Answer:
a) v=1.81 m/s; B) v=4.95 m/s
Explanation:
using momentum conservation
m1v1+m2v1=m1v2+m2v2
A)
5.2*672+700*0=5.2*428+700v2
The initial velocity of the block is 0, solving for v2
v2=1.81 m/s
B) Now both the bullet and the block travel together
m1v1+m2v1=(m1+m2)v2
5.2*672+700*0=(5.2+700)v
v=4.95 m/s
a. To find the resulting speed of the block after the bullet strikes it, we can use the law of conservation of momentum. b. The speed of the bullet-block center of mass can be found by calculating the weighted average of the speeds of the bullet and block.
Explanation:a. To find the resulting speed of the block, we can use the law of conservation of momentum. The momentum before the collision is equal to the momentum after the collision.
The initial momentum of the bullet is given by m1 * v1, and the final momentum is given by (m1 + m2) * vf, where m1 is the mass of the bullet, v1 is the initial velocity of the bullet, m2 is the mass of the block, and vf is the final velocity of the bullet and block combined.
Using these values, we can solve for vf and then find the resulting speed of the block.
b. The speed of the bullet-block center of mass can be found by calculating the weighted average of the speeds of the bullet and block. Since the mass of the bullet is much smaller than the mass of the block, the center of mass will be closer to the speed of the block.
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A cylinder is 0.10 m in radius and 0.20 in length. Its rotational inertia, about the cylinder axis on which it is mounted, is 0.020 kg m2. A string is wound around the cylinder and pulled with a force of 1.0 N. The angular acceleration of the cylinder is:
Answer:[tex]5 rad/s^2[/tex]
Explanation:
Given
Radius of cylinder r=0.1 m
Length L=0.2 in.
Moment of inertia I=0.020 kg-m^2
Force F=1 N
We Know Torque is given by
[tex]Torque =I\alpha =F\cdot r[/tex]
where [tex]\alpha =angular\ acceleration[/tex]
[tex]I\alpha =F\cdot r[/tex]
[tex]0.02\cdot \alpha =1\cdot 0.1[/tex]
[tex]\alpha =5 rad/s^2[/tex]
The angular acceleration of the cylinder is 5 rad/s².
To calculate the angular acceleration of the cylinder, we use the formula of torque below.
What is torque?Torque is the force that causes a body to rotate about an axis.
Formula:
Iα = Fr........... Equation 1Where:
I = rotational initialα = Angular accelerationr = Radius of the cylinderF = Forceθ = Angle.
Make α the subject of the equation
α = RFsinθ/I............. Equation 2From the question,
Given:
r = 0.1 mF = 1.0 NI = 0.020 kgm²θ = 90° ( Since the rope is tangental to the side of the cylinder).Substitute these values into equation 2
α = (1.0×0.1×sin90°)/(0.02)α = 5 rad/s²Hence, The angular acceleration of the cylinder is 5 rad/s².
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An electron that has an instantaneous velocity of ???? = 2.0 × 106 m ???? ???? + 3.0 × 106 m ???? ???? is moving through the uniform magnetic field ???? = 0.030T ???? − 0.15T ???? . (a) Find the force on the electron due to the magnetic field (b) Repeat your calculation for a proton having the same velocity.
Explanation:
It is given that,
Velocity of the electron, [tex]v=(2\times 10^6i+3\times 10^6j)\ m/s[/tex]
Magnetic field, [tex]B=(0.030i-0.15j)\ T[/tex]
Charge of electron, [tex]q_e=-1.6\times 10^{-19}\ C[/tex]
(a) Let [tex]F_e[/tex] is the force on the electron due to the magnetic field. The magnetic force acting on it is given by :
[tex]F_e=q_e(v\times B)[/tex]
[tex]F_e=1.6\times 10^{-19}\times [(2\times 10^6i+3\times 10^6j)\times (0.030i-0.15j)][/tex]
[tex]F_e=-1.6\times 10^{-19}\times (-390000)(k)[/tex]
[tex]F_e=6.24\times 10^{-14}k\ N[/tex]
(b) The charge of electron, [tex]q_p=1.6\times 10^{-19}\ C[/tex]
The force acting on the proton is same as force on electron but in opposite direction i.e (-k). Hence, this is the required solution.
Mimas, a moon of Saturn, has an orbital radius of 1.62 × 108 m and an orbital period of about 23.21 h. Use Newton’s version of Kepler’s third law and these data to find the mass of Saturn. Answer in units of kg.
Answer:
[tex]3.60432\times 10^{26}\ kg[/tex]
Explanation:
a = Orbital radius = [tex]1.62\times 10^8\ m[/tex]
T = Orbital period = 23.21 hours
G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²
From Kepler's third law we get
[tex]M=\frac{4\pi^2a^3}{GT^2}\\\Rightarrow M=\frac{4\pi^2\times (1.62\times 10^8)^3}{6.67\times 10^{-11}\times (23.21\times 3600)^2}\\\Rightarrow M=3.60432\times 10^{26}\ kg[/tex]
From the given data the mass of Saturn is [tex]3.60432\times 10^{26}\ kg[/tex]
Final answer:
To find the mass of Saturn, we can use Newton's version of Kepler's third law. Given the orbital radius and period of Saturn's moon Mimas, we can use the equation p² = a³ to calculate the mass of Saturn. Using the provided data, the mass of Saturn is approximately 5.69 x 10^26 kg.
Explanation:
To find the mass of Saturn, we can use Newton's version of Kepler's third law. Kepler's third law states that the square of the orbital period of a planet or moon is proportional to the cube of its semimajor axis. We can use the equation p² = a³ to solve for the mass of Saturn. Given that the orbital radius of Mimas, a moon of Saturn, is 1.62 × 108 m and its orbital period is about 23.21 h, we can use this data to find the mass of Saturn.
First, convert the orbital period to seconds: 23.21 h = 23.21 x 60 x 60 s = 83,556 s. Next, plug the values into the equation p² = a³: (83,556 s)² = (1.62 × 10^8 m)³. Solving for a³, we get a³ = (83,556 s)² / (1.62 × 10^8 m)³ = 38.3 x 10^18 s²/m³.
Now, rearrange the equation to solve for the mass of Saturn: mass of Saturn = (a³ x 4π²) / (G), where G is the gravitational constant. Plug in the known values: mass of Saturn = (38.3 x 10^18 s²/m³ x 4π²) / (6.67 x 10^-11 N m²/kg²). Calculating this, we get a mass of Saturn of approximately 5.69 x 10^26 kg.
A playground merry-go-round has a radius of 3.0 m and arotational inertia of 600 kg ⋅ m2. It
is initially spinning at 0.80 rad/s when a 20-kg child crawlsfrom the center to the rim. When
the child reaches the rim the angular velocity of themerry-go-round is:_________
A) 0.61 rad/s
B) 0.73 rad/s
C) 0.80 rad/s
D) 0.89 rad/s
E) 1.1rad/s
Answer:
Final angular velocity, [tex]\omega_f=0.61\ rad/s[/tex]
Explanation:
Given that,
Radius of merry- go- round, r = 3 m
Inertia of the merry- go- round, [tex]I=600\ kg-m^2[/tex]
Angular speed, [tex]\omega=0.8\ rad/s[/tex]
Mass, m = 20 kg
Let I' is the new rotational inertia of the merry- go- round. Here, the angular momentum of the system remains conserved. So,
[tex]L_f=L_o[/tex]
[tex]I_f\omega_f=I_o\omega_o[/tex]
[tex]\omega_f=(\dfrac{I}{I+mr^2})\omega_o[/tex]
[tex]\omega_f=(\dfrac{600}{600+20\times 3^2})\times 0.8[/tex]
[tex]\omega_f=0.61\ rad/s[/tex]
So, the angular velocity of the merry-go-round is 0.61 rad/s. So, the correct option is (A). Hence, this is the required solution.
What is evaporation? Condensation? Drag the terms on the left to the appropriate blanks on the right to complete the sentences. ResetHelp liquid gaseous solid Evaporation is the physical change in which a substance converts from its state to its state. Condensation is the physical change in which a substance converts from its state to its state.
Answer:
Evaporation is the physical change in which a substance converts from its liquid state to its gaseous state. Condensation is the physical change in which a substance converts from its gaseous state to its liquid state.
Explanation:
Evaporation and condensation are opposite processes to each other. Evaporation changes a liquid to a gas and condensation is the reverse.
You have been hired to check the technical correctness of an upcoming made-for-TV murder mystery that takes place in a space shuttle. In one scene, an astronaut's safety line is cut while on a spacewalk. The astronaut, who is 200 meters from the shuttle and not moving with respect to it, finds that the suit's thruster pack has also been damaged and no longer works and that only 4 minutes of air remains. To get back to the shuttle, the astronaut unstraps a 10-kg tool kit and throws it away with a speed of 8 m/s. In the script, the astronaut, who has a mass of 80 kg without the toolkit, survives, but is this correct?
Answer:
The astronaut who has a mass of 80 kg without the toolkit do survive with 40 seconds of remaining air
Explanation:
Due the astronaut throws the 10-kg tool kit away with a speed of 8 m/s, it gives a momentum equivalent but in the other direction, so [tex]I=mv=(10Kg)(8m/s)=80kg*m/s[/tex], then we can find the speed that the astronaut reaches due to its weight we get, [tex]v=\frac{I}{m} =\frac{80kg*m/s}{80Kg} =1m/s[/tex].
Finally, as the distance to the space shuttle is 200m, the time taken to the astronaut to reach it at the given speed will be [tex]t=\frac{d}{v}=\frac{200m}{1m/s}=200s[/tex], as the remaining air time is 4 min or 240 seconds, The astronaut who has a mass of 80 kg without the toolkit do survive with 40 seconds of remaining air.
1. The ___________ was used to find a Jupiter-sized planet through careful measurements of the changing position of a star in the sky.2. Discovering planets through the __________ requires obtaining and studying many spectra of the same star.3. The________________ successfully discovered thousands of extrasolar planets with a spacecraft that searched for transits among some 100,000 stars.4. The ____________ is used to find extrasolar planets by carefully monitoring changes in a star's brightness with time.5. Compared to the planets of our solar system, the composition of a _________________ most resembles the compositions of Uranus and Neptune.6. Observations indicating that other planetary systems often have jovian planets orbiting close to their stars are best explained by what we call______.7. An extrasolar planet that is rocky and larger than Earth is often called a_____.
Answer:
(1) The ____astrometric method_______ was used to find a Jupiter-sized planet through careful measurements of the changing position of a star in the sky.
(2) Discovering planets through the ____doppler method ______ requires obtaining and studying many spectra of the same star.
(3) The____kepler mission ____________ successfully discovered thousands of extrasolar planets with a spacecraft that searched for transits among some 100,000 stars.
