Answers
The mass of the solid cylindrical wheel is 71.14 kg .
What is mass?
Measuring inertia, a property of all stuff that cannot be changed. What a body of matter is in essence is the resistance it provides to a change in its speed or position brought about by the application of a force.
What is acceleration ?
Acceleration is the term for the rate of change in speed. In most cases, but not always, acceleration denotes a shift in speed. An object is still moving in a circular path, but the direction of its velocity is changing, thus it is still gaining speed.
final angular speed of wheel is
w = 50 rev/m/n = 50 * 2π/ 60 = 1rev/ min = 2π/ 60 rad/s
w = 10π/6 rad/s
work done w = 590 J
radius of wheel is r = 1.10 m
as we know that work done in rotating of wheel is
W = 1/2 I W²,
Where I is moment of inertia of the wheel
moment of inertia of a solid cylindrical wheel is
I = 1/2 mr², where m is mass
∴ W = Hw/ r²w²
m= 4*590/ (1.10)² ( 10π/6) ² = 4* 590* 36/ 1.21* 100 π²
m= 71.14 kg
Therefore, mass of the solid cylindrical wheel is 71.14 kg .
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Related Questions
A cobalt-60 source with activity 2.60×10-4 Ci is embedded in a tumor that has
mas 0.20 kg. The source emits gamma photons with average energy 1.25 MeV.
Half the photons are absorbed in the tumor, and half escape.
i. What energy is delivered to the tumor per second? [4 marks]
ii. What absorbed dose, in rad, is delivered per second? [2 marks]
iii. What equivalent dose, in rem, is delivered per second if the RBE for
these gamma rays is 0.70? [2 marks]
Page 6 of 7
iv. What exposure time is required for an equivalent dose of 200 rem? [2
marks]
B. A laser with power output of 2.0 mW at a wavelength of 400 nm is projected
onto a Calcium metal. The binding energy is 2.31 eV.
i. How many electrons per second are ejected? [6 marks]
ii. What power is carried away by the electrons? [4 marks]
C. A hypodermic needle of diameter 1.19 mm and length 50 mm is used to
withdraw blood from a patient? How long would it take for 500 ml of blood to be
taken? Assume a blood viscosity of 0.0027 Pa.s and a pressure in the vein of
1,900 Pa. [10 marks]
D. A person with lymphoma receives a dose of 35 gray in the form of gamma
radiation during a course of radiotherapy. Most of this dose is absorbed in 18
grams of cancerous lymphatic tissue.
i. How much energy is absorbed by the cancerous tissue? [2 marks]
ii. If this treatment consists of five 15-minute sessions per week over the
course of 5 weeks and just one percent of the gamma photons in the
gamma ray beam are absorbed, what is the power of the gamma ray
beam? [4 marks]
iii. If the gamma ray beam consists of just 0.5 percent of the photons
emitted by the gamma source, each of which has an energy of 0.03
MeV, what is the activity, in Curies, of the gamma ray source? [4 marks]
E. A water heater that is connected across the terminals of a 15.0 V power supply
is able to heat 250 ml of water from room temperature of 25°C to boiling point
in 45.0 secs. What is the resistance of the heater? The density of water is 1,000
kg/m2 and the specific heat capacity of water is 4,200 J/kg/°C. [10 marks]
Answers
Answer:
A i. E = 9.62 × 10⁻⁷ J/s
ii. The absorbed dose is 4.81 × 10⁻⁶ Gy
iii. The equivalent dose is 3.37 × 10⁻⁴ rem/s
iv. t = 593471.81 seconds
B. i. 4.025 × 10¹⁵/s
ii. 0.512 mW
C. 7218092.2 seconds
D. i. 6.3 × 10⁻¹ J
ii. 1.4 × 10⁻² W
iii. 1.57 × 10³ Curie
E. 0.129 Ω
Explanation:
The given parameters are;
Mass of tumor = 0.20 kg
Activity of Cobalt-60 = 2.60 × 10⁻⁴ Ci
Photon energy = 1.25 MeV
(i) The energy, E, delivered to the tumor is given by the relation;
\(E = \frac{1}{2}\left (Number \, of \, decay / seconds \right )\times \left (Energy \, of \, photon \right )\)
\(E = \frac{1}{2}\left (2.6\times 10^{-4}Ci )\times \left (\frac{3.70\times 10^{10}decays/s}{1 Ci} \right )\times 1.25\times 10^{6}eV\times \frac{1.6\times 10^{-19}J}{1eV}\)
E = 9.62 × 10⁻⁷ J/s
(ii) The equation for absorbed dose is given as follows;
Absorbed dose, D, in Grays Gy = (Energy Absorbed Joules J)/Mass kg
Therefore, absorbed dose = (9.62 × 10⁻⁷ J/s)/( kg) = 4.81 × 10⁻⁶ Gy
1 Gray = 100 rad
4.81 × 10⁻⁷ Gy = 100 × 4.81 × 10⁻⁶ = 4.81 × 10⁻⁴ rad/s
(iii) Equivalent dose, H, is given by the relation;
H = D × Radiation factor, \(w_R\)
∴ H = 0.7 × 4.81 × 10⁻⁴ rad/s = 3.37 × 10⁻⁴ Sv = 3.37 × 10⁻⁴ rem/s
(iv) The exposure time required for an equivalent dose of 200 rem is given as follows;
\(\dot{H} = \dfrac{H}{t}\)
Therefore;
\(t= \dfrac{200}{{3.37 \times 10^{-4}} } = 593471.81 \, s\)
∴ t = 6.9 days
B. The number of electrons ejected is given by the relation;
\(N = \frac{P}{E} = \frac{P \times \lambda}{hc}\)
\(N = \dfrac{2.0 \times 10^{-3} \times 400 \times 10^{-9}}{6.626 \times 10^{-34} \times 3 \times 10^8} = 4.025 \times 10^{15}/s\)
(ii) The power carried by the electron
The energy carried away by the electrons is given by the relation;
\(KE_e = hv - \Phi\)
\(KE_e = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{400 \times 10^{-9}} - 2.31 \times \frac{1.6 \times 10 ^{-19} }{1}\)
\(KE_e = 4.9695 \times 10^{-19} - 3.696 \times 10 ^{-19} = 1.2735 \times 10^{-19} J\)
Power, P\(_e\), carried away by the electron = 4.025 × 10¹⁵ × 1.2735 × 10⁻¹⁹ = 0.512 mW
C. The given parameters are;
d = 1.19 mm, ∴ r = 1.19/2 = 0.595 × 10⁻³ m
l = 50 mm = 5 × 10⁻³ m
V = 500 ml = 5 × 10⁻⁴ m³
η = 0.0027 Pa
p = 1,900 Pa.
\(\dfrac{V}{t} = \dfrac{\pi }{8} \times \dfrac{P/l}{\eta } \times r^4\)
\(t = \dfrac{8\times \eta\times V\times l }{\pi \times P \times r^4}\)
\(t = \dfrac{8\times 0.0027 \times 5 \times 10^{-4} \times 5 \times 10^{-2} }{\pi \times 1900 \times (0.595 \times 10^{-4} )^4}\)
t = 7218092.2 seconds
D) i. Energy absorbed is given by the relation;
E = m×D
Where:
D = 35 Gray = 35 J/kg
m = 18 g = 18 × 10⁻³ kg
∴ E = 35 × 18 × 10⁻³ = 6.3 × 10⁻¹ J
ii. Total time for treatment = 15 × 5 = 75 minutes
Energy absorbed = 6.3 × 10⁻¹ × 100 = 63 J
Power = Energy(in Joules)/Time (in seconds)
∴ Power = 63/(75×60) = 1.4 × 10⁻² W
iii. Whereby the power is provided by 0.5% of the photons emitted by the source, we have;
\(P_{source}= \frac{P_{beam}}{0.005} =\frac{0.0014}{0.005} =0.28 \, W\)
1 MeV = 1.60218 × 10⁻¹³ J
0.03 MeV = 0.03 × 1.60218 × 10⁻¹³ J = 4.80654 × 10⁻¹⁵ J/photon
Therefore, the number of disintegration per second = 0.28 J/s ÷ 4.80654 × 10⁻¹⁵ J/photon = 5.83 × 10¹³ disintegrations per second
1 Curie = 3.7 × 10¹⁰ disintegrations per second
Hence, 5.83 × 10¹³ disintegrations per second = (5.83 × 10¹³)/(3.7 × 10¹⁰) Curie
= 1.57 × 10³ Curie
E. The parameters given are;
Density of water = 1000 kg/m³
Volume of water = 250 ml = 0.00025 m³
Initial temperature, T₁, = 25°C
Final temperature, T₂, = 100°C
Change in temperature, ΔT = 100 - 25 = 75°
Specific heat capacity of the water = 4200 J/kg/°C
Mass of water = Density × Volume = 1000 × 0.00025 = 0.25 kg
∴ Heat supplied = 4200 × 0.25 × 75 = 78,750 J
Time to heat the water = 45.0 sec
Therefore, power = Energy/time = 78750/45 = 1750 W
The formula for electrical power = I²R =VI = V²/R
Therefore, where V = 15.0 V, we have;
15²/R = 1750
R = 15²/1750 = 0.129 Ω.
The resistance of the heater = 0.129 Ω.
An object has mass M (uppercase) and is located at the origin of the coordinate system. A second object has mass m (lowercase) and is located at a distance r from the origin. This exercise explores the potential energy of the two-mass system. A spherical coordinate system is appropriate, but only the radial direction, r^, needs to be considered.
Enter a vector expression for the force, F⃗ , acting on mass m . The expression must be valid for all values of the radial coordinate, 0 0, and the zero of potential energy is taken as limr→∞UG=0.
Answers
Mass of the second object located at a distance r from the origin, r^ is the unit vector in the radial direction, and the negative sign indicates that the force
What is a system ?The System can refer to a set of interacting or interdependent components forming an integrated whole. The term can be applied to various fields, including physics, engineering, biology, and social sciences, among others. In physics, a system typically refers to a collection of objects or particles that are studied together, often with the goal of understanding the behavior of the system as a whole. In engineering, a system can refer to a group of components that work together to perform a specific function, such as an electrical power grid or an automobile engine. In biology, a system can refer to an organism or group of organisms that interact with their environment, such as an ecosystem or the human body.
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The inner cylinder of a long, cylindrical capacitor has radius r and linear charge density +λ. It is surrounded by a coaxial cylindrical conducting shell with inner radius r and linear charge density -λ. (a) What is the energy density in the region between the conductors at a distance r from the axis?b) Integrate the energy density calculated in part (a) over the volume between the conductors in a length L of the capacitor to obtain the total electric-field energy per unit length
Answers
Hi there!
a)
We can begin by using the equation for energy density.
\(U = \frac{1}{2}\epsilon_0 E^2\)
U = Energy (J)
ε₀ = permittivity of free space
E = electric field (V/m)
First, derive the equation for the electric field using Gauss's Law:
\(\Phi _E = \oint E \cdot dA = \frac{Q_{encl}}{\epsilon_0}\)
Creating a Gaussian surface being the lateral surface area of a cylinder:
\(A = 2\pi rL\\\\E \cdot 2\pi rL = \frac{Q_{encl}}{\epsilon_0}\\\\Q = \lambda L\\\\E \cdot 2\pi rL = \frac{\lambda L}{\epsilon_0}\\\\E = \frac{\lambda }{2\pi r \epsilon_0}\)
Now, we can calculate the energy density using the equation:
\(U = \frac{1}{2} \epsilon_0 E^2\)
Plug in the expression for the electric field and solve.
\(U = \frac{1}{2}\epsilon_0 (\frac{\lambda}{2\pi r \epsilon_0})^2\\\\U = \frac{\lambda^2}{8\pi^2r^2\epsilon_0}\)
b)
Now, we can integrate over the volume with respect to the radius.
Recall:
\(V = \pi r^2L \\\\dV = 2\pi rLdr\)
Now, we can take the integral of the above expression. Let:
\(r_i\) = inner cylinder radius
\(r_o\) = outer cylindrical shell inner radius
Total energy-field energy:
\(U = \int\limits^{r_o}_{r_i} {U_D} \, dV = \int\limits^{r_o}_{r_i} {2\pi rL *U_D} \, dr\)
Plug in the equation for the electric field energy density and solve.
\(U = \int\limits^{r_o}_{r_i} {2\pi rL *\frac{\lambda^2}{8\pi^2r^2\epsilon_0}} \, dr\\\\U = \int\limits^{r_o}_{r_i} { L *\frac{\lambda^2}{4\pi r\epsilon_0}} \, dr\\\)
Bring constants in front and integrate. Recall the following integration rule:
\(\int {\frac{1}{x}} \, dx = ln(x) + C\)
Now, we can solve!
\(U = \frac{\lambda^2 L}{4\pi \epsilon_0}\int\limits^{r_o}_{r_i} { \frac{1}{r}} \, dr\\\\\\U = \frac{\lambda^2 L}{4\pi \epsilon_0} ln(r)\left \| {{r_o} \atop {r_i}} \right. \\\\U = \frac{\lambda^2 L}{4\pi \epsilon_0} (ln(r_o) - ln(r_i))\\\\U = \frac{\lambda^2 L}{4\pi \epsilon_0} ln(\frac{r_o}{r_i})\)
To find the total electric field energy per unit length, we can simply divide by the length, 'L'.
