Answer:
The maximum error is [tex]\Delta \eta = 2032.9[/tex]
Explanation:
From the question we are told that
The length is [tex]l = 1\ m[/tex]
The radius is [tex]r = 0.002 \pm 0.0002 \ m[/tex]
The pressure is [tex]P = 4 *10^{5} \ \pm 1750[/tex]
The rate is [tex]v = 0.5*10^{-9} \ m^3 /t[/tex]
The viscosity is [tex]\eta = \frac{\pi}{8} * \frac{P * r^4}{v}[/tex]
The error in the viscosity is mathematically represented as
[tex]\Delta \eta = | \frac{\delta \eta}{\delta P}| * \Delta P + |\frac{\delta \eta}{\delta r} |* \Delta r + |\frac{\delta \eta}{\delta v} |* \Delta v[/tex]
Where [tex]\frac{\delta \eta }{\delta P} = \frac{\pi}{8} * \frac{r^4}{v}[/tex]
and [tex]\frac{\delta \eta }{\delta r} = \frac{\pi}{8} * \frac{4* Pr^3}{v}[/tex]
and [tex]\frac{\delta \eta }{\delta v} = - \frac{\pi}{8} * \frac{Pr^4}{v^2}[/tex]
So
[tex]\Delta \eta = \frac{\pi}{8} [ |\frac{r^4}{v} | * \Delta P + | \frac{4 * P * r^3}{v} |* \Delta r + |-\frac{P* r^4}{v^2} |* \Delta v][/tex]
substituting values
[tex]\Delta \eta = \frac{\pi}{8} [ |\frac{(0.002)^4}{0.5*10^{-9}} | * 1750 + | \frac{4 * 4 *10^{5} * (0.002)^3}{0.5*10^{-9}} |* 0.0002 + |-\frac{ 4*10^{5}* (0.002)^4}{(0.5*10^{-9})^2} |* 0 ][/tex]
[tex]\Delta \eta = \frac{\pi}{8} [56 + 5120 ][/tex]
[tex]\Delta \eta = 647 \pi[/tex]
[tex]\Delta \eta = 2032.9[/tex]
please help In a video game, a ball moving at 0.6 meter/second collides with a wall. After the collision, the velocity of the ball changed to -0.4 meter/second. The collision takes 0.2 seconds to occur. What’s the acceleration of the ball during the collision? Use . a= v-u/t
Answer:
the acceleration during the collision is: - 5 [tex]\frac{m}{s^2}[/tex]
Explanation:
Using the formula:
[tex]a=\frac{\Delta\,v}{\Delta\,t}[/tex]
we get:
[tex]a=\frac{-0.4-0.6}{0.2} \,\frac{m}{s^2} =\frac{-1}{0.2} \,\frac{m}{s^2} =-5\,\,\frac{m}{s^2}[/tex]
Briefly describe the relationship between an equipotential surface and an electric field, and use this to explain why we will plot equipotential lines.
Answer:
E = - dV/dx
Explanation:
Las superficies equipòtenciales son superficie donde el potencial eléctrico es constante por lo cual nos podemos desplazaren ella sin realizar nigun trabajo.
El campo electrico es el campo que existen algún punto en el espacio creado por alguna ddistribucion de carga.
De los antes expuesto las dos magnitudes están relacionadas
E = - dV/dx
por lo cual el potenical es el gradiente del potencial eléctrico.
Como el campo eléctrico sobre un superficie equipotenciales constante, podemos colocar una punta de prueba con un potencial dado y seguir la linea que de una diferencia de potencial constar, lo cual permite visualizar las forma de cada linea equipotencial
Consider a conducting rod of length 31 cm moving along a pair of rails, and a magnetic field pointing perpendicular to the plane of the rails. At what speed (in m /s) must the sliding rod move to produce an emf of 0.75 V in a 1.75 T field?
Answer:
The speed of the rod is 1.383 m/s
Explanation:
Given;
length of the conducting rod, L = 31 cm = 0.31 m
induced emf on the rod, emf = 0.75V
magnetic field around the rod, B = 1.75 T
Apply the following Faraday's equation for electromagnetic induction in a moving rod to determine the speed of the rod.
emef = BLv
where;
B is the magnetic field
L is length of the rod
v is the speed of the rod
v = emf / BL
v = (0.75) / (1.75 x 0.31)
v = 1.383 m/s
Therefore, the speed of the rod is 1.383 m/s
The electric field at the surface of a charged, solid, copper sphere with radius 0.220 mm is 4200 N/CN/C, directed toward the center of the sphere. What is the potential at the center of the sphere, if we take the potential to be zero infinitely far from the sphere?
