 - 0)
Q = 209.2 × 60
Q = 12552 J
Hence, the amount of heat energy used to heat the water is 12552 J or 3000 calories
an explanation can be consistent but may not be the best explanation for a phenomenon
True
False
Answer:
true
Explanation:
The pharyngeal tonsil is found in the __________________________ .
Answer:
The pharyngeal tonsils are located near the opening of the nasal cavity into the pharynx. When these tonsils become enlarged they may interfere with breathing and are called adenoids.
An engineer is designing a runway for an airport. Of the planes that will use the airport, the
lowest acceleration rate is likely to be 3 m/s2
. The takeoff speed for this plane will be 70 m/s.
Assuming this minimum acceleration, what is the minimum allowed length for the runway?
Answer:
816m
Explanation:
70^2m/s=2(3m/ss)(x)
x=816
A 5 kg ball is moving at a velocity of +2 m/s when it
speeds up to +5 m/s in 7 seconds.
Calculate the acceleration.
Heya!!
For calculate aceleration, lets applicate formula:
[tex]\boxed{a=\frac{V-V_o}{t} }[/tex]
Δ Being Δ
V = Final Velocity = 5 m/s
Vo = Initial Velocity = 2 m/s
a = Aceleration = ? m/s²
t = Time = 7 s
⇒ Let's replace according the formula:
[tex]\boxed{a=\frac{5\ m/s - 2 \ m/s}{7\ s} }[/tex]
⇒ Resolving
[tex]\boxed{ a=0,428\ m/s^{2}}[/tex]
Result:
The aceleration of the object is 0,428 m/s²
Good Luck!!
Answer:
Ksasdasd
Explanation: