Answer:
Explanation:
Work done on a spring is expressed as [tex]W = 1/2 ke^{2}[/tex]
k is the elastic constant
e is the extension of the material
If 4 J of work is needed to stretch a spring from its natural length of 36 cm to a length of 47 cm, then;
Work done = 4J and the extension e = 47 cm - 36 cm; e = 11 cm
11cm = 0.11m
Substituting the given values into the equation above to get the elastic constant;
[tex]W = 1/2 ke^{2}\\4 = 1/2k(0.11)^{2} \\8 = 0.0121k\\k = 8/0.0121\\k = 661.16N/m[/tex]
a) In order to determine the amount of work needed work is needed to stretch the spring from 41 cm to 45 cm, wre will use the same formula as above.
[tex]W = 1/2ke^{2} \\e = 0.45 - 0.41\\e = 0.04 m\\ k = 661.16N/m[/tex]
[tex]W = 1/2 * 661.16 * 0.04^{2} \\W = 330.58*0.0016\\W = 0.53J (to\ 2d.p)[/tex]
b) According to hooke's law, F = ke where F is the applied force
We are to get the extension when a force of 15N is applied to the original length of the material.
e = F/k
e = 15/661.16
e = 0.02 m (to 1 d.p)
This means that the natural length of the spring will be stretched by 0.02 m when a force of 15N is applied to it.
3. Identify the mathematical relationship that exists between pressure and volume, when temperature and quantity are held constant, as being directly proportional or inversely proportional. Explain your answer and write an equation that relates pressure and volume to a constant, using variables
Answer:
P = cte / V
therefore pressure and volume are inversely proportional
Explanation:
For this exercise we can join the ideal gases equation
PV = n R T
they indicate that the amount of matter and the temperature are constant, therefore
PV = cte
P = cte / V
therefore pressure and volume are inversely proportional
What portion of the difference in the angular speed before and after you increased the mass can be accounted for by frictional losses
Answer:
As the mass increases, the moment of inertia(I) increases, therefore, the angular momentum(L) increases too.
Explanation:
friction can be defined as resistance in motion of bodies in relative to one another
momentum is the product of mass and velocity
torque is the time rate of change in momentum
τ = [tex]\frac{dL}{dt}[/tex]
where L = Iω = mvr
I = moment of inertia
ω= angular frequency
if there is no external force(torque) acting on the system, then
[tex]\frac{dL}{dt}[/tex] = 0
dL = 0 = constant
moment of inertia I depends on the distribution of mass on the axis of rotation.
as the mass increases, the angular momentum(L) increases
angular frequency, ω, remains constant
Waves from two slits are in phase at the slits and travel to a distant screen to produce the second minimum of the interference pattern. The difference in the distance traveled by the wave is:
Answer:
Three halves of a wavelength I.e 7lambda/2
Explanation:
See attached file pls
A single slit of width 0.3 mm is illuminated by a mercury light of wavelength 254 nm. Find the intensity at an 11° angle to the axis in terms of the intensity of the central maximum.
Answer:
The the intensity at an 11° angle to the axis in terms of the intensity of the central maximum is
[tex]I_c = \frac{I}{I_o} =8.48 *10^{-8}[/tex]
Explanation:
From the question we are told that
The width of the slit is [tex]D = 0.3 \ mm = 0.3 *10^{-3} \ m[/tex]
The wavelength is [tex]\lambda = 254 \ nm = 254 *10^{-9} \ m[/tex]
The angle is [tex]\theta = 11^o[/tex]
The intensity of at [tex]11^o[/tex] to the axis in terms of the intensity of the central maximum. is mathematically represented as
[tex]I_c = \frac{I}{I_o} = [ \frac{sin \beta }{\beta }] ^2[/tex]
Where [tex]\beta[/tex] is mathematically represented as
[tex]\beta = \frac{D sin (\theta ) * \pi}{\lambda }[/tex]
substituting values
[tex]\beta = \frac{0.3 *10^{-3} sin (11 ) * 3.142}{254 *10^{-9} }[/tex]
[tex]\beta = 708.1 \ rad[/tex]
So
[tex]I_c = \frac{I}{I_o} = [ \frac{sin (708.1) }{(708.1)}] ^2[/tex]
[tex]I_c = \frac{I}{I_o} =8.48 *10^{-8}[/tex]
When static equilibrium is established for a charged conductor, the electric field just inside the surface of the conductor is
Answer:
The electric field just inside the charged conductor is zero.
