Explanation:
Let l is the length of the wire. such that, side of square is l/4. Area of square loop is, [tex]A_s=\dfrac{l^2}{16}[/tex]
Radius of the circular loop,
[tex]l=2\pi r\\\\r=\dfrac{l}{2\pi}[/tex]
Area of the circular loop,
[tex]A_a=\pi r^2\\\\A_a=\pi \times (\dfrac{l}{2\pi})^2\\\\A=\dfrac{l^2}{4\pi}[/tex]
Torque in magnetic field is given by :
[tex]\tau=NIAB\sin\theta[/tex]
It is clear that, [tex]\tau\propto A[/tex]
So,
[tex]\dfrac{\tau_s}{\tau_a}=\dfrac{A_s}{A_a}\\\\\dfrac{\tau_s}{\tau_a}=\dfrac{\dfrac{l^2}{16}}{\dfrac{l^2}{4\pi}}\\\\\dfrac{\tau_s}{\tau_a}=\dfrac{\pi}{4}[/tex]
So, the ratio of maximum torque on the square loop to the maximum torque on the circular loop is [tex]\pi :4[/tex].
A block of mass 0.464 kg is hung from a vertical spring and allowed to reach equilibrium at rest. As a result, the spring is stretched by 0.646 m. Find the spring constant. N/m The block is then pulled down an additional 0.338 m and released from rest. Assuming no damping, what is its period of oscillation? s How high above the point of release does the block reach as it oscillates?
Explanation:
Mass, m = 0.464 kg
Compression in the spring, x = 0.646 m
(a) The net force acting on the spring is given by :
[tex]kx=mg[/tex]
k is spring constant
[tex]k=\dfrac{mg}{x}\\\\k=\dfrac{0.464\times 9.8}{0.646 }\\\\k=7.03\ N/m[/tex]
(b) The angular frequency of the spring mass system is given by :
[tex]\omega=\sqrt{\dfrac{k}{m}} \\\\\omega=\sqrt{\dfrac{7.03}{0.464 }} \\\\\omega=3.89\ rad/s[/tex]
The period of oscillation is :
[tex]T=\dfrac{2\pi}{\omega}\\\\T=\dfrac{2\pi}{3.89}\\\\T=1.61\ m/s[/tex]
(c) As the spring oscillates, its will reach to a height of 2(0.338) = 0.676 m
The spring constant is calculated to be 7.03 N/m using Hooke's Law. The period of oscillation for the block-spring system is found to be 1.74s utilizing the formula for simple harmonic motion. The height reached above the point of release by the block is 0.338m, based on the conservation of energy principle.
Explanation:To find the spring constant, which can be denoted by k, we can use Hooke's Law (F = -kx). The force, F, in this case is the weight of the block (mass x gravity), mg = 0.464 kg x 9.8 m/s² = 4.5472 N. This force causes the spring to stretch 0.646 m. Therefore, k is calculated as F/x = 4.5472 N / 0.646 m = 7.03 N/m.
The period of oscillation, T, for a block-spring system executing simple harmonic motion is given by T = 2π√(m/k), where m is the mass and k is the spring constant. Substituting the given values, T = 2π√(0.464 kg / 7.03 N/m) = 1.74s.
For the height above the point of release that the block will reach, again we come back to conservation of energy. At the point of release, the block possesses only elastic potential energy, 1/2kA², where A is the amplitude. This energy will be entirely converted into gravitational potential energy, mgh, at the block's highest point. Therefore, h = (1/2kA²)/mg. So, h = (1/2 * 7.03 N/m * (0.338 m) ²) / (0.464 kg * 9.8 m/s²) = 0.338 m, above the point of release.
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Neutron stars, such as the one at the center of the Crab Nebula, have about the same mass as our sun but a much smaller diameter. Part A If you weigh 665 N on the earth, what would be your weight on the surface of a neutron star that has the same mass as our sun and a diameter of 16.0 km ? Take the mass of the sun to be ms = 1.99×1030 kg , the gravitational constant to be G = 6.67×10−11 N⋅m2/kg2 , and the free-fall acceleration at the earth's surface to be g = 9.8 m/s2 . Express your weight wstar in newtons. View Available Hint(s)
Answer:
1.41*10^14 N
Explanation:
Given that
Weight of the person on earth, W = 665
Radius of the neutron star, r = 8 km
Mass of the sun, M = 1.99*10^30 kg
Gravitational constant, G = 6.67*10^-11 Nm²/kg²
Acceleration due to gravity in earth, g = 9.8 m/s²
Weight on earth is given by
W = mg, so that, mass m would be
m = W/g
m = 665 / 9.8
m = 67.86 kg
The mass on earth is 67.86 kg
Weight on the neutron star is then
W = F = GmM/r²
F = (6.67*10^-11 * 67.86 * 1.99*10^30) / 8000²
F = 9*10^21 / 6.4*10^7
F = 1.41*10^14 N
Thus, the the weight in the neutron star is 1.41*10^14 N
Calculate the energy released by the electron-capture decay of 5727Co. Consider only the energy of the nuclei (ignore the energy of the surrounding electrons). The following masses are given:5727Co: 56.936296u5726Fe: 56.935399uExpress your answer in millions of electron volts (1u=931.5MeV/c2) to three significant figures.
Answer:
Explanation:
The mass difference is
Δm = m2 - m1
56.936296 u - 56.935399 u
= 0.000897 u
The energy released by the electron-capture decay
E = Δmc²
( 0.000897 u) c² ( 931. 5 MeV /c²÷ 1 u)
= 0.8355555 MeV
An AC generator with an output rms voltage of 42.0 V at a frequency of 60.0 Hz is connected across a 20.0-?F capacitor. Find the following. (a) capacitive reactance ? (b) rms current A (c) maximum current in the circuit A (d) Does the capacitor have its maximum charge when the current takes its maximum value? Yes No
Explanation:
answer and explanation is in the picture
5.19 Identify the force present and explain whether work is being performed in the following cases: (a) You lift a pencil off the top of a desk. (b) A spring is compressed to half its normal length.
Work in physics is done when a force causes displacement in the direction of the force. Lifting a pencil off a desk and compressing a spring are both examples of work being done because in both cases there is a force and a displacement.
