Answer:
A
Explanation:
potential energy is stored energy so the ball has potential energy to bounce or roll which would then have been converted to kinetic but the rest are in motion meaning energy is no longer stored but used in motion and therefore cannot be potential energy
Two objects attract each other with a gravitational force of magnitude 1.02 10-8 N when separated by 19.7 cm. If the total mass of the two objects is 5.14 kg, what is the mass of each
Answer:
The two masses are 3.39 Kg and 1.75 Kg
Explanation:
The gravitational force of attraction between two bodies is given by the formula;
F = Gm₁m₂/d²
where G is the gravitational force constant = 6.67 * 10⁻¹¹ Nm²Kg⁻²
m₁ = mass of first object; m₂ = mass of second object; d = distance of separation between the objects
Further calculations are provided in the attachment below
When a nerve cell depolarizes, charge is transferred across the cell membrane, changing the potential difference. For a typical nerve cell, 9.0 pC of charge flows in a time of 0.50 ms. You may want to review (Page 706) . Part A What is the average current
Answer:
I = 18 x 10⁻⁹ A = 18 nA
Explanation:
The current is defined as the flow of charge per unit time. Therefore,
I = q/t
where,
I = Average Current passing through nerve cell
q = Total flow of charges through nerve cell
t = time period of flow of charges
Here, in our case:
I = ?
q = (9 pC)(1 x 10⁻¹² C/1 pC) = 9 x 10⁻¹² C
t = (0.5 ms)(1 x 10⁻³ s/1 ms) = 5 x 10⁻⁴ s
Therefore,
I = (9 x 10⁻¹² C)/(5 x 10⁻⁴ s)
I = 18 x 10⁻⁹ A = 18 nA
The potential energy function
U(x,y)=A[(1/x2) + (1/y2)] describes a conservative force, where A>0.
Derive an expression for the force in terms of unit vectors i and j.
Answer:
[tex]F=-2A[\frac{1}{x^3}\hat{i}+\frac{1}{y^3}\hat{j}][/tex]
Explanation:
You have the following potential energy function:
[tex]U(x,y)=A[\frac{1}{x^2}+\frac{1}{y^2}}][/tex] (1)
A > 0 constant
In order to find the force in terms of the unit vectors, you use the gradient of the potential function:
[tex]\vec{F}=\bigtriangledown U(x,y)=\frac{\partial}{\partial x}U\hat{i}+\frac{\partial}{\partial y}U\hat{j}[/tex] (2)
Then, you replace the expression (1) into the expression (2) and calculate the partial derivatives:
[tex]\vec{F}=A\frac{\partial}{\partial x}[\frac{1}{x^2}+\frac{1}{y^2}]} \hat{i}+A\frac{\partial}{\partial x}[\frac{1}{x^2}+\frac{1}{y^2}]\hat{j}\\\\\vec{F}=A(-2x^{-3})\hat{i}+A(-2y^{-3})\hat{j}\\\\F=-2A[\frac{1}{x^3}\hat{i}+\frac{1}{y^3}\hat{j}][/tex](3)
The result obtained in (3) is the force expressed in terms of the unit vectors, for the potential energy function U(x,y).
The center of gravity of an ax is on the centerline of the handle, close to the head. Assume you saw across the handle through the center of gravity and weigh the two parts. What will you discover?
Answer:
I believe it is they will weigh the same
Explanation:
Center of gravity is the axis on which the mass rotates evenly if I remember correctly from AP Physics
The head side is heavier than the handle side. - this will be discovered.
What is center of gravity of a object?Theoretically, the body's center of gravity is where all of the weight is believed to be concentrated. Knowing the centre of gravity is crucial because it may be used to forecast how a moving object will behave when subjected to the effects of gravity. In designing immobile constructions like buildings and bridges, it is also helpful.
We know that center of gravity is close to some particular point refers the mass of the point is greater then others. It is given that: The center of gravity of an ax is on the centerline of the handle, close to the head.
So, we can conclude that the head side of the ax is heavier than the handle side of it.