\(\frac{U}{L} = \frac{\lambda^2 L}{4\pi \epsilon_0} ln(\frac{r_o}{r_i})\frac{1}{L} \\\\\frac{U}{L} = \boxed{\frac{\lambda^2 }{4\pi \epsilon_0} ln(\frac{r_o}{r_i})}\)
And here's our equation!
Work Energy Theorem QUESTION: A 1200kg automobile is moving at 25m/s along level ground. What is the initial KE of the automobile? What is the final KE of the automobile? What is the change in KE of the automobile?What is the work done?
Answers
(a) The initial kinetic energy (KE) of the automobile is 375,000 J
(b) The final KE will also be 375,000 J.
(c) The work done on the automobile is zero
What is the initial kinetic energy?The initial kinetic energy (KE) of the automobile can be found using the formula:
KE = 1/2mv²
where;
m is the mass of the automobile and v is its velocity.KE = 1/2 x 1200 kg x (25 m/s)²
KE = 375,000 J
The final KE of the automobile will be the same as the initial KE if the velocity remains constant. However, if there is a change in velocity, the final KE can be found using the same formula as above.
The change in KE can be found by subtracting the initial KE from the final KE, or by using the work-energy theorem:
ΔKE = W
where;
ΔKE is the change in kinetic energy and W is the work done.Assuming there is no external work done on the automobile, the change in KE will be zero.
Therefore, the final KE will also be 375,000 J.
The work done on the automobile can be found using the work-energy theorem:
W = ΔKE = 0 J (since there is no change in KE)
Therefore, the work done on the automobile is zero.
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which diagram best shows the field lines around two bar magnets that repel each other?
Answers
Answer:
Explanation:
I hope this helped
PLEASE HELP IM HEING TIMED
Answers
Answer:
Uh, I could be wrong but doesn’t it mean that the wave and particle are reacting together to make light? I think it’s something like that... I hope this helps!
Calculate The pressure produced by a force of 392 N acting on an area of 8.0 m^2
Answers
Answer:
Explanation:
500-0.05=499.95
500-100+5=405
100+100=10000
A skier starts from rest and slides 9.0 m down a slope in 3.0 s. Assuming constant acceleration, in
what time after starting will the skier acquire a speed of 24 ms?
Answers
Answer:
3 seconds
Explanation:
initial velocity = 0ms
final velocity = 24ms
time(1) = 3s
acceleration = ?
time (2) =?
solution:-
to find acceleration,
a =( vf - vi ) / t
= 24-0 /3
= 24/ 3
a= 8ms^-2
To find time(2)
t= (vf - vi )/ a
= 24- 0 / 8
= 24/ 8
= 3seconds
A well-coated structure is defined as A) 95% or better B) 90% or better C) 99% or better D) 93% or better
Answers
Answer and Explanation:
A well-coated structure is defined as having a coating that meets a certain standard of quality. The answer to this particular question depends on the specific criteria being used to evaluate the coating. This would typically require a coating coverage of 90% or better, if not higher.
However, in general, a well-coated structure would typically refer to a surface that has been thoroughly and evenly covered with a coating material such as paint or varnish. This ensures that the underlying material is protected from environmental factors such as moisture and UV radiation. In addition, a well-coated structure can also improve the overall appearance of the surface, making it more aesthetically pleasing. Regarding the options provided in the question, the answer would depend on the specific criteria being used to evaluate the coating. However, it is safe to say that a well-coated structure would require a high level of coating coverage, with minimal areas left uncovered or with an uneven application. This would typically require a coating coverage of 90% or better, if not higher. Ultimately, the specific answer would depend on the standards and expectations set by the evaluating body.
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A well-coated structure is defined as having a coating that meets a certain standard of quality. The answer to this particular question depends on the specific criteria being used to evaluate the coating. This would typically require a coating coverage of 90% or better, if not higher.
However, in general, a well-coated structure would typically refer to a surface that has been thoroughly and evenly covered with a coating material such as paint or varnish. This ensures that the underlying material is protected from environmental factors such as moisture and UV radiation. In addition, a well-coated structure can also improve the overall appearance of the surface, making it more aesthetically pleasing.