Explanation:In physics, work is defined as the product of force, displacement, and the cosine of the angle between them. It's represented by the formula W = Fd cos(θ). Let's apply this definition to two scenarios:
Lifting a pencil: When you lift a pencil off the top of a desk, the force you apply is against gravity, causing an upward displacement. Work is being performed since there is a displacement in the direction of the force.Compressing a spring: When a spring is compressed to half its normal length, work is done on the spring. This work is negative because the force applied is in the opposite direction of the spring's natural expansion.Considering the examples from the given reference, we understand that work requires both a force and a displacement. No work is done if either the force or the displacement is zero, or if the force is perpendicular to the direction of the displacement.
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The point is at the edge of the disk and the component bodies are:
a. A uniform disk of radius and mass .
b. A uniform rod of length and mass .
c. A uniform rectangle with side lengths and , and mass .
d. A point mass at with mass .
What is the moment of inertia about the axis through the point?
Answer:
a. A uniform disk of radius and mass .
Explanation:
The moment of inertia I of an object depends on a chosen axis and the mass of the object. Given the axis through the point, the inertia will be drawn from the uniform disc having a radius and the mass.
.
Lindsay is standing in the middle of the aisle of a bus that’s at rest at a stop light. The light turns green and the bus speeds up. Without grabbing or hanging on to anything, Lindsay manages to remain stationary with respect to the floor. While the bus is speeding up, the net force on Lindsay points in which direction?
a. Toward the front of the bus b. There is no net force on Lindsay c. Toward the back of the bus
Answer:
b. There is no net force on Lindsay.
Explanation:
Since Lindsay manages to stay stationary then this means that either there is no force acting on Lindsay or all the forces acting on Lindsay are balanced and produces no net force on Lindsay with respect to the bus.
Answer:
No net force acting on Lindsay.
Explanation:
Solution:-
- Lets assume the mass of the bus to be = M
- Assume mass of Lindsay = m
- The initial speed of bus and lindsay was vi = 0 m/s
- We will consider the system ( Lindsay + Bus ) to be isolated with no fictitious unbalanced forces acting on the system.
- Since, the system can be isolated the application of principle of conservation of linear momentum is completely valid.
- We will assume the direction of bus moving to be positive.
- The principle of conservation of momentum states:
Pi = Pf
vi*(m+M) = m*vL + M*vb
Where,
vL : The velocity of Lindsay after bus starts moving
vb : The velocity with which bus moves
Since, vi = 0 m/s.
0 = m*vL + M*vb
vL = - ( M / m )*vb
- The lindsay moves in opposite direction of motion of bus. We will apply the Newton's second law of motion on Lindsay:
Impulse = F*t = m*( vL - vi )
Impulse = F*t = m*( - ( M / m )*vb - 0 )
Impulse = F*t = -M*vb
F = -M*vb/t
- We see that a force ( F ) acts on Lindsay as soon as the bus starts moving. The negative sign shows the direction of force which is opposite to the motion of bus. So the force acts on Lindsay towards the back of the bus.
- However, if we consider the system of ( bus + Lindsay ), the net force exerted on Lindsay remains zero as the impulse force ( F ) becomes an internal force of the system and is combated by the friction force ( Ff ) between Lindsay's shoes and the aisle floor.
- The magnitude of Friction force ( Ff ) is equivalent to the impulse created by force ( F ) by Newton's third Law of motion. Every action has its equal but opposite reaction. Hence,
Fnet = F + Ff
Fnet = F - F
Fnet = 0
- However, The force ( F ) created by the rate of change of momentum of bus must be less than Ff which is equivalent to weight of Lindsay:
Ff = m*g*us
Where,
us: The coefficient of static friction.
- So for forces to remain balanced, with no resultant force then:
Ff ≥ F
m*g*us ≥ M*ab
us ≥ (M/m)*(ab/g)
- The coefficient of static friction should be enough to combat the acceleration of bus ( ab ).
The equation of a transverse wave traveling along a very long string is y 6.0 sin(0.020px 4.0pt), where x and y are expressed in centimeters and t is in seconds. Determine (a) the amplitude, (b) the wavelength, (c) the frequency, (d) the speed, (e) the direction of propagation of the wave, and (f) the maximum transverse speed of a particle in the string
Answer:
given
y=6.0sin(0.020px + 4.0pt)
the general wave equation moving in the positive directionis
y(x,t) = ymsin(kx -?t)
a) the amplitude is
ym = 6.0cm
b)
we have the angular wave number as
k = 2p /?
or
? = 2p / 0.020p
=1.0*102cm
c)
the frequency is
f = ?/2p
= 4p/2p
= 2.0 Hz
d)
the wave speed is
v = f?
= (100cm)(2.0Hz)
= 2.0*102cm/s
e)
since the trignometric function is (kx -?t) , sothe wave propagates in th -x direction
f)
the maximum transverse speed is
umax =2pfym
= 2p(2.0Hz)(6.0cm)
= 75cm/s
g)
we have
y(3.5cm ,0.26s) = 6.0cmsin[0.020p(3.5) +4.0p(0.26)]
= -2.0cm
The amplitude of the wave is 6.0 cm, the wavelength is 100π cm, the frequency is approximately 0.637 Hz, the speed is approximately 200π cm/s, the direction of propagation is the positive x-direction, and the maximum transverse speed of a particle in the string is 120π cm/s.