Learn more about center of gravity here:
https://brainly.com/question/17409320
#SPJ5
A hot air balloon competition requires a balloonist to drop a ribbon onto a target on the ground. Initially the hot air balloon is 50 meters above the ground and 100 meters from the target. The wind is blowing the balloon at v = 15 meters/sec on a course to travel directly over the target. The ribbon is heavy enough that any effects of the air slowing the vertical velocity of the ribbon are negligible. How long should the balloonist wait to drop the ribbon so that it will hit the target?
time =
Answer:
The waiting time is [tex]t_w = 3 .47 \ s[/tex]
Explanation:
From the question we are told that
The height of the hot air balloon above the ground is [tex]d = 50 \ m[/tex]
The distance of the balloon from the target is [tex]l = 100 \ m[/tex]
The velocity of the balloon is [tex]v = 15 \ m/s[/tex]
Generally the time it will take to reach the ground is
[tex]t = \sqrt{2 * \frac{d}{g} }[/tex]
substituting values
[tex]t = \sqrt{2 * \frac{50}{9.8} }[/tex]
[tex]t = 3.2 \ s[/tex]
The distance that is covered at time with the given velocity is mathematically evaluated as
[tex]z = v * t[/tex]
substituting values
[tex]z = 15 * 3.2[/tex]
[tex]z = 48 \ m[/tex]
This implies that for the balloon moving at a velocity (v) to hit the target it must be dropped at this distance (z)
Now the distance the balloonist has to wait before dropping in order to hit the target is
[tex]A = d - z[/tex]
substituting values
[tex]A = 100 - 48[/tex]
[tex]A = 52 \ m[/tex]
This implies that the time the balloonist has to wait is
[tex]t_w = \frac{A}{v}[/tex]
substituting values
[tex]t_w = \frac{52}{15}[/tex]
[tex]t_w = 3 .47 \ s[/tex]
g At some point the road makes a right turn with a radius of 117 m. If the posted speed limit along this part of the highway is 25.1 m/s, how much should Raquel bank the turn so that a vehicle traveling at the posted speed limit can make the turn without relying on the frictional force between the tires and the road
Answer:
Ф = 28.9°
Explanation:
given:
radius (r) = 117m
velocity (v) = 25.1 m/s
required: angle Ф
Ф = inv tan (v² / (r * g)) we know that g = 9.8
Ф = inv tan (25.1² / (117 * 9.8))
Ф = 28.9°
the density of gold is 19 300kg/m^3. what is the mass of gold cube with the length 0.2015m?
Answer:
The mass is [tex]157.87m^3[/tex]Explanation:
Given data
length of cube= 0.2015 m
density = 19300 kg/m^3.
But the volume of cube is given as [tex]l*l*l= l^3[/tex]
[tex]volume -of- cube= 0.2015*0.2015*0.2015= 0.00818 m^3[/tex]
The density is expressed as = mass/volume
[tex]mass=19300*0.00818= 157.87m^3[/tex]
A force of 44 N will stretch a rubber band 88 cm (0.080.08 m). Assuming that Hooke's law applies, how far will aa 11-N force stretch the rubber band? How much work does it take to stretch the rubber band this far?
Answer:
The rubber band will be stretched 0.02 m.
The work done in stretching is 0.11 J.
Explanation:
Force 1 = 44 N
extension of rubber band = 0.080 m
Force 2 = 11 N
extension = ?
According to Hooke's Law, force applied is proportional to the extension provided elastic limit is not extended.
F = ke
where k = constant of elasticity
e = extension of the material
F = force applied.
For the first case,
44 = 0.080K
K = 44/0.080 = 550 N/m
For the second situation involving the same rubber band
Force = 11 N
e = 550 N/m
11 = 550e
extension e = 11/550 = 0.02 m
The work done to stretch the rubber band this far is equal to the potential energy stored within the rubber due to the stretch. This is in line with energy conservation.
potential energy stored = [tex]\frac{1}{2}ke^{2}[/tex]
==> [tex]\frac{1}{2}* 550* 0.02^{2}[/tex] = 0.11 J
1. As you pass a freight truck with a trailer on a highway, you notice that its trailer is bouncing up and down slowly. Is it more likely that the trailer is heavily loaded or nearly empty
Answer:
It's more likely that the trailer is heavily loaded
Explanation:
Due to the fact that the frequency is proportional to the square root of the force constant and inversely proportional to the square root of the mass, it is very likely that the truck would be heavily loaded because the force constant would be the same whether the truck is empty or heavily loaded.
