Answer: 13 cm
Explanation:
Given
Inductance of the solenoid, L = 3.01•10⁻⁶ H
Width of the wire, x = 1.13 m
Length of the tube, z = ?
Now, we know that
L = μ₀N²A/z, where
N = x/2πr, making r subject of formula,
r = x/2πN
Also,
A = πr², on substituting for A, we have
A = πx²/4π²N²
Now finally, we substitute in the initial equation and solve
L = μ₀N²A/z
L = μ₀N²πx²/4π²N²z
L = μ₀x²/4z, making z subject of formula, we have
z = μ₀x²/4L
z = 4π*10⁻⁷ * 1.13² /4 * 3.01*10⁻⁶
z = (4π*10⁻⁷ * 1.2769) / (4 * 3.01*10⁻⁶)
z = 1.6*10^-6 / 1.2*10^-5
z = 0.13 m
Therefore, the length of the tube should be 13 cm
The length of the tube should be 469 centimeters.
To find the required length of the tube for Tia's solenoid, we can use the formula for the inductance of a solenoid:
[tex]\[ L = \frac{{\mu_0 \cdot N^2 \cdot A}}{l} \][/tex]
Where:
L = inductance of the solenoid (3.01 μH)
[tex]\( \mu_0 \)[/tex] = permeability of free space (4π × 10^-7 T*m/A)
N = number of turns of wire
A = cross-sectional area of the solenoid
[tex]\( l \)[/tex] = length of the solenoid
We're given [tex]\( L \)[/tex] and [tex]\( N \cdot l \)[/tex] (the total length of wire wound around the tube). We need to find [tex]\( l \)[/tex], the length of the tube.
First, let's rearrange the formula to solve for [tex]\( l \)[/tex]:
[tex]\[ l = \frac{{\mu_0 \cdot N^2 \cdot A}}{{L}} \][/tex]
We know the cross-sectional area ( A ) can be represented as [tex]\( A = \pi r^2 \)[/tex], where ( r ) is the radius of the tube. Since the wire is uniformly wound, we can calculate ( A ) using the length of the wire and the number of turns.
[tex]\[ A = \frac{{\text{{Length of wire}}}}{{\text{{Number of turns}}}} \][/tex]
Given that the wire length is 1.13 m and there are [tex]\( N \)[/tex] turns, [tex]\( A = \frac{{1.13}}{{N}} \)[/tex].
Substituting this into the equation for [tex]\( l \)[/tex], we get:
[tex]\[ l = \frac{{\mu_0 \cdot N^2 \cdot \left(\frac{{1.13}}{{N}}\right)}}{{L}} \][/tex]
[tex]\[ l = \frac{{\mu_0 \cdot 1.13}}{{L}} \][/tex]
Now, plug in the known values:
[tex]\[ l = \frac{{4\pi \times 10^{-7} \cdot 1.13}}{{3.01 \times 10^{-6}}} \][/tex]
[tex]\[ l = \frac{{4\pi \times 1.13}}{{3.01}} \][/tex]
[tex]\[ l = \frac{{4 \times 3.1416 \times 1.13}}{{3.01}} \][/tex]
[tex]\[ l = \frac{{14.1376}}{{3.01}} \][/tex]
[tex]\[ l = 4.69 \, \text{m} \][/tex]
Finally, convert this length to centimeters:
[tex]\[ l = 469 \, \text{cm} \][/tex]
So, the length of the tube should be 469 centimeters.
A block of mass m attached to the end of a spring of spring constant k undergoes simple harmonic motion with amplitude A and angular frequency ω. The position of the block is described by a cosine function with an initial phase angle ϕ = 0. Which of the variables m, k, A, or ϕ should you increase if you want to increase the frequency of oscillation?
Answer:
To increase the frequency the spring constant k should be increased.
Explanation:
In simple harmonic motion, the frequency is expressed as;
f = (1/2π)√(k/m)
where;
k is spring constant
m is mass of the object
From the frequency formula i wrote, It’s obvious that frequency depends on only the spring constant and mass while Amplitude(A) and phase angle(ϕ) are not related and would have no effect on the frequency. Thus, by inspection, an increase in spring constant increases frequency while an increase in mass decreases frequency
Which statement best explains why objects are pulled towards earth's center
Answer:
I didn't see any statement. But here is the answer:--
Explanation:
But objects are pulled towards earth's center by gravity. classical mechanics can not define the gravity but general relativity describe it as a curve of space-time by a concentrated energy or mass. As the mass of an object increases the more space time will be curved in the direction of positive energy so the center of the earth is the minimum point of the curved space-time by earth.That's why smaller objects than earth will fall to the center of the earth.
The statement best explains why objects are pulled toward Earth’s center is : Earth has a much greater mass than objects on its surface .Gravity is the attractive force acting between masses. This force is why objects fall to the ground when dropped.
This phenomenon is explained by gravity. As Newton's Law of Universal Gravitation says, every object with mass attracts every other object with mass. The Earth, having a much greater mass compared to objects on its surface, exerts a significant gravitational force pulling objects towards its center. This is why when you drop an object, it falls to the ground. The gravitational force of Earth keeps us grounded and is responsible for attracting objects towards its center.
Complete Question
Which statement best explains why objects are pulled toward Earth’s center?
Earth has a magnetic force that is strongest at its core.
Earth has a much greater mass than objects on its surface.
The weight of the atmosphere pushes objects toward Earth's surface.
The strength of the Sun’s gravity pushes down on objects at Earth's surface.
A sanding disk with rotational inertia 2.0 x 10-3 kg·m2 is attached to an electric drill whose motor delivers a torque of magnitude 11 N·m about the central axis of the disk. About that axis and with torque applied for 19 ms, what is the magnitude of the (a) angular momentum and (b) angular velocity of the disk?
Answer
Given,
Rotational inertia = 2.0 x 10-3 kg·m²
Torque = 11 N.m
time, t = 19 ms
a) Angular momentum
[tex]\tau = \dfrac{\Delta L}{\Delta t}[/tex]
L is angular momentum
[tex]\Delta L = \tau \Delta t[/tex]
[tex]\Delta L = 11\times 19 \times 10^{-3}[/tex]
[tex]\Delta L = 0.209\ Kg m^2/s[/tex]
b) Angular velocity
We know.
[tex]L = I \omega[/tex]
[tex]\omega = \dfrac{L}{I}[/tex]
[tex]\omega = \dfrac{0.209}{2\times 10^{-3}}[/tex]
[tex]\omega = 104.5\ rad/s[/tex]
A truck with a mass of 1400 kg and moving with a speed of 14.0 m/s rear-ends a 641-kg car stopped at an intersection. The collision is approximately elastic since the car is in neutral, the brakes are off, the metal bumpers line up well and do not get damaged. Find the speed of both vehicles after the collision.
After collision, truck remains at 14.0 m/s, car stops (0 m/s) due to elastic collision.
To find the velocities of both vehicles after the collision, we can use the principle of conservation of momentum, which states that the total momentum before the collision is equal to the total momentum after the collision.
The momentum [tex]\( p \) of an object is given by its mass \( m \) multiplied by its velocity \( v \)[/tex]:
[tex]\[ p = mv \][/tex]
Before the collision, the total momentum is the sum of the momentum of the truck and the momentum of the car. After the collision, the total momentum remains the same.
Let [tex]\( v_t \) be the velocity of the truck after the collision, and \( v_c \)[/tex] be the velocity of the car after the collision.
Before the collision:
[tex]\[ \text{Total momentum} = \text{momentum of truck} + \text{momentum of car} \][/tex]
[tex]\[ (1400 \, \text{kg}) \times (14.0 \, \text{m/s}) + (641 \, \text{kg}) \times (0 \, \text{m/s}) = (1400 \, \text{kg}) \times (v_t) + (641 \, \text{kg}) \times (v_c) \][/tex]
[tex]\[ 1400 \times 14.0 = 1400v_t + 641v_c \][/tex]
[tex]\[ 19600 = 1400v_t + 641v_c \][/tex]
Since the collision is approximately elastic, we also have the condition that the relative velocity of approach equals the relative velocity of separation.
[tex]\[ v_t - v_c = 14.0 \, \text{m/s} \][/tex]
We now have a system of two equations:
[tex]\[ 19600 = 1400v_t + 641v_c \][/tex]
[tex]\[ v_t - v_c = 14.0 \][/tex]
We can solve this system of equations to find the values of [tex]\( v_t \)[/tex] and [tex]\( v_c \)[/tex].
[tex]\[ v_t = \frac{19600 - 641v_c}{1400} \][/tex]
Substitute this expression for [tex]\( v_t \)[/tex] into the second equation:
[tex]\[ \frac{19600 - 641v_c}{1400} - v_c = 14.0 \][/tex]
Multiply both sides by 1400 to eliminate the denominator:
[tex]\[ 19600 - 641v_c - 1400v_c = 14.0 \times 1400 \][/tex]
[tex]\[ 19600 - 2041v_c = 19600 \][/tex]
Now, isolate [tex]\( v_c \)[/tex] :
[tex]\[ -2041v_c = 0 \][/tex]
[tex]\[ v_c = 0 \][/tex]
Substitute [tex]\( v_c = 0 \)[/tex] into [tex]\( v_t = \frac{19600 - 641v_c}{1400} \)[/tex] :
[tex]\[ v_t = \frac{19600 - 641 \times 0}{1400} \][/tex]
[tex]\[ v_t = \frac{19600}{1400} \][/tex]
[tex]\[ v_t = 14.0 \, \text{m/s} \][/tex]
So, after the collision, the truck's speed remains 14.0 m/s, and the car's speed becomes 0 m/s.
The speed of the truck and car after the collision is approximately 9.47 m/s
Use the principle of conservation of momentum to solve this problem. The total momentum before the collision is equal to the total momentum after the collision. We can express this as:
[tex]$$ m_{\text{truck}} \cdot v_{\text{truck, initial}} + m_{\text{car}} \cdot v_{\text{car, initial}} = (m_{\text{truck}} + m_{\text{car}}) \cdot v_{\text{final}} $$[/tex]
Given the following information:
- Mass of the truck, [tex]\(m_{\text{truck}}\)[/tex]: 1400 kg
- Initial velocity of the truck, [tex]\(v_{\text{truck, initial}}\)[/tex]: 14.0 m/s
- Mass of the car, [tex]\(m_{\text{car}}\):[/tex] 641 kg
- Initial velocity of the car, [tex]\(v_{\text{car, initial}}\)[/tex]: 0 m/s (since the car is stopped)
We can solve for the final velocity, [tex]\(v_{\text{final}}\):[/tex]
[tex]$$ 1400 \cdot 14.0 + 641 \cdot 0 = (1400 + 641) \cdot v_{\text{final}} $$[/tex]
Solving for [tex]\(v_{\text{final}}\):[/tex]
[tex]$$ v_{\text{final}} = \frac{1400 \cdot 14.0}{1400 + 641} $$[/tex]
Calculating the value:
[tex]$$ v_{\text{final}} \approx 9.47 \, \text{m/s} $$[/tex]
Therefore, the speed of the truck and car after the collision is approximately 9.47 m/s.