Explanation:
Electric field is defined as the region where electrical force is experienced by an electric charge usually as a result of the presence of another electric charge. A charged conductor is said to be in electrostatic equilibrium when it is in an electrostatically balanced state. This simply means a state in which the free electrical charges in the charged conductor have stopped moving.
For any charged conductor that has attained electrostatic equilibrium, the electric field at any point below the surface of the charged conductor falls to zero. Hence the electric field just inside the charged conductor is zero.
The velocity selector in in a mass spectrometer consists of a uniform magnetic field oriented at 90 degrees to a uniform electric field so that a charge particle entering the region perpendicular to both fields will experience an electric force and a magnetic force that are oppositely directed. If the uniform magnetic field has a magnitude of 37.8 ~\text{mT}37.8 mT, then calculate the magnitude of the electric field that will cause a proton entering the velocity selector at 40.640.6 km/s to be undeflected. Give your answer in units of kV/m.
Answer:
50k/h is the answer to iy
Suppose a proton moves to the right and enters a uniform magnetic field into the page. It follows trajectory B with radius rp. An alpha particle (twice the charge and 4 times the mass) enters the same magnetic field in the same way and with the same velocity as the proton. Which path best represents the alpha particle’s trajectory?
Answer:
R = r_protón / 2
Explanation:
The alpha particle when entering the magnetic field experiences a force and with Newton's second law we can describe its movement
F = m a
Since the magnetic force is perpendicular, the acceleration is centripetal.
a = v² / R
the magnetic force is
F = q v x B = q v B sin θ
the field and the speed are perpendicular so the sin 90 = 1
we substitute
qv B = m v² / R
R = q v B / m v²
in the exercise they indicate
the charge q = 2 e
the mass m = 4 m_protón
R = 2e v B / 4m_protón v²
we refer the result to the movement of the proton
R = (e v B / m_proton) 1/2
the data in parentheses correspond to the radius of the proton's orbit
R = r_protón / 2
A cylindrical capacitor is made of two thin-walled concentric cylinders. The inner cylinder has radius 5 mm , and the outer one a radius 11 mm . The common length of the cylinders is 160 m . What is the potential energy stored in this capacitor when a potential difference 6 V is
Answer:
The potential energy is [tex]PE = 2.031 *10^{-7} \ J[/tex]
Explanation:
From the question we are told that
The inner radius is [tex]r_i = 5 \ mm = 0.005 \ m[/tex]
The outer radius is [tex]r_o = 11 \ mm = 0.011 \ m[/tex]
The common length is [tex]l = 160 \ m[/tex]
The potential difference is [tex]V = 6 \ V[/tex]
Generally the capacitance of the cylindrical capacitor is mathematically represented as
[tex]C = \frac{2 \pi * k * \epsilon_o }{ ln \frac{ r_o }{r_i} } * l[/tex]
Where [tex]\epsilon _o[/tex] is the permitivity of free space with the values [tex]\epsilon _o = 8.85*10^{-12} \ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2[/tex]
and k is the dielectric constant of the dielectric material here the dielectric material is free space so k = 1
Substituting values
[tex]C = \frac{2* 3.142 * 1 * 8.85*10^{-12} }{ ln \frac{ 0.011}{0.005} } * 160[/tex]
[tex]C = 1.129 *10^{-8} \ F[/tex]
The potential energy stored is mathematically represented as
[tex]PE = \frac{1}{2} * C * V ^2[/tex]
substituting values
[tex]PE = 0.5 * 1.129 *10^{-8} * (6)^2[/tex]
[tex]PE = 2.031 *10^{-7} \ J[/tex]
what is the mass of an oil drop having two extra electrons that is suspended motionless by the field between the plates
Answer:
m = 3,265 10⁻²⁰ E
Explanation:
For this exercise we can use Newton's second law applied to our system, which consists of a capacitor that creates the uniform electric field and the drop of oil with two extra electrons.