A 150m race is run on a 300m circular track of circumference. Runners start running from the north and turn west until reaching the south. What is the magnitude of the displacement made by the runners?
Answer:
95.5 m
Explanation:
The displacement is the position of the ending point relative to the starting point.
In this case, the magnitude of the displacement is the diameter of the circular track.
d = 300 m / π
d ≈ 95.5 m
A charge of 87.6 pC is uniformly distributed on the surface of a thin sheet of insulating material that has a total area of 65.2 cm^2. A Gaussian surface encloses a portion of the sheet of charge. If the flux through the Gaussian surface is 9.20 N⋅m^2/C, what area of the sheet is enclosed by the Gaussian surface?
Answer:
60.8 cm²
Explanation:
The charge density, σ on the surface is σ = Q/A where q = charge = 87.6 pC = 87.6 × 10⁻¹² C and A = area = 65.2 cm² = 65.2 × 10⁻⁴ m².
σ = Q/A = 87.6 × 10⁻¹² C/65.2 × 10⁻⁴ m² = 1.34 × 10⁻⁸ C/m²
Now, the charge through the Gaussian surface is q = σA' where A' is the charge in the Gaussian surface.
Since the flux, Ф = 9.20 Nm²/C and Ф = q/ε₀ for a closed Gaussian surface
So, q = ε₀Ф = σA'
ε₀Ф = σA'
making A' the area of the Gaussian surface the subject of the formula, we have
A' = ε₀Ф/σ
A' = 8.854 × 10⁻¹² F/m × 9.20 Nm²/C ÷ 1.34 × 10⁻⁸ C/m²
A' = 81.4568/1.34 × 10⁻⁴ m²
A' = 60.79 × 10⁻⁴ m²
A' ≅ 60.8 cm²
The flux through the Gaussian surface is 9.20 N⋅m^2/C then the surface area of the Gaussian Sheet is 60.76 square cm.
Charge and Charge DensityA certain amount of electrons in excess or defect is called a charge. Charge density is the amount of charge distributed over per unit of volume.
Given that, for a thin sheet of insulating material, the charge Q is 87.6 pC and surface area A is 65.2 square cm. Then the charge density for a thin sheet is given below.
[tex]\sigma = \dfrac {Q}{A}[/tex]
[tex]\sigma = \dfrac {87.6\times 10^{-12}}{65;.2\times 10^{-4}}[/tex]
[tex]\sigma = 1.34\times 10^{-8} \;\rm C/m^2[/tex]
Thus the charge density for a thin sheet of insulating material is [tex]1.34\times 10^{-8} \;\rm C/m^2[/tex].
Now, the flux through the Gaussian surface is 9.20 N⋅m^2/C. The charge over the Gaussian Surface is given as below.
[tex]Q' = \sigma A'[/tex]
Where Q' is the charge at the Gaussian Surface, A' is the surface area of the Gaussian surface and [tex]\sigma[/tex] is the charge density.
For the closed Gaussian Surface, Flux is given below.
[tex]\phi = \dfrac {Q'}{\epsilon_\circ}[/tex]
Hence the charge can be written as,
[tex]Q' = \phi\epsilon_\circ[/tex]
So the charge can be given as below.
[tex]Q' = \phi\epsilon_\circ = \sigma A'[/tex]
Then the surface area of the Gaussian surface is given below.
[tex]A' = \dfrac {\phi\epsilon_\circ}{\sigma}[/tex]
Substituting the values in the above equation,
[tex]A' = \dfrac {9.20 \times 8.85\times 10^{-12}}{1.38\times 10^{-8}}[/tex]
[tex]A' =0.006076\;\rm m^2[/tex]
[tex]A' = 60.76 \;\rm cm^2[/tex]
Hence we can conclude that the area of the Gaussian Surface is 60.76 square cm.
To know more about the charge and charge density, follow the link given below.
https://brainly.com/question/8532098.
A very long, solid cylinder with radius R has positive charge uniformly distributed throughout it, with charge per unit volume \rhorho.
(a) Derive the expression for the electric field inside the volume at a distance r from the axis of the cylinder in terms of the charge density \rhorho.