Explanation:The given equation representing a transverse wave along a very long string is y = 6.0 sin(0.020px - 4.0pt). We can extract various information from this equation:
The amplitude of the wave is 6.0 cm. The amplitude represents the maximum displacement of the particles in the string from their equilibrium position.The wavelength of the wave can be determined by comparing the equation to its standard form y = A sin(kx - ωt), where k represents the wave number. From the equation given, we can find k = 0.020p. The wave number is related to the wavelength by the equation λ = (2π)/k. Substituting the value of k, we get λ = (2π)/(0.020p), which simplifies to 100π cm.The frequency of the wave can be determined from the equation ω = 2πf, where ω is the angular frequency. In the given equation, comparing it to the standard form, we find ω = 4.0 pt. Solving for f, we get f = (ω/2π) = 4.0 pt/(2π) ≈ 0.637 Hz.The speed of the wave can be determined by the equation v = fλ, where v is the speed. Substituting the frequency and wavelength values, we get v = (0.637 Hz)(100π cm) ≈ 200π cm/s.The direction of propagation of the wave can be determined from the sign of the coefficient of x in the equation. In this case, the coefficient is positive, indicating that the wave is propagating in the positive x-direction.The maximum transverse speed of a particle in the string can be determined by taking the derivative of the displacement equation with respect to time, which gives the velocity equation. Taking the derivative of the given equation, we get v = -120πe^(-0.020px) cos(0.020px - 4.0pt). The maximum transverse speed will occur when the cos(0.020px - 4.0pt) term is equal to 1, resulting in a maximum speed of 120π cm/s.Learn more about Properties of Transverse Waves here:https://brainly.com/question/32780857
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A new metal alloy is found to have a specific heat capacity of 0.321 J/(g⋅∘C). First, 44.5 g of the new alloy is heated to 100. ∘C. Then, it is placed in an ideal constant-pressure calorimeter containing 175 g of water (Vs,water=4.184 J/(g⋅∘C)) at an initial temperature of 20.0 ∘C. What will the final temperature of the mixture be after it attains thermal equilibrium
Answer:
The final temperature is 21.531°c
Explanation:
Heat gained by the water = heat lost by the metal
Heat gained = (m)(c)(∆tita)
Where m = mass
C = specific heat capacity
∆tita = temperature change
X = equilibrium temperature
So ....
Heat gained by water
= 175*4.185*(x-20)
= 732.2x - 14644
Heat lost by metal
= 44.5*0.321*(100-x)
=1428.45 -14.2845x
So....
1428.45 - 14.2845x = 732.2x - 14644
1428.45+14644= 732.2x + 14.2845x
16072.45= 746.4845x
16072.45/746.4845= x
21.531 = x
The final temperature is 21.531°c
Two charged particles, Q 1 and Q 2, are a distance r apart with Q 2 = 5 Q 1. Compare the forces they exert on one another when is the force Q 2 exerts on Q 1 and is the force Q 1 exerts on Q 2.
Describe a situation in which an electron will be affected by an external electric field but will not be affected by an external magnetic field. Is it possible that an electron is affected by an external magnetic field but not by an external electric field?
Answer:
Explanation:
Situations in which an electron will be affected by an external electric field but will not be affected by an external magnetic field
a ) When an electron is stationary in the electric field and magnetic field , he will be affected by electric field but not by magnetic field. Magnetic field can exert force only on mobile charges.
b ) When the electron is moving parallel to electric field and magnetic field . In this case also electric field will exert force on electron but magnetic field field will not exert force on electrons . Magnetic field can exert force only on the perpendicular component of the velocity of charged particles.
Situations when electron is affected by an external magnetic field but not by an external electric field
There is no such situation in which electric field will not affect an electron . It will always affect an electron .
Final answer:
In certain conditions, an electron is affected by an external electric field but not by an external magnetic field. Conversely, an electron can be influenced by an external magnetic field but not by an external electric field.
Explanation:
An electron will be affected by an external electric field but will not be affected by an external magnetic field when it is at rest in a magnetic field as it experiences no force.
It is possible for an electron to be affected by an external magnetic field but not by an external electric field, as seen in the case of a moving charged particle forming a magnetic field around wires carrying electrical currents.
Describe how the horizontal position of The baseball varies overtime is the change in position constant variable? Why might this be so?
Answer:
When you hit a baseball with a bat, you are applying a force only in the time while the bat is in contact with the ball, after that, the horizontal speed of the ball is constant. (if you ignore the friction of the air)
This happens because of the second Newton's law:
Force is equal to mass times the acceleration or:
F = m*a
If we do not have a force, then we do not have acceleration.
This means that the change in position for a fixed amount of time is constant, so horizontal speed is constant.
the coefficient of static friction between the bed of a truck and a box resting on it is 0.67. The truck is traveling along a straight horizontal road at 30 m/s. what is the least distance in which the truck can stop if the box is not to slide/
Answer:
68.46 m.
Explanation:
Given,
coefficient of friction,μ = 0.67
speed of truck, v = 30 m/s
distance travel by the truck to stop = ?
Now,
Calculation of acceleration
we know,
f = m a
and also
f = μ N = μ mg
Equating both the forces equation
m a = μ mg
a = μ g
a = 0.67 x 9.81
a = 6.57 m/s²
Now, using equation of kinematics
v² = u² + 2 a s
0 = 30² - 2 x 6.57 x s
s = 68.46 m
Hence, the minimum distance travel by the truck is equal to 68.46 m.
The truck would need to stop in a minimum distance of approximately 68.5 meters to prevent the box from sliding off, based on the given parameters and considering the law of inertia and the role of static friction.
To determine the minimum distance the truck can stop without the box sliding off, we need to consider the law of inertia and the role of static friction.
The force must meet or exceed the force exerted by the truck's deceleration to keep the box from sliding.
Given that the initial velocity of the truck is 30 m/s and the box has a mass of 500 kg, we can use the formula for static friction, which is F = μN, where F is the force of friction, μ is the coefficient of static friction, and N is the normal force.
In this case, N equates to the gravitational force on the box, so N = mg = 500 kg * 9.8 m/s² = 4900 N. Substituting these values in the formula for static friction, we get F = 0.67 * 4900 N = 3283 N.
The force of friction, in terms of motion, is also equivalent to mass times acceleration (F = ma), so we can set this equal to the static friction we calculated and solve for acceleration: 3283 N = 500 kg * a. Solving for a gives an acceleration of approximately 6.57 m/s².
Using motion equations, specifically v² = u² + 2as where v is the final velocity (0 m/s), u is the initial velocity (30 m/s), a is the acceleration (-6.57 m/s² because it's deceleration), and s is the distance, we can compute for the distance s which gives a value of approximately 68.5 meters.
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The probable question is :-
A truck is moving along a straight horizontal road at 30 m/s. A box is placed on the bed of the truck, and the coefficient of static friction between the bed and the box is 0.67. What is the least distance in which the truck can come to a stop without causing the box to slide? Consider the deceleration of the truck due to braking.