∑ F = 0
[tex]F_{e}[/tex] - W = 0
the electric force is
F_{e} = q E
as they indicate that the charge is two electrons
F_{e} = 2e E
The weight is given by the relationship
W = mg
we substitute in the first equation
2e E = m g
m = 2e E / g
let's put the value of the constants
m = (2 1.6 10⁻¹⁹ / 9.80) E
m = 3,265 10⁻²⁰ E
The value of the electric field if it is a theoretical problem must be given and if it is an experiment it can be calculated with measures of the spacing between plates and the applied voltage, so that the system is in equilibrium
An air-filled parallel-plate capacitor is connected to a battery and allowed to charge up. Now a slab of dielectric material is placed between the plates of the capacitor while the capacitor is still connected to the battery. After this is done, we find that
Answer:
The voltage across the capacitor will remain constant
The capacitance of the capacitor will increase
The electric field between the plates will remain constant
The charge on the plates will increase
The energy stored in the capacitor will increase
Explanation:
First of all, if a capacitor is connected to a voltage source, the voltage or potential difference across the capacitor will remain constant. The electric field across the capacitor will stay constant since the voltage is constant, because the electric field is proportional to the voltage applied. Inserting a dielectric material into the capacitor increases the charge on the capacitor.
The charge on the capacitor is equal to
Q = CV
Since the voltage is constant, and the charge increases, the capacitance will also increase.
The energy in a capacitor is given as
E = [tex]\frac{1}{2}CV^{2}[/tex]
since the capacitance has increased, the energy stored will also increase.
You indicate that a symbol
is a vector by drawing
A. through the symbol.
B. over the symbol.
c. under the symbol.
D. before the symbol.
Answer:
B. over the symbol.
Explanation:
vectors are represented with a symbol carrying an arrow head with also indicates direction
The refractive index n of transparent acrylic plastic (full name Poly(methyl methacrylate)) depends on the color (wavelength) of the light passing through it. At wavelength 486.1 nm (blue, designated with letter F) -> nF=1.497, and at wavelength 656.3 nm (red, designated with letter C) -> nC=1.488. Two beams (one of each wavelength) are prepared to coincide, and enter the flat polished surface of an acrylic block at angle of 45 arc degree measured from the normal to the surface. What is the angle between the blue beam and the red beam in the acrylic block?
Answer:
The angle between the blue beam and the red beam in the acrylic block is
[tex]\theta _d =0.19 ^o[/tex]
Explanation:
From the question we are told that
The refractive index of the transparent acrylic plastic for blue light is [tex]n_F = 1.497[/tex]
The wavelength of the blue light is [tex]F = 486.1 nm = 486.1 *10^{-9} \ m[/tex]
The refractive index of the transparent acrylic plastic for red light is [tex]n_C = 1.488[/tex]
The wavelength of the red light is [tex]C = 656.3 nm = 656.3 *10^{-9} \ m[/tex]
The incidence angle is [tex]i = 45^o[/tex]
Generally from Snell's law the angle of refraction of the blue light in the acrylic block is mathematically represented as
[tex]r_F = sin ^{-1}[\frac{sin(i) * n_a }{n_F} ][/tex]
Where [tex]n_a[/tex] is the refractive index of air which have a value of[tex]n_a = 1[/tex]
So
[tex]r_F = sin ^{-1}[\frac{sin(45) * 1 }{ 1.497} ][/tex]
[tex]r_F = 28.18^o[/tex]
Generally from Snell's law the angle of refraction of the red light in the acrylic block is mathematically represented as
[tex]r_C = sin ^{-1}[\frac{sin(i) * n_a }{n_C} ][/tex]
Where [tex]n_a[/tex] is the refractive index of air which have a value of[tex]n_a = 1[/tex]
So
[tex]r_C = sin ^{-1}[\frac{sin(45) * 1 }{ 1.488} ][/tex]
[tex]r_F = 28.37^o[/tex]
The angle between the blue beam and the red beam in the acrylic block
[tex]\theta _d = r_C - r_F[/tex]
substituting values
[tex]\theta _d = 28.37 - 28.18[/tex]
[tex]\theta _d =0.19 ^o[/tex]
An alternating current is supplied to an electronic component with a warning that the voltage across it should never exceed 12 V. What is the highest rms voltage that can be supplied to this component while staying below the voltage limit in the warning?