(b) What is the electric field at a point outside the volume in terms of the charge per unit length \lambdaλ in the cylinder?
(c) Compare the answers to parts (a) and (b) for r = R.
(d) Graph the electric-field magnitude as a function of r from r = 0 to r = 3R.
Answer:
the answers are provided in the attachments below
Explanation:
Gauss law state that the net electric field coming out of a closed surface is directly proportional to the charge enclosed inside the closed surface
Applying Gauss law to the long solid cylinder
A) E ( electric field ) = p*r / 2 * [tex]e_{0}[/tex]
B) E = 2K λ / r
C) Answers from parts a and b are the same
D) attached below
Applying Gauss's law which states that the net electric field in an enclosed surface is directly ∝ to the charge found in the enclosed surface.
A ) The expression for the electric field inside the volume at a distance r
Gauss law : E. A = [tex]\frac{q}{e_{0} }[/tex] ----- ( 1 )
where : A = surface area = 2πrL , q = p(πr²L)
back to equation ( 1 )
E ( electric field ) = p*r / 2 * [tex]e_{0}[/tex]
B) Electric field at point Outside the volume in terms of charge per unit length λ
Given that: linear charge density = area * volume charge density
λ = πR²P
from Gauss's law : E ( 2πrL) = [tex]\frac{q}{e_{0} }[/tex]
∴ E = [tex]\frac{\pi R^{2}P }{2e_{0}r\pi }[/tex] ----- ( 2 )
where : πR²P = λ
Back to equation ( 2 )
E = λ / 2e₀π*r where : k = 1 / 4πe₀
∴ The electric field ( E ) at point outside the volume in terms of charge per unit Length λ
E = 2K λ / r
C) Comparing answers A and B
Answers to part A and B are similar
Hence we can conclude that Applying Gauss law to the long solid cylinder
E ( electric field ) = p*r / 2 * [tex]e_{0}[/tex], E = 2K λ / r also Answers from parts a and b are the same.
Learn more about Gauss's Law : https://brainly.com/question/15175106
The rock and meterstick balance at the 25-cm mark, as shown in the sketch. The meterstick has a mass of 1 kg. What must be the mass of the rock? (Show work).
Answer:
1 kgExplanation:
Check the diagram attached below for the diagram.
Let the weight of the rock be W and the mass of the meter stick be M. Note that the mass of the meter stick will be placed at the middle of the meter stick i.e at the 50cm mark
Using the principle of moment to calculate the weight of the rock. It states that the sum of clockwise moments is equal to the sum of anti clockwise moment.
Moment = Force * perpendicular distance
The meterstick acts in the clockwise direction while the rock acys in the anti clockwise direction
Clockwise moment = 1kg * 25 = 25kg/cm
Anticlockwise moment = W * 25cm = 25W kg/cm
Equating both moments of forces
25W = 25
W = 25/23
W = 1 kg
The mass of the rock is also 1 kg
The objective lens of a microscope has a focal length of 5.5mm. Part A What eyepiece focal length will give the microscope an overall angular magnification of 300
Complete Question
The distance between the objective and eyepiece lenses in a microscope is 19 cm . The objective lens has a focal length of 5.5 mm .
What eyepiece focal length will give the microscope an overall angular magnification of 300?
Answer:
The eyepiece focal length is [tex]f_e = 0.027 \ m[/tex]
Explanation:
From the question we are told that
The focal length is [tex]f_o = 5.5 \ mm = -0.0055 \ m[/tex]
This negative sign shows the the microscope is diverging light
The angular magnification is [tex]m = 300[/tex]
The distance between the objective and the eyepieces lenses is [tex]Z = 19 \ cm = 0.19 \ m[/tex]
Generally the magnification is mathematically represented as
[tex]m = [\frac{Z - f_e }{f_e}] [\frac{0.25}{f_0} ][/tex]
Where [tex]f_e[/tex] is the eyepiece focal length of the microscope
Now making [tex]f_e[/tex] the subject of the formula
[tex]f_e = \frac{Z}{1 - [\frac{M * f_o }{0.25}] }[/tex]
substituting values
[tex]f_e = \frac{ 0.19 }{1 - [\frac{300 * -0.0055 }{0.25}] }[/tex]
[tex]f_e = 0.027 \ m[/tex]
When a potential difference of 12 V is applied to a wire 7.2 m long and 0.35 mm in diameter the result is an electric current of 2.1 A. What is the resistivity of the wire?