A bucket of water of mass 20 kg is pulled at constant velocity up to a platform 32 meters above the ground. This takes 8 minutes, during which time 6 kg of water drips out at a steady rate through a hole in the bottom. Find the work needed to raise the bucket to the platform. Assume g
Answer:
W=-1881.6J
Explanation:
we have that the change in the mass is
[tex]\frac{dm}{dt}=-c\\m(0)=60kg\\m(8)=60kg-6kg=54kg[/tex]
by solving the differential equation and applying the initial conditions we have
[tex]\int dm=-c\int dt\\m=-ct+d\\m(0)=-c(0) + d=60 \\m(8)=-8c+d=54[/tex]
by solving for c and d
d=60
c=0.75
The work needed is
W = m(t) gh
by integrating we have
[tex]dW_T= gh\int dm \\\\W_T=gh\int_0^8 -0.75dt\\\\W_T=(9.8\frac{m}{s^2})(32m)(-0.75(8))=-1881.6J[/tex]
hope this helps!!
Near the earth the intensity of radiation from the sun is 1.35 kW/m2. What volume of space in this region contains 1.0 J of electromagnetic energy? (c = 3.0 x 108 m/s, μ0 = 4π × 10-7 T ∙ m/A, ε0 = 8.85 × 10-12 C2/N ∙ m2)
Answer:
The volume is 2.22x10⁵m³
Explanation:
the solution is in the attached Word file
To calculate the volume of space containing 1.0 J of electromagnetic energy near the Earth, use the formula Volume = Energy / Intensity after converting the intensity from kW/m² to W/m².
The volume of space that contains 1.0 J of electromagnetic energy can be found using the formula:
Volume = Energy / Intensity
Given that the intensity is 1.35 kW/m², you need to convert it to W/m² (1 kW = 1000 W).
After converting the intensity, you can then calculate the volume using the given energy of 1.0 J.
The volume of space near Earth containing 1.0 J of electromagnetic energy can be found by first calculating the energy density using the given intensity and speed of light, and then dividing the amount of energy by energy density, resulting in a volume of 2.22 imes 10^5 m^3.
Finding the Volume of Space Containing 1.0 J of Electromagnetic Energy
To find the volume of space near Earth that contains 1.0 J of electromagnetic energy given the intensity (I) from the sun as 1.35 kW/m2, we can use the formula for energy density (u), which is given by the equation u = I/c, where c is the speed of light. Now, to find the volume (V) that contains energy (E), we use the equation V = E/u.
First, let's convert the intensity to watts per square meter: 1.35 kW/m2 = 1350 W/m2. Next, calculate the energy density (u):
u = 1350 W/m2 / (3.0 * 108 m/s) = 4.5 * 10-6 J/m3.
With the energy density known, we can calculate the volume (V) that contains 1.0 J of energy:
V = 1.0 J / (4.5 * 10-6 J/m3) = 2.22 * 105 m3.
Therefore, a volume of 2.22 * 105 m3 near Earth contains 1.0 J of electromagnetic energy.
9. A 50 kg physics i student trudges from the 1st floor to the 3rd floor, going up a total height of 13 m.
Calculate the work done by the student
Answer:
6370 J
Explanation:
By the law of energy conservation, the work done by the student would be the change in potential enegy from 1st floor to 3rd floor, or a change of 13 m
[tex]W = E_p = mgh [/tex]
where m = 50kg is the mass of the student, g = 9.8 m/s2 is the gravitational constant and h = 13 m is the height difference
[tex]W = 50*9.8*13 = 6370 J[/tex]
Which units are used to measure force?
newtons
feet
miles
grams
Newton is used to measure force
The most used unit for the force measurement is "newton" which is the SI unit of the force
Answer:
newtons
Explanation:
A square metal plate 2.5m on each side is pivoted about an axis though point O at its center and perpendicular to the plate. Calculate the net torque due to the three forces if the magnitudes of the forces are F1=18N, F2=20N, and F3=11N..
Complete Question
The diagram for this question is shown on the first uploaded image
Answer:
The net torque [tex]\tau = 21.95N \cdot m[/tex]
Explanation:
From the question we are told that
The length of each side is [tex]L = 2.5 m[/tex]
The first force is [tex]F_1 = 18N[/tex]
The second force is [tex]F_2 = 20 N[/tex]
The third force is [tex]F_3 = 11N[/tex]
The free body diagram for the question is shown on the second uploaded image
Generally torque is mathematically represented as
[tex]\tau = r * F[/tex]
Where [tex]\tau[/tex] is the torque
r is the length from the rotating point to the point the force is applied, this is also the radius of the circular path made
F is the force causing the rotation.