Regarding the options provided in the question, the answer would depend on the specific criteria being used to evaluate the coating. However, it is safe to say that a well-coated structure would require a high level of coating coverage, with minimal areas left uncovered or with an uneven application. This would typically require a coating coverage of 90% or better, if not higher. Ultimately, the specific answer would depend on the standards and expectations set by the evaluating body
Ebo throws the ball again at an initial velocity of 18 m/s but this time at an angle of 53° to the ground. a) Work out the velocity in component vector form. b) Why will the ball spend a longer time in the air than it did before. c) Calculate the range of the ball.
Answers
The velocity is 16.52 m/s, and the component vector form of the velocity is vcosθ.
What is an easy way to define velocity?A moving object's position and speed are primarily determined by its velocity. It can be defined as the amount of space an object covers in a given amount of time. Velocity is the measure of how far an object travels in a predetermined amount of time.
velocity in vector form= vcosθ
vector form= vcosθ
vector form=18cos53
v=16.52 m/s
How quickly an object changes its location is quantified by a vector number called velocity. Consider a person who moves quickly while taking two steps forward and two steps back while remaining in one location. Velocity is a vector quantity. Therefore, velocity is cognizant of direction. The direction must be taken into account when determining an object's velocity. A speed of 55 mph offers insufficient information. The direction must be used to properly describe the object's velocity. Simply put, the direction of the velocity vector indicates the motion of an object.
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what is a U.S state you can visit a desert in
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What evidence did Wegener NOT use to support his idea of continental drift?
A. Mountain ranges on different continents lined up when coastlines were matched up.
B. Fossils of plants and animals in climates where their survival would have been impossible. C. The thickness of layers of ice in the Antarctic.
D. Rock strata on different continents lined up when coastlines were matched up.
Answers
The evidence that Wegener did NOT use to support his idea of continental drift is "the thickness of layers of ice in the Antarctic.
option C
What is Wegener's primary evidence for continental drift?Wegener's primary evidence for continental drift included the fit of the coastlines of different continents, the distribution of fossils across different continents, and the alignment of rock strata on different continents.
So the thickness of layers of ice in the Antarctic, was not used by Wegener to support his idea of continental drift. While this evidence is important for supporting the theory of glaciation, it is not relevant to the theory of continental drift.
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If Maggie is walking at 1.4 m/s and accelerates at 0.20 m/s^2. What is her final velocity at the end of 100.0m
Answers
Answer:
6.478m/s
Explanation:
There are some assumptions to be made. To answer this problem, I assume that Maggie is walking in a straight line and that she initially starts off with a velocity of 1.4m/s.
Maggie initially walks at a speed of 1.4m/s. We are also told that the acceleration is 0.2m/s^2 and that she walked a total of 100m. Therefore our parameters are:
\(v_i=1.4\frac{m}{s}\\a=0.2\frac{m}{s^2}\\d=100m\\v_f=?\)
Looking up kinematic equations that contains these parameters are
\(v_f^2=v_i^2+2ad\\v_f^2=(1.4\frac{m}{s})^2+2*(0.2\frac{m}{s^2})(100m)\\v_f^2=41.96\frac{m^2}{s^2}\\v_f=6.478\frac{m}{s}\)
Fill in the blanks using the following words: Solid, Liquids, Gasses, More, Less, Gas, Fluids, Higher, Lower, Sun, Radiation, Conductors
● Radiation transfers heat best through _________ because there is _______ space between the particles.
● Conduction transfers heat best through _______ because there is space between the particles.
● Convection transfers heat best through _______ which includes _______ and ______.
● Heat always moves from _________ temperature to _________ temperature.
● Heat from ______ travels to earth by ___________.
● Solids that transfer heat well ate known as _____________.
Answers
Answer:
Blank 1: Gasses
Blank 2: More
Blank 3: Solids
Blank 4: Fluids
Blank 5: Liquid
Blank 6: Gas
Blank 7: Higher
Blank 8: Lower
Blank 9: Sun
Blank 10: Radiation
Blank 11: Conductors
P.S. order of answers does not matter between Blank 5 and 6.
The nuchal ligament in a horse supports the weight of the horse's head. This ligament is much more elastic than a typical ligament, stretching from 15% to 45% longer than its resting length as a horse's head moves up and down while it runs This stretch of the ligament stores energy, making locomotion more efficient. Measurements on a segment of ligament show a linear stress-versus-strain relationship until the strain approaches 0.80. Smoothed data for the stretch are shown in (Figure 1).
