Answer:
L = 8.66 m
Explanation:
The length measured by the moving observer is related to the true length is given by
L = L₀ √1 - (v²/c²)
Where L₀ is the length of the spaceship as measured by person A, v is the speed of spaceship of person A and c is the speed of light c = 3.8x10ᵃ m/s
L = 10√1 - (0.5c)²/c²
L = 10√1 - (0.5*3.8x10ᵃ)²/3.8x10ᵃ ²
L = 8.66 m
Therefore, the Person B would measure the spaceship length to be 8.66 m
A 640 kg automobile slides across an icy street at a speed of 63.9 km/h and collides with a parked car which has a mass of 816 kg. The two cars lock up and slide together. What is the speed of the two cars just after they collide?
Answer:
V3 = 7.802 m/s
Explanation:
m1 = 640 Kg, M2 = 816 kg, V1 = 63.9 Km/h = 17.75 m/s, V2 =0 m/s
Let V3 is the combine velocity after collision.
According to the law of conservation of momentum
m1 v1 + m2 v2 = (m1 + m2) v3
⇒ V3 =( m1 v1 + m2 v2 ) / (m1 + M2)
V3 = ( 640 Kg × 17.75 m/s + 816 kg × 0m/s) / (640 Kg + 816 kg)
V3 = 7.802 m/s
Answer:
28.088 km/h or 7.802 m/s
Explanation:
From the law of conservation of momentum,
Total momentum before collision = Total momentum after collision
mu+m'u' = V(m+m')....................... Equation 1
Where m = mass of the automobile, m' = mass of the car, u = initial velocity of the car, u' = initial velocity of the car, V = velocity of the two car after collision
Make V the the subject of the equation
V = (mu+m'u')/(m+m')................ Equation 2
Given: m = 640 kg, m' = 816 kg, u = 63.9 km/h, u' = 0 m/s (parked).
Substitute into equation 2
V = (640×63.9+816×0)/(640+816)
V = 40896/1456
V = 28.088 km/h = 28.088(1000/3600) m/s = 7.802 m/s
Hence the speed of the two cars after they collide = 28.088 km/h or 7.802 m/s
Point charges of 4.75 µC and −1.50 µC are placed 0.350 m apart. (Assume the negative charge is located to the right of the positive charge. Include the sign of the value in your answers.)
(a) Where can a third charge be placed so that the net force on it is zero?
_______ m to the right of the −1.50 µC charge
(b) What if both charges are positive?
_______ m to the right of the 1.50 µC charge
Answer:
Explanation:
Let the position of balance point be x distance from 1.5 µC on the right side . Balance point can not be on the left side or between the charges because in that case electric field by both the charges will be in the same direction .
For equilibrium of field
k x 4.75 µC / ( .35 +x )² - k x 1.5 µC / x ² = 0
4.75 / ( .35 +x )² = 1.5 / x²
( .35 +x )² / x² = 4.75 / 1.5
( .35 +x )² / x² = 3.167
( .35 +x ) / x = 1.78
.35 + x = 1.78 x
.78 x = .35
x = .45 m ( + ve ) to the right of 1.50 µC
b ) When both the charges are positive , balance point will lie in between them because , electric field will be in opposite direction .
For equilibrium of field
k x 4.75 µC / ( .35 - x )² = k x 1.5 µC / x ²
4.75 / ( .35 -x )² = 1.5 / x²
( .35 -x )² / x² = 4.75 / 1.5
( .35 -x )² / x² = 3.167
( .35 -x ) / x = 1.78
.35 -x = 1.78 x
2.78 x = .35
x = .126 m ( - ve ) to the right of 1.50 µC
Be sure to answer all parts. To improve conductivity in the electroplating of automobile bumpers, a thin coating of copper separates the steel from a heavy coating of chromium. (a) What mass of Cu is deposited on an automobile trim piece if plating continues for 1.25 h at a current of 5.1 A?
Answer:
Explanation:
Total charge passed
= 1.25 x 60 x 60 x 5.1 C
= 22950 C
Equivalent mass of copper
= 63.5 / 2
= 31.75 g
96500 coulomb is required to obtain 31.75 g of copper
22950 C will release (31.75 / 96500) x 22950
= 7.55 g of copper .
Consider a platinum wire (σ= 1.0 × 107 Ω-1·m-1) with a cross-sectional area of 1 mm2 (similar to your connecting wires) and carrying 0.3 amperes of current, which is about what you get in a circuit with a round bulb and two batteries in series. Calculate the strength of the very small electric field required to drive this current through the wire.
Answer: 0.03 N/C
Explanation:
We use the current density formula to solve this question.
I/A = σ * E
Where,
I = current flowing in the circuit = 0.3 A
A = cross sectional area of the wire = 1 mm²
σ = resistivity of the wire = 1*10^7 Ω^-1·m^-1
E = strength of the electric field required
I/A = σ *E
E = I/(A * σ)
First we convert area from mm to m, so that, 1*10^-3 mm = 1*10^-6 m
E = 0.3 A / (1*10^-6 m * 1*10^7 Ω^-1·m^-1)
E = 0.3 A / 10 Ω^-1
E = 0.03 N/C
The required electric field to drive given current through the wire is 0.03 V/m.
To calculate the strength of the electric field required to drive a current of 0.3 amperes through a platinum wire, we can use the relationship between current density, electrical conductivity, and electric field.
The current density (J) is given by:
J = I / A
where I = 0.3 A and A = 1 mm² = 1 × 10⁻⁶ m².
Substituting the values, we get:
J = 0.3 A / (1 × 10⁻⁶ m²) = 3 × 10⁵A/m²
Next, we use Ohm's law in the form that relates current density to the electric field:
J = σE
where σ is the electrical conductivity of platinum, σ = 1.0 × 10⁷ S/m (S = 1/Ω), and E is the electric field.
Rearranging for the electric field, we get:
E = J / σ
Substituting the values, we get:
E = (3 × 10⁵ A/m²) / (1.0 × 10⁷ S/m) = 3 × 10⁻² V/m
Therefore, the strength of the very small electric field required to drive the current through the wire is 0.03 V/m.