looking at the free body diagram we can deduce that L is the diameter of the circular path made as a result of toque
Now for the torque due to force [tex]F_1[/tex]
[tex]\tau_1 = - F_1 * r_1[/tex]
The negative sign is because the direction of [tex]F_1[/tex] is clockwise
=> [tex]\tau_1 = - F_1 * \frac{L}{2}[/tex]
Substituting value
[tex]\tau_1 = - 18 * \frac{2.5}{2 }[/tex]
[tex]\tau_1 = - 22.5 N \cdot m[/tex]
The torque as a result of the second force is mathematically evaluated as
[tex]\tau_2 = F_2 * r_2[/tex]
[tex]\tau_2 = F_2 * \frac{L}{2}[/tex]
[tex]= 20 * \frac{2.5}{2}[/tex]
[tex]\tau_ 2 = 25 \ N \cdot m[/tex]
The torque as a result of the third force is mathematically evaluated as
[tex]\tau_3 =r_3 (F_3 sin \theta + F_3 cos \theta )[/tex]
[tex]\tau_3 = \frac{L}{2} (F_3 sin \theta + F_3 cos \theta )[/tex]
Where the free body diagram [tex]\theta = 45^o[/tex]
[tex]\tau_3 = \frac{2.5}{2} (11 * sin (45) +11 cos (45) )[/tex]
[tex]\tau_ 3 = 19.45 \ N \cdot m[/tex]
The net torque the mathematically
[tex]\tau = \tau_1 + \tau_2 + \tau_3[/tex]
substituting value
[tex]\tau = -22.5 + 25 + 19.45[/tex]
[tex]\tau = 21.95N \cdot m[/tex]
A 1000 kg elevator accelerates upward at 1.0 m/s2 for 10 m, starting from rest. a. How much work does gravity do on the elevator? b. How much work does the tension in the elevator cable do on the elevator? c. What is the elevator’s kineti
Answer:
a)= 98kJ
b)=108kJ
c) = 10kJ
Explanation:
a. The work that is done by gravity on the elevator is:
Work = force * distance
= mass * gravity * distance
= 1000 * 9.81 * 10
= 98,000 J
= 98kJ
b)The net force equation in the cable
T - mg = ma
T = m(g+a)
T = 1000(9.8 + 10)
T = 10800N
The work done by the cable is
W = T × d
= 10800N × 10
= 108000
=108kJ
c) PE at 10m = 1000 * 9.81 * 10 = 98,100 J
Work done by cable = PE +KE
108,100 J = KE + 98,100 J
KE = 10,000 J
= 10kJ
=
Answer:
A)Work done by gravity = -98 Kj
B) W_tension = 108 Kj
C) Final kinetic energy Kf = 10 Kj
Explanation:
We are given;
mass; m = 1000kg
Upward acceleration; a = 1 m/s²
Distance; d = 10m
I've attached a free body diagram to show what is happening with the elevator.
A) From the image i attached, the elevators weight acting down is mg.
While T is the tension of the cable pulling the elevator upwards.
Now, we know that,
Work done = Force x distance
Thus, W = F•d
We want to calculate the work done by gravity;
From the diagram I've drawn, gravity is acting downwards and in am opposite direction to the motion having an upward acceleration.
Thus, Force of gravity = - mg
So, Work done by gravity;
W_grav = - mgd
W_grav = -1000 x 9.8 x 10 = -98000 J = -98 Kj
B) Again, W = F.d
W_tension = T•d
Let's find T by summation of forces in the vertical y direction
Thus, Σfy = ma
So, T - mg = ma
Thus, T = ma + mg
T = m(a + g)
Plugging in values,
T = 1000(1 + 9.8)
T = 1000 x 10.8 = 10800 N
So, W_tension = T•d = 10800 N x 10m = 108000 J = 108 Kj
C) From work energy theorem,
Net work = change in kinetic energy
Thus, W_net = Kf - Ki
Where Kf is final kinetic energy and Ki is initial kinetic energy.
Now since the elevator started from rest, Ki = 0 because velocity at that point is zero.
Thus, W_net = Kf - 0
W_net = Kf
Now, W_net is the sum of work done due to gravity and work done due to another force.
Thus, in this case,
W_net = W_grav + W_tension
W_net = -98000 J + 108000 J
W_net = 10000J = 10Kj
So,since W_net = Kf
Thus, Final kinetic energy Kf = 1000J
Find the mass of the solid cylinder Dequals{(r,theta,z): 0less than or equalsrless than or equals2, 0less than or equalszless than or equals10} with density rho(r,theta,z)equals1plusStartFraction z Over 2 EndFraction . Set up the triple integral using cylindrical coordinates that should be used to find the mass of the solid cylinder as efficiently as possible. Use increasing limits of integration. Integral from 0 to nothing Integral from nothing to nothing Integral from nothing to nothing (nothing )font size decreased by 3 dz font size decreased by 3 dr font size decreased by 3 d theta
The mass of the solid cylinder is calculated by setting up a triple integral over the volume of the cylinder using cylindrical coordinates with the given density function. The integral is calculated from the inside out, starting with the integral over z (height), followed by r (radius), and finally θ (angular measure).
Explanation:The mass of a solid object in three-dimensions using a cylindrical coordinate system (r,θ,z) can be expressed as an integral over the volume of the object multiplied by the density function. In your case, given that the solid cylinder D equals {(r, θ, z) : 0 ≤ r ≤ 2, 0 ≤ z ≤ 10} with density function ρ(r, θ, z) = 1+ z/2, we calculate mass M as follows:
M=∫02π∫02∫010(1+ z/2)rdzdrdθ
Here, we start by calculating the innermost integral (∫010(1+ z/2)dz) first, then move to the middle integral (∫02dr) and finally the outermost integral (∫02πdθ). Each integral works on the outcome of the next inner integral until the entire mass of the cylinder is calculated.
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The mass of the solid cylinder is [tex]\(140\pi\).[/tex]
The mass of the solid cylinder is given by the triple integral
[tex]\[ \int_{0}^{2\pi} \int_{0}^{2} \int_{0}^{10} \left(1 + \frac{z}{2}\right) r \, dz \, dr \, d\theta. \][/tex]
To find the mass of the solid cylinder, we need to evaluate the triple integral of the density function over the volume of the cylinder in cylindrical coordinates. The density function is given by [tex]\(\rho(r, \theta, z) = 1 + \frac{z}{2}\).[/tex]
The solid cylinder is defined by the set [tex]\(D = \{(r, \theta, z) : 0 \leq r \leq 2, 0 \leq z \leq 10\}\)[/tex]. The limits of integration for[tex]\(r\), \(z\), and \(\theta\)[/tex] are determined by the geometry of the cylinder:
- For[tex]\(r\)[/tex], the radial distance from the z-axis, the limits are from 0 to 2, since the radius of the cylinder is 2 units.
- For[tex]\(z\),[/tex] the height of the cylinder, the limits are from 0 to 10, since the height is 10 units.
- For [tex]\(\theta\),[/tex] the angle around the z-axis, the limits are from 0 to [tex]\(2\pi\),[/tex] since a full rotation around the z-axis is [tex]\(2\pi\)[/tex] radians.
The differential volume element in cylindrical coordinates is [tex]\(r \, dz \, dr \, d\theta\),[/tex] where the [tex]\(r\)[/tex] factor accounts for the Jacobian of the transformation from Cartesian to cylindrical coordinates.