Answer:
The highest rms voltage will be 8.485 V
Explanation:
For alternating electric current, rms (root means square) is equal to the value of the direct current that would produce the same average power dissipation in a resistive load
If the peak or maximum voltage should not exceed 12 V, then from the relationship
[tex]V_{rms} = \frac{V_{p} }{\sqrt{2} }[/tex]
where [tex]V_{rms}[/tex] is the rms voltage
[tex]V_{p}[/tex] is the peak or maximum voltage
substituting values into the equation, we'll have
[tex]V_{rms} = \frac{12}{\sqrt{2} }[/tex] = 8.485 V
A jet transport with a landing speed of 200 km/h reduces its speed to 60 km/h with a negative thrust R from its jet thrust reversers in a distance of 425 m along the runway with constant deceleration. The total mass of the aircraft is 140 Mg with mass center at G. Compute the reaction N under the nose wheel B toward the end of the braking interval and prior to the application of mechanical braking. At lower speed, aerodynamic forces on the aircraft are small and may be neglected.
Answer:
257 kN.
Explanation:
So, we are given the following data or parameters or information in the following questions;
=> "A jet transport with a landing speed
= 200 km/h reduces its speed to = 60 km/h with a negative thrust R from its jet thrust reversers"
= > The distance = 425 m along the runway with constant deceleration."
=> "The total mass of the aircraft is 140 Mg with mass center at G. "
We are also give that the "aerodynamic forces on the aircraft are small and may be neglected at lower speed"
Step one: determine the acceleration;
=> Acceleration = 1/ (2 × distance along runway with constant deceleration) × { (landing speed A)^2 - (landing speed B)^2 × 1/(3.6)^2.
=> Acceleration = 1/ (2 × 425) × (200^2 - 60^2) × 1/(3.6)^2 = 3.3 m/s^2.
Thus, "the reaction N under the nose wheel B toward the end of the braking interval and prior to the application of mechanical braking" = The total mass of the aircraft × acceleration × 1.2 = 15N - (9.8 × 2.4 × 140).
= 140 × 3.3× 1.2 = 15N - (9.8 × 2.4 × 140).
= 257 kN.
The reaction N under the nose wheel B towards the end of the braking interval = 257 kN
Given data :
Landing speed of Jet = 200 km/h
Distance = 425 m
Total mass of aircraft = 140 Mg with mass center at G
Determine the reaction N under the nose of wheel B First step : calculate the value of the Jet accelerationJet acceleration = 1 / (2 *425) * (200² - 60² ) * 1 / (3.6)²
= 3.3 m/s²
Next step : determine the reaction N under the nose of WheelReaction N = Total mass of aircraft * jet acceleration* 1.2 = 15N - (9.8*2.4* 140). ----- ( 1 )
∴ Reaction N = 140 * 3.3 * 1.2 = 15 N - ( 9.8*2.4* 140 )
Hence Reaction N = 257 KN
We can conclude that the The reaction N under the nose wheel B towards the end of the braking interval = 257 kN
Learn more about : https://brainly.com/question/15776281
A wave travels at a consent speed. how does the frequency change if the wavelength is reduced by a factor of 4?
Answer:
The frequency increases by 4 because it is inversely proportional to the wavelength.
Two positive charges are located at x = 0, y = 0.3m and x = 0, y = -.3m respectively. Third point charge q3 = 4.0 μC is located at x = 0.4 m, y = 0.