(4) The ___transit method_________ is used to find extrasolar planets by carefully monitoring changes in a star's brightness with time.
(5) Compared to the planets of our solar system, the composition of a ____water world _____________ most resembles the compositions of Uranus and Neptune.
(6) Observations indicating that other planetary systems often have jovian planets orbiting close to their stars are best explained by what we call__migration ____.
(7) An extrasolar planet that is rocky and larger than Earth is often called a__super-Earth ___.
The Doppler technique and transit technique are used to discover extrasolar planets such as Jupiter-sized ones and those larger than Earth respectively. The Kepler mission has been successful in identifying thousands of these. The composition of extrasolar planets can differ, some resembling Uranus and Neptune, while the concept of 'planet migration' explains the closeness of large planets to their stars.
Explanation:1. The Doppler technique was used to find a Jupiter-sized planet through careful measurements of the changing position of a star in the sky.
2. Discovering planets through the spectroscopy requires obtaining and studying many spectra of the same star.
3. The Kepler mission successfully discovered thousands of extrasolar planets with a spacecraft that searched for transits among some 100,000 stars.
4. The transit technique is used to find extrasolar planets by carefully monitoring changes in a star's brightness with time.
5. Compared to the planets of our solar system, the composition of a mini-Neptune most resembles the compositions of Uranus and Neptune.
6. Observations indicating that other planetary systems often have jovian planets orbiting close to their stars are best explained by what we call planet migration.
7. An extrasolar planet that is rocky and larger than Earth is often called a super-Earth.
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A vertical spring withk= 245N/m oscillates with an amplitude of 19.2cm when 0.457kg hangs from it. The mass posses through the equilibrium point (y= 0) with a negative velocity at t= 0.Assume that downward is the positive direction of motion. What equation describes this motion as a function of time?
Answer:
y = -19.2 sin (23.15t) cm
Explanation:
The spring mass system is an oscillatory movement that is described by the equation
y = yo cos (wt + φ)
Let's look for the terms of this equation the amplitude I
y₀ = 19.2 cm
Angular velocity is
w = √ (k / m)
w = √ (245 / 0.457
w = 23.15 rad / s
The φ phase is determined for the initial condition t = 0 s , the velocity is negative v (0) = -vo
The speed of the equation is obtained by the derivative with respect to time
v = dy / dt
v = - y₀ w sin (wt + φ)
For t = 0
-vo = -yo w sin φ
The angular and linear velocity are related v = w r
v₀ = w r₀
v₀ = v₀ sinφ
sinφ = 1
φ = sin⁻¹ 1
φ = π / 4 rad
Let's build the equation
y = 19.2 cos (23.15 t + π/ 4)
Let's use the trigonometric ratio π/ 4 = 90º
Cos (a +90) = cos a cos90 - sin a sin sin 90 = 0 - sin a
y = -19.2 sin (23.15t) cm
1. A pair of oppositely charged parallel plates is separated by 5.51 mm. A potential difference of 614 V exists between the plates. What is the strength of the electric field between the plates? The fundamental charge is 1.602 × 10?19 . Answer in units of V/m.
2. What is the magnitude of the force on an electron between the plates? Answer in units of N.
3. How much work must be done on the electron to move it to the negative plate if it is initially positioned 2.7 mm from the positive plate? Answer in units of J.
Answer:
Part a)
[tex]E = 1.11 \times 10^5 N/C[/tex]
Part 2)
[tex]F = 1.78 \times 10^{-14} N[/tex]
Part 3)
[tex]W = 5 \times 10^{-17} J[/tex]
Explanation:
Part 1)
As we know that electric field and potential difference related to each other as
[tex]E = \frac{\Delta V}{x}[/tex]
so we will have
[tex]\Delta V = 614 V[/tex]
[tex]x = 5.51 mm[/tex]
so we have
[tex]E = \frac{614}{5.51 \times 10^{-3}}[/tex]
[tex]E = 1.11 \times 10^5 N/C[/tex]
Part 2)
Charge of an electron
[tex]e = 1.6 \times 10^{-19} C[/tex]
now force is given as
[tex]F = qE[/tex]
[tex]F = (1.6 \times 10^{-19})(1.11 \times 10^5)[/tex]
[tex]F = 1.78 \times 10^{-14} N[/tex]
Part 3)
Work done to move the electron
[tex]W = F.d[/tex]
[tex]W = (1.78 \times 10^{-14})(5.51 - 2.7) \times 10^{-3}[/tex]
[tex]W = 5 \times 10^{-17} J[/tex]
An empty rubber balloon has a mass of 12.5 g. The balloon is filled with helium at a density of 0.181 kg/m3. At this density the balloon has a radius of 0.498 m. If the filled balloon is fastened to a vertical line, what is the tension in the line? The density of air is 1.29 kg/m3.
Answer:
5.5 NExplanation:
mass of balloon (m) = 12.5 g = 0.0125 kg
density of helium = 0.181 kg/m^{3}
radius of the baloon (r) = 0.498 m
density of air = 1.29 kg/m^{3}
acceleration due to gravity (g) = 1.29 m/s^{2}
find the tension in the line
the tension in the line is the sum of all forces acting on the line
Tension =buoyant force + force by helium + force of weight of rubber
force = mass x acceleration
from density = \frac{mass}{volume} , mass = density x volume
buoyant force = density x volume x accelerationwhere density is the density of air for the buoyant force
buoyant force = 1.29 x (\frac{4]{3} x π x 0.498^{3}) x 9.8 = 6.54 N
force by helium = density x volume x accelerationforce by helium = 0.181 x (\frac{4]{3} x π x 0.498^{3}) x 9.8 = 0.917 N
force of its weight = mass of rubber x accelerationforce of its weight = 0.0125 x 9.8 = 0.1225 N
Tension = buoyant force + force by helium + force of weight of rubberthe force of weight of rubber and of helium act downwards, so they
carry a negative sign.