Now, we can set up the triple integral as follows:
[tex]\[ \int_{0}^{2\pi} \int_{0}^{2} \int_{0}^{10} \left(1 + \frac{z}{2}\right) r \, dz \, dr \, d\theta. \][/tex]
To evaluate this integral, we would first integrate with respect to [tex]\(z\)[/tex] from 0 to 10, then with respect to [tex]\(r\)[/tex] from 0 to 2, and finally with respect to [tex]\(\theta\)[/tex] from 0 to[tex]\(2\pi\)[/tex] . The integral can be computed as follows:
1. Integrate with respect to [tex]\(z\)[/tex] first:
[tex]\[ \int_{0}^{10} \left(1 + \frac{z}{2}\right) dz = \left[z + \frac{z^2}{4}\right]_{0}^{10} = 10 + \frac{100}{4} = 10 + 25 = 35. \][/tex]
2. Next, integrate with respect to[tex]\(r\):[/tex]
[tex]\[ \int_{0}^{2} 35r \, dr = \left[\frac{35r^2}{2}\right]_{0}^{2} = \frac{35 \cdot 4}{2} = 70. \][/tex]
3. Finally, integrate with respect to [tex]\(\theta\):[/tex]
[tex]\[ \int_{0}^{2\pi} 70 \, d\theta = \left[70\theta\right]_{0}^{2\pi} = 70 \cdot 2\pi = 140\pi. \][/tex]
Which appliances uses the larger current, a toaster or a desk lamp?
Answer:
DeskLamp
Explanation:
You're average toaster uses so much more electricity than the average desk lamp. 2 slice toasters I believe uses about 1000 watts? And 4 slice uses 1600. The brightest lightbulbs tend to run on 100 watts so just imagine a lightbulb with 1600 watts. (There's a video of a man cranking the watts to a 1000 watt lightbulb up to millions until it pops) Lightbulbs have a very far range in the amounts of energy they can take in. Here is a simplified answer-
On average- the toaster would use the most watts in terms of use
In other circumstances- The lightbulb would be the case.
So yeah,
It's Desklamps. They're really powerful.
You push your physics book 1.50 m along a horizontal tabletop with a horizontal push of 2.40 N while the opposing force of friction is 0.600 N. How much work does each of the following forces do on the book: (a) your 2.40-N push, (b) the friction force, (c) the normal force from the tabletop, and (d) gravity? (e) What is the net work done on the book?
Answer:
a) 3.6 J
b) 0.9 J
c) 0 J
d) 0 J
e) 2.7 J
Explanation:
Work = force F x diaplament in direction of force d
Given that,
Displacement in direction of push force d = 1.5 N
Push force = 2.4 N
Friction = 0.6 N
a) work done by push force = 2.4 x 1.5 = 3.6 J
b) word done by friction = -0.6 x 15 = -0.9 J (minus sign shows the work is opposite that done by the push force)
c) work done by normal force from table top ( force acting upwards to oppose gravity) = 0 J since there is no vertical displacement
d) work done by gravity (doward force) = 0 J since there is no vertical displacement
e) net work done on book = 3.6 J + (-0.9J) + 0 J + 0 J = 2.7 J
The work done by each force is calculated using the formula Work = Force × Distance. The net work done on the book is the sum of the work done by each force. (a) 3.6 J (b) 0.9 J (c) 0 J (d) 0 J (e) 4.5 J
Explanation:To calculate the work done by each force, we use the formula:
Work = Force × Distance
(a) The work done by your push is:
Work = 2.40 N × 1.50 m = 3.6 J
(b) The work done by the friction force is:
Work = 0.600 N × 1.50 m = 0.9 J
(c) The normal force from the tabletop does no work because it is perpendicular to the direction of motion.
(d) The work done by gravity is zero since the book doesn't move vertically and gravity acts vertically.
(e) The net work done on the book is the sum of the work done by all forces:
Net Work = Work by your push + Work by friction + Work by normal force + Work by gravity = 3.6 J + 0.9 J + 0 J + 0 J = 4.5 J
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Two equal mass carts approach each other with velocities equal in magnitude but opposite in direction. Friction can be neglected. If the carts collide completely inelastically, what will be the final velocity of the combined system?
Answer: their final velocity will be zero
Explanation:
Since they have equal masses m,
And their velocity is equal but in opposite direction u and -u, then,
mu + (-mu) = 2mv
0 = 2mv
Which implies that V = 0
Arc welding uses electric current to make an extremely hot electric arc that can melt metal. The arc emits ultraviolet light that can cause sunburn and eye damage if a welder is not wearing protective gear. Why does the arc give off ultraviolet light
Answer: The arc gives off ultraviolet light as a result of the high temperature of the arc which can be as high as above 3000°c.
Explanation: arc welding employs or uses electric current to generate heat for the purpose of joining metals. Metals being joined usually have high melting points above 3000°c, This high melting points of metals means the welding arc needs to attain a higher temperature to be able to join the metals. In the process of attaining that temperature needed to join the metals the arc gives of ultraviolet light (UV).
The reaction converting glycerol to glycerol-3-phosphate is energetically unfavorable by 9.2 kJkJ. The reaction of ATP with water to yield ADP, hydrogen phosphate ion, and hydrogen ion is energetically favorable by 30.5 kJkJ. These reactions are coupled so that the overall process is energetically favorable. What is ΔGΔGDelta G for the overall process?
Answer:
- 21.3 kJ
Explanation:
For the favorable reaction, ΔG₁ = + 9.2 kJ (ΔG > 0 for unfavorable reaction)
For the favorable reaction, ΔG₂ = - 30.5 kJ (ΔG < 0 for favorable reaction)
For the overall reaction, ΔG = ΔG₁ + ΔG₂ = + 9.2 kJ + (- 30.5 kJ)
ΔG = + 9.2 kJ - 30.5 kJ = - 21.3 kJ
Two sheets of polarizing material are placed such that their polarizing axes are 90° to each other and no light passes through the combination. Can you place a third sheet of polarizing material between the two and cause light to pass through this three-sheet combination? Explain why if this is possible.