A) Make a careful sketch of decent size that illustrates all force vectors with directions and magnitudes.
B) What is the resulting vector of the total force on charge q1 exerted by the other two charges using vector algebra?
Answer:
0.46N
Explanation:
See attached file
The medical profession divides the ultraviolet region of the electromagnetic spectrum into three bands: UVA (320-420 nm), UVB (290-320 nm), and UVC (100-290 nm). UVA and UVB promote skin cancer and premature skin aging; UVB also causes sunburn, but helpfully fosters production of vitamin D. Ozone in Earth's atmosphere blocks most of the more dangerous UVC. Find the frequency range associated with UVB radiation.
Answer:
υ = 9.375 x 10¹⁴ Hz to 10.34 x 10¹⁴ Hz
Explanation:
The frequency of an electromagnetic radiation can be given by the following formula:
υ = c/λ
where,
υ = frequency of electromagnetic wave = ?
c = speed of light = 3 x 10⁸ m/s
λ = wavelength of electromagnetic wave = 290 nm to 320 nm
FOR LOWER LIMIT OF FREQUENCY:
λ = 320 nm = 3.2 x 10⁻⁷ m
Therefore,
υ = (3 x 10⁸ m/s)/(3.2 x 10⁻⁷ m)
υ = 9.375 x 10¹⁴ Hz
FOR UPPER LIMIT OF FREQUENCY:
λ = 290 nm = 3.2 x 10⁻⁷ m
Therefore,
υ = (3 x 10⁸ m/s)/(2.9 x 10⁻⁷ m)
υ = 10.34 x 10¹⁴ Hz
Therefore, the frequency range for UVB radiations is:
υ = 9.375 x 10¹⁴ Hz to 10.34 x 10¹⁴ Hz
Suppose a 225 kg motorcycle is heading toward a hill at a speed of 29 m/s. The two wheels weigh 12 kg each and are each annular rings with an inner radius of 0.280 m and an outer radius of 0.330 m. How high can it coast up the hill, if you neglect friction in m?
a) m = 180 kg
b) v = 29 m/s
c) h = 32 m
Answer:
It can coast uphill 6.2m
Explanation:
See attached file pls
Find an article online or application in your daily life involving rotating objects and physics.
Answer:
the planet Earth is a good example
A circular loop in the plane of a paper lies in a 0.75 T magnetic field pointing into the paper. The loop's diameter changes from 18.0 cm to 6.8 cm in 0.46 s.
A) Determine the direction of the induced current and justify your answer.
B) Determine the magnitude of the average induced emf.
C) If the coil resistance is 2.5 Ω, what is the average induced current?
Answer:
Explanation:
A.the direction of induced current will be clockwise
B: Changing 18cm and 6.8cm into 0.18m and 0.68
2.5
Divide them both by 2 to find the radius . Now we have 0.09 and .034m.
Now use Φ=(π*0.09^2)(.75 T)cos0 and the 0.019wb
(π*0.034^2)(.75 T)cos0 and the 0.00272wb
ow use ε=-N(ΔΦ/Δt)
For ΔΦ, 0.091-0.0027=0.0883
C.
To find the current, use I=ε/R
0.0883/2.5= 0.035A
An electron and a proton each have a thermal kinetic energy of 3kBT/2. Calculate the de Broglie wavelength of each particle at a temperature of 1950 K. (kb is Boltzmann's constant, 1.38x10-23 J/K).
Answer:
The de Broglie wavelength of electron βe = 2.443422 × 10⁻⁹ m
The de Broglie wavelength of proton βp = 5.70 × 10⁻¹¹ m
Explanation:
Thermal kinetic energy of electron or proton = KE
∴ KE = 3kbT/2
given that; kb = 1.38 x 10⁻²³ J/K , T = 1950 K
so we substitute
KE = ( 3 × 1.38 x 10⁻²³ × 1950 ) / 2
kE = 4.0365 × 10⁻²⁰ ( is the kinetic energy for both electron and proton at temperature T )
Now we know that
mass of electron M'e = 9.109 × 10⁻³¹
mass of proton M'p = 1.6726 × 10⁻²⁷
We also know that
KE = p₂ / 2m
from the equation, p = √ (2mKE)
{ p is momentum, m is mass }
de Broglie wavelength = β
so β = h / p = h / √ (2mKE)
h = Planck's constant = 6.626 × 10⁻³⁴
∴ βe = h / √ (2m'e × KE)
βe = 6.626 × 10⁻³⁴ / √ (2 × 9.109 × 10⁻³¹ × 4.0365 × 10⁻²⁰ )
βe = 6.626 × 10⁻³⁴ / √ 7.3536957 × 10⁻⁵⁰
βe = 6.626 × 10⁻³⁴ / 2.71176984642871 × 10⁻²⁵
βe = 2.443422 × 10⁻⁹ m
βp = h / √ (2m'p ×KE)
βp = 6.626 × 10⁻³⁴ / √ (2 × 1.6726 × 10⁻²⁷ × 4.0365 × 10⁻²⁰ )
βp = 6.626 × 10⁻³⁴ / √ 1.35028998 × 10⁻⁴⁶
βp = 6.626 × 10⁻³⁴ / 1.16201978468527 × 10⁻²³
βp = 5.702140 × 10⁻¹¹ m
Question 5
A fidget spinner that is 4 inches in diameter is spinning clockwise. The spinner spins at 3000
revolutions per minute.
At t = 0, consider the point A on the outer edge of the spinner that is along the positive horizontal
axis. Let h(t) be the vertical position of A in inches. Suppose t is measured in minutes. Find a
sinusoidal function that models h(t).
h(t) =
Can someone please help me for this question?!!!!! ASAP?!!!!
Answer:
h = 4 sin (314.15 t)
Explanation:
This is a kinematics exercise, as the system is rotating at a constant speed.
w = θ / t
θ = w t
in angular motion all angles are measured in radians, which is defined
θ = s / R
we substitute
s / R = w t
s = w R t
let's reduce the magnitude to the SI system
w = 3000 rev / min (2π rad / 1rev) (1min / 60 s) = 314.16 rad / s
let's calculate
s = 314.16 4 t
s = 1,256.6 t
this is the value of the arc
Let's find the function of this system, let's use trigonometry to find the projection on the x axis
cos θ = x / R
x = R cos θ
x = R cos wt
projection onto the y-axis is
sin θ = y / R
how is a uniform movement
θ = w t
y = R sin wt
In the case y = h
h = R sin wt
h = 4 sin (314.15 t)
A solid cylinder has a diameter of 17.4 mm and a length of 50.3mm. It's mass is 49g . What is its density of the cylinder in metric tonnes per cubic metre? Give your answer to 1 significant figure.
Answer:
4 tonne/m³
Explanation:
ρ = m / V
ρ = 49 g / (π (17.4 mm / 2)² (50.3 mm))
ρ = 0.0041 g/mm³
Converting to tonnes/m³:
ρ = 0.0041 g/mm³ (1 kg / 1000 g) (1 tonne / 1000 kg) (1000 mm / m)³
ρ = 4.1 tonne/m³
Rounding to one significant figure, the density is 4 tonne/m³.
A charged particle is moving with speed v perpendicular to a uniform magnetic field. A second identical charged particle is moving with speed 2v perpendicular to the same magnetic field. If the frequency of revolution of the first particle is f, the frequency of revolution of the second particle is
Answer:
the frequency of revolution of the second particle is f
Explanation:
centripetal force is balanced by the magnetic force for object under magnetic field is given as
Mv²/r= qvB
But v= omega x r
Omega= 2pi x f
f= qB/2pi x M
So since frequency does not depend on the velocity.therefore the frequency of revolution of the second particle remains the same and its equal to f
The specific heat of a certain type of cooking oil is 1.75 J/(g⋅°C). How much heat energy is needed to raise the temperature of 2.01 kg of this oil from 23 °C to 191 °C?