Tension = 6.54 - 0.917 - 0.1225 = 5.5 NCalculating the tension on a string tied to a helium-filled balloon involves principles of buoyancy and weight. The difference between the buoyant force exerted by the displaced air and the sum of the weight of the empty balloon and weight of the helium gives the tension on the line.
Explanation:The tension on the vertical line tied to the helium-filled balloon can be found using Archimedes' Principle and Buoyancy. First, the volume (V) of the helium-filled balloon can be calculated using the formula for the volume of a sphere; V = 4/3 x π x (0.498 m)³ = ~0.516 m³. The mass (m) of the helium in the balloon is calculated by multiplying this volume by the density of helium, which is m = 0.516 m³ x 0.181 kg/m³ = ~0.093 kg.
Next, the weight (w) of air displaced by the balloon is calculated. This weight is equal to the product of the volume of the balloon, the density of the air, and gravitational acceleration; w = 0.516 m³ x 1.29 kg/m³ x 9.8 m/s² = ~6.5 N. This is the buoyant force that lifts the balloon.
The weight (W) of the filled balloon is the sum of the weight of the empty balloon and the helium; W = (0.0125 kg + 0.093 kg) x 9.8 m/s² = ~1.03 N. The tension (T) in the line is then the difference between the buoyant force and this weight (i.e., T = 6.5 N - 1.03 N = ~5.47 N).
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A 2.35 kg ball is attached to a ceiling by a3.53 m long string. The height of the room is5.03 m .The acceleration of gravity is 9.8 m/s2.
What is the gravitational potential energyassociated with the ball relative to the ceiling?Answer in units of J.
Answer:-81.29 J
Explanation:
Given
mass of ball [tex]m=2.35 kg[/tex]
Length of string [tex]L=3.53 m[/tex]
height of Room [tex]h=5.03 m[/tex]
Gravitational Potential Energy is given by
[tex]P.E.=mgh[/tex]
where h=distance between datum and object
here Reference is ceiling
therefore [tex]h=-3.53 m[/tex]
Potential Energy of ball w.r.t ceiling
[tex]P.E.=2.35\times 9.8\times (-3.53)=-81.29 J[/tex]
i.e. 81.29 J of Energy is required to lift a ball of mass 2.35 kg to the ceiling
Which of the following is true about the speed of light?
A.) It is a constant when the light is traveling in a vacuum.
B.) It speeds up or slows down depending on the observer.
C.) It is slowest in a vacuum.
D.) It varies depending on the color of the light.
Answer:
It is a constant when the light is traveling in a vacuum.
Explanation:
A wood-burning stove (emissivity = 0.900 and surface area = 3.75 m2) is being used to heat a room. The fire keeps the stove surface at a constant 182 °C (455 K) and the room at a constant 22 °C (295 K). Determine the net radiant power generated by the stove.
Answer:
Radiated power will be [tex]675241.7175watt[/tex]
Explanation:
We have given emissivity [tex]\epsilon =0.9[/tex]
Stove temperature T = 455 K
Room temperature [tex]T_C=295K[/tex]
We know the Stephan's constant [tex]\sigma =5.67\times 10^{-6}w/m^2K^4[/tex]
We know that radiated power is given by [tex]P=\epsilon \sigma A(T^4-T_C^4)=0.9\times5.67\times 10^{-6}\times 3.75\times (455^4-295^4)=675241.7175watt[/tex]
The net radiant power generated by the wood-burning stove can be determined by using the Stefan-Boltzmann law. This law combines the factors of Stefan-Boltzmann constant, emissivity of the body, surface area, and the difference of the fourth powers of the absolute temperatures.
Explanation:The rate of radiant power generated by the stove can be found using the Stefan-Boltzmann law, which describes the rate of heat transfer by emitted radiation. This law can be stated as P = σeAT⁴, where P is the radiant power, σ is the Stefan-Boltzmann constant (5.67 × 10⁻⁸ J/s.m².K⁴), e is the emissivity of the body, A is the surface area, and T is the absolute temperature in Kelvins.
For the net rate of heat transfer by radiation (or the net radiant power) for the wood-burning stove, you would use the Stefan-Boltzmann law as Q-net = σeA(T₂⁴ - T₁⁴), where T₂ is the absolute temperature of the stove and T₁ the absolute temperature of the room.
Plugging in the stove's values from the problem: emissivity = 0.900, surface area = 3.75 m², T₂ = 455 K (stove temperature), and T₁ = 295 K (room temperature), we can calculate the net radiant power allocated by the stove.
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A cabin has a 0.159-m thick wooden floor [k = 0.141 W/(m · C°)] with an area of 13.4 m2. A roaring fire keeps the interior of the cabin at a comfortable 18.0 °C while the air temperature in the crawl space below the cabin is –16.4°C. What is the rate of heat conduction through the wooden floor?