Answer:
Explanation:
Polarization In this case angle of incidence is not equal to angle of polarization, hence reflected light is partially polarized and transmitted light is also partially polarized. by reflection is explained by Brewster's law,
According to this when unpolarized light incident on glass plate at an angle is called as angle of polarizing the reflected light is plane polarized, and transmitted light is partially polarized. The plane of vibration of polarized light is having plane of vibrations perpendicular to plane of incidence.
When two sheets of polarizing material with perpendicular axes are placed together, no light passes through. However, if a third sheet of polarizing material is placed between the first two sheets at an angle, some light can pass through the combination.
Explanation:When two sheets of polarizing material are placed with their polarizing axes at right angles to each other, no light passes through the combination because the second sheet blocks the light that is polarized by the first sheet. However, if a third sheet of polarizing material is placed between the first two sheets, at an angle between 0° and 90° to the axis of the first sheet, some light can pass through the three-sheet combination.
This is because the third sheet allows a fraction of the previously blocked light to pass through. As the angle between the first and third sheets approaches 90°, more light is transmitted by the combination.
For example, in the case of Figure 27.41, when the axes of the first and second filters are aligned (parallel), all of the polarized light passed by the first filter is also passed by the second filter. When the second filter is rotated to make the axes perpendicular, no light is passed by the second filter.
Two hollow cylinders have the same inner and outer diameters and the same mass, but they have different lengths because one is made of low density wood and the other of high density brass. Which cylinder has the greater moment of inertia about its cylindrical center axis?
Answer:
Both cylinders will have the same moment of inertia.
Explanation:
The moment of inertia of a rigid body depends on the distribution of mass around the axis of rotation, i.e at what radius from the axis how much mass is located. What the moment of of inertia DOES NOT depend in is the distribution of mass parallel to the axis of rotation. This means that two hollow cylinders of the same mass but with different lengths will have the same moment of inertia!
For completeness, the momentum of inertia of a hollow cylinder is
[tex]I \approx mR^2[/tex]
from which we clearly see that [tex]I[/tex] only depends on the mass and the radius of the hollow cylinder and not on its height; Hence, both the wooden and the brass cylinders will have the same moment of inertia.
The two (2) hollow cylinders will have the same moment of inertia about their cylindrical center axis.
What is moment of inertia?Moment of inertia is also referred to as the mass moment of inertia and it can be defined as a measure of the rotational inertia of an object or its resistance to angular acceleration about a reference axis.
Mathematically, the moment of inertia of a hollow cylinder is given by the formula:
[tex]I=mr^2[/tex]
Where:
I is the moment of inertia.m is the mass.r is the radius.From the above formula, we can deduce that the moment of inertia of a hollow cylinder is highly dependent on the distribution of mass around its axis of rotation (radius) rather than its height.
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A proton with a speed of 2 x 105 m/s falls through a potential difference V and thereby increases its speed to 6 x 105 m/s. Through what potential difference did the proton fall
16.7 x 10²V
Explanation:The work-energy theorem states that the change in kinetic energy of a particle, Δ[tex]K_{E}[/tex], results in work done by the particle, W. i.e;
Δ[tex]K_{E}[/tex] = W ------------------(i)
But:
Δ[tex]K_{E}[/tex] = [tex]\frac{1}{2}[/tex]mv² -
Δ[tex]K_{E}[/tex] = [tex]\frac{1}{2}[/tex]m [v² - u²] ------------------(ii)
Where;
m = mass of particle (proton in this case) = 1.67 x 10⁻²⁷kg
v = final velocity of the particle = 6 x 10⁵m/s
u = initial velocity of the particle = 2 x 10⁵m/s
Substitute these values into equation (ii) as follows;
Δ[tex]K_{E}[/tex] = [tex]\frac{1}{2}[/tex](1.67 x 10⁻²⁷) [(6 x 10⁵)² - (2 x 10⁵)²]
Δ[tex]K_{E}[/tex] = [tex]\frac{1}{2}[/tex](1.67 x 10⁻²⁷) [(36 x 10¹⁰) - (4 x 10¹⁰)]
Δ[tex]K_{E}[/tex] = [tex]\frac{1}{2}[/tex](1.67 x 10⁻²⁷) [(36 - 4) x 10¹⁰]
Δ[tex]K_{E}[/tex] = [tex]\frac{1}{2}[/tex](1.67 x 10⁻²⁷) [32 x 10¹⁰]
Δ[tex]K_{E}[/tex] = (1.67 x 10⁻²⁷) [16 x 10¹⁰]
Δ[tex]K_{E}[/tex] = 26.72 x 10⁻¹⁷J
Also:
W = qΔV ----------------(iii)
Where;
q = charge of the particle (proton) = 1.6 x 10⁻¹⁹C
ΔV = change in potential difference of the particle = V (from the question)
Substitute these values into equation (iii) as follows;
W = 1.6 x 10⁻¹⁹ x V
W = 1.6 x 10⁻¹⁹V -------------------(iv)
Now:
Substitute the values of Δ[tex]K_{E}[/tex] = 26.72 x 10⁻¹⁷ and W in equation(iv) into equation (i)
26.72 x 10⁻¹⁷ = 1.6 x 10⁻¹⁹V
Solve for V;
V = (26.72 x 10⁻¹⁷) / (1.6 x 10⁻¹⁹)
V = 16.7 x 10²V
Therefore, the potential difference through which the proton fell was 16.7 x 10²V
The potential contrast between different locations represents the work or energy dissipated in the transmission of the unit amount of voltage from one point to another.
Following are the calculation of the potential difference:
Change in energy [tex], e^{v}=\frac{1}{2} \ m(v_2^2-v_1^2)[/tex]
[tex]1.602 \times ^{-19}=\frac{1}{2} \times 1.67 \times 10^{-27} (6^2-2^2)\times 10^{10}\\\\[/tex]
[tex]\to v \times 1.602 \times 10^{-19}=0.835 \times 10^{-17}\times 32\\\\[/tex]
[tex]\to v = \frac{0.835 \times 10^{-17}\times 32}{1.602 \times 10^{-19}}\\\\[/tex]
[tex]= \frac{0.835 \times 10^{2}\times 32}{1.602 }\\\\ = \frac{26.72 \times 10^{2} }{1.602 }\\\\=16.67\times 10^{2}[/tex]
Therefore the final answer is "[tex]\bold{16.67 \times 10^2}[/tex]".
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You are explaining to friends why astronauts feel weightless orbiting in the space shuttle, and they respond that they thought gravity was just a lot weaker up there. Convince them and yourself that it isn't so by calculating how much weaker gravity is 450 km above the Earth's surface. Express your answer using two significant figures.
Answer:
Only 9% weaker
Explanation:
Because this is where most stuff that people do in space takes place. So, um, here we're at a radius of the earth plus 300 kilometers. You may already be seeing why this isn't going to have much effect if this were except the 6.68 times, 10 to the sixth meters. And so the value of Gout here. You know, Newton's gravitational constant times, the mass of the Earth divided by R squared for the location we're looking at. And so this works out to be 8.924 meters per second squared, which is certainly less than it is at the surface of the earth. However, this is only 9% less than acceleration for gravity at the surface. So the decrease in the gravity gravitational acceleration of nine percent not really going toe produces a sensation of weightlessness.
Answer:
The gravity at the altitude of 450 km above the earth's surface is (0.87 × the normal acceleration due to gravity); that is, 0.87g.
This shows that the gravity at this altitude isn't as weak as initially thought; it is still 87% as strong as the gravity on the surface of the earth.
The gravity is only 13% weaker at an altitude of 450 km above the earth's surface.
Explanation:
The force due to gravity that is usually talked about is basically the force of attraction between the planetary bodies such as the earth and objects on its surface.
The force is given through the Newton's gravitational law which explains that the force of attraction between two bodies is directly proportional to the product of the masses of both bodies and inversely proportional to the square of their distance apart .
Let the force of attraction between a body of mass, m on the surface of the earth and the earth itself, with mass M
F ∝ (Mm/r²)
F = (GMm/r²)
G = Gravitational constant (the constant of proportionality)
r = distance between the body and the earth, and this is equal.to the radius of the earth.
This force is what is now translated to force of gravity or weight of a body.
F = mg
where g = acceleration due to gravity = 9.8 m/s²
F = (GMm/r²) = (mg)
g = (GM/r²) = 9.8 m/s²
So, for a body at a distance that is 450 km above the earth's atmosphere, the distance between that body and the centre of the earth is (r + 450,000) m
Let that be equal to R.
R = (r + 450,000) m
Note that earth's radius is approximately 6400 km
r = 6400 km = 6,400,000 m
R = 6400 + 450 = 6850 km = 6,850,000 m
Normal acceleration due to gravity = 9.8 m/s²
9.8 = (GM/6,400,000²)
GM = 9.8 × 6,400,000²
Acceleration due to gravity at a point 450 km above the earth's surface
a = (GM/R²)
a = (GM/6,850,000²)
Note that GM = 9.8 × 6,400,000²
a = (9.8 × 6,400,000²) ÷ (6,850,000²)
a = 9.8 × 0.873 = 8.56 m/s²
a = 87% of g.
The gravity at this altitude above the earth's surface is (0.87 × the normal acceleration due to gravity)
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A long solenoid has n turns per unit length, has a radius R1, and carries a current R2. A circular coil with radius R2 and with N total turns is coaxial with the solenoid and equidistant from its ends. (a) Find the magnetic flux through the coil ifR2 > R1 (b) Find the magnetic flux through the coil if R2 < R1.
Answer:
a) Фm = μ₀*I*n*π*N*R₁²
b) Фm = μ₀*I*n*π*N*R₂²
Explanation:
Given
n = number of turns per unit length of the solenoid
R₁ = radius of the solenoid
I = current passing through the solenoid
R₂ = radius of the circular coil
N = number of total turns on the coil
a) Фm = ? If R₂ > R₁
b) Фm = ? If R₂ < R₁.
We can use the formula
Фm = N*B*S*Cos θ
where
N is the number of turns
B is the magnetic field
S is the area perpendicular to magnetic field B.
The magnetic field outside the solenoid can be approximated to zero. Therefore, the flux through the coil is the flux in the core of the solenoid.
We know that the magnetic field inside the solenoid is uniform. Thus, the flux through the circular coil is given by the same expression with R2 replacing R1 (from the area).
a) Then, the flux through the large circular loop outside the solenoid (R₂ > R₁) is obtained as follows:
Фm = N*B*S*Cos θ
where
B = μ₀*I*n
S = π*R₁²
θ = 0°
⇒ Фm = (N)*(μ₀*I*n)*(π*R₁²)*Cos 0°
⇒ Фm = μ₀*I*n*π*N*R₁²
b) The flux through the coil when R₂ < R₁ is
Фm = (N)*(μ₀*I*n)*(π*R₂²)*Cos 0°
⇒ Фm = μ₀*I*n*π*N*R₂²
Answer:
a) ∅ [tex]= u_0*n*I*N*pi*(R_1)^2[/tex]
b) ∅ [tex]= u_0 * n*I*N * pi * (R_2)^2 [/tex]
Explanation:
The magnetic field inside the solenoid is uniform and the flux in the coil is the same as the flux in the solenoid.
The magnetic field outside can be said to be zero which also means that the flux through the coil is the same as flux in the solenoid.
Thus, we can say
R2 = R1
Where:
N = total turns on coil
n = number of turns per unit length of solenoid
I = current
R1 = radius of solenoid
R2 = circular coil radius
a) The magnetic flux through the coil if R2 > R1 is giben as:
∅ [tex]= N(u_0 *n*I)(pi*(R_1 )^2) Cos 0[/tex]
Solving further, we have:
∅ [tex]= u_0*n*I*N*pi*(R_1)^2[/tex]
Here,
[tex] u_0*n*I [/tex] is the magnetic field
(Pi*R1²) is the area pependicular to magnetic field
N is the number of turns
b) The magnetic flux through the coil if R2 < R1
∅ [tex] = N(u_0*n*I)(pi*(R_2)^2)Cos 0 [/tex]
∅ [tex]= u_0 * n*I*N * pi * (R_2)^2 [/tex]
Here,
[tex] u_0*n*I [/tex] is the magnetic field
(Pi*R2²) is the area pependicular to magnetic field
N is the number of turns