Answer:
The potential at the center of the sphere is -924 V
Explanation:
Given;
radius of the sphere, R = 0.22 m
electric field at the surface of the sphere, E = 4200 N/C
Since the electric field is directed towards the center of the sphere, the charge is negative.
The Potential is the same at every point in the sphere, and it is given as;
[tex]V = \frac{1}{4 \pi \epsilon_o} \frac{q}{R}[/tex] -------equation (1)
The electric field on the sphere is also given as;
[tex]E = \frac{1}{4 \pi \epsilon _o} \frac{|q|}{R^2}[/tex]
[tex]|q |= 4 \pi \epsilon _o} R^2E[/tex]
Substitute in the value of q in equation (1)
[tex]V = \frac{1}{4 \pi \epsilon_o} \frac{-(4 \pi \epsilon _o R^2E)}{R} \ \ \ \ q \ is \ negative\ because \ E \ is\ directed \ toward \ the \ center\\\\V = -RE\\\\V = -(0.22* 4200)\\\\V = -924 \ V[/tex]
Therefore, the potential at the center of the sphere is -924 V
A bucket of water with total mass 23 kg is attached to a rope, which in turn is wound around a 0.050-m radius pulley at the top of a well. The bucket is raised to the top of the well and released from rest. The bucket is falling for 2 s and has a speed of 8.0 m/s upon hitting the water surface in the well. What is the moment of inertia of the pulley?
Answer:
[tex]I = 0.083 kg m^2[/tex]
Explanation:
Mass of the bucket, m = 23 kg
Radius of the pulley, r = 0.050 m
The bucket is released from rest, u = 0 m/s
The time taken to fall, t = 2 s
Speed, v = 8.0 m/s
Moment of Inertia of the pulley, I = ?
Using the equation of motion:
v = u + at
8 = 0 + 2a
a = 8/2
a = 4 m/s²
The relationship between the linear and angular accelerations is given by the equation:
[tex]a = \alpha r[/tex]
Angular acceleration, [tex]\alpha = a/r[/tex]
[tex]\alpha = 4/0.050\\\alpha = 80 rad/s^2[/tex]
Since the bucket is falling, it can be modeled by the equation:
mg - T = ma
T = mg - ma = m(g-a)
T = 23(9.8 - 4)
The tension, T = 133.4 N
The equation for the pulley can be modeled by:
[tex]T* r = I * \alpha\\133.4 * 0.050 = I * 80\\6.67 = 80 I\\I = 6.67/80\\I = 0.083 kg m^2[/tex]
An elastic band is hung on a hook and a mass is hung on the lower end of the band. When the mass is pulled downward and then released, it vibrates vertically. The equation of motion is s = 9 cos(t) + 9 sin(t), t ≥ 0, where s is measured in centimeters and t in seconds. (Take the positive direction to be downward.) (a) Find the velocity and acceleration at time t.
Answer:
v(t) = s′(t) = −9sin(t)+9cos(t)
a(t) = v′(t) = −9cos(t) −9sin(t)
Explanation:
Given that
s = 9 cos(t) + 9 sin(t), t ≥ 0
Then acceleration and velocity is
v(t) = s′(t) = −9sin(t)+9cos(t)
a(t) = v′(t) = −9cos(t) −9sin(t)
The compressor of an air conditioner draws an electric current of 16.2 A when it starts up. If the start-up time is 1.45 s long, then how much electric charge passes through the circuit during this period
Answer:
Q = 23.49 C
Explanation:
We have,
Electric current drawn by the air conditioner is 16.2 A
Time, t = 1.45 s
It is required to find the electric charge passes through the circuit during this period. We know that electric current is defined as the electric charge flowing per unit time. So,
[tex]I=\dfrac{q}{t}\\\\q=It\\\\q=16.2\times 1.45\\\\q=23.49\ C[/tex]
So, the charge of 23.49 C is passing through the circuit during this period.
find the value of k for which the given pair of vectors are not equal
2ki +3j and 8i + 4kj
Answer:
5
Explanation:
The speed of a sound wave in air is 343m/s. If the density of the air is 1.2kg/m3, find the bulk modulus.
Answer:
141178.8
Explanation:
use : density x velocity²
1.2 x 343² = 141178.8 pa
g At 4 ft. from a point source of gamma radiation the intensity is 158 mRem/hr. How far away would you have to move to decrease the intensity to 17 mRem/hr.? [I1/I2 = (D2)2/(D1)2] I1(D1)2 = I2(D2)2
Answer:
You will have to move 12.2 ft to decrease the initial intensity to 17 mRem/h.
Explanation:
The distance can be calculated using the following equation:
[tex] \frac{I_{1}}{I_{2}} = \frac{D_{2}^{2}}{D_{1}^{2}} [/tex]
Where:
I₁ = 158 mRem/h
I₂ = 17 mRem/h
D₁ = 4 ft.
D₂=?
Hence, the distance D₂ is:
[tex] D_{2} =\sqrt{\frac{I_{1}}{I_{2}}*D_{1}^{2}} = \sqrt{\frac{158 mRem/h}{17 mRem/h}*(4 ft)^{2}} = 12.2 ft [/tex]
Therefore, you will have to move 12.2 ft to decrease the initial intensity to 17 mRem/h.
I hope it helps you!
A centrifuge rotor is accelerated from rest to 20000 rpm in 30s a) what is its average angular acceleration b) through how many revolutions has the centrifuge rotor turned during it's acceleration period, assuming constant angular acceleration
Answer:
a. 70 rad/s²
b. 5000 rev
Explanation:
As we know,
[tex]\omega = 20000\frac{rev}{min}\frac{2 \pi rad}{1 \ rev}\frac{1 \ min}{60 \ sec}[/tex]
then,
[tex]\omega=2100 \ rad/s[/tex]
a...
⇒ [tex]\bar{\alpha}=\frac{\omega-\omega_{0}}{\Delta t}[/tex]
On substituting the values, we get
⇒ [tex]=\frac{2100}{30}[/tex]
⇒ [tex]=70 \ rad/s^2[/tex]
b...
⇒ [tex]\theta=\theta_{0}=\omega_{0}t+\frac{1}{2}\alpha t^2[/tex]
[tex]=\frac{1}{2}\alpha t^2[/tex]
[tex]=\frac{1}{2}\times 70\times (30)^2[/tex]
[tex]=31500 \ rad[/tex]
[tex]=31500 \ rad\frac{1 \ rev}{2\pi rad}[/tex]
[tex]=5000 \ rev[/tex]
(a) The average angular acceleration will be 70 rad/s².
(b) 20063.69 revolutions have the centrifuge rotor turned during its acceleration period.
What is angular acceleration?Angular acceleration is defined as the pace of change of angular velocity with reference to time. It is denoted by α. Its unit is rad/s².
The given data in the problem is;
n is the revolution of centrifugal rotor = 20000 rpm
t is the time interval= the 30s
(α) is the Angular acceleration=?
n₁ is the revolution when the acceleration is constant =?
(a) The average angular acceleration will be 70 rad/s².
The value of the angular velocity is given by
[tex]\rm \omega_f = \frac{2\pi N}{60} } \\\\ \rm \omega_f = \frac{2 \times 3.14 \times 20000}{60} \\\\ \rm \omega_f= 2100\ rad/sec.[/tex]
The formula for angular acceleration is guven by;
[tex]\rm \alpha =\frac{ \omega_f-\omega_i}{dt} \\\\ \rm \alpha =\frac{ 2100-0}{3}\\\\ \rm \alpha =70\ rad/sec^2[/tex]
Hence the average angular acceleration will be 70 rad/s².
(b 5000 revolutions have the centrifuge rotor turned during its acceleration period.