A long, thin straight wire with linear charge density λ runs down the center of a thin, hollow metal cylinder of radius R. The cylinder has a net linear charge density 2λ. Assume λ is positive. Part A Find expressions for the magnitude of the electric field strength inside the cylinder, r
Answer:
[tex]E=\frac{\lambda}{2\pi r\epsilon_0}[/tex]
Explanation:
We are given that
Linear charge density of wire=[tex]\lambda[/tex]
Radius of hollow cylinder=R
Net linear charge density of cylinder=[tex]2\lambda[/tex]
We have to find the expression for the magnitude of the electric field strength inside the cylinder r<R
By Gauss theorem
[tex]\oint E.dS=\frac{q}{\epsilon_0}[/tex]
[tex]q=\lambda L[/tex]
[tex]E(2\pi rL)=\frac{L\lambda}{\epsilon_0}[/tex]
Where surface area of cylinder=[tex]2\pi rL[/tex]
[tex]E=\frac{\lambda}{2\pi r\epsilon_0}[/tex]
A proton is located at the origin and an electron is located at (1.0, 1.0) mm:
(a) Determine the electric dipole moment of these two particles in unit vector notation.
(b) If we put this electric dipole moment in an external electric field vector(E) = 300vector(i) (N/C), calculate the work done by the electric field to rotate the dipole so that it becomes in the same direction as the field.
Answer:
(a). [tex]{\vec{p} =(1.6*10^{-22}\bold{i}+1.6*10^{-22}\bold{j})m \cdot C.}[/tex]
(b). [tex]U = 4.8*10^{-20}J.[/tex]
Explanation:
(a).
The electric dipole moment of the charges is
[tex]\vec{p} = q \vec{r}[/tex]
In our case
[tex]\vec{r} = (1.0*10^{-3}\bold{i}+1.0*10^{-3}\bold{j})m[/tex]
and
[tex]q =1.6*10^{-19}C[/tex];
therefore, the dipole moment is
[tex]\vec{p} =1.6*10^{-19}C *(1.0*10^{-3}\bold{i}+1.0*10^{-3}\bold{j})m[/tex]
[tex]\boxed{\vec{p} =(1.6*10^{-22}\bold{i}+1.6*10^{-22}\bold{j})m \cdot C.}[/tex]
(b).
The work done [tex]U[/tex] by an external electric field [tex]\vec{E}[/tex] is
[tex]U = -\vec{p}\cdot \vec{E}[/tex]
[tex]U = [1.6*10^{-22}\bold{i}+1.6*10^{-22}\bold{j}] \cdot[300\bold{i}][/tex]
[tex]\boxed{U = 4.8*10^{-20}J.}[/tex]
If the entire apparatus were submerged in water, would the width of the central peak change? View Available Hint(s) If the entire apparatus were submerged in water, would the width of the central peak change? The width would increase. The width would decrease. The width would not change.
Answer:
The width would decrease
Explanation:
The width of central maximum of screen is proportional to wavelength of light and wavelength of light in water is less than that of air
If the apparatus were submerged in water, the width of the central peak as a part of the light diffraction pattern would decrease. This is because water slows down light more than air does - reducing the wavelength and therefore narrowing the central peak.
Explanation:If the entire apparatus were submerged in water, the width of the central peak would indeed change. This is due to the phenomenon of diffraction, which describes the way waves spread out when they pass through an opening. The central peak referred to in the question is a part of the diffraction pattern observed when light passes through a slit.
When diffraction occurs, a pattern of bright and dark spots, or 'fringes', is created. The width of these fringes is determined by the wavelength of the light and the size of the opening. If we were to submerge the entire apparatus in water, the diffraction pattern would change. This is because water has a higher refractive index than air, meaning it slows down the light more. As a result, the wavelength of the light in water becomes smaller compared to that in air. According to the formula for diffraction, the smaller the wavelength, the narrower the central peak. So in conclusion, if the apparatus were submerged in water, the width of the central peak would decrease.
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Alligators and other reptiles don't use enough metabolic energy to keep their body temperatures constant. They cool off at night and must warm up in the sun in the morning. Suppose a 300 kg kg alligator with an early-morning body temperature of 25 ∘ C ∘C is absorbing radiation from the sun at a rate of 1200 W How long will the alligator need to warm up to a more favorable 30 C?
Answer:
Explanation:
Amount of heat required can be found from the following relation
Q = mcΔT
m is mass of the body , c is specific heat and ΔTis rise in temperature .
Here m = 300 kg
c = 3350 J /kg k
ΔT = 30 - 25
= 5 °C
Putting the values in the expression above
Q = 300 x 3350 x 5
= 5025000 J
Rate at which energy is absorbed = 1200 J /s
Time required
= 5025000 / 1200
= 4187.5 S
= 69.8 minute
= 1 hour 9.8 mimutes.
Consider 500 g of silver (atomic mass 107 g/mol). Assume that the temperature is high enough that equipartition applies. We heat the silver by immersing it in water; by measuring the temperature change of the water we compute that 300 J of heat energy entered the silver.