Answer:
[/tex] 408.8[/tex]
Explanation:
[tex]t[/tex] = thickness of the floor = 0.159 m
[tex]k[/tex] = thermal conductivity of wooden floor = 0.141 Wm⁻¹ ⁰C⁻¹
[tex]T_{i}[/tex] = Temperature of the interior of the cabin = 18.0 ⁰C
[tex]T_{o}[/tex] = Temperature of the air below the cabin = - 16.4 ⁰C
Difference in temperature is given as
[tex]\Delta T[/tex] = Difference in temperature = [tex]T_{i} - T_{o} = 18 - (- 16.4) = 34.4[/tex]⁰C
[tex]A[/tex] = Area of the floor = 13.4 m²
[tex]Q[/tex] = Rate of heat conduction
Rate of heat conduction is given as
[tex]Q = \frac{kA \Delta T}{t}[/tex]
[tex]Q = \frac{(0.141) (13.4) (34.4)}{0.159}\\Q = 408.8[/tex] W
Waves on a swimming pool propagate at 0.720 m/s. You splash the water at one end of the pool and observe the wave go to the opposite end, reflect, and return in 31.0 s. How far away is the other end of the pool?
Answer:
11.16 m
Explanation:
total distance traveled by the wave = speed * time
= 0.720 m/s * 31.0 s
= 22.32 m
Since this is an echo back to the starting point then the length of the swimming pool must be half of the distance traveled.
length of swimming pool = 22.32 m÷2 = 11.16 m
Which of the following would constitute evidence against Turcotte's model?
A) the success of gradualist models explaining the surface of Mars
B) an even more detailed map of the surface of Venus
C) an even longer river of lava on Io, a moon of Jupiter
D) a few active volcanoes on Ishtar Terra, a continent on Venus
E) a volcano on Earth releasing a massive burst of thermal energy all at once
Answer: answer is option A
Explanation: gradually model ascertained the features of the surface of mass is explaining it as gradual incremental changes continuously over a long period of time which negates the catastrophic model nature of turtles model
A football player kicks a 0.41-kg football initially at rest; and the ball flies through the air. If the kicker's foot was in contact with the ball for 0.051 s and the ball's initial speed after the collision is 21 m/s, what was the magnitude of the average force on the football?
Answer:
Average force on the football = 168.82 N
Explanation:
Force = Mass x Acceleration
F = ma
Mass, m = 0.41 kg
We have equation of motion, v = u + at
Initial velocity, u = 0 m/s
Final velocity, v = 21 m/s
Time, t = 0.051 s
Substituting
21 = 0 + a x 0.051
a = 411.76 m/s²
Substituting in force equation,
F = ma = 0.41 x 411.76 = 168.82 N
Average force on the football = 168.82 N
A spring (70 N/m ) has an equilibrium length of 1.00 m. The spring is compressed to a length of 0.50 m and a mass of 2.0 kg is placed at its free end on a frictionless slope which makes an angle of 41 ∘ with respect to the horizontal. The spring is then released
A)If the mass is not attached to the spring, how far up the slope from the compressed point will the mass move before coming to rest?
B)If the mass is attached to the spring, how far up the slope from the compressed point will the mass move before coming to rest?
Answer:
Part a)
[tex]L = 0.68 m[/tex]
Part b)
[tex]L = 0.63 m[/tex]
Explanation:
Part a)
As we know that there is no friction in the path
So here we can use energy conservation to find the distance moved by the mass
Initial spring energy = final gravitational potential energy
so we will have
[tex]\frac{1}{2}kx^2 = mgL sin\theta[/tex]
[tex]\frac{1}{2}70(0.5)^2 = 2(9.81)(L) sin41[/tex]
[tex]8.75 = 12.87 L[/tex]
[tex]L = 0.68 m[/tex]
Part b)
Now if spring is connected to the block then again we can use energy conservation
so we will have
[tex]\frac{1}{2}kx^2 = mg(x + x')sin\theta + \frac{1}{2}kx'^2[/tex]
so we will have
[tex]\frac{1}{2}(70)(0.5^2) = 2(9.81) (0.50 + x') sin41 + \frac{1}{2}(70)x'^2[/tex]
[tex]8.75 = 6.43 + 12.87 x' + 35 x'^2[/tex]
[tex]x' = 0.13 m[/tex]
so total distance moved upwards is
[tex]L = 0.5 + 0.13 = 0.63 m[/tex]
If the spring is not attached, the mass will move 0.68 meters. however, if the spring is attached, the mass will move by 0.63 meters.
How can we arrive at these results?We will first calculate the motion of the mass if it is not attached to the spring. This will be done with the following formula:[tex]\frac{1}{2} kx^2=m*g*L*sin41[/tex]
Within this formula, the symbol "k" represents 70 N/m. The "x" represents 0.50m. The "m" is 2.0 kg, the "g" represents gravity and the L is the value we need to find.
Therefore, we can substitute the values of the formula as follows:
[tex]\frac{1}{2}70(0.50)^2= 2*9.81 *L*sin41\\8.75=12.87L\\L= \frac{8.75}{12.87} = 0.67 ------- 0.68m[/tex]
Then we can calculate the distance traveled by the mass attached to the spring. For this, we will use the formula:[tex]\frac{1}{2} kx^2= m*g*(x+x')*sin41+\frac{1}{2} 70x'^2[/tex]
By substituting the values, we can answer the equation as follows:
[tex]\frac{1}{2} 70*0.50^2=2*9.81*(0.50+x')*sin41+\frac{1}{2} 70x'^2\\8.75=6.43+12.87x'+35x'\\X'=0.13m \\\\L=0.5+0.13=0.63 m[/tex]
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A 50.1 kg diver steps off a diving board and drops straight down into the water. The water provides an average net force of resistance of 1598 N to the diver’s fall. If the diver comes to rest 5.2 m below the water’s surface, what is the total distance between the diving board and the diver’s stopping point underwater? The acceleration due to gravity is 9.81 m/s 2. Answer in units of m.