Answer:
7.63 x 10^-8ohmm
Explanation:
resistivity of the wire = 7.63 x 10^-8ohmm
When a particular wire is vibrating with a frequency of 6.3 Hz, a transverse wave of wavelength 53.3 cm is produced. Determine the speed of wave pulses along the wire.
Answer:
335.79cm/s
Explanation:
When a transverse wave of wavelength λ is produced during the vibration of a wire, the frequency(f), and the speed(v) of the wave pulses are related to the wavelength as follows;
v = fλ ------------------(ii)
From the question;
f = 6.3Hz
λ = 53.3cm
Substitute these values into equation (i) as follows;
v = 6.3 x 53.3
v = 335.79cm/s
Therefore, the speed of the wave pulses along the wire is 335.79cm/s
An ice skater spinning with outstretched arms has an angular speed of 5.0 rad/s . She tucks in her arms, decreasing her moment of inertia by 11 % . By what factor does the skater's kinetic energy change? (Neglect any frictional effects.)
Answer:
K_{f} / K₀ =1.12
Explanation:
This problem must work using the conservation of angular momentum (L), so that the moment is conserved in the system all the forces must be internal and therefore the torque is internal and the moment is conserved.
Initial moment. With arms outstretched
L₀ = I₀ w₀
the wo value is 5.0 rad / s
final moment. After he shrugs his arms
[tex]L_{f}[/tex] = I_{f} w_{f}
indicate that the moment of inertia decreases by 11%
I_{f} = I₀ - 0.11 I₀ = 0.89 I₀
L_{f} = L₀
I_{f} w_{f} = I₀ w₀
w_{f} = I₀ /I_{f} w₀
let's calculate
w_{f} = I₀ / 0.89 I₀ 5.0
w_{f} = 5.62 rad / s
Having these values we can calculate the change in kinetic energy
[tex]K_{f}[/tex] / K₀ = ½ I_{f} w_{f}² (½ I₀ w₀²)
K_{f} / K₀ = 0.89 I₀ / I₀ (5.62 / 5)²
K_{f} / K₀ =1.12
what is the value of the tropic of cancer
Answer: The latitudinal value of tropic of cancer is 23.5° N on June 21, when the sun is directly up above the head at noon. The Equator is the circle at which sun is straight above the head.
Explanation:
Objects A and B are both positively charged. Both have a mass of 900 g, but A has twice the charge of B. When A and B are placed 30.0 cm apart, B experiences an electric force of 0.870 N.
How large is the force on A?
What is the charge on qA and qB?
If the objects are released, what is the initial acceleration of A?
Answer:
- Force on A = 0.870N
- charge of the object B = q = 2.1 μC
charge of the object A = 2q = 4.2 μC
- a = 0.966 m/s^2
Explanation:
- In order to determine the force on the object A, you take into account the third Newton law, which states that the force experienced by A has the same magnitude of the force experienced by B, but with an opposite direction.
Then, the force on A is 0.870N
- In order to calculate the charge of both objects, you use the following formula:
[tex]F_e=k\frac{q_Aq_B}{r^2}[/tex] (1)
k: Coulomb's constant = 8.98*10^9 Nm^2/C^2
r: distance between the objects = 30.0cm = 0.30m
A has twice the charge of B. If the charge of B is qB=q, then the charge of A is qA=2qB = 2q.
You replace the expression for qA and qB into the equation (1), solve for q, and replace the values of the parameters.
[tex]F_e=k\frac{(2q)(q)}{r^2}=2k\frac{q^2}{r^2}\\\\q=\sqrt{\frac{r^2Fe}{2k}}\\\\q=\sqrt{\frac{(0.30m)^2(0.870N)}{2(8.98*10^9Nm^2/C^2)}}=2.1*10^{-6}C\\\\q=2.1\mu C[/tex]
Then, you have:
charge of the object B = q = 2.1 μC
charge of the object A = 2q = 4.2 μC
- In order to calculate the acceleration of A, you use the second Newton law with the electric force, as follow:
[tex]F_e=ma\\\\a=\frac{F_e}{m}[/tex]
m: mass of the object A = 900g = 0.900kg
[tex]a=\frac{0.870N}{0.900kg}=0.966\frac{m}{s^2}[/tex]
The acceleration of A is 0.966m/s^2
Use Coulomb’s law to derive the dimension for the permittivity of free space.