[tex]\rm \theta= \theta_0+\frac{1}{2} \alpha t^2 \\\\ \rm \theta= \frac{1}{2} \times 70 \times (30)^2 \\\\ \rm \theta=31500\ rad[/tex]
As we know that the angular velocity is given by
[tex]\rm \omega = \frac{\theta}{t} \\\\ \rm \omega = \frac{31500}{30} \\\\ \rm \omega = 1050 \ rad/sec[/tex]
The relation of angular velocity and revolution will be
[tex]\rm n= \frac{ \omega \times 60}{2\pi} \\\\ \rm n= \frac{ 2100 \times 60}{2\times 3.14 } \\\\ \rm n = 20063.69 \ rev[/tex]
Hence 20063.69 revolutions have the centrifuge rotor turned during its acceleration period.
To learn more about angular acceleration refer to the link ;
https://brainly.com/question/408236
An automobile accelerates from zero to 30 m/s in 6 s. The wheels have a diameter of 0.4 m. What is the angular acceleration of each wheel
Answer:
12.5 rad/s²
Explanation:
Angular Acceleration: This can be defined as the ratio of linear acceleration and radius. The S.I unit is rad/s²
From the question,
a = αr................... Equation 1
Where a = linear acceleration, α = angular acceleration, r = radius.
But,
a = (v-u)/t.............. Equation 2
Where v = final velocity, u = initial velocity, t = time.
Substitute equation 2 into equation 1
(v-u)/t = αr
make α the subject of the equation
α = (v-u)/tr................. Equation 3
Given: v = 30 m/s, u = 0 m/s, t = 6 s, r = 0.4 m
Substitute into equation 3
α = (30-0)/(0.4×6)
α = 30/2.4
α = 12.5 rad/s²
a 5.0 charge is placed at the 0 cm mark of a meterstick and a -4.0 charge is placed at the 50 cm mark. what is the electric field at the 30 cm mark
Answer:
-1748*10^N/C
Explanation:
See attached file
In a series RC circuit, the resistor voltage is 124 V and the capacitor voltage is 167 V. What is the total voltage
Answer:
208 V
Explanation:
resistor voltage = Vr = 124 V
capacitor voltage Vc = 167 V
the total voltage in the RC circuit is the resultant voltage of the resistor and the capacitor
total voltage i= [tex]\sqrt{Vr^{2} + Vc^{2} }[/tex]
==> [tex]\sqrt{124^{2} + 167^{2} } =[/tex] 208 V
Particle A has charge qA and particle B has charge qB. When they are separated by a distance ri, they experience an attractive force Fi. The particles are moved without altering their charges. Now they experience an attractive force with a magnitude of 36Fi. Find an expression for their new separation.
Answer:
[tex]r_f=\frac{1}{6}r_i[/tex]
Explanation:
To find the new separation of the charges, you first take into account the formula for the electric force, when the force are separated a distance of ri.
You use the following expression:
[tex]F_i=k\frac{q_Aq_B}{r_i^2}[/tex] (1)
k: Coulomb's constant
qA: charge of A particle
qB: charge of B particle
When the charges are separated to a new distance rf, the new force is 36Fi, if the charges have not changed, you have:
[tex]F_f=36F_i=k\frac{q_Aq_B}{r_f^2}[/tex] (2)
To find the new separation you replace the expression for Fi of the equation (1) into the equation (2) and solve for rf in terms of ri:
[tex]36F_i=36k\frac{q_Aq_B}{r_i^2}=k\frac{q_Aq_B}{r_f^2}\\\\\frac{36}{r_i^2}=\frac{1}{r_f^2}\\\\r_f=\frac{1}{6}r_i[/tex]
The new separation of the charges is 1/6 times of the initial separation
What is the force of gravity acting on a 1-kg m mass? (g = 9.8 m/s ^ 2)
Answer: Use this F=Ma.
Explanation: So your answer will be
F=1 Kg+9.8 ms-2
So the answer will be
F=9.8N
How'd I do this?
I just used Newton's second law of motion.
I'll also put the derivation just in case.
Applied force α (Not its alpha, proportionality symbol) change in momentum
Δp α p final- p initial
Δp α mv-mu (v=final velocity, u=initial velocity and p=v*m)
or then
F α m(v-u)/t
So, as we know v=final velocity & u= initial velocity and v-u/t =a.
So F α ma, we now remove the proportionality symbol so we'll add a proportionality constant to make the RHS & LHS equal.
So, F=kma (where k is the proportionality constant)
k is 1 so you can ignore it.
So, our equation becomes F=ma
A machinist is required to manufacture a circular metal disk with area 1300 cm2. (a) What radius produces such a disk
Answer:
Radius r = 20.34 cm
The radius that can produces such a disk is 20.34 cm
Explanation:
Area of a circle;
A = πr^2
A = area
r = radius
Making r the subject of formula;
r = √(A/π) ........1
Given;
A = 1300 cm^2
Substituting into the equation 1;
r = √(1300/π)
r = 20.34214472564 cm
r = 20.34 cm
The radius that can produces such a disk is 20.34 cm
Using the ideas of electric field and force, explain what would happen to an electron if released from rest at r
Given that,
Distance = r
Electric field :
Electric field is equal to the multiplication of electric constant and charge divided by square of distance.
In mathematically form,
[tex]E=\dfrac{kq}{r^2}[/tex]
[tex]E=\dfrac{1}{4\pi\epsilon_{0}}\dfrac{q}{r^2}[/tex]
Electric force :
Electric force is equal to the multiplication of electric constant and both charges divided by square of distance.
In mathematically form,
[tex]E=\dfrac{kqQ}{r^2}[/tex]
[tex]E=\dfrac{1}{4\pi\epsilon_{0}}\dfrac{qQ}{r^2}[/tex]
We know that,
The relation between electric field and electric force is
[tex]E=\dfrac{F}{q}[/tex]
According to question,
If electron is released from rest then electron move towards the source point due to attractive force.
Hence, The electron move towards the source point due to attractive force.
what is the difference between a good conductor and a good insulator?
Answer:
Explanation:
In a conductor, electric current can flow freely, in an insulator it cannot.
Metals such as copper typify conductors, while most non-metallic solids are said to be good insulators, having extremely high resistance to the flow of charge through them.
Most atoms hold on to their electrons tightly and are insulators.
Three wires meet at a junction. Wire 1 has a current of 0.40 A into the junction. The current of wire 2 is 0.75 A out of the junction. How many electrons per second move past a point in wire 3?
Answer:
number of electrons = 2.18*10^18 e
Explanation:
In order to calculate the number of electrons that move trough the second wire, you take into account one of the Kirchoff's laws. All the current that goes inside the junction, has to go out the junction.
Then, if you assume that the current of the wire 1 and 3 go inside the junction, then, all this current have to go out trough the second junction:
[tex]i_1+i_3=i_2[/tex] (1)
i1 = 0.40 A
i2 = 0.75 A
you solve the equation i3 from the equation (1):
[tex]i_3=i_2-i_1=0.75A-0.40A=0.35A[/tex]
Next, you take into account that 1A = 1C/s = 6.24*10^18
Then, you have:
[tex]0.35A=0.35\frac{C}{s}=0.35*\frac{6.24*10^{18}e}{s}=2.18*10^{18}\frac{e}{s}[/tex]
The number of electrons that trough the wire 3 is 2.18*10^18 e/s
When the pivot point of a balance is not at the center of mass of the balance, how is the net torque on the balance calculated
Answer:
It is calculated as Force × perpendicular distance.
Explanation:
Torque is a rotational force and twisting force that can cause an object to rotate in it's axis. This cause angular rotation.
The torque due to gravity on a body about its centre of mass is zero because the centre of mass is the that point of the body at which the force acts by the gravity that is mg.
But if the pivot point of a balance is not at the centre of mass of the balance, it will be FORCE × PERPENDICULAR DISTANCE because at that point, there is no centre in which the force act on a body by gravity. The distance and force will be use to calculate.
A parallel-plate capacitor C is charged up to a potential V0 with a charge of magnitude Q0 on each plate. It is then disconnected from the battery, and the plates are pulled apart to twice their original separation.
A. What is the new capacitance in terms of C?
B. How much charge is now on the plates in terms of Q_0 ?
C. What is the potential difference across the plates in terms of V_0
Answer:
A. The new capacitance is half in term of C
B. The charge remains the same in terms of [tex]Q_{0}[/tex]
C. The potential difference is double in terms of [tex]V_{0}[/tex]
Explanation:
The battery with a voltage of [tex]V_{0}[/tex] is used to charge the plates, giving it a capacitance of C.
The charging process leaves a charge of magnitude [tex]Q_{0}[/tex] on the plate
The battery is disconnected (this will leave it with a constant charge [tex]Q_{0}[/tex])
the relationship between the charge, voltage and capacitance of the plate is
[tex]Q_{0}[/tex] = C[tex]V_{0}[/tex] ......... equ 1
A. The relationship between capacitance and the distance of the plate is given as
C = Aε/d ......... equ 2
where A is the area of the plate,
ε is the permeability of free space,
d is the distance between the plates
The area of the plate does not change, and permeability of free space is a constant. The combination of all these means that if the distance is doubled, then the capacitance will be halved. This is from equ 2 when the distance becomes 2d, then we have
C' = Aε/2d
==> C' = C/2
B. Since the battery is disconnected, and the capacitor is not discharged, the charge on the plate will remains the same as [tex]Q_{0}[/tex]. This is due to the conservation of charges.
C. Since the charge remains constant, and the capacitance is halved, then from equ 1, the new potential difference V will become double of the initial potential difference [tex]V_{0}[/tex]
==> V = 2[tex]V_{0}[/tex]
6. Two forces of 50 N and 30 N, respectively, are acting on an object. Find the net force (in
N) on the object if
the forces are acting in the same direction
b. the forces are acting in opposite directions.
Answer:
same direction = 80 (n)
opposite direction = 20 (n) going one direction
Explanation:
same direction means they are added to each other
and opposite means acting on eachother
Calculate the ideal banking angle in degrees for a gentle turn of 1.88 km radius on a highway with a 136.3 km/hr speed limit, assuming everyone travels at the speed limit.
Answer:
Ф = 4.4°Explanation:
given:
radius (r) = 1.88 km
velocity (v) = 136.3 km/hr
required:
banking angle ∡ ?
first:
convert 1.88 km to m = 1.88km * 1000m / 1km
r = 1880 m
convert velocity v = 136.3 km/hr to m/s = 136.3 km/hr * (1000 m/ 3600s)
v = 37.86 m/s
now.. calculate the angle
Ф = inv tan (v² / r * g) we know that gravity = 9.8 m/s²
Ф = inv tan (37.86² / (1880 * 9.8))
Ф = 4.4°
A force of 30 N is required to hold a spring that has been stretched from its natural length of 20 cm to a length of 35 cm. How much work is done in stretching the spring from 35 cm to 40 cm
Answer:
0.25 J
Explanation:
Given that
Force on the spring, F = 30 N
Natural length of the spring, l1 = 20 cm = 0.2 m
Final length of the spring, l2 = 35 cm = 0.35 m
Extension of the spring, x = 0.35 - 0.2 = 0.15 m
The other extension is 40 - 35 cm = 5 cm = 0.05 m
Work done = ?
Considering Hooke's Law of Elasticity.
We're told that the spring stretches through 15 cm, and thereafter asked to find the work done in stretching it through 5 cm
The question is solved in the attachment below
Alternating Current In Europe, the voltage of the alternating current coming through an electrical outlet can be modeled by the function V 230 sin (100t), where tis measured in seconds and Vin volts.What is the frequency of the voltage
Answer:
[tex]\frac{50}{\pi }[/tex]Hz
Explanation:
In alternating current (AC) circuits, voltage (V) oscillates in a sine wave pattern and has a general equation as a function of time (t) as follows;
V(t) = V sin (ωt + Ф) -----------------(i)
Where;
V = amplitude value of the voltage
ω = angular frequency = 2 π f [f = cyclic frequency or simply, frequency]
Ф = phase difference between voltage and current.
Now,
From the question,
V(t) = 230 sin (100t) ---------------(ii)
By comparing equations (i) and (ii) the following holds;
V = 230
ω = 100
Ф = 0
But;
ω = 2 π f = 100
2 π f = 100 [divide both sides by 2]
π f = 50
f = [tex]\frac{50}{\pi }[/tex]Hz
Therefore, the frequency of the voltage is [tex]\frac{50}{\pi }[/tex]Hz
A merry-go-round is spinning at a rate of 3.0 revolutions per minute. Cora is sitting 1.0 m from the center of the merry-go-round and Cameron is sitting right on the edge, 2.0 m from the center.
1. What is the relationship between the rotational speed of the two children?
a. Cameron's rotational speed is double coral rotational speed.
b. Cameron's rotational speed is four times as much as coral rotational speed.
c. Cora rotational speed is double Cameron's rotational speed.
d. Cora rotational speed is the same as Cameron's rotational speed.
e. Cora rotational speed is four times as much as Cameron's rotational speed
2. What is the relationship between the tangential speed of the two children?
a. Cora tangential speed is four times as much as Cameron's rotational speed.
b. Camerons tangential speed is four times coral tangential speed.
c. Cora tangential speed is the same as Cameron's tangential speed.
d. Cora tangential speed is double Cameron's tangential speed.
e. Cameron's tangential speed is double coral tangential speed.
Answer:
Explanation:
1 )
angular velocity of merry go round = 2π n
= 2π x 3 / 60
ω = .1 x π radian / s
This will be the rotational speed of the whole system including that of Cora and Cameron . It will not depend upon their relative position with respect to
the centre of the merry go round .
So rotational speed of Cora = Rotational speed of Cameron
option ( d ) is correct .
2 )
Tangential speed v = ω R where R is the distance from the centre of merry go round .
Tangential speed of Cora = .1 x π x 1
= .314 m /s
Tangential speed of Cameron = .1 x π x 2
= .628 m /s .
So tangential speed of Cameron is twice that of Cora .
Option ( e ) is correct .
Answer:
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Explanation:
A body is sent out in space. Which of the following statements is true of this body as it moves away from Earth?
A. The body's mass and weight remain equal.
B. The body's mass remains constant, and its weight decreases.
C. The body's mass decreases, and its weight remains constant.
Cell membranes (the walled enclosure around a cell) are typically about d = 7.1 nm thick. They are partially permeable to allow charged material to pass in and out, as needed. Equal but opposite charge densities build up on the inside and outside faces of such a membrane, and these charges prevent additional charges from passing through the cell wall. We can model a cell membrane as a parallel-plate capacitor, with the membrane itself containing proteins embedded in an organic material to give the membrane a dielectric constant of about 12.
Required:
a. What is the capacitance per square centimeter of such a cell wall?
b. In its normal resting state, a cell has a potential difference of 85mV across its membrane. What is the electric field inside this membrane?
Answer:
A. 1.24micro farad
B. 1.20x10^7V/m
Explanation:
See attached file
1.24micro farad is the capacitance per square centimeter of such a cell wall
In its normal resting state, a cell has a potential difference of 85mV across its membrane. the electric field inside this membrane is 1.20x10^7V/m.
what are the function of cell membrane ?The cell membrane is a selectively semi-permeable membrane surrounds the cytoplasm, the primary function of the membrane is to provide protection the integrity of the interior of the cell.
The cell membrane is called as multifaceted membrane which envelopes a cell's cytoplasm and help the cell to maintain the its shape.
Cell membrane is composed of Proteins and lipids where the mix or ratio of proteins and lipids can vary depending on the function of a specific cell.
Phospholipids are the crucial components of cell membranes which are arranged to form a lipid bilayer and certain substances can diffuse through the membrane to the cell's interior.
some cell organelles are surrounded by cell membranes, The nucleus and mitochondria have its own membrane.
For more details regarding cell membrane, visit
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The density of a sample of metal was measured to be 8.91 g/cm3. An X-ray diffraction experiment measures the edge of a face-centered cubic cell as 352.4 pm. Part APart complete What is the atomic weight of the metal
Answer:
The atomic weight of the metal is 58.7 g/mol
Explanation:
Given;
density of the metal sample, ρ = 8.91 g/cm³
edge length of the face centered cubic cell, α = 352.4 pm = 352.4 x 10⁻¹⁰ cm
Volume of the unit cell of the metal;
V = α³
V = (352.4 x 10⁻¹⁰ cm)³
V = 4.376 x 10⁻²³ cm³
Mass of the metal in unit cell
mass = density x volume
mass = 8.91 g/cm³ x 4.376 x 10⁻²³ cm³
mass = 3.899 x 10⁻²² g
Atomic weight, based on 4 atoms per unit cell;
4 atoms = 3.899 x 10⁻²² g
6.022 x 10²³ atoms = ?
= (6.022 x 10²³atoms x 3.899 x 10⁻²² g) / (4 atoms)
= 58.699 g/mol
= 58.7 g/mol (this metal is Nickel)
Theerefore, the atomic weight of the metal is 58.7 g/mol