What was the change in temperature of the silver?
A. 2.57 K
B. 0.6 K
C. 4.99 K
D. 169 K
E. 0.0466 K
Answer:
A. 2.57 K
Explanation:
From specific heat capacity,
Q = cmΔT........................ Equation 1
Where Q = Amount of heat that entered into silver, m = mass of silver, c = specific heat capacity of silver, ΔT = change in temperature of the silver.
make ΔT the subject of the equation
ΔT = Q/cm................... Equation 2
Given: Q = 300 J, m = 500 g = 0.5 kg
Constant: c = 233 J/kg.K
Substitute into equation 2
ΔT = 300/(0.5×233)
ΔT = 300/116.5
ΔT = 2.57 K
Hence the right option is A. 2.57 K
The change in the temperature of the silver should be considered as the 2.57 K.
Change in temperature:
From specific heat capacity,
Q = cmΔT........................ Equation 1
Here
Q = Amount of heat that entered into silver,
m = mass of silver,
c = specific heat capacity of silver,
ΔT = change in temperature of the silver.
Now
ΔT = Q/cm................... Equation 2
Since
Q = 300 J, m = 500 g = 0.5 kg
Constant: c = 233 J/kg.K
So,
ΔT = 300/(0.5×233)
ΔT = 300/116.5
ΔT = 2.57 K
hence, The change in the temperature of the silver should be considered as the 2.57 K.
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CI. Design challenge: Your goal is to build a compound microscope than can at least double the size of the object. We will constrain ourselves to values that will work later in a simulation, which has the unrealistic- for microscopes - focal lengths in meters. (Imagine instead that those values are in cm) You need to choose two converging lenses. You have available lenses with focal lengths of 2, 4, 8 and 12 m. First choose an objective lens focal length, and a distance to place the object from the lens. Then choose an eyepiece lens focal length and a distance to place the lens from the objective lens. C2. Draw your microscope design including a ray diagram. Your diagram should include: 1. Location and properties (real/virtual, upright/inverted, bigger/smaller) of the image formed by the objective. 2. Location and properties (real/virtual, upright/inverted, bigger/smaller) of the image formed by the eyepiece. C3. Calculate the magnification of your microscope
Answer:
2. 17.7cm
3. -7 that is magnification
Explanation:
See attached handwritten document
Two astronauts are 1.40 m apart in their spaceship. One speaks to the other. The conversation is transmitted to earth via electromagnetic waves. The time it takes for sound waves to travel at 328 m/s through the air between the astronauts equals the time it takes for the electromagnetic waves to travel to the earth. How far away from the earth is the spaceship?
Answer:
The spaceship is placed at a distance of [tex]1.28\times 10^6\[/tex] meters from the Earth.
Explanation:
Distance between two astronauts, d = 1.4 m
The time it takes for sound waves to travel at 328 m/s through the air between the astronauts equals the time it takes for the electromagnetic waves to travel to the earth.
For sound, time taken is given by :
[tex]t=\dfrac{1.4}{328}\ ........(1)[/tex]
For electromagnetic wave, time taken to travel the Earth is :
[tex]t=\dfrac{d}{3\times 10^8}\ ........(2)[/tex]
d is the distance from the Earth to the spaceship.
As time are same :
[tex]\dfrac{d}{3\times 10^8}=\dfrac{1.4}{328}[/tex]
[tex]d=\dfrac{1.4}{328}\times 3\times 10^8\\\\d=1.28\times 10^6\ m[/tex]
So, the spaceship is placed at a distance of [tex]1.28\times 10^6[/tex] meters from the Earth.
Is it possible for an inverted pyramid to exist, like the one show here? Explain your answer.
The inverted pyramid of energy contradicts the unidirectional flow of energy in ecosystems, violating thermodynamic principles. Natural energy pyramids depict decreasing energy availability at higher trophic levels, reflecting ecological dynamics and energy conservation.
The concept of an inverted pyramid of energy contradicts the fundamental principles of energy flow in ecosystems. In ecological systems, energy is transferred through trophic levels in a unidirectional manner, following the second law of thermodynamics. Primary producers, such as plants, harness solar energy and convert it into chemical energy through photosynthesis. Herbivores then consume these plants, transferring some of the energy to the next trophic level. Carnivores, in turn, consume herbivores, and so on.
An inverted pyramid of energy would imply a scenario where the energy at higher trophic levels exceeds that at lower levels, contrary to the established flow. This goes against the principles of energy conservation and the inefficiencies inherent in energy transfer between trophic levels, where energy is lost as heat in each transfer.
The inverted pyramid concept is not observed in natural ecosystems due to the energy losses associated with metabolism, heat production, and other inefficiencies. The pyramid shape represents the decreasing energy availability at higher trophic levels, a pattern consistent with the laws of thermodynamics and ecological dynamics.
In summary, an inverted pyramid of energy is not feasible within ecological systems, as it contradicts established principles of energy transfer and conservation in ecosystems.
The question probable may be:
Is it possible to have an inverted pyramid of energy?? why or why not??
A capacitor charging circuit consists of a battery, an uncharged 20 μF capacitor, and a 6.0 kΩ resistor. At t = 0 s the switch is closed; 0.15 s later, the current is 0.46 mA . Part A What is the battery's emf?