Answer:
Coulomb's law is:
[tex]F = \frac{1}{4*pi*e0} *(q1*q2)/r^2[/tex]
First, force has units of Newtons, the charges have units of Coulombs, and r, the distance, has units of meters, then, working only with the units we have:
N = (1/{e0})*C^2/m^2
then we have:
{e0} = C^2/(m^2*N)
And we know that N = kg*m/s^2
then the dimensions of e0 are:
{e0} = C^2*s^2/(m^3)
(current square per time square over cubed distance)
And knowing that a Faraday is:
F = C^2*S^2/m^2
The units of e0 are:
{e0} = F/m.
Pulling out of a dive, the pilot of an airplane guides his plane into a vertical circle with a radius of 600 m. At the bottom of the dive, the speed of the airplane is 150 m/s. What is the apparent weight of the 70.0-kg pilot at that point?
Answer:
The apparent weight of the pilot is 3311 N
Explanation:
Given;
radius of the vertical circle, r = 600 m
speed of the plane, v = 150 m/s
mass of the pilot, m = 70 kg
Weight of the pilot due to his circular motion;
[tex]W= F_v\\\\F_v = \frac{mv^2}{r} \\\\F_v = \frac{70*150^2}{600} \\\\F_v = 2625 \ N[/tex]
Real weight of the pilot;
[tex]W_R = mg\\\\W_R = 70 *9.8\\\\W_R = 686 \ N[/tex]
Apparent weight - Real weight of pilot = weight due to centripetal force
[tex]F_N - mg = \frac{mv^2}{r} \\\\F_N = \frac{mv^2}{r} + mg\\\\F_N = 2625 \ N + 686 \ N\\\\F_N = 3311\ N[/tex]
Therefore, the apparent weight of the pilot is 3311 N
Two space ships collide in deep space. Spaceship P, the projectile, has a mass of 4M,
while the target spaceship T has a mass of M. Spaceship T is initially at rest and the
collision is elastic. If the final velocity of Tis 8.1 m/s, what was the initial velocity of
P?
Answer:
The initial velocity of spaceship P was u₁ = 5.06 m/s
Explanation:
In an elastic collision between two bodies the expression for the final velocity of the second body is given as follows:
[tex]V_{2} = \frac{(m_{2}-m_{1}) }{(m_{1}+m_{2})}u_{2} + \frac{2m_{1} }{(m_{1}+m_{2})}u_{1}[/tex]
Here, subscript 1 is used for spaceship P and subscript 2 is used for spaceship T. In this equation:
V₂ = Final Speed of Spaceship T = 8.1 m/s
m₁ = mass of spaceship P = 4 M
m₂ = mass of spaceship T = M
u₁ = Initial Speed of Spaceship P = ?
u₂ = Initial Speed of Spaceship T = 0 m/s
Using these values in the given equation, we get:
[tex]8.1 m/s = \frac{M-4M }{4M+M}(0 m/s) + \frac{2(4M) }{4M+M}u_{1}[/tex]
8.1 m/s = (8 M/5 M)u₁
u₁ = (5/8)(8.1 m/s)
u₁ = 5.06 m/s
how do a proton and neutron compare?
Answer:
c.they have opposite charges.
Explanation:
because the protons have a positive charge and the neutrons have no charge.
A spherical shell is rolling without slipping at constant speed on a level floor. What percentage of the shell's total kinetic energy is translational kinetic energy
Answer:
41.667 per cent of the total kinetic energy is translational kinetic energy.
Explanation:
As the spherical shell is rolling without slipping at constant speed, the system can be considered as conservative due to the absence of non-conservative forces (i.e. drag, friction) and energy equation can be expressed only by the Principle of Energy Conservation, whose total energy is equal to the sum of rotational and translational kinetic energies. That is to say:
[tex]E = K_{t} + K_{r}[/tex]
Where:
[tex]E[/tex] - Total energy, measured in joules.