Answer:
The total distance is 16.9 m
Explanation:
We understand work in physics as certain force exerted through certain distance. To reach that point below the water, the work done by the diver must be equal to the work done by the water's force of resistance. Therefore, we determine both work expressions and we solve the equation for the diver distance, which is the total distance between the diving board and the stopping point underwater.
[tex]W_{d}: work done by diver\\W_{R}: work due the force of resistance\\W_{d} = W_{R}\\F_{d}\times d_{d}=F_R \times d_{R}\\d_{d}= \frac{F_R \times d_R}{F_d}= \frac{F_R \times d_R}{m_d \times g}= \frac {1598 N \times 5.2 m}{50.1 kg \times 9.81 m/s^2}\\d_{d}= 16. 91 m[/tex]
When a stone is dropped into a pond, ripples form and move in concentric circles outward from the location of the stone impact. As the ripples move outward, they lose waves lose amplitude. If we neglect friction, why would the wave amplitude decrease the further they move outward?
Explanation:
When a stone is dropped into a pond, ripples form and move in concentric circles outward. This can be treated as a point source emitting waves. And for point source emitting waves
Intensity of the wave I = Source power Ps/Area
⇒[tex]I= \frac{Ps}{\pi r^2}[/tex]
⇒I∝1/r^2
Therefore, on increasing the distance r, the intensity of the wave decreases, this also means that the amplitude also decreases.
A uniform rectangular plate of length B = 15 cm and height A = 24 cm has a rectangular corner cut out of it of length D = 6 cm and height C = 8 cm. The plate is made of a material of area mass density σ. For this problem we set the origin at the lower left corner of the plate with the x-axis horizontal pointing right and the y-axis vertical pointing up.
a. Calculate the value of the y-coordinate, in centimeters, for the center of mass of the plate.
b. Calculate the value of the x-coordinate, in centimeters, for the center of mass of the plate.
The center of mass for a rectangle with a corner cut out can be calculated using the centroid formula by considering the plate and the cut-out as separate figures and then determining the average weighted by area.
Explanation:This problem concerns the calculation of the center of mass for 2-dimensional figures. The center of mass represents the point at which we could balance the object in a uniform gravitational field and is calculated by integrating the mass distributed in the object.
First, consider that area mass density σ is constant. Thus, this is equivalent to finding the centroid of the shape. The location of the centroid C (x,y) is given by:
x = (A1x1 + A2x2)/(A1 + A2)
and
y = (A1y1 + A2y2)/(A1 + A2)
where Ax and Ay represent the area and centroid coordinate of the individual figure, respectively. For this problem, A1 is the whole rectangular plate and A2 is the cutout. The centroid of a rectangle is at the midpoint of its diagonals, so the x and y coordinates for A1 are (B/2, A/2) and for A2 are (D, D/2).
Then, the x-coordinate for the center of mass becomes, x = [(AB*(B/2) - CD*(D/2)]/(AB - CD) and y-coordinate for center of mass becomes, y = [(AB*(A/2) - CD*(C/2)]/(AB - CD) Calculate them to get numerical value.
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A manufacturer of chocolate chips would like to know whether its bag filling machine works correctly at the 420 gram setting. It is believed that the machine is underfilling the bags. A 24 bag sample had a mean of 414 grams with a standard deviation of 15. Assume the population is normally distributed. A level of significance of 0.1 will be used. Find the P-value of the test statistic.
Answer:
P-value of the test statistic is 2.499
Explanation:
given data:
size sample is 24
sample mean is 414 gm
standard deviation is 15
Null hypotheis is
[tex]H_0: \mu = 420 gm[/tex]
[tex]H_1 \mu< 420[/tex]
level of significance is 0.01
from test statics
[tex]t = \frac{\hat x - \mu}{\frac{s}{\sqrt{n}}}[/tex]
degree od freedom is = n -1
df = 24 -1 = 13
[tex]t = \frac{414 - 420}{\frac{15}{\sqrt{24}}}[/tex]
t = -1.959
from t table critical value of t at 0.1 significane level and 23 degree of freedom is 2.499
A fielder tosses a 0.15 kg baseball at 26 m/s at a 36 ∘ angle to the horizontal.Part AWhat is the ball's kinetic energy at the start of its motion?Express your answer with the appropriate units.Part BWhat is the kinetic energy at the highest point of its arc?Express your answer with the appropriate units.
Answer:
A. 50.7 J
B. 33.18 J
Explanation:
At the start of the motion the kinetic energy of the ball would be
[tex]E_k = \frac{mv^2}{2} = \frac{0.15*26^2}{2} = 50.7 J[/tex]
When the ball gets to the highest point, vertical velocity must be 0, only horizontal velocity contributes to the kinetic energy. Since air resistant can be neglected, horizontal energy is preserved since the start
[tex]v_h = vcos(36^o) = 26*0.81 = 21.03 m/s[/tex]
So the kinetic energy at this point is
[tex]E = \frac{mv_h^2}{2} = \frac{0.15*21.03^2}{2} = 33.18J[/tex]
Given the data from the question, the kinetic energy at the start and at the highest point are:
A. The kinetic energy at the start of the motion is 50.7 J
B. The kinetic energy at the highest point is 0 J
What is kinetic energy?This is the energy possessed by an object in motion. Mathematically, it can be expressed as:
KE = ½mv²
Where
KE is the kinetic energy m is the mass v is the velocity A. How to determine the KE at the start Mass (m) = 0.15 KgVelocity (v) = 26 m/sKinetic energy (KE) =?KE = ½mv²
KE = ½ × 0.15 × 26²
KE = 50.7 J
B. How to determine the KE at the highest point Mass (m) = 0.15 KgVelocity (v) = 0 m/s (at highest point) Kinetic energy (KE) =?KE = ½mv²
KE = ½ × 0.15 × 0
KE = 0 J
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A 340 g bird flying along at 6.0 m/s sees a 13 g insect heading straight toward it with a speed of 30 m/s. The bird opens its mouth wide and enjoys a nice lunch. What is the bird's speed immediately after swallowing?
_____m/s
Answer:
V = 4.67 m/s
Explanation:
given,
mass of the flying bird (M)= 340 g = 0.34 Kg
Speed of bird(u₁) = 6 m/s
mass of insect(m) = 13 g = 0.013 Kg
speed of insect(u₂) = 30 m/s
Speed of bird after eating the insect(V) =?
Using conservation of momentum
M u₁ + m u₂ = (M+m)V
[tex]V = \dfrac{M u_1+ m u_2}{(M+m)}[/tex]
[tex]V = \dfrac{0.34\times 6 - 0.013 \times 30}{0.34+0.013}[/tex]
[tex]V = \dfrac{1.65}{0.353}[/tex]
V = 4.67 m/s
the bird's speed immediately after swallowing V = 4.67 m/s
A 1.5-cm object is placed 0.50 m to the left of a diverging lens with a focal length of 0.20 m. A converging lens with a focal length of 0.17 m is located 0.08 m to the right of the diverging lens. What is the height and orientation with respect to the original object of the final image?
Answer:
The object for the converging lens is upright and 0.429 cm tall, the image of this converging lens is inverted and 1.375 cm high
Explanation:
Let
[tex]d_{o}=distance of object[tex]\\f=focal length\\d_{i}=distance of image\\I_{h}=Image height[/tex]
For diverging lens:
[tex]d_{o} = 0.50\\f = -0.20\\\frac{1}{d_{o}}+\frac{1}{-0.20}\\\frac{1}{d_{i}}=\frac{1}{-0.20}-\frac{1}{0.50}=-7\\d_{o}=-\frac{1}{7}[/tex]
Magnification = [tex]\frac{d_{i}}{d_{o}}= -\frac{1}{7}÷ 0.5 = -0.286[/tex]
Image height [tex]= -0.286 * 1.5 = -0.429 cm[/tex] (negative sign means the image is virtual, inverted.
This image is [tex]\frac{1}{7}[/tex] meter to left of the center of the diverging lens.
The converging lens is located 0.08 m to the right of the diverging lens
The distance between the image of the diverging lens and center of the converging lens = [tex]\frac{1}{7} + 0.08 = 0.229 m[/tex]
The image of the diverging lens becomes the object of the converging lens.
[tex]d_{o} = 0.223\\f = 0.17\\\frac{1}{d_{i}}=\frac{1}{0.17}-\frac{1}{0.223}=0.715\\d_{i}=0.715m to the right of the converging lens[/tex]
[tex]Magnification =\frac{d_{i}}{d_{o}} = \frac{0.715}{0.223}=3.206\\image height=3.206 * 0.429 = 1.375 cm.[/tex]
A mass of 4.8 kg is dropped from a height of 4.84 meters above a vertical spring anchored at its lower end to the floor. If the spring has a height of 82 cm and a constant of 24 N/cm, how far, to the nearest tenth of a cm, is the spring compressed?
Answer:
45.6 cm
Explanation:
Let x (m) be the length that the spring is compressed. We know that when we drop the mass from 4.84 m above and compress the springi, ts gravitational energy shall be converted to spring potential energy due to the law of energy conservation
[tex]E_g = E_p[/tex]
[tex]mgh = 0.5kx^2[/tex]
where h = 4.84 + x is the distance from the dropping point the the compressed point, and k = 24N/cm = 2400N/m is the spring constant, g = 9.81 m/s2 is the gravitational acceleration constant. And m = 4.8 kg is the object mass.
[tex]4.8*9.81(4.84 + x) = 0.5 * 2400 * x^2[/tex]
[tex]47.088x + 227.906 = 1200x^2[/tex]
[tex]1200x^2 - 47.088x - 227.906 = 0[/tex]
[tex]x = 0.456m[/tex] or 45.6 cm
A hand pump is being used to inflate a bicycle tire that has a gauge pressure of 59.0 lb/in2. If the pump is a cylinder of length 17.4 in. with a cross-sectional area of 3.00 in.2, how far down must the piston be pushed before air will flow into the tire? Assume the temperature is constant.
Answer must be in inches
Answer:
∴The air cannot be made to flow in with the given pump at the given conditions.