Answer:
Battery voltage will be equal to 9.65 volt
Explanation:
We have given capacitance [tex]C=20\mu F=20\times 10^{-6}F[/tex]
Resistance [tex]R=6kohm=6000ohm[/tex]
Time constant of RC circuit is
[tex]\tau =RC=20\times 10^{-6}\times 6000=0.12sec[/tex]
Time is given t = 0.15 sec
Current i = 0.46 mA
Current in RC circuit is given by
[tex]i=\frac{V}{R}e^{\frac{-t}{\tau }}[/tex]
[tex]0.00046=\frac{V}{6000}e^{\frac{-0.15}{0.12 }}[/tex]
[tex]0.00046=\frac{V}{6000}\times 0.286[/tex]
V = 9.65 volt
So battery emf will be equal to 9.65 volt
To find the emf, use Ohm's Law and the charging formula for a capacitor. With the current, resistance, and capacitor values, calculate the voltage across the resistor. Then, use that to solve for emf with the formula V = emf(1 - e^-t/RC).
Explanation:To determine the electromotive force (emf) of the battery, we can use the relationship between current, resistance, and emf in the circuit containing a resistor and a capacitor. The relevant formula derived from the natural charging process of a capacitor through a resistor is:
V = emf(1 - e-t/RC) (charging)
Since the question provides the current (I) 0.15 seconds after the circuit is closed, we know that:
I = V/R, where V is the voltage across the resistor at time t, R is the resistance, and I is the current.
Let's solve for emf using the given information:
Identify the given values: R = 6.0 kΩ and C = 20 μF.Adjust units for computations: R = 6000 Ω and C = 20 x 10-6 F.Calculate the voltage (V) across the resistor using Ohm's law: V = I * R = 0.46 mA * 6000 Ω.Convert the current to amperes: I = 0.46 x 10-3 A.Substitute V and I into the equation to find emf: 0.46 x 10-3 A * 6000 Ω = emf(1 - e-0.15/RC).Calculate RC: RC = R * C = 6000 Ω * 20 x 10-6 F = 0.12 s.Substitute RC into the exponent: emf = V/(1 - e-0.15/0.12).After calculations, the emf can be found to be approximately the voltage drop across the resistor divided by the remaining fraction given by the exponential term. This result is the battery's emf.Learn more about Capacitor Charging Circuit here:https://brainly.com/question/31415514
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Boltzmann’s constant is 1.38066×10^−23 J/K, and the universal gas constant is 8.31451 J/K · mol. If 0.8 mol of a gas is confined to a 6.7 L vessel at a pressure of 10.8 atm. What is the average kinetic energy of a gas molecule?
Answer: The average kinetic energy is 2.3 × 10^-20 J
Explanation: Please see the attachments below
A 5.20 kg chunk of ice is sliding at 13.5 m/s on the floor of an ice-covered valley when it collides with and sticks to another 5.20 kg chunk of ice that is initially at rest. Since the valley is icy, there is no friction. After the collision, the blocks slide partially up a hillside and then slide back down.
how high above the valley floor will the combined chunks go?
Answer:
The chunk went as high as
2.32m above the valley floor
Explanation:
This type of collision between both ice is an example of inelastic collision, kinetic energy is conserved after the ice stuck together.
Applying the principle of energy conservation for the two ice we have based on the scenery
Momentum before impact = momentum after impact
M1U1+M2U2=(M1+M2)V
Given data
Mass of ice 1 M1= 5.20kg
Mass of ice 2 M2= 5.20kg
velocity of ice 1 before impact U1= 13.5 m/s
velocity of ice 2 before impact U2= 0m/s
Velocity of both ice after impact V=?
Inputting our data into the energy conservation formula to solve for V
5.2*13.5+5.2*0=(10.4)V
70.2+0=10.4V
V=70.2/10.4
V=6.75m/s
Therefore the common velocity of both ice is 6.75m/s
Now after impact the chunk slide up a hill to solve for the height it climbs
Let us use the equation of motion
v²=u²-2gh
The negative sign indicates that the chunk moved against gravity
And assuming g=9.81m/s
Initial velocity of the chunk u=0m/s
Substituting we have
6.75²= 0²-2*9.81*h
45.56=19.62h
h=45.56/19.62
h=2.32m
An Arrow (0.5 kg) travels with velocity 60 m/s to the right when it pierces an apple (1 kg) which is initially at rest. After the collision, the arrow and the apple are stuck together. Assume that no external forces are present and therefore the momentum for the system is conserved. What is the final velocity (in m/s) of apple and arrow after the collision
Answer:
Velocity after collision will be 20 m/sec
Explanation:
We have given mass of arrow [tex]m_1=0.5kg[/tex]
Mass of arrow [tex]v_1=60m/sec[/tex]
Mass of apple [tex]m_2=1kg[/tex]
Apple is at rest so velocity of apple [tex]v_2=0m/sec[/tex]
According to conservation of momentum
Momentum before collision is equal to momentum after collision
[tex]m_1v_1+m_2v_2=(m_1+m_2)v[/tex]
[tex]0.5\times 60+1\times 0=(0.5+1)v[/tex]
[tex]1.5v=30[/tex]
v = 20 m/sec
A star ending its life with a mass of four to eight times the Sun's mass is expected to collapse and then undergo a supernova event. In the remnant that is not carried away by the supernova explosion, protons and electrons combine to form a neutron star with approximately twice the mass of the Sun. Such a star can be thought of as a gigantic atomic nucleus. Assume r-aA1/3. If a star of mass 3.88 x 1030 kg is composed entirely of neutrons (mn 1.67 x 1027 kg), what would its radius be?
Answer:
r = 16 Km
Explanation:
given
m_n= 1.67 x 10^-27 Kg
M_star = 3.88 x 10^30 Kg
A= M_star/m_n
A= 3.88*10^30/1.67 x 10^-27
A=2.28 *10^57 neutrons A = The number of neutrons
we use the number of neutrons as a mass number because the star has only neutrons. = 1.2 x 10-15 m
r = r_o*A^1/3
r = 1.2*10^-15*2.28 *10^57^1/3
r = 16 Km
Consider a ball in projectile motion under gravitational force so that it is has motion in both the vertical and horizontal directions. The effects of air friction may be ignored. When the ball reaches the highest point in its trajectory, what can be said about the magnitudes of the ball's velocity and acceleration vectors? Select the correct answer
a. The horizontal components of both the velocity and acceleration are zero.
b. The magnitude of velocity is zero, but the acceleration is a constant
c. The magnitudes of both velocity and acceleration are zero
d. The magnitude of acceleration is zero but the ball has a positive velocity
e. The magnitude of velocity is at its minimum nonzero value, but the magnitude of acceleration is a constant
At the highest point in the ball's trajectory, the magnitude of its velocity vector is at its minimum nonzero value, while the magnitude of its acceleration vector is a constant.
Explanation:When the ball reaches the highest point in its trajectory, the magnitude of its velocity vector is at its minimum nonzero value, while the magnitude of its acceleration vector is a constant. Since the ball is at the highest point, it momentarily comes to rest in the vertical direction, resulting in its velocity magnitude being zero in that direction. However, the ball still has a horizontal velocity component, which is constant throughout its motion.
A double-slit diffraction pattern is formed on a distant screen. If the separation between the slits decreases, what happens to the distance between interference fringes? Assume the angles involved remain small. A double-slit diffraction pattern is formed on a distant screen. If the separation between the slits decreases, what happens to the distance between interference fringes? Assume the angles involved remain small. The distance between interference fringes remains the same. The distance between interference fringes also decreases. The distance between interference fringes increases. The effect cannot be determined unless the distance between the slits and the screen is known.
Answer:
Option C. The distance between interference fringes increases.
Explanation:
The distance between interference fringes for small angles can be given by the formula:
[tex]y = \frac{\lambda D}{d}[/tex]..........(1)
Where D = Distance between the slits and the screen
[tex]\lambda =[/tex] the wavelength of light
d = separation between the two slits
from the formula given in equation (1)
[tex]y \alpha 1/d[/tex]
It is obvious from the relationship above that the distance between interference fringes is inversely proportional to the separation between the slits.
Therefore, if the separation between slits is increased, the distance between interference fringes is increased.
Light striking a metal surface causes electrons to be emitted from the metal via the photoelectric effect.The intensity of the incident light and the temperature of the metal are held constant. Assuming that the initial light incident on the metal surface causes electrons to be ejected from the metal, what happens if the frequency of the incident light is increased? Check all that apply.A. The work function of the metal increases.B. The number of electrons emitted from the metal per second increases.C. The maximum speed of the emitted electrons increases.D. The stopping potential increases.
Answer:
C. The maximum speed of emitted electrons increase
D. The stopping potential increases
Explanation:
Albert Einstein provided a successful explanation of the photoelectric effect on the basis of quantum theory. He proposed that,
“An electron either absorbs one whole photon or it absorbs none. After absorbing a photon, an electron either leaves the surface of metal or dissipate its energy within the metal in such a short time interval that it has almost no chance to absorb a second photon. An increase in intensity of light source simply increase the number of photons and thus, the number of electrons, but the energy of electron remains same. However, increase in frequency of light increases the energy of photons and hence, the energy of electrons too.”
Therefore, due to this increase in energy of photons, the kinetic energy of emitted electrons also increase. And the increase in kinetic energy follows with the increase in velocity.
Now, the stopping potential is directly proportional to Kinetic Energy of the electrons. Thus, the increase in Kinetic Energy, results in an increase in stopping potential.
Therefore, two options apply here:
C. The maximum speed of emitted electrons increases
D. The stopping potential increases
If the frequency of the incident light in the photoelectric effect is increased, C. The maximum speed of the emitted electrons increases and D. The stopping potential increases; however, the work function does not change and the number of electrons emitted per second remains the same at a constant light intensity.
For the question about what happens if the frequency of the incident light in the photoelectric effect is increased, the correct options are:
The maximum speed of the emitted electrons increases.The stopping potential increases.The work function of the metal (Option A) is a material property and does not change with the frequency of the incident light. Therefore, this option is incorrect. The number of electrons emitted from the metal per second (Option B) depends on the intensity of the light, not its frequency, so increasing the frequency at constant intensity does not increase the emission rate. Thus, this option is also incorrect. When the frequency of the incident light is increased, the kinetic energy of the emitted electrons increases (as per Einstein's equation for the photoelectric effect), which correlates with Option C. The stopping potential, which is the electric potential needed to stop the fastest photoelectrons, also increases with the electron's kinetic energy, confirming Option D as correct.
Electrons are allowed "in between" quantized energy levels, and, thus, only specific lines are observed. The energies of atoms are not quantized. The energies of atoms are quantized. Electrons are not allowed "in between" quantized energy levels, and, thus, only specific lines are observed. When an electron moves from one energy level to another during absorption, a specific wavelength of light (with specific energy) is emitted. When an electron moves from one energy level to another during emission, a specific wavelength of light (with specific energy) is emitted.
Answer:
This is because The energies of atoms are quantized.
Electrons are allowed "in between" quantized energy levels, and, thus, only specific lines are observed
A 500 pF capacitor is charged up so that it has 10μC of charge on its plates. The capacitor is then quicklyconnected to a 10 H inductor. Calculate themaximum energystored in the magnetic field of the inductoras the circuit oscillates. What is the current through the inductor when the maximum energy stored in theinductor is reached?
Answer:
Explanation:
Given that,
A capacitor of capacitance
C = 500pF
Charge on capacitor is
Q = 10μC
Capacitor is then connected to an inductor of inductance 10H
L = 10H
Since we want to calculate the maximum energy stored by the inductor, then, we will assume all the energy from the capacitor is transfer to the inductor
So energy stored in capacitor can be determined by using
U = ½CV²
Then, Q = CV
Therefore V = Q/C
U = ½ C • (Q/C)² = ½ C × Q²/C²
U = ½Q² / C
Then,
U = ½ × (10 × 10^-6)² / (500 × 10^-9)
U = 1 × 10^-4 J
U = 0.1 mJ
So the energy stored in this capacitor is transfers to the inductor.
So, energy stored in the inductor is 0.1mJ
B. Current through the inductor
Energy in the inductor is given as
U = ½Li²
1 × 10^-4 = ½ × 10 × i²
1 × 10^-4 = 5× i²
i² = 1 × 10^-4 / 5
i² = 2 × 10^-5
I = √(2×10^-5)
I = 4.47 × 10^-3 Amps
Then,
I = 4.47 mA
A student sits on a rotating stool holding two 3.0-kg objects. When his arms are extended horizontally, the objects are 1.0 m from the axis of rotation and he rotates with an angular speed of 0.75 rad/s. The moment of inertia of the student plus stool is 3.0 kg⋅m^2 and is assumed to be constant. The student then pulls in the objects horizontally to 0.30 m from the rotation axis. Find the kinetic energy of the student before and after the objects are pulled in.
Answer:
2.86J
Explanation:
M = 3.0kg
R₁ = 1.0m
R₂ = 0.3m
I₁ = I(mass) + I(student + stool)
I₁ = 2mr₁² + I(student + stool)
I₁ = 2*(3*1²) + 3.0
I₁ = 9.0kgm²
the initial moment of inertia of the system = 9.0kgm²
I₂ = 2mr₂² + I(student + stool)
I₂ = 2*(3 * 0.3²) + 3.0
I₂ = 0.54 + 3.0
I₂ = 3.54kgm²
the final moment of inertia of the system is 3.54kgm²
From conservation of angular momentum
I₁ω₁ = I₂ω₂
ω₂ = (I₁ * ω₁) / I₂
ω₂ = (0.75 * 9) / 3.54
ω₂ = 1.09rad/s
kinetic energy of rotation (k.e) =½ Iω²
K.E = (K.E)₂ - (K.E)₁
k.e = [½ * 3.54 * (1.90)²] - [½ * 9.0 * 0.75²]
K.E = 6.3897 - 2.53125
K.E = 2.85845
K.E = 2.86J
A flat coil of wire consisting of 15 turns, each with an area of 40 cm 2, is positioned perpendicularly to a uniform magnetic field that increases its magnitude at a constant rate from 1.5 T to 5.1 T in 2.0 s. If the coil has a total resistance of 0.20 Ω, what is the magnitude of the induced current?
Answer:
0.54 A
Explanation:
Parameters given:
Number of turns, N = 15
Area of coil, A = 40 cm² = 0.004 m²
Change in magnetic field, ΔB = 5.1 - 1.5 = 3.6 T
Time interval, Δt = 2 secs
Resistance of the coil, R = 0.2 ohms
To get the magnitude of the current, we have to first find the magnitude of the EMF induced in the coil:
|V| = |(-N * ΔB * A) /Δt)
|V| = | (-15 * 3.6 * 0.004) / 2 |
|V| = 0.108 V
According to Ohm's law:
|V| = |I| * R
|I| = |V| / R
|I| = 0.108 / 0.2
|I| = 0.54 A
The magnitude of the current in the coil of wire is 0.54 A
Overload refers to: A. Performing a weight-lifting exercise with the resistance (load) held overhead B. Using a demand (load) above the normal demand faced by a muscle C. The principle that strength will be best developed when the resistance (load) exceeds the individual's physical abilities (e.g., 8 reps of the 5-RM) D. The principle that a resistance (load) must be presented repeatedly in order to elicit any adaptations E. All of the answers are correct
Answer:
E. All of the answers are correct
Explanation:
Overload principle in fitness training is associated with a gradual development of an athlete's abilities by progressively increasing the athlete's load and training.
In order to do this, the athlete's limit must be surpassed albeit gradually at first then picked up later over time.
Final answer:
Overload refers to using a demand (load) above the normal demand faced by a muscle, fundamental for muscle growth and strength increase. It involves progressively increasing the workload to stimulate muscle adaptation and improvements. Overload, progression, and specificity are key principles in effective training programs.
Explanation:
Overload refers to B. Using a demand (load) above the normal demand faced by a muscle. This is a fundamental concept in strength training and physical conditioning. Overload is the principle that for muscles to grow stronger or for endurance to increase, they must be challenged with a demand that is greater than what they are accustomed to. This concept involves progressively increasing the workload on the muscle to stimulate adaptation and improvements in muscle strength, endurance, and size.
The principle of overload is based on the body's ability to adapt to increased demands. When you perform exercises that are more challenging than your muscles are used to, your body adapts by making those muscles stronger, which in turn makes them capable of handling the increased load. This can be achieved through various methods such as increasing the weight lifted, the number of repetitions performed, or the intensity of the exercise.
It is also important to note that overload, along with progression and specificity, form the three foundational principles of training that guide the development of effective strength and conditioning programs. These principles ensure that exercises not only challenge the body but do so in a way that promotes optimal growth, strength, and performance enhancements over time.
A partially evacuated vertical cylindrical container is covered by a circular lid that makes an airtight seal. The pressure in the room is 1.01 x 105 Pa and the pressure inside the container is 0.41 x 105 Pa. What other two quantities would you need to know in order to calculate the minimum upward applied force required to lift the lid? Select two answers.
The pressure difference (∆P) would be the atmospheric pressure in the room (1.01 x 10⁵ Pa) minus the pressure inside the container (0.41 x 10⁵ Pa).
To calculate the minimum upward applied force required to lift the lid of a partially evacuated vertical cylindrical container, you would need to know two additional quantities: the radius of the lid, and consequently the surface area of the lid.
The force required to lift the lid can be calculated using the equation F = P x A, where F is the force, P is the difference in pressure (external minus internal), and A is the surface area of the lid. Recall that the area of the circular lid would be calculated using the formula A = πr², where r is the radius of the circular lid.
In this case, the pressure difference (∆P) would be the atmospheric pressure in the room (1.01 x 10⁵ Pa) minus the pressure inside the container (0.41 x 10⁵ Pa).
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A cat walks along a plank with mass M= 6.00 kg. The plank is supported by two sawhorses. The center of mass of the plank is a distance d1= 0.850 m to the left of sawhorse B. When the cat is a distance d2= 1.11 m to the right of sawhorse B, the plank just begins to tip.
If the cat has a mass of 2.9 kg, how far to the right of sawhorse B can it walk before the plank begins to tip?
Answer:
d₂ = 1.466 m
Explanation:
In this case we must use the rotational equilibrium equations
Στ = 0
τ = F r
we must set a reference system, we use with origin at the easel B and an axis parallel to the plank , we will use that the counterclockwise ratio is positive
+ W d₁ - w_cat d₂ = 0
d₂ = W / w d₁
d₂ = M /m d₁
d₂ = 5.00 /2.9 0.850
d₂ = 1.466 m
A person is pushing a fully loaded 21.60 kg wheelbarrow at constant velocity along a plank that makes an angle ????=20.0∘ with the horizontal. The load is distributed such that the center of mass of the wheelbarrow is exactly halfway along its length L. What is the magnitude of the total force Fx the person must apply so that the wheelbarrow is parallel to that plank? You may neglect the radius of the wheel in your analysis. The gravitational acceleration is g=9.81 m/s2.
Answer:
Explanation:
The weight of the wheelbarrow will act downwards . Its component parallel to the plank will act downwards along the plank .
The value of component = mgsinθ
21.6 x 9.8 x sin20
= 72.4 N
The person shall have to apply force equal to this component so that the barrow moves with uniform velocity.
Force required = 72.4 N .
Your group has been asked to examine collisions in which the pucks stick together after the collision. To get the pucks to stick together, your group has wrapped the circumference of each puck with velcro (the regulation puck has the velcro hooks and the practice puck has the velcro loops) with the hooks or loops outward in each case. To get consistent puck speeds before the collision, your group has constructed a pair of rubber-band launchers, in which a very strong rubber band is stretched tightly between two posts and then used like a slingshot. A few trials are sufficient to determine that these launchers, if used carefully, can launch the pucks with predictable speeds and with essentially no spin?
Answer:
If necessary, readjust the f-stop of the CCD camera until only the LED’s from the pucks
are visible. You may find that a vertical stripe appears associated with the LED. This
is called streaking or ‘blooming’ and is a limitation of CCD technology in the presence of
localized bright spots. Some amount of streaking is acceptable, and can be compensated for
later in the computer analysis.
Once the exposure level of the camera has been set you should capture a ‘dark frame’.
For this, first remove all pucks, hands, etc., from the air table and click on the box to the
right of Background Frame. The dark frame can later be subtracted from your images of
collisions to suppress any constant background such as the edge of the table.
A thin block of soft wood with a mass of 0.0720 kg rests on a horizontal frictionless surface. A bullet with a mass of 4.67 g is fired with a speed of 556 m/s at a block of wood and passes completely through it. The speed of the block is 18.0 m/s immediately after the bullet exits the block. (a) Determine the speed of the bullet as it exits the block.
Answer:
Explanation:
Let the mass of the thin block be M = .072 kg .
mass of bullet m = 4.67 x 10⁻³ kg.
speed of bullet v = 556 m/s.
speed of block after billet exits the block = 18 m/s.
We shall apply law of conservation of momentum
momentum of bullet block system
.072 x 0 + 4.67 x 10⁻³ x 556 = .072 x 18 + 4.67 x 10⁻³ x V here V is velocity of bullet after exit
= 2.59652 = 1.296 + 4.67 x 10⁻³ x V
4.67 x 10⁻³ x V = 1.30052
V = .278 x 10³ m /s
= 278 m /s