The segment of ligament tested has a resting length of 40 mm.
How long is the ligament at a strain of 0.60?
Answers
Answer:
Explanation:The nuchal ligament in a horse supports the weight of the horse's head. This ligament is much more elastic than a typical ligament, stretching from 15% to 45% longer than its resting length as a horse's head moves up and down while it runs This stretch of the ligament stores energy, making locomotion more efficient. Measurements on a segment of ligament show a linear stress-versus-strain relationship until the strain approaches 0.80. Smoothed data for the stretch are shown in (Figure 1).
The segment of ligament tested has a resting length of 40 mm.
How long is the ligament at a strain of 0.60?
Unfortunately, there is no image or data provided for "Figure 1" in the given question. Therefore, it is impossible to determine the length of the ligament at a strain of 0.60. Can you please provide more information or context for this question?
Hyper Hamza
The nuchal ligament in a horse supports the weight of the horse's head. This ligament is much more elastic than a typical ligament, stretching from 15% to 45% longer than its resting length as a horse's head moves up and down while it runs This stretch of the ligament stores energy, making locomotion more efficient. Measurements on a segment of ligament show a linear stress-versus-strain relationship until the strain approaches 0.80. Smoothed data for the stretch are shown
The segment of ligament tested has a resting length of 40 mm.
How long is the ligament at a strain of 0.60?
Assuming we have the data for the stress-strain relationship of the nuchal ligament in a horse, we can use the linear relationship between stress and strain to determine the length of the ligament at a strain of 0.60.
Let's say that the stress-strain data for the nuchal ligament is given by the equation:
stress = m * strain + c
where m is the slope of the line and c is the y-intercept.
Since the stress-strain relationship is linear, we can determine the slope and y-intercept using two points on the line. We can use the resting length of the ligament (40 mm) and the strain at which the linear relationship breaks down (0.80) to determine the slope and y-intercept.
At a strain of 0, the length of the ligament is the resting length (40 mm). At a strain of 0.80, the length of the ligament is:
Length at strain 0.80 = resting length * (1 + strain) = 40 mm * (1 + 0.80) = 72 mm
So we have two points on the line: (0, 40) and (0.80, 72). Using these two points, we can calculate the slope:
m = (y2 - y1) / (x2 - x1) = (72 - 40) / (0.80 - 0) = 90
And the y-intercept:
c = y1 - m * x1 = 40 - 90 * 0 = 40
Now we can use the equation of the line to find the length of the ligament at a strain of 0.60:
stress = m * strain + c
stress = 90 * 0.60 + 40
stress = 94
At a strain of 0.60, the stress in the ligament is 94. We can use this stress value and the slope of the line to find the corresponding length of the ligament:
stress = m * strain + c
94 = 90 * 0.60 + 40
94 = 94
Therefore, the length of the ligament at a strain of 0.60 is 64 mm.
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A mass of 0.24 kg is attached to a spring and set into oscillation on a horizontal frictionless surface. The simple harmonic motion of the mass is described by x) (0.46 m)cos (12 rad/s)t. Determine the following (a) Amplitude of oscillation for the oscillating mass (b) Period of the oscillation for the oscillating mass (c) Force constant (spring constant) for the spring. (d) Position of the mass after it has been oscillating for one half a period. (e) Time, it takes the mass to get to the position x=-010 m.
Answers
a) Amplitude A of oscillation for the oscillating mass = 0.46 m
b) Period of the oscillation for the oscillating mass is 0.52 s
c) Spring constant k is 34.56 N/m
d) The mass's position is zero after oscillating for half a time.
e) Time, it takes the mass to get to the position x= - 0.10 m is 0.1125 s.
The mass attached to the spring is 0.24 kg.
The simple harmonic motion of the mass is given as x = (0.46 m) [cos(12 rad/s)t]
The general equation of x(t) is known to be,
x(t) = A cos ωt
where,
A is amplitude
ω is angular frequency
t is time
Following that, let's compare the presented equation to the general equation:
(a) The amplitude of oscillation:
A = 0.46 m
(b) The period of oscillation is,
T = 2π √(m/k) = 2π √(0.24/34.56) = 2π √(0.0069) = 2π ×0.08 = 0.52 s
(c) Spring constant k is given by,
ω = √(k/m)
ω² = k/m
So, k = m ω² = 0.24 × 12² = 34.56 N/m
(d) After one half of the period, position is
x(t) = 0.46 cos(π/2) = 0.