Answer:
Q = 590,940 J
Explanation:
Given:
Specific heat (c) = 1.75 J/(g⋅°C)
Mass(m) = 2.01 kg = 2,010
Change in temperature (ΔT) = 191 - 23 = 168°C
Find:
Heat required (Q)
Computation:
Q = mcΔT
Q = (2,010)(1.75)(168)
Q = 590,940 J
Q = 590.94 kJ
if a speed sound in air at o°c is 331m/s. what will be its value at 35 °c
Answer:
please brainliest!!!
Explanation:
V1/√T1 =V2/√T2
V1 = 331m/s
T1 = 0°C = 273k
V2 = ?
T2 = 35°c = 308k
331/√273 = V2/√308331/16.5 = V2/17.520.06 = V2/17.5V2 = 20.06 x 17.5 V2 = 351.05m/sA spherical capacitor contains a charge of 3.40 nC when connected to a potential difference of 240.0 V. Its plates are separated by vacuum and the inner radius of the outer shell is 4.10 cm.
Calculate:
a. The capacitance
b. The radius of the inner sphere.
c. The electric field just outside the surface of the inner sphere.
Answer:
A) 1.4167 × 10^(-11) F
B) r_a = 0.031 m
C) E = 3.181 × 10⁴ N/C
Explanation:
We are given;
Charge;Q = 3.40 nC = 3.4 × 10^(-9) C
Potential difference;V = 240 V
Inner radius of outer sphere;r_b = 4.1 cm = 0.041 m
A) The formula for capacitance is given by;
C = Q/V
C = (3.4 × 10^(-9))/240
C = 1.4167 × 10^(-11) F
B) To find the radius of the inner sphere,we will make use of the formula for capacitance of spherical coordinates.
C = (4πε_o)/(1/r_a - 1/r_b)
Rearranging, we have;
(1/r_a - 1/r_b) = (4πε_o)/C
ε_o is a constant with a value of 8.85 × 10^(−12) C²/N.m
Plugging in the relevant values, we have;
(1/r_a - 1/0.041) = (4π × 8.85 × 10^(−12) )/(1.4167 × 10^(-11))
(1/r_a) - 24.3902 = 7.8501
1/r_a = 7.8501 + 24.3902
1/r_a = 32.2403
r_a = 1/32.2403
r_a = 0.031 m
C) Formula for Electric field just outside the surface of the inner sphere is given by;
E = kQ/r_a²
Where k is a constant value of 8.99 × 10^(9) Nm²/C²
Thus;
E = (8.99 × 10^(9) × 3.4 × 10^(-9))/0.031²
E = 3.181 × 10⁴ N/C
WILL MARK BRAINLIEST!!An igneous rock has large red, black, and green crystals. How else can this rock be accurately described?
O fine texture
O cooled quickly
O intrusive origin
O created by lava
Answer:
D
Explanation:
A 0.400-kg ice puck, moving east with a speed of 5.86 m/s , has a head-on collision with a 0.900-kg puck initially at rest.
A. Assume that the collision is perfectly elastic, what will be the speed of the 0.300 kg object after the collision?
B. What will be the direction of the 0.300 kg object after the collision?
C. What will be the speed of the 0.900 kg object after the collision?
Answer:
a) The final speed of the 0.400-kg puck after the collision is 2.254 meters per second, b) The negative sign of the solution found in part a) indicates that 0.400-kg puck is moving westwards, c) The speed of the 0.900-kg puck after the collision is 3.606 meters per second eastwards.
Explanation:
a) Since collision is perfectly elastic and there are no external forces exerted on pucks system, the phenomenon must be modelled after the Principles of Momentum and Energy Conservation. Changes in gravitational potential energy can be neglected. That is:
Momentum
[tex]m_{1}\cdot v_{1,o} + m_{2}\cdot v_{2,o} = m_{1}\cdot v_{1,f} + m_{2}\cdot v_{2,f}[/tex]
Energy
[tex]\frac{1}{2}\cdot (m_{1}\cdot v_{1,o}^{2}+ m_{2}\cdot v_{2,o}^{2})=\frac{1}{2}\cdot (m_{1}\cdot v_{1,f}^{2}+ m_{2}\cdot v_{2,f}^{2})[/tex]
[tex]m_{1}\cdot v_{1,o}^{2} + m_{2}\cdot v_{2,o}^{2} = m_{1}\cdot v_{1,f}^{2} + m_{2}\cdot v_{2,f}^{2}[/tex]
Where:
[tex]m_{1}[/tex], [tex]m_{2}[/tex] - Masses of the 0.400-kg and 0.900-kg pucks, measured in kilograms.