Explanation:
Given:
gauge pressure of bicycle tyre, [tex]P_g=59\ lb.in^{-2}[/tex]length of cylinder of the pump, [tex]l=17.4\ in[/tex]area of the the cylinder of the pump, [tex]a= 3\ in^2[/tex]we have the density of air at STP, [tex]\rho=4.4256\times 10^{-5}\ lb.in^{-2}[/tex]The piston must be pushed more than the pressure inside the tyre:
[tex]P_g=\rho\times V\div a[/tex]
[tex]59=4.4256\times 10^{-5}\times a\times h\div a[/tex]
[tex]h=13.33\times 10^5\ in[/tex]
∴The air cannot be made to flow in with the given pump at the given conditions.
What is the asthenosphere and describe its characteristics?
Answer:
Layer of the interior of the Earth that extends approximately between 50 and 150 km deep, formed mainly by partially molten plastic rocks that can deform: the asthenosphere is located between the lithosphere and the mesosphere.
Explanation:
The asthenosphere is composed of ductile silicate materials, in solid state and partially or totally semi-molten (according to their depth and / or proximity to magma bags), which allow continental drift and isostasis. On it the tectonic plates move.
Characteristics of the asthenosphere
The asthenosphere is a part of the upper mantle, just below the lithosphere that is involved in tectonic plate movements and isostatic adjustments.
Despite its heat, the pressure keeps it like plastic and has a relatively low density. Seismic waves pass relatively slowly through the asthenosphere, compared to the mantle that covers the lithosphere, which is why it has been called the low-speed zone.
This was the observation that originally alerted the seismologists to their presence and gave some information about their physical properties, such as the speed of the seismic waves that decreases as stiffness decreases. Under the thin oceanic plates the asthenosphere is generally much closer to the surface of the seabed, and rises a few kilometers from the ocean floor.
Due to the temperature and pressure conditions in the asthenosphere, the rock becomes ductile, moving at a deformation rate that is measured in cm / year at linear distances that finally reach thousands of kilometers. In this way, it flows like a convection current, radiating heat outward from inside the Earth. Above the asthenosphere, at the same rate of deformation, the rock behaves elastically and, if fragile, can break, causing failures. The rigid lithosphere is believed to "float" over the asthenosphere from which it flows slowly, creating the movement of tectonic plates.
Answer:
The conistency is solid layer of rock
Explanation:
A steel cylinder at sea level contains air at a high pressure. Attached to the tank are two gauges, one that reads absolute pressure and one that reads gauge pressure. The tank is then brought to the top of a mountain.Which statement is true for the gauge that reads gauge pressure?
Select the correct answer
Both gauges will change by the same amount.
The pressure reading increases.
The pressure reading stays the same.
The pressure reading decreases.
Final answer:
The gauge pressure reading stays the same when a steel cylinder is moved to the top of a mountain because it measures pressure relative to the external atmospheric pressure, which does not influence the pressure inside the cylinder.
Explanation:
The correct statement for the gauge that reads gauge pressure when a steel cylinder is brought to the top of a mountain is that the pressure reading stays the same. Gauge pressure is the pressure relative to atmospheric pressure, which means it measures the excess pressure in the tank over the external atmospheric pressure. Since the pressure inside the steel cylinder is not dependent on external atmospheric pressure changes, the reading on the gauge pressure meter would remain constant, even if the cylinder is taken to a different altitude.
Calculate the specific heat of a metal from the following data. A container made of the metal has a mass of 3.9 kg and contains 11 kg of water. A 2.0 kg piece of the metal initially at a temperature of 189°C is dropped into the water. The container and water initially have a temperature of 16.0°C, and the final temperature of the entire system is 18.0°C.
Answer:
[tex]c_m =272.365\frac{J}{KgK}[/tex]
Explanation:
1) Notation and data given
[tex]m_c[/tex]= mass of the container = 3.9 kg
[tex]m_w[/tex]= mass of the water inside of the container= 11 kg
[tex]m_m[/tex]=mass of the metal= 2 kg
[tex]T_{im}=189\degree C[/tex] initital temperature of the metal
[tex]T_{iw}=16\degree C[/tex] initital temperature of the water
[tex]T_{ic}=16\degree C[/tex] initital temperature of the container
[tex]T_{f}=18\degree C[/tex] final equilibrium temperature
2) Concepts and formulas
Since the inital temperature for the pice of metal is higher than the temperature for the container and the water inside, the piece of metal will transfer heat, and this transferred heat is the same amount absorbed by the container and the water reaching the equilibrium. So then we have this formula:
[tex]Q_m =Q_w + Q_c[/tex]
We don't have any change of phase so then we just have the presence of sensible heat, and using this we got that:
[tex]m_p c_m \Delta T_m =m_w c_w \Delta T +m_c c_c \Delta T[/tex]
On this case the container is made by the same metal so then [tex]c_c=c_m[/tex]
[tex]m_p c_m \Delta T_m =m_w c_w \Delta T +m_c c_m \Delta T[/tex]
And solving for [tex]c_m[/tex] from the last expression we got:
[tex]m_p c_m \Delta T_m -m_c c_m \Delta T_c =m_w c_w \Delta T[/tex]
[tex]c_m [m_p \Delta T_m -m_c \Delta T_c] =m_w c_w \Delta T[/tex]
[tex]c_m=\frac{m_w c_w \Delta T}{m_p \Delta T_m -m_c \Delta T_c}[/tex]
Replacing the values we have:
[tex]c_m=\frac{11kg(4187\frac{J}{KgK}(18\degree C-16\degree C)}{2kg(189\degree C -16 \degree C) -3.9Kg(18\degree C -16\degree C)}= 272.365\frac{J}{Kg K}[/tex]