(e) Time required at x = - 0.10 m
-0.10 = 0.46 cos 12t
cos 12t = -0.217
12t = cos⁻¹(-0.217)
t = 1.35/12 = 0.1125 s
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The force of attraction between things that have mass called_____. Someone please help
Answers
Why do astronomers use spectroscopes to analyze light from distant objects?
Answers
Water has a heat capacity of 4.184 J/g °C. If 50 g of water has a temperature of 30ºC and a piece of hot copper is added to the water causing the temperature to increase to 70ºC. What is the amount of heat absorbed by the water?
Answers
The amount of heat absorbed by the water will be 8368 J.
What are heat gain and heat loss?Heat gain is defined as the amount of heat required to increase the temperature of a substance by some degree of Celsius. While heat loss is inverse to heat gain.
It is given by the formula as ;
\(\rm Q= mcdt\)
The given data in the problem is;
Equilibrium temperature = 30°C.
mass of water = 50 g ,
Temperature change = 70ºC
The specific heat of water =4.184 J//g °C
The amount of heat absorbed by the water is;
\(\rm Q= mcdt \\\\ Q=50 \times 4.184 \times (70^0 -30^0)C\\\\ Q= 8368 J\)
Hence, the amount of heat absorbed by the water will be 8368 J.
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In a race, Usain Bolt accelerates at
1.99 m/s2 for the first 60.0 m, then
decelerates at -0.266 m/s2 for the final
40.0 m. What was his final velocity? (Unit = m/s)
Answers
HELP!!! 100 POINTSSS!!!
Answers
Answer:
3
Explanation:
From the question given above, the following data were obtained:
Length (L) = 3 m
Height (H) = 1 m
Mechanical advantage (MA) =.?
The ideal mechanical advantage for the system can be obtained as follow:
MA = L/H
MA = 3/1
MA = 3
Therefore, the ideal mechanical advantage for the system is 3
Answer:
Explanation:
Mechanical Advantage is the ratio of Input Distance / Output Distance.
In this case, the output distance is to bring up the cart by 1m.
The input distance is to push the cart by 3m.
Assume an ideal machine, MA = 3/1 = 3
While in Earth’s orbit, an 80-kg astronaut carrying a 20-kgtool kit is initially drifting toward a stationary (to her) spaceshuttle at a speed of 2 m/s. What is her final speed if shethrows the tool kit toward the shuttle with a speed of 6 m/s asseen from the shuttle.
Please show a correct solution to this problem so that I canrate you a lifesaver.
Answers
The final speed of the astronaut in this scenario will be 1 m/s toward the shuttle.
This can be calculated by using the conservation of momentum - the momentum of the astronaut plus the momentum of the tool kit before the throw must equal the momentum of the astronaut plus the tool kit after the throw.
Momentum before the throw = (80 kg)(2 m/s) + (20 kg)(0 m/s) = (160 kg m/s) Momentum after the throw = (80 kg)(v) + (20 kg)(6 m/s)
Solving for v gives us v = 1 m/s. Therefore, the astronaut's final speed is 1 m/s toward the shuttle.
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There are 9,000 houses in Elizabeth's town. Last summer, 7,020 of the houses were for sale. What percentage of the houses in the town were for sale last summer?
Answers
78% of the houses in Elizabeth's town were for sale last summer.
To find the percentage of houses that were for sale last summer in Elizabeth's town, we need to divide the number of houses for sale by the total number of houses and then multiply by 100.
The total number of houses in Elizabeth's town is given as 9,000. Last summer, 7,020 houses were for sale. To calculate the percentage, we divide 7,020 by 9,000:
Percentage = (7,020 / 9,000) * 100
Simplifying the calculation:
Percentage = 0.780 * 100
Percentage = 78.0%
To explain further, we take the number of houses for sale (7,020) and divide it by the total number of houses (9,000) to get the ratio of houses for sale to the total. Multiplying this ratio by 100 gives us the percentage. In this case, 78.0% indicates that a significant portion of the houses in Elizabeth's town were available for sale during the summer.
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What is the concentration of molecular oxygen (O2) in mol/L on a June day in Toronto when atmospheric pressure is 1.0 atm and the temperature is 25 oC? (Remember O2 is present at a concentration of 21% (or 21pph) in the atmosphere).
Answers
Answer:
The concentration of mole evil at oxygen on that day is 0.00858 mol/L
Explanation:
Here, we want to calculate the concentration of molecular oxygen
The pressure on that day is 1.0 atm
Since oxygen is at a concentration of 21%, the pressure of oxygen will be 21/100 * 1 = 0.21 atm
Now let’s calculate the concentration;
From Ideal gas law;
PV = nRT
This can be written as;
P/RT = n/V
The term n/V refers to concentration;
Let’s make substitutions now;
P = pressure = 0.21 atm
R = molar gas constant = 0.0821 L•atm/mol•k
T = temperature = 25 = 25 + 273.15 = 298.15 K
Substituting these values, we have;
n/V = C = 0.21/(0.0821 * 298.15) = 0.00858 mol/L
The current in a lightning bolt is 2.6 x 105
A.
How much charge passes through a cross-
sectional area of the lightning bolt in 4.1 s?
Answer in units of C.
Answers
Answer:
The right solution is "\(10.66\times 10^5 \ C\)". A further explanation is given below.
Explanation:
The given values are:
Current,
I = \(2.6\times 10^5\)
Time,
t = 4.1 s
As we know,
⇒ \(Q=I\times t\)
On substituting the given values, we get
⇒ \(=2.6\times 10^5\times 4.1\)
⇒ \(=1,066,000 \ C\)
⇒ \(=10.66\times 10^5 \ C\)
HELP!!! how does gravity affect how objects fall to the ground
Answers
Answer:
c
Explanation:
Answer:
When objects fall to the ground, gravity causes them to accelerate. Acceleration is a change in velocity, and velocity, in turn, is a measure of the speed and direction of motion. Gravity causes an object to fall toward the ground at a faster and faster velocity the longer the object falls.
Explanation:
A solid ball (I = 2/5 MR^2) has a mass of 87.0 g and a diameter of 17.0 cm. It is initially at reston a long inclined plane that makes an angle of 25° with respect to the horizontal. If theball rolls down the slope without slipping, calculate the speed of the ball after it has rolled3.00 m down the slope.
Answers
Given,
The moment of inertia of the solid ball, I=(2/5)MR²
The mass of the ball, m=87.0 g=0.087 kg
The diameter of the ball, d=17.0 cm=0.17 m
The initial velocity of the ball, u=0 m/s
The angle of inclination of the slope, θ=25°
The distance covered by the ball, s=3.00 m
The acceleration of the ball is given by,
\(a=g\sin \theta\)Where g is the acceleration due to gravity.
On substituting the known values,
\(\begin{gathered} a=9.8\times\sin 25^{\circ} \\ =4.14\text{ m/s}^2 \end{gathered}\)From the equation of the motion,
\(v^2=u^2+2as\)Where v is the speed of the ball after it has rolled 3.00 m
On substituting the known values.
\(\begin{gathered} v^2=0+2\times4.14\times3.00 \\ \Rightarrow v=\sqrt[]{24.84} \\ v=4.98\text{ m/s} \end{gathered}\)Thus the speed of the ball after it rolls 3.00 m down the slope is 4.98 m/s
The speed of light is 3×10^8 meters per second, which means that light can travel 300 million meters in just one second. How far can light travel in one minute?
Answers
Answer:
(1.8 × 10^9) meters in one minute
Explanation:
To determine how far light can travel in one minute, we need to multiply its speed by the number of seconds in a minute.
The speed of light is 3 × 10^8 meters per second.
There are 60 seconds in a minute.
Therefore, the distance light can travel in one minute is:
Distance = Speed × Time
Distance = (3 × 10^8 meters per second) × (60 seconds)
Calculating this, we get:
Distance = 3 × 10^8 meters/second × 60 seconds
Distance = 18 × 10^8 meters
Distance = 1.8 × 10^9 meters
So, light can travel approximately 1.8 billion (1.8 × 10^9) meters in one minute.
How is newton's third law different from newton's first law
Answers
Third law states that for every action there is an equal and opposite reaction
Hope that’s helps
Answer : Newton's first law of motion (also known as Law of Inertia) describes about the tendency of an object to resist change from its state of rest / motion / direction (inertia).
Whereas, Newton's third law of motion tells us that forces exists in pairs. It talks about the action & reaction forces, which are equal in magnitude but opposite in direction.
In this way, Newton's third law of motion is different from Newton's first law of motion.
Energy can be transferred from one _______ to another and one __________ to another. *
Answers
Answer:
Source, form
Explanation:
Answer:
Explanation:
Energy can be transferred from one _source_ to another and one __form____ to another.