[tex]K_{r}[/tex] - Rotational kinetic energy, measured in joules.
[tex]K_{t}[/tex] - Translational kinetic energy, measured in joules.
The spherical shell can be considered as a rigid body, since there is no information of any deformation due to the motion. Then, rotational and translational components of kinetic energy are described by the following equations:
Rotational kinetic energy
[tex]K_{r} = \frac{1}{2}\cdot I_{g}\cdot \omega^{2}[/tex]
Translational kinetic energy
[tex]K_{t} = \frac{1}{2}\cdot m \cdot R^{2}\cdot \omega^{2}[/tex]
Where:
[tex]I_{g}[/tex] - Moment of inertia of the spherical shell with respect to its center of mass, measured in [tex]kg\cdot m^{2}[/tex].
[tex]\omega[/tex] - Angular speed of the spherical shell, measured in radians per second.
[tex]R[/tex] - Radius of the spherical shell, measured in meters.
After replacing each component and simplifying algebraically, the total energy of the spherical shell is equal to:
[tex]E = \frac{1}{2}\cdot (I_{g} + m\cdot R^{2})\cdot \omega^{2}[/tex]
In addition, the moment of inertia of a spherical shell is equal to:
[tex]I_{g} = \frac{2}{3}\cdot m\cdot R^{2}[/tex]
Then, total energy is reduced to this expression:
[tex]E = \frac{5}{6}\cdot m \cdot R^{2}\cdot \omega^{2}[/tex]
The fraction of the total kinetic energy that is translational in percentage is given by the following expression:
[tex]\%K_{t} = \frac{K_{t}}{E}\times 100\,\%[/tex]
[tex]\%K_{t} = \frac{\frac{1}{2}\cdot m \cdot R^{2}\cdot \omega^{2} }{\frac{5}{6}\cdot m \cdot R^{2}\cdot \omega^{2} } \times 100\,\%[/tex]
[tex]\%K_{t} = \frac{5}{12}\times 100\,\%[/tex]
[tex]\%K_{t} = 41.667\,\%[/tex]
41.667 per cent of the total kinetic energy is translational kinetic energy.
An asteroid that has an orbit with a semi-major axis of 4 AU will have an orbital period of about ______ years.
Answer:
16 years.
Explanation:
Using Kepler's third Law.
P2=D^3
P=√d^3
Where P is the orbital period and d is the distance from the sun.
From the question the semi major axis of the asteroid is 4 AU= distance. The distance is always express in astronomical units.
P=?
P= √4^3
P= √256
P= 16 years.
Orbital period is 16 years.
An ice skater is in a fast spin with her arms held tightly to her body. When she extends her arms, which of the following statements in NOT true?
A. Het total angular momentum has decreased
B. She increases her moment of inertia
C. She decreases her angular speed
D. Her moment of inertia changes
Answer:
A. Her total angular momentum has decreased
Explanation:
Total angular momentum is the product of her moment of inertia and angular velocity. In this scenario it doesn’t decrease but rather remains constant as the movement of the arms doesn’t have any effect on the total angular momentum.
The movement of the arm under certain conditions however has varying effects and changes on parameters such as the moment of inertia and the angular speed.
Which of the following statements is valid:_______.A) If the potential throughtout a particular region is constant, the field throughout the region must be zero.B) If the field throughout a particular region is constant, the potential throughout that region must be zero.C) If the field at a particular point is zero, the potential at that point must be zero.D) If the potential at a particular point is zero, the field at that point must be zero.
Answer:
Option A is the only valid statement.
Explanation:
A)The electric field intensity is defined by the relationship:
E= -ΔV/Δr.
Now, according to the relationship above, the electric field would be the negative gradient of electric potential. Now, if the electric potential is constant throughout the given region of space, then the change in electric potential would be ΔV=0.
Thu,E= 0.
So the answer is that, E will be zero in this case.
So, the statement is valid.
B) Statement not valid because the field is the gradient of the potential. Hence, the field would be zero in any region where the potential is constant. However, constant does not necessarily mean a value of zero. With that being said, we can always change the definition of the potential function by adding a constant, to thus make it zero there. But then the potential will no longer be zero at infinity or in any different “flat” regions.