What is Pascal's Law? When gasses aren't constrained they tend to expand or contract, which depends on the pressure. You can squeeze air into tighter spaces by pressing the molecules together. In a contained incompressible fluid, any external pressure applied at one point will raise pressure equally at every point. An object placed in water is buoyed up with a force equal to the weight of the fluid the object displaces.
Answer: In a contained incompressible fluid, any external pressure applied at one point will raise pressure equally at every point.
Explanation: This law was put forward by Blaise Pascal, a French mathematician in 1648. Pascal's Law states that in a contained incompressible fluid, any external pressure applied at one point will raise pressure equally at every point. Pascal's law has been used in fluid mechanics for different applications these includes:
- the hydraulic jack used in automobile listings,
- most automobile braking systems,
-water towers, and dams.
A laboratory dish, 20 cm in diameter, is half filled with water. One at a time, 0.49 μL drops of oil from a micropipette are dropped onto the surface of the water, where they spread out into a uniform thin film. After the first drop is added, the intensity of 640 nm light reflected from the surface is very low. As more drops are added, the reflected intensity increases, then decreases again to a minimum after a total of 13 drops have been added. What is the index of refraction of the oil?
Explanation:
Formula for path difference is as follows.
x = 2tn
and, refractive index (n) = [tex]\frac{\lambda}{2t}[/tex]
Thickness is calculated as follows.
Thickness (t) = [tex]\frac{volume}{area}[/tex]
Area = [tex]\pi r^{2}[/tex]
= [tex]\pi \times (\frac{d}{2})^{2}[/tex]
= [tex]\frac{0.49 \times 10^{-6}}{3.14 \times 0.01 m}[/tex]
= [tex]1.56 \times 10^{-8}[/tex] m
Now, the refractive index will be calculated as follows.
For drop, n = [tex]\frac{\lambda}{2t}[/tex]
For B drop, n = [tex]\frac{\lambda}{26t}[/tex]
So, n = [tex]\frac{640 \times 10^{-9}}{26 \times 1.56 \times 10^{-8}}[/tex]
= [tex]\frac{640 \times 10^{-9}}{40.56 \times 10^{-8}}[/tex]
= 1.5
Thus, we can conclude that index of refraction of the oil is 1.5.
The GPS (Global Positioning System) satellites are approximately 5.18 mm across and transmit two low-power signals, one of which is at 1575.42 MHz (in the UHF band). In a series of laboratory tests on the satellite, you put two 1575.42 MHz UHF transmitters at opposite ends of the satellite. These broadcast in phase uniformly in all directions. You measure the intensity at points on a circle that is several hundred meters in radius and centered on the satellite. You measure angles on this circle relative to a point that lies along the centerline of the satellite (that is, the perpendicular bisector of a line which extends from one transmitter to the other). At this point on the circle, the measured intensity is 2.00 W/m^2.
Required:
What is the intensity at a point on the circle at an angle of 4.45° from the centerline?
Answer:
intensity at a point on the circle at an angle of 4.45° from the center line is 1.77 W/m².
Explanation:
See attached picture.
The intensity at a point on the circle at an angle of 4.45° from the centerline is approximately 4.00 W/m².
To find the intensity at a point on the circle at an angle of 4.45° from the centerline, we need to consider the interference pattern created by the two transmitters.
1. Wavelength Calculation:
The wavelength [tex](\(\lambda\))[/tex] of the transmitted signal can be calculated using the frequency [tex](\(f\))[/tex] and the speed of light c.
[tex]\[ \lambda = \frac{c}{f} \][/tex]
Given:
- [tex]\( c = 3 \times 10^8 \)[/tex] m/s (speed of light)
- [tex]\( f = 1575.42 \times 10^6 \)[/tex] Hz (frequency)
[tex]\[ \lambda = \frac{3 \times 10^8}{1575.42 \times 10^6} \approx 0.1902 \text{ meters} \][/tex]
2. Phase Difference Calculation:
The phase difference [tex](\(\Delta \phi\))[/tex] between the two waves at a point on the circle is determined by the path difference [tex](\(\Delta L\))[/tex] travelled by the waves.
The path difference is given by:
[tex]\[ \Delta L = d \sin(\theta) \][/tex]
where:
- d is the distance between the two transmitters (5.18 mm = 0.00518 m)
- [tex]\(\theta\)[/tex] is the angle from the centerline (4.45°)
[tex]\[ \Delta L = 0.00518 \sin(4.45^\circ) \approx 0.00518 \sin(0.0776) \approx 0.00518 \times 0.0775 \approx 0.0004019 \text{ meters} \][/tex]
The phase difference is:
[tex]\[ \Delta \phi = \frac{2 \pi \Delta L}{\lambda} = \frac{2 \pi \times 0.0004019}{0.1902} \approx 0.0133 \text{ radians} \][/tex]
3. Intensity Calculation:
The intensity at a given point due to two sources interfering constructively or destructively can be found using:
[tex]\[ I = I_0 \left(1 + \cos(\Delta \phi)\right) \][/tex]
Here, [tex]\(I_0\)[/tex] is the intensity when both waves interfere constructively at the centerline [tex](\(\theta = 0^\circ\))[/tex], which is given as 2.00 W/m^2.
[tex]\[ I = 2.00 \left(1 + \cos(0.0133)\right) \][/tex]
Since [tex]\(\cos(0.0133) \approx 0.9999\)[/tex]:
[tex]\(\cos(0.0133) \approx 0.9999\)[/tex]
Thus, the intensity at a point on the circle at an angle of 4.45° from the centerline is approximately 4.00 W/m².
Problem 1 An object with m1 = 5kg is attached to a spring of negligible mass. This mass/spring combination is then slid horizontally on a frictionless surface with a velocity of 5 m/s towards a stationary object with m2 = 6kg. Upon impact, the spring compresses, then we examine two cases. First, find the velocities of the two objects assuming the spring completely relaxes again after the interaction. Second, assume that m2, after they separate, slides up a frictionless incline. (a)What is the relative speed of the masses when the spring is maximally compressed?
Answer and Explanation:
The answer is attached below
The velocity of both the objects after the collision is [tex]\frac{25}{11} m/s[/tex].
(a) the relative velocity of the masses is zero.
(b) the compression of the spring is 0.185m.
Inelastic collision:
The given case is an example of an inelastic collision in which two objects after the collision move together. The momentum of the system is conserved, therefore,
[tex]m_1u_1+m_2u_2=(m_1+m_2)v[/tex]
here, m₁ = 5kg , m₂ = 6kg
u₁ = 5 m/s , initial velocity of mass m₁, and
u₂= 0, initial speed of the mass m₂
[tex]5\times5+0=(5+6)v\\\\v=\frac{25}{11}m/s[/tex]
(a) When the spring is fully compressed both the masses move with the same velocity, therefore the relative speed of the masses is zero.
(b) from the law of conservation of energy:
the initial kinetic energy of the masses is converted into final kinetic energies and the potential energy of the spring:
[tex]\frac{1}{2} m_1u_1^2+\frac{1}{2} m_2u^2_2=\frac{1}{2} (m_1+m_2)v^2+\frac{1}{2}kx^2[/tex]
where x is the compression of the spring
[tex]\frac{1}{2}\times5\times5^2+0=\frac{1}{2}(5+6)\times(\frac{25}{11})^2=\frac{1}{2}\times2000x^2\\\\x=0.185m[/tex]
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What is the ideal banking angle (in degrees) for a gentle turn of 1.75 km radius on a highway with a 120 km/h speed limit (about 75 mi/h), assuming everyone travels at the limit?
Answer:
Ф = 4.98 °
Explanation:
Given:
R = 1.75 km = 1.75 * 10^3
g = 9.8 m/s^2
v = 120 km/h ==> 33.33 m/s
required:
Ф
solution:
The angle(Ф) is given by
Ф = arctan(v^2/R*g)
= arctan(33.33^2/9.8*1.75 * 10^3)
= 4.98 °
note:
there maybe error in calculation but method is correct.
A backcountry skier weighing 700 N skis down a steep slope, unknowingly crossing a snow bridge that spans a deep, hidden crevasse. The bridge can support 470 N − meaning that's the maximum normal force it can sustain without collapsing.
Answer:
The minimum slope angle for which the skier can safely traverse the snow bridge is 47.82°
Explanation:
Given that,
Weight = 700 N
Normal force = 470 N
Suppose, Find the minimum slope angle for which the skier can safely traverse the snow bridge.
We need to calculate the minimum slope angle
Using balance equation
[tex]F_{N}=mg\cos\theta[/tex]
[tex]\theta=\cos^{-1}(\dfrac{F_{N}}{mg})[/tex]
Put the value into the formula
[tex]\theta=\cos^{-1}(\dfrac{470}{700})[/tex]
[tex]\theta=47.82^{\circ}[/tex]
Hence, The minimum slope angle for which the skier can safely traverse the snow bridge is 47.82°
When a 700 N skier crosses a snow bridge that can only support 470 N, the bridge will collapse due to the skier's weight exceeding the bridge's maximum normal force. Newton's second law can be used to understand this situation, which states that the net force acting on an object is equal to its mass multiplied by its acceleration. The relationship between weight, mass, and the gravitational force can also be used to determine that the normal force should be equal to the skier's weight for the bridge to be safe.
Explanation:The maximum normal force that the snow bridge can sustain is 470 N. The backcountry skier weighs 700 N, so the bridge will collapse under their weight.
To understand why, we can use Newton's second law, which states that the net force acting on an object is equal to its mass multiplied by its acceleration. The skier's weight is acting downwards, while the normal force of the snow bridge is acting upwards. When the skier crosses the bridge, the normal force should be equal to their weight for the bridge to be in equilibrium. However, since the normal force is less than the skier's weight, the bridge cannot sustain the weight and will collapse.
This situation can be described using the formula: Weight = mass x gravitational force. So, we can calculate the mass of the skier by dividing their weight by the gravitational force. With a weight of 700 N and a gravitational force of 9.8 m/s², the mass is approximately 71.4 kg. Therefore, the normal force supporting the skier on the bridge should be equal to 700 N for it to be safe.
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A polarized light is incident on several polarizing disks whose planes are parallel and centered on common axis. Suppose that the transmission axis of the first polarizer is rotated 20° relative to the axis of polarization of the incident light, and that the transmission axis of each additional analyzer is rotated 20° relative to the transmission axis of the previous one. What is the maximum number of polarizer needed (whole number), so the transmitted light through all polarizing sheets has an intensity that is equal at least 12% that striking the first polarizer?
Answer:
The number of polarizer needed so transmitted light has at least 12% intensity = 17
Explanation:
Given :
Angle between incident light and optic axis of polarizer = 20°
Given that, the transmission axis of each additional analyzer is rotated 20° relative to the transmission axis of the previous one
According to the malus law,
The intensity of the transmitted light passes through the polarizer is proportional to the square of the cosine of angle between the transmission axis to the optic axis.
⇒ [tex]I = I_{o} cos^{2} \alpha[/tex]
Where, [tex]I =[/tex] transmitted intensity through polarizer, [tex]I_{o} =[/tex] incident intensity of the light.
Given in question, all the time [tex]\alpha =[/tex] 20°
By calculation ∴ [tex]cos^{2} 20 = 0.883[/tex]
After 1st polarizer,
∴ [tex]I_{1} = 0.883I_{o}[/tex]
Now we need to multiply all the time 0.883 until we get 0.12 (relative 20° angle given in question)
After 17th polarizer we get 0.1205 ≅ 0.12
[tex]I_{17} = 0.883^{17} = 0.1205 \times 100 = 12[/tex]% [tex]I_{o}[/tex]
Means we get 12% intensity after 17th polarizing disk.
An ideal battery would produce an extraordinarily large current if "shorted" by connecting the positive and negative terminals with a short wire of very low resistance. Real batteries do not. The current of a real battery is limited by the fact that the battery itself has resistance.
What is the resistance of a 9.0 V battery that produces a 17 A current when shorted by a wire of negligible resistance?
R=____
Answer:
Resistance, [tex]R=0.529\ \Omega[/tex]
Explanation:
Given that,
Voltage of the battery, V = 9 volts
Current produced in the circuit, I = 17 A
We need to find the resistance when shorted by a wire of negligible resistance. It is a case of Ohm's law. The voltage is given by :
[tex]V=IR[/tex]
[tex]R=\dfrac{V}{I}[/tex]
[tex]R=\dfrac{9\ V}{17\ A}[/tex]
[tex]R=0.529\ \Omega[/tex]
So, the resistance in the circuit is 0.529 ohms. Hence, this is the required solution.
The radius of Venus (from the center to just above the atmosphere) is 6050 km (6050✕103 m), and its mass is 4.9✕1024 kg. An object is launched straight up from just above the atmosphere of Venus. (a) What initial speed is needed so that when the object is far from Venus its final speed is 8000 m/s? vinitial = m/s (b) What initial speed is needed so that when the object is far from Venus its final speed is 0 m/s? (This is called the "escape speed.") vescape = m/s
Answer:
(a) The initial speed required is 13116 m/s
(b) The escape speed is 10394 m/s
This problem involves the application of newtons laws of gravitation. The forces in action here are conservative and as a result mechanical energy is conserved.
The full calculation can be found in the attachment below.
Explanation:
In both parts (a) and (b) the energy conservation equation were used. Assumption was made that when the object is very far from the planet the distance from the planet's center approaches infinity and the gravitational potential energy approaches zero.
The calculation can be found below.
Suppose a person riding on top of a freight car shines a searchlight beam in the direction in which the train is traveling. Compare the speed of the light beam relative to the ground when:
a. The train is at rest.
b. The train is moving.
How does the behavior of the light beam differ from the behavior of a bullet fired in the same direction from the top of the freight car?
Answer:
Explanation:
a ) The speed of light will be 3 x 10⁸ m /s
b ) Speed of light will again be 3 x 10⁸ m /s . for an observer on the ground. It is so because speed of light is independent of moving frame of reference . It is absolute . It can not be changed by changing frame of reference .
The behavior of light differs because the speed of bullet will be less in case of stationary train . For moving train , speed of bullet will be increased by amount equal to speed of train for an observer on the ground.
There has long been an interest in using the vast quantities of thermal energy in the oceans to run heat engines. A heat engine needs a temperature difference, a hot side and a cold side. Conveniently, the ocean surface waters are warmer than the deep ocean waters. Suppose you build a floating power plant in the tropics where the surface water temperature is ≈≈ 30 ∘C∘C. This would be the hot reservoir of the engine. For the cold reservoir, water would be pumped up from the ocean bottom where it is always ≈≈ 5 ∘C∘C.What is the maximum possible efficiency of such a power plant? In%.
Answer:
[tex]\eta_{th} = 8.247\%[/tex]
Explanation:
The maximum possible efficiency for the floating power plant is given by the Carnot's Efficiency:
[tex]\eta_{th} = \left(1-\frac{278.15\,K}{303.15\,K} \right)\times 100\%[/tex]
[tex]\eta_{th} = 8.247\%[/tex]
What pressure gradient along the streamline, dp/ds, is required to accelerate air at standard temperature and pressure in a horizontal pipe at a rate of 274 ft/s2?
Answer:
The answer to this question is attached.
The pressure gradient required to accelerate the air inside a pipe can be calculated using Euler's simplified equation for fluid motion. At standard temperature and pressure, you would need a pressure gradient of approximately -102.3 Pa/m to accelerate air at 274 ft/s².
Explanation:To calculate the pressure gradient, or dp/ds, required to accelerate the air inside a pipe, we can use Euler's equation for fluid motion, which in its simplified form is dp/ds = -ρa, where ρ represents the density of the fluid (in this case, air) and a is the acceleration. For standard temperature and pressure, the density of the air is approximately 1.225 kg/m³. It is important to note that the acceleration provided is in feet per second squared, so we would need to convert that to meters per second squared to match units with the density. The conversion factor is 1 ft/s² = 0.3048 m/s², resulting in an acceleration of approximately 83.515 m/s². By substituting these values into the equation, we get dp/ds = -(1.225 kg/m³)(83.515 m/s²) = -102.3 kg/(m·s²), or -102.3 Pa/m, since a Pascal (Pa) is equivalent to a kg/(m·s²).
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Four forces of the same magnitude but differing directions act at and tangent to the rim of a uniform wheel free to spin about its center of mass (CM). Which statement about the wheel is correct?
A. None of these
B. Neither the net force on the wheel nor the net torque on the wheel about the CM
is zero.
C. The net force on the wheel is not zero but the net torque about the CM is zero.
D. The net force on the wheel and the net torque on the wheel about the CM are zero.
E. The net force on the wheel is zero but the net torque about the CM is not zero.
Final answer:
The correct statement about a wheel with four forces of the same magnitude but different directions acting tangentially at its rim is that both the net force on the wheel and the net torque about the CM are zero.
Explanation:
When four forces of the same magnitude but differing directions act tangentially at the rim of a uniform wheel free to spin about its center, the net force acting on the wheel may be zero if the forces cancel each other out. However, the direction and point of application of these forces are critical in determining the net torque. If the forces are applied in such a way that they create rotational effects that cancel each other, the net torque about the center of mass (CM) will also be zero. Therefore, the correct statement about the wheel is:
D. The net force on the wheel and the net torque on the wheel about the CM are zero.
To put it simply, the forces are arranged in a way that causes them to balance out, leading to no linear acceleration (net force is zero), and they are positioned symmetrically around the center of mass, producing no rotational acceleration (net torque is zero).
The desperate contestants on a TV survival show are very hungry. The only food they can see is some fruit hanging on a branch high in a tree. Fortunately, they have a spring they can use to launch a rock. The spring constant is 1000 N/m, and they can compress the spring a maximum of 30 cm. All the rocks on the island seem to have a mass of 400 g.a. With what speed does the rock leave the spring?b. If the fruit hangs 15 m above the ground, will they feast or go hungry?
Answer:
(a) v = 15m/a
(b) No they won't feast because the rock can only rise to a height of 11.5m which is less than 15m.
Explanation:
Please see the attachment below for film solution.
The rock does not reach the fruit because the maximum height it achieves is 11.48 meters, which is below the required 15 meters. Therefore, the contestants will go hungry.
Let's solve the problem step by step. First, we'll determine the speed with which the rock leaves the spring, and then we'll check if this speed is enough to reach the fruit hanging in the tree.
Part (a): Speed of the rock leaving the spring
Using the principle of conservation of energy, the potential energy stored in the spring when compressed will be converted into the kinetic energy of the rock when the spring is released.
The potential energy stored in the spring [tex]\( E_{\text{spring}} \)[/tex] is given by:
[tex]\[E_{\text{spring}} = \frac{1}{2} k x^2\][/tex]
Substituting the given values:
[tex]\[E_{\text{spring}} = \frac{1}{2} \times 1000 \, \text{N/m} \times (0.30 \, \text{m})^2\][/tex]
[tex]\[E_{\text{spring}} = \frac{1}{2} \times 1000 \times 0.09\][/tex]
[tex]\[E_{\text{spring}} = 45 \, \text{J}\][/tex]
This energy will be converted into kinetic energy [tex]\( E_{\text{kinetic}} \)[/tex] of the rock:
[tex]\[E_{\text{kinetic}} = \frac{1}{2} m v^2\][/tex]
Equating the spring potential energy to the kinetic energy:
[tex]\[45 \, \text{J} = \frac{1}{2} \times 0.4 \, \text{kg} \times v^2\][/tex]
Solving for ( v ):
[tex]\[45 = 0.2 \times v^2\][/tex]
[tex]\[v^2 = \frac{45}{0.2}\][/tex]
[tex]\[v^2 = 225\][/tex]
[tex]\[v = \sqrt{225}\][/tex]
[tex]\[v = 15 \, \text{m/s}\][/tex]
So, the speed of the rock as it leaves the spring is [tex]\( 15 \, \text{m/s} \).[/tex]
Part (b): Checking if the rock can reach the fruit
To determine if the rock can reach the height of 15 m, we use the following kinematic equation for vertical motion:
[tex]\[h = v_0 t - \frac{1}{2} g t^2\][/tex]
First, determine the time ( t ) it takes to reach the maximum height where the vertical velocity becomes zero:
[tex]\[v = v_0 - g t\][/tex]
At the maximum height, [tex]\( v = 0 \):[/tex]
[tex]\[0 = 15 - 9.81 t\][/tex]
[tex]\[t = \frac{15}{9.81}\][/tex]
[tex]\[t \approx 1.53 \, \text{s}\][/tex]
Now, calculate the maximum height [tex]\( h_{\text{max}} \):[/tex]
[tex]\[h_{\text{max}} = v_0 t - \frac{1}{2} g t^2\][/tex]
[tex]\[h_{\text{max}} = 15 \times 1.53 - \frac{1}{2} \times 9.81 \times (1.53)^2\][/tex]
[tex]\[h_{\text{max}} = 22.95 - 11.47\][/tex]
[tex]\[h_{\text{max}} = 11.48 \, \text{m}\][/tex]
The maximum height reached by the rock is approximately 11.48 m, which is less than the 15 m needed to reach the fruit.
A phone cord is 4.67 m long. The cord has a mass of 0.192 kg. A transverse wave pulse is produced by plucking one end of the taunt cord. The pulse makes four trips down and back along the cord in 0.794 s.
What is the tension in the cord?
Answer:
91.017N
Explanation:
Parameters
L=4.67m, m=0.192kg, t = 0.794s, The pulse makes four trips down and back along the cord, we have 4 +4 =8 trips( to and fro)
so N= no of trips = 8, From Wave speed(V) = N *L/t , we have :
V= 8*4.67/0.794 = 47.0529 m/s.
We compute the cords mass per length, Let it be P
P = M/L = 0.192/4.67 = 0.04111 kg/m
From T = P * V^2 where T = Tension, we have
T = 0.04111 * (47.0529)^2
T = 91.017N.
The tension in the cord is 91.017N
A 925 kg car rounds an unbanked curve at a speed of 25 m/s. If the radius of the curve is 72, what is the minimum coefficient of friction between the car and the road required so that the car does not skid?
Answer:
[tex]\mu_s^{min}=0.885[/tex]
Explanation:
The centripetal force is provided by the static friction between the car and the road, and always have to comply with [tex]f\leq\mu_sN[/tex], so we have:
[tex]ma_{cp}=f\leq\mu_sN=\mu_smg[/tex]
Which means:
[tex]a_{cp}=\frac{v^2}{r}\leq\mu_sg[/tex]
So we have:
[tex]\frac{v^2}{gr}\leq\mu_s[/tex]
Which means that [tex]\frac{v^2}{gr}[/tex] is the minimum value the coefficient of static friction can have, which for our values is:
[tex]\mu_s^{min}=\frac{v^2}{gr}=\frac{(25m/s)^2}{(9.81m/s^2)(72m)}=0.885[/tex]
To keep the car from skidding, the frictional force between the car and the road must provide the necessary centripetal force to keep the car moving in a circle. The minimum coefficient of friction required would be about 0.86.
Explanation:To determine this, we must understand that the friction force between the car and the road surface is what keeps the car in a curved path. If the speed of the car or the radius of the curve is too large, a greater force is needed to keep the car from skidding. This force must be supplied as an increased friction force, which implies a larger coefficient of friction.
The centripetal force needed to keep a car on the road as it rounds a turn is provided by the frictional force between the road and the car's tires. We can use the formula for centripetal force where Fc=m*v^2/r. Here, Fc is the centripetal force, m is the mass of the car, v is the speed of the car, and r is the radius of the curve. In this case, friction provides the centripetal force, so Fc is also equal to the force of static friction, which is less than or equal to the coefficient of static friction times the normal force (µ*m*g).
Setting these equal and solving for µ gives µ=v^2/(g*r). Plugging in the given values (v=25 m/s, g=9.8 m^2/s, and r=72 m), we find that the minimum coefficient of friction required is about 0.86. This is reasonable; a typical car with good tires on dry concrete requires a minimum coefficient of friction of about 0.7 to keep from skidding in a curve.
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An inventor claims to have devised a cyclical engine for use in space vehicles that operates with a nuclear-fuelgenerated energy source whose temperature is 920 R and a sink at 490 R that radiates waste heat to deep space. He also claims that this engine produces 4.5 hp while rejecting heat at a rate of 15,000 Btu/h. Is this claim valid?
Answer:
Valid
Explanation:
to determine if the claim is valid, we compare the efficiency of the device to that of a Carnot engine.
The following data were given
High Temperature = 920R,
Low temperature =490R
work=4.5hp =4.5*2544.5=11450.25Btu/h
low heat Ql heat= 15000Btu/h
High heat Qh=work +Ql=11450.25Btu/h+15000Btu/h=26450.25Btu/h
Next we cal calculate the efficiency of the Carnot engine
[tex]E_Carnot=1-\frac{T_L}{T_H}\\ E_Carnot=1-\frac{490}{920}\\ E_Carnot=0.467[/tex]
Hence the maximum efficiency at the given temperature is 47%
Next we calculate the efficiency of the device
[tex]E_device=\frac{work}{Q_H} \\E_device=\frac{11450.25}{26450.25} \\E_device=0.433\\[/tex]
which is 43%
since the maximum efficiency of 47% is not exceeded, we can conclude that the claim is valid
While describing a circular orbit 300 mi above the earth a space vehicle launches a 6000-lb communications satellite. Determine the additional energy required to place the satellite in a geosynchronous orbit at an altitude of 22,000 mi
Answer:
[tex]\Delta U = 2.2126039 x 10^{12} J[/tex]
Explanation:
While the satellite is in the space vehicle, it has the next potential energy
[tex]U = -\frac{GmMe}{r}[/tex]
where G is the gravitational constant
m is the satellite's mass in kilograms
Me is the earth's mass
r is the orbit's radius from to the earth's center in meters
[tex]U = - \frac{6.67x10^{-11}*2721.554*5.972x10^{24} }{482803}[/tex]
[tex]U = -2.2423x10^{12} J[/tex]
The additional energy required is the difference between this energy and the energy that the satellite would have in an orbit with an altitude of 22000 mi
[tex]U = -\frac{6.67x10^{-11}*2721.554*5.792x10^{24} }{35405568}[/tex]
[tex]U = -29696124610.3 J[/tex]
Then
[tex]\Delta U = 2.2126039 x 10^{12} J[/tex]
When driving straight down the highway at a constant velocity you have to give the engine a little gas (which means an added external force of tires pushing on the road) to maintain your uniform motion. Explain why this does not violate Newton’s 1st law. What are the other forces acting on the vehicle?"
Answer:
Explanation:
This does not violate Newton's 1st law because the net force would still be 0 in order to produce uniform motion (aka constant velocity). The other forces acting on the vehicles is air resistance which is non-zero. So we need car internal force to counter balance this force, which require extra gas for the car.
Applying extra gas to a car moving at a constant velocity does not violate Newton's 1st Law as the additional force is required to counteract other forces, such as air resistance and road friction, allowing the car to maintain its velocity. This is, in fact, a confirmation of Newton's 1st Law.
Even while driving at a constant velocity on a straight highway, one must occasionally apply a bit of extra gas, which means applying an extra external force.
However, this does not violate Newton's 1st Law, or the law of inertia, which states that an object in motion tends to stay in motion with the same speed and in the same direction unless acted upon by an unbalanced force.
The reason for this is the existence of other forces acting on the car that work against the forward motion. These can include air resistance (or drag), friction from the road, and uphill forces if the road is inclined.
The extra force (gas) provided helps overcome these forces, allowing the car to maintain a constant velocity. So, this activity is actually a confirmation of Newton's first Law, not a violation.
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A piece of Nichrome wire has a radius of 6.5 104 m. It is used in a laboratory to make a heater that uses 4.00 102 W of power when connected to a voltage source of 120 V. Ignoring the effect of temperature on resistance, estimate the necessary length of wire.
Answer:
[tex]L=4.8*10^{17}m[/tex]
Explanation:
Given data
Power P=4.00×10²W
Radius r=6.5×10⁴m
Voltage V=120V
To find
Length of wire L
Solution
We know that resistance of wire can be obtained from
[tex]P=\frac{V_{2}}{R}\\ R=\frac{V_{2}}{P}[/tex]
We also know that R=pL/A solving the length noting that A=πr²
and using p=100×10⁻⁸Ω.m we find that
So
[tex]L=\frac{RA}{p}\\ L=\frac{\frac{(V^{2})}{P}(\pi r^{2}) }{p} \\L=\frac{V^{2}(\pi r^{2})}{pP}\\ L=\frac{(120V)^{2}\pi (6.5*10^{4} m)^{2} }{100*10^{-8}(4.00*10^{2} W) }\\ L=4.8*10^{17}m[/tex]
Final answer:
To calculate the length of Nichrome wire needed for a heating element using 400 W at 120 V, one determines the resistance and then uses it with the wire's cross-sectional area and resistivity. Approximately 41.45 meters of wire is required.
Explanation:
To estimate the necessary length of Nichrome wire for a laboratory heater, we start by calculating the resistance using the power and voltage supplied. The formula for power (P) in terms of voltage (V) and resistance (R) is P = V2 / R. Given P = 400 W and V = 120 V, the resistance can be calculated as follows:
R = V2 / P = (1202) / 400 = 36 Ω.
Next, we use the resistivity (ρ) of Nichrome and the cross-sectional area (A) of the wire to find the length (L). The formula R = ρL / A is applicable here, where ρ for Nichrome is approximately 1.10×10-6 Ω·m and A = πr2. The radius (r) given is 6.5×10-4 m, so:
A = π(6.5×10-4)2 = 1.33×10-6 m2.
Substituting the values into the formula, we get:
L = (R · A) / ρ = (36 · 1.33×10-6) / (1.10×10-6) ≠ 41.45 m.
Thus, approximately 41.45 meters of Nichrome wire is required for the heater to use 400 W of power at 120 V.
You are a member of a citizen's committee investigating safety in the high school sports program. You are interested in knee damage to athletes participating in the long jump (sometimes called the broad jump). The coach has her best long jumper demonstrate the event for you. He runs down the track and, at the take-off point, jumps into the air at an angle of 30 degrees from the horizontal. He comes down in a sand pit at the same level as the track 26 feet away from his take-off point. With what velocity (both magnitude and direction) did he hit the ground?
The velocity at which the long jumper landed can be computed by separating the jump into horizontal and vertical components, using the initial take-off angle and the horizontal distance. The horizontal and vertical velocities are combined to find the magnitude of the total landing velocity, and the arctangent of their ratio gives the direction.
Explanation:To determine the velocity at which the long jumper landed, we need to consider the projectile motion of the jump. There are two components of velocity to consider: the horizontal (vx) and the vertical (vy) at the point of landing.
First, the horizontal velocity (vx) can be found by dividing the total horizontal distance by the time of flight (t). The equation of horizontal motion is:
vx = d / t, where d is the horizontal distance.
The vertical velocity (vy) of the jumper when he lands will be the same magnitude but opposite in direction to the vertical velocity at take-off due to symmetry in projectile motion, assuming no air resistance. The vertical velocity at take-off can be calculated using the initial jump angle (Ө) and the initial velocity (v0).
Using the initial angle of 30°, the vertical component of the initial velocity at take-off (v0y) is:
v0y = v0 * sin(Ө).
Since the vertical motion is subject only to acceleration due to gravity (g), the final vertical velocity at landing (vy) will be v0y (but in the opposite direction).
The total landing velocity (v) is then found by combining these two components using the Pythagorean theorem:
v = √(vx^2 + vy^2).
The direction of the landing velocity is given by the angle made with the horizontal, found using the arctangent of the ratio of vy over vx:
θ = arctan(vy / vx).
These calculations would provide the magnitude and direction of the landing velocity.
Problem 14.45 Two small spheres A and B, of mass 2.5 kg and 1 kg, respectively, are connected by a rigid rod of negligible weight. The two spheres are resting on a horizontal, frictionless surface when A is suddenly given the velocity Determine 0 v (3.5 m/s) .i (a) the linear momentum of the system and its angular momentum about its mass center G, (b) the velocities of A and B after the rod AB has rotated through 180 .
Answer:
(a) linear momentum = 8.75. angular momentum =0. (b) 2.5m/s
Explanation:
A and B are on horizontal and so is the velocity given to A = 3.5m/s i (in i direction which is x direction ), this means that there would be motion only in x direction.
let's first calculate linear momentum.
(A)
p(total)= P_A+P_B = (2.5Kgx3.5m/s)+(1kgx0m/s) = 8.75m/s.
Angular momentum.
L=m*v x r. as the motion is only in x direction and there is no rotation in the system, there fore m*v x r = 0
(B)
Velocity of A and B come from the fact that total linear momentum is conserved.
p_final = P_A+P_B = (mA=mB)v_new = mA*V1+mB*vB.
second term on right is = 0 because B has 0 velocity.
solving for V_new and with the values of all unknown substituted in gives
V_new = (mA*VA)/(mA+mB)= 8.75/3.5= 2.5m/s
The diagram showing the 2 spheres is missing, so i have attached it.
Answer:
A) Linear momentum =(8.75 kg/m.s)i
Angular momentum = (-0.5kg.m²/s)k
B) Va' = (1.5m/s)i and Vb' = (5 m/s)j
Explanation:
First of all, let's find the position of the mass centre;
y' = Σ(mi.yi)/mi = [2.5(0) + 1(0.2)]/(2.5+1) = 0.2/3.5 =0.057143 m
A) Linear momentum is given as;
L = m(a) x v(o) = (2.5 x 3.5)i = (8.75 kg/m.s)i
Angular momentum is given as;
HG = Vector GA x m(a) x v(o)
Where vector GA is the position of the mass centre;
Thus;
HG = 0.057143j x (2.5 x 3.5)i = (-0.5 kg.m²/s) k
B) from conservation of linear momentum;
MaVo = Ma Va' + Mb Vb
2.5 x 3.5 = 2.5Va' + 1 Vb'
8.75 = 2.5Va' + 1 Vb' - - - - eq(1)
Also, for conservation of angular momentum ;
raMaVo = - raMa Va' + rbMb Vb'
from the diagram attached, ra + rb = 0.2
Now, ra is the same as value as that of the centre of the mass.
Thus, ra = 0.057143.
rb = 0.2 - ra = 0.2 - 0.057143 = 0.14286
Thus;
0.057143 x 8.75 = -(0.057143 x 2.5)Va' + (0.14286 x 1) Vb'
0.5 = - 0.14286Va' + 0.14286Vb' - - - eq 2
Solving eq 1 and 2 simultaneously, we get; Va' = 1.5m/s and Vb' = 5 m/s
An RLC circuit with a resistor of R\:=\:1 R = 1 k, capacitor ofC\:=\:3 C = 3 F, and inductor ofL\:=\:2 L = 2 H reaches a maximum current through the inductor of 7 mA. When all of the energy stored in the circuit is in the inductor, what is the magnetic energy stored? (Express your answer in micro-Joules) In the same circuit explained above, if all the energy is then transferred into the capacitor, what voltage drop will there be across the capacitor? (Express your answer in Volts to the hundredths place)
Answer:
Magnetic energy stored in the inductor when all of the energy in the circuit is in the inductor = 0.049 mJ
If all the energy is then transferred into the capacitor, the voltage drop across the capacitor = 0.00572 V = 0.01 V (expressed to the hundredths value)
Explanation:
In an RLC circuit with maximum current of 7mA = 0.007 A
When all of the energy is stored in the inductor, maximum current will flow through it,
Hence E = (1/2) LI²
L = inductance of the inductor = 2 H
E = (1/2) (2)(0.007²) = 0.000049 J = 0.049 mJ
When all the energy in the circuit is in the capacitor, this energy will be equal to the energy calculated above.
And for a capacitor, energy is given as
E = (1/2) CV²
E = 0.000049 J, C = 3 F, V = ?
0.000049 = (1/2)(3)(V²)
V = 0.00572 V = 0.01 V
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A vaulter is holding a horizontal 3.00-kg pole, 4.50 m long. His front arm lifts straight up on the pole, 0.750 from the end, and his back arm pushes straight down on the end of the pole. How much force does his back arm exert on the pole?
(Unit= N)
NEED HELP FAST
Answer:
58.8 N
Explanation:
Let 'F₁' be the force by front arm and 'F₂' be the force by back arm.
Given:
Mass of the rod (m) = 3.00 kg
Length of the pole (L) = 4.50 m
Acceleration due to gravity (g) = 9.8 m/s²
Distance of 'F₁' from one end of pole (d₁) = 0.750 m
'F₂' acts on the end. So, distance between 'F₁' and 'F₂' = 0.750 m
Now, weight of the pole acts at the center of pole.
Now, distance of center of pole from 'F₁' is given as:
d₂ = (L ÷ 2) - d₁
[tex]d_2=\frac{4.50}{2}-0.75=1.5\ m[/tex]
Now, as the pole is held horizontally straight, the moment about the point of application of force 'F₁' is zero for equilibrium of the pole.
So, Anticlockwise moment = clockwise moment
[tex]F_2\times d_1=mg\times d_2\\\\F_2=\frac{mg\times d_2}{d_1}[/tex]
Plug in the given values and solve for 'F₂'. This gives,
[tex]F_2=\frac{3.00\ kg\times 9.8\ m/s^2\times 1.5\ m}{0.75\ m}\\\\F_2=\frac{44.1}{0.75}=58.8\ N[/tex]
Therefore, the force exerted by the back arm on the pole is 58.8 N vertically down.
Reasons for which petroleum is the chosen fuel for transportation in the U.S. include
1) its energy value per unit volume
2) its ability to quickly start or stop providing energy
3) its low amount of pollution produced per joule
Answer:
The Reasons for which petroleum is the chosen fuel for transportation in the U.S. include number 1) and 2).
Explanation:
The main reasons why oil is the most used fuel in the USA is because of its energy potential with a high combustion power and a better octane rating compared to other fuels. Thus, fossil fuel derived from oil has a high energy value per unit volume and an ability to quickly start or stop providing energy.
calorimeter has aluminum inner cup of mass 120 gram containing 100 ml water at temperature 20 degree Celsius. Brass piece with mass 100 gram is heated to 100 degree Celsius, and then immersed in the calorimeter. Calculate the final temperature of the system. The specific heat of brass is 0.09 cal/(gramXdeg.C) . The additional necessary data are provided in the text.
Answer:
the final temperature of the system of the system is 25.32°C
Explanation:
We are not given specific capacity of water and aluminium, so we use their standard values, also we are not given the density if water so we assume the standard vale of density of water
The aluminium calorimeter has a mass Mc= 120g
Volume of water in calorimeter = 100ml at θc =20°C
Density of water is
1000Kg/m³ = 1g/mL
Then, density = mass/ volume
Mass=density ×volume
Mass=1g/mL×100mL
Mass=100gram
Then, the mass of water is
Mw = 100gram
Mass of brass is Mb = 100gram
The temperature of brass is θb=100°C
The specific heat capacity of water is Cw= 1cal/g°C
The specific heat capacity of aluminum Ca=0.22cal/g°C
We are looking for final temperature θf=?
Given that the specific heat capacity of brass is Cb=0.09Cal/g°C
Using the principle of calorimeter;
The principle of calorimetry states that if there is no loss of heat in surrounding the total heat loss by hot body equals to total heat gained by a cold body.
So, the calorimeter gained heat and the liquid in the calorimeter gain heat too
Heat gain by calorimeter(Hc) = Mc•Ca•∆θ
Where Mc is mass of calorimeter,
Ca is Specific Heat capacity of Calorimeter
∆θ=(θf-θc)
Hc=Mc•Ca•∆θ
Hc=120•0.22•(θf-20)
Hc=26.4(θf-20)
Hc=26.4θf-528
Also, heat gain by the water
Heat gain by wayer(Hw) = Mw•Cw•∆θ
Where Mw is mass of water,
Cw is Specific Heat capacity of water
∆θ=(θf-θw),
Note that the temperature of the water and the calorimeter are the same at the beginning i.e. θc=θw=20°C
Hw=Mw•Cw•∆θ
Hw=100•1•(θf-20)
Hw=100(θf-20)
Hw=100θf-2000
Also heat loss by the brass is given by
heat loss by brass
Heat loss by brass(Hb)= Mb•Cb•∆θ
Where Mb is mass of brass,
Cb is Specific Heat capacity of brass
∆θ=(θb-θf)
Therefore,
Hb=Mb•Cb•∆θ
Hb=100•0.09•(100-θf)
Hb=9(100-θf)
Hb=900-9θf
Applying the principle of calorimeter
Heat gain = Heat loss
Hc+Hw=Hb
26.4θf-528 + 100θf-2000=900-9θf
26.4θf+100θf+9θf=900+2000+528
135.4θf=3428
Then, θf=3428/133.4
θf=25.32°C
The final temperature of the system is 25.32 degree Celsius.
Given data:
The mass of aluminum cup is, m = 120 g .
The mass of brass piece is, m' = 100 g.
The volume of water in aluminum cup is, V = 100 ml.
The temperature of water is, T = 20 degree Celsius.
The specific heat of brass is, [tex]c''=0.09 \;\rm cal/g ^\circ C[/tex].
The temperature of brass is, T'' = 100 degree Celsius.
The principle of calorimetry states that if there is no loss of heat in surrounding the total heat loss by hot body equals to total heat gained by a cold body.
So, first calculate heat gain by calorimeter,
H = mc (T' - T)
Here, c is the specific heat of aluminum and its value is, [tex]0.22 \;\rm cal/g^\circ C[/tex]. Solving as,
[tex]H = 120 \times 0.22 (T' - 20)\\H = 26.4T' - 528[/tex]
And, heat gain by water is,
[tex]H'=m'c'(T'-T)[/tex]
Here, c' is the specific heat of water. ( c' =1 )
Solving as,
[tex]H'=100 \times 1 \times (T'-20)\\H' = 100T' -2000[/tex]
Now, heat loss by brass is,
[tex]H'' = m'c'' (T''-T')\\\\H'' = 100 \times 0.09 \times (100-T')\\\\H'' = 900-9T'[/tex]
Applying the principle of calorimeter
Heat gain = Heat loss
H + H' = H''
(26.4T' - 528) + (100T' - 2000) = (900 - 9T')
135.4 T' = 3428
T ' = 25.32 degree Celsius
Thus, we can conclude that the final temperature of the system is 25.32 degree Celsius.
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