[tex]v_{1,o}[/tex], [tex]v_{2,o}[/tex] - Initial speeds of the 0.400-kg and 0.900-kg pucks, measured in meters per second.
[tex]v_{1}[/tex], [tex]v_{2}[/tex] - Final speeds of the 0.400-kg and 0.900-kg pucks, measured in meters per second.
If [tex]m_{1} = 0.400\,kg[/tex], [tex]m_{2} = 0.900\,kg[/tex], [tex]v_{1,o} = +5.86\,\frac{m}{s}[/tex], [tex]v_{2,o} = 0\,\frac{m}{s}[/tex], the system of equation is simplified as follows:
[tex]2.344\,\frac{kg\cdot m}{s} = 0.4\cdot v_{1,f} + 0.9\cdot v_{2,f}[/tex]
[tex]13.736\,J = 0.4\cdot v_{1,f}^{2}+0.9\cdot v_{2,f}^{2}[/tex]
Let is clear [tex]v_{1,f}[/tex] in first equation:
[tex]0.4\cdot v_{1,f} = 2.344 - 0.9\cdot v_{2,f}[/tex]
[tex]v_{1,f} = 5.86-2.25\cdot v_{2,f}[/tex]
Now, the same variable is substituted in second equation and resulting expression is simplified and solved afterwards:
[tex]13.736 = 0.4\cdot (5.86-2.25\cdot v_{2,f})^{2}+0.9\cdot v_{2,f}^{2}[/tex]
[tex]13.736 = 0.4\cdot (34.340-26.37\cdot v_{2,f}+5.063\cdot v_{2,f}^{2})+0.9\cdot v_{2,f}^{2}[/tex]
[tex]13.736 = 13.736-10.548\cdot v_{2,f} +2.925\cdot v_{2,f}^{2}[/tex]
[tex]2.925\cdot v_{2,f}^{2}-10.548\cdot v_{2,f} = 0[/tex]
[tex]2.925\cdot v_{2,f}\cdot (v_{2,f}-3.606) = 0[/tex]
There are two solutions:
[tex]v_{2,f} = 0\,\frac{m}{s}[/tex] or [tex]v_{2,f} = 3.606\,\frac{m}{s}[/tex]
The first root coincides with the conditions before collision and the second one represents a physically reasonable solution.
Now, the final speed of the 0.400-kg puck is: ([tex]v_{2,f} = 3.606\,\frac{m}{s}[/tex])
[tex]v_{1,f} = 5.86-2.25\cdot (3.606)[/tex]
[tex]v_{1,f} = -2.254\,\frac{m}{s}[/tex]
The final speed of the 0.400-kg puck after the collision is 2.254 meters per second.
b) The negative sign of the solution found in part a) indicates that 0.400-kg puck is moving westwards.
c) The speed of the 0.900-kg puck after the collision is 3.606 meters per second eastwards.
A 2.0 kg handbag is released from the top of the Leaning Tower of Pisa, and 55 m before reaching the ground, it carries a speed of 29 m / s. What was the average force of air resistance?
Answer:
4.31 N
Explanation:
Given:
Δy = -55 m
v₀ = 0 m/s
v = -29 m/s
Find: a
v² = v₀² + 2aΔy
(-29 m/s)² = (0 m/s)² + 2a (-55 m)
a = -7.65 m/s²
Sum of forces in the y direction:
∑F = ma
R − mg = ma
R = m (g + a)
R = (2.0 kg) (9.8 m/s² − 7.65 m/s²)
R = 4.31 N