C) Statement not valid because, for the fact that electric field is zero at a particular point, it doesn't necessarily
imply that the electric potential is zero at that point. A good example would be the case of two identical charges which are separated by some distance. At the midpoint between the charges, the
electric field due to the charges would be zero. However, the electric potential due to the charges at that same point would not be zero. Thus, the potential will either have two positive contributions, if the charges are positive, or two negative contributions, if the charges are negative.
(D) Statement is not valid because, for the fact that electric potential is zero at a particular point, it does not necessarily imply that the electric field is zero at that point. A good example would be the case of a dipole, which
has two charges of the same magnitude, but opposite sign, and are separated by some distance. At
the midpoint between the charges, the electric potential due to the charges would be zero, but the electric field due to the charges at that same point would not be zero.
The interference of two sound waves of similar amplitude but slightly different frequencies produces a loud-soft-loud oscillation we call __________.
a. the Doppler effect
b. vibrato
c. constructive and destructive interference
d. beats
Answer:
the correct answer is d Beats
Explanation:
when two sound waves interfere time has different frequencies, the result is the sum of the waves is
y = 2A cos 2π (f₁-f₂)/2 cos 2π (f₁ + f₂)/2
where in this expression the first part represents the envelope and the second part represents the pulse or beatings of the wave.
When examining the correct answer is d Beats
A 0.500-kg mass suspended from a spring oscillates with a period of 1.50 s. How much mass must be added to the object to change the period to 2.00 s
Answer:
389 kg
Explanation:
The computation of mass is shown below:-
[tex]T = 2\pi \sqrt{\frac{m}{k} }[/tex]
Where K indicates spring constant
m indicates mass
For the new time period
[tex]T^' = 2\pi \sqrt{\frac{m'}{k} }[/tex]
Now, we will take 2 ratios of the time period
[tex]\frac{T}{T'} = \sqrt{\frac{m}{m'} }[/tex]
[tex]\frac{1.50}{2.00} = \sqrt{\frac{0.500}{m'} }[/tex]
[tex]0.5625 = \sqrt{\frac{0.500}{m'} }[/tex]
[tex]m' = \frac{0.500}{0.5625}[/tex]
= 0.889 kg
Since mass to be sum that is
= 0.889 - 0.500
0.389 kg
or
= 389 kg
Therefore for computing the mass we simply applied the above formula.
The mass added to the object to change the period to 2.00 s is 0.389 kg and this can be determined by using the formula of the time period.
Given :
A 0.500-kg mass suspended from a spring oscillates with a period of 1.50 s.
The formula of the time period is given by:
[tex]\rm T = 2\pi\sqrt{\dfrac{m}{K}}[/tex] ---- (1)
where m is the mass and K is the spring constant.
The new time period is given by:
[tex]\rm T'=2\pi\sqrt{\dfrac{m'}{K}}[/tex] ---- (2)
where m' is the total mass after the addition and K is the spring constant.
Now, divide equation (1) by equation (2).
[tex]\rm \dfrac{T}{T'}=\sqrt{\dfrac{m}{m'}}[/tex]
Now, substitute the known terms in the above expression.
[tex]\rm \dfrac{1.50}{2}=\sqrt{\dfrac{0.5}{m'}}[/tex]
Simplify the above expression in order to determine the value of m'.
[tex]\rm m'=\dfrac{0.5}{0.5625}[/tex]
m' = 0.889 Kg
Now, the mass added to the object to change the period to 2.00 s is given by:
m" = 0.889 - 0.500
m" = 0.389 Kg
For more information, refer to the link given below:
https://brainly.com/question/2144584
When looking at the chemical symbol, the charge of the ion is displayed as the
-superscript
-subscript
-coefficient
-product
Answer:
superscript
Explanation:
When looking at the chemical symbol, the charge of the ion is displayed as the Superscript. This is because the charge of ions is usually written up on the chemical symbol while the atom/molecule is usually written down the chemical symbol. The superscript refers to what is written up on the formula while the subscript is written down on the formula.
An example is H2O . The 2 present represents two molecule of oxygen and its written as the subscript while Fe2+ in which the 2+ is written up is known as the superscript.
Answer:
superscript
Explanation: