Answer:
a) v = 517.99 m / s, b) θ = 296.3º
Explanation:
This is an exercise in kinematics, we are going to solve each axis independently
X axis
the acceleration is aₓ = 5.10 1 / S², they are on for t = 670 s and reaches a speed of vₓ= 3670 m / s, let's use the relation
vₓ = v₀ₓ + aₓ t
v₀ₓ = vₓ - aₓ t
v₀ₓ = 3670 - 5.10 670
v₀ₓ = 253 m / s
Y axis
the acceleration is ay = 7.30 m / s², with a velocity of 4378 m / s after
t = 670 s
v_y = v_{oy} + a_y t
v_{oy} = v_y - a_y t
v_oy} = 4378 - 7.30 670
v_{oy} = -513 m / s
to find the velocity modulus we use the Pythagorean theorem
v = [tex]\sqrt{v_o_x^2 + v_o_y^2}[/tex]
v = [tex]\sqrt{253^2 +513^2}[/tex]
v = 517.99 m / s
to find the direction we use trigonometry
tan θ ’= [tex]\frac{v_o_y}{v_o_x}[/tex]
θ'= tan⁻¹ [tex]\frac{voy}{voy}[/tex]
θ'= tan⁻¹ (-513/253)
tea '= -63.7
the negative sign indicates that it is below the ax axis, in the fourth quadrant
to give this angle from the positive side of the axis ax
θ = 360 - θ
θ = 360 - 63.7
θ = 296.3º
What is the biggest planet in the solar system
Answer:
Jupiter
Explanation:
Answer:
The answer is Jupiter.
Explanation:
Jupiter is an orange/yellow colored planet.
As a roller coaster car crosses the top of a 48.01-m-diameter loop-the-loop, its apparent weight is the same as its true weight. What is the car's speed at the top?
Answer:
The speed of the car, v = 21.69 m/s
Explanation:
The diameter is = 48.01 m
Therefore, the radius of the loop R = 24.005 m
Weight at the top is n = mv^2/R - mg
Since the apparent weight is equal to the real weight.
So, mv^2/R - mg = mg
v = √(2Rg)
v = √[2(24.005 m)(9.8 m/s^2)]
The speed of the car, v = 21.69 m/s
Answer:
The speed is 15.34 m/s.
Explanation:
Diameter, d = 48.01 m
Radius, R = 24.005 m
Let the speed is v and the mass is m.
Here, the weight of the car is balanced by the centripetal force.
According to the question
[tex]m g = \frac{mv^2}{R}\\\\v =\sqrt{24.005\times9.8}\\\\v = 15.34 m/s[/tex]
Find the magnitude and direction of a force between a 25.0 coulomb charge and a 40.0coulomb charge when they are separated by a distance of 30.0cm
Answer:
95.0 colomb
Explanation:
Make sure to understand the concept
vector A has a magnitude of 8 unit make an angle of 45° with posetive x axis vector B also has the same magnitude of 8 unit along negative x axis find the magnitude of A+B?
Answer:
45 × 8 units = A + B as formular
Give an example of a substance with an amorphous structure.
Answer:
Tempered glass
Explanation:
When warmed, an amorphous substance has a non-crystalline architecture that differentiates from its isochemical liquid, but this does not go through structural breakdown or the glass transition.
A ball on a frictionless plane is swung around in a circle at constant speed. The acceleration points in the same direction as the velocity vector.
a. True
b. False
Answer:
False
Explanation:
You have a circle so think back to circular motion. Theres 2 directions, centripetal and tangential. The problem tells you there's a constant tangential speed so tangential acceleration is 0. However there is a centripetal acceleration acting on the ball that holds it in its circular motion (i.e. tension, or gravity). Since centripetal is perpendicular to the tangential direction, acceleration and velocity are in different directions.
An electron is pushed into an electric field where it acquires a 1-V electrical potential. Suppose instead that two electrons are pushed the same distance into the same electric field (but far enough apart that they don't effect eachother). What is the electrical potential of one of the electrons now?
Answer:
0.5 V
Explanation:
The electric potential distance between different locations in an electric field area is unaffected by the charge that is transferred between them. It is solely dependent on the distance. Thus, for two electrons pushed together at the same distance into the same field, the electric potential will remain at 1 V. However, the electric potential of one of the two electrons will be half the value of the electric potential for the two electrons.
A block of mass M is connected by a string and pulley to a hanging mass m. The coefficient of kinetic friction between block M and the table is 0.2, and also, M = 20 kg, m = 10 kg. How far will block m drop in the first seconds after the system is released?
How long will block M move during above time?
At the time, calculate the velocity of block M
Find out the deceleration of the block M, if the connected string is
removal by cutting after the first second. Then, calculate the time
taken to contact block M and pulley.
Answer:
a) y = 0.98 t², t=1s y= 0.98 m,
b) he two blocks must move the same distance
c) v = 1.96 m / s, d) a = -1.96 m / s², e) x = 0.98 m
Explanation:
For this exercise we can use Newton's second law
Big Block
Y axis
N-W = 0
N = M g
X axis
T- fr = Ma
the friction force has the expression
fr = μ N
fr = μ Mg
small block
w- T = m a
we write the system of equations
T - fr = M a
mg - T = m a
we add and resolved
mg- μ Mg = (M + m) a
a = [tex]g \ \frac{m - \mu M}{m+M}[/tex]
a = [tex]9.8 \ \frac{10- 0.2 \ 20}{ 10 \ +\ 20}[/tex]
a = 9.8 (6/30)
a = 1.96 m / s²
a) now we can use the kinematic relations
y = v₀ t + ½ a t²
the blocks come out of rest so their initial velocity is zero
y = ½ a t²
y = ½ 1.96 t²
y = 0.98 t²
for t = 1s y = 0.98 m
t = 2s y = 1.96 m
b) Time is a scale that is the same for the entire system, the question should be oriented to how far the big block will move.
As the curda is in tension the two blocks must move the same distance
c) the velocity of the block M
v = vo + a t
v = 0 + 1.96 t
for t = 1 s v = 1.96 m / s
t = 2 s v = 3.92 m / s
d) the deceleration if the chain is cut
when removing the chain the tension becomes zero
-fr = M a
- μ M g = M a
a = - μ g
a = - 0.2 9.8
a = -1.96 m / s²
e) the distance to stop the block is
v² = vo² - 2 a x
0 = vo² - 2a x
x = vo² / 2a
x = 1.96² / 2 1.96
x = 0.98 m
the time to travel this distance is
v = vo - a t
t = vo / a
t = 1.96 /1.96
t = 1 s
Two objects attract each other with a gravitational force of magnitude 1.00 3 1028 N when separated by 20.0 cm. If the total mass of the two objects is 5.00 kg, what is the mass of each
Answer:
The mass of each object is 2kg and 3 kg.
Explanation:
Given that,
Gravitational force,[tex]F=1\times 10^{-8}\ N[/tex]
The distance between masses, d = 20 cm = 0.2 m
The total mass of the two objects, M + m = 5 kg
M = 5-m
The formula for the gravitational force is :
[tex]F=G\dfrac{Mm}{d^2}\\\\1\times 10^{-8}=6.67\times 10^{-11}\times \dfrac{(5-m)m}{(0.2)^2}\\\\\frac{1\times10^{-8}}{6.67\times10^{-11}}=\frac{(5-x)x}{(0.2)^{2}}\\\\\frac{1\times10^{-8}}{6.67\times10^{-11}}\cdot(0.2)^{2}\\\\5.99=(5-x)x\\\\x=2\ kg\ and\ 3 \ kg[/tex]
So, the mass of each object is 2kg and 3 kg.
A wave pulse travels along a stretched string at a speed of 200 cm/s. What will be the speed if:
a. The string's tension is doubled?
b. The string's mass is quadrupled (but its length is unchanged)?
c. The string's length is quadrupled (but its mass is unchanged)?
d. The string's mass and length are both quadrupled?
Answer:
a. 282.84 cm/s b. 100 cm/s c. 400 cm/s d. 200 cm/s
Explanation:
The speed of the wave v = √(T/μ) where T = tension and μ = mass per unit length = m/l where m = mass of string and l = length of string.
So, v = √(T/μ)
v = √(T/m/l)
v = √(Tl/m)
a. The string's tension is doubled?
If the tension is doubled, T' = 2T the new speed is
v' = √(T'l/m)
v' = √(2Tl/m)
v' = √2(√Tl/m)
v' = √2v
v' = √2 × 200 cm/s
v' = 282.84 cm/s
b. The string's mass is quadrupled (but its length is unchanged)?
If the mass is quadrupled, m' = 4m the new speed is
v' = √(Tl/m')
v' = √(Tl/4m)
v' = (1/√4)(√Tl/m)
v' = v/2
v' = 200/2 cm/s
v' = 100 cm/s
c. The string's length is quadrupled (but its mass is unchanged)?
If the length is quadrupled, l' = 4l the new speed is
v' = √(Tl'/m)
v' = √(T(4l)/m)
v' = √4)(√Tl/m)
v' = 2v
v' = 200 × 2 cm/s
v' = 400 cm/s
d. The string's mass and length are both quadrupled?
If the length is quadrupled, l' = 4l and mass quadrupled, m' = 4m, the new speed is
v' = √(Tl'/m')
v' = √(T(4l)/4m)
v' = √(Tl/m)
v' = v
v' = 200 cm/s
what is the dimensional formula of force and torque
Answer:
Units. Torque has the dimension of force times distance, symbolically T−2L2M. Although those fundamental dimensions are the same as that for energy or work, official SI literature suggests using the unit newton metre (N⋅m) and never the joule. The unit newton metre is properly denoted N⋅m.
Dimension: M L2T−2
In SI base units: kg⋅m2⋅s−2
Other units: pound-force-feet, lbf⋅inch, ozf⋅in
Answer:
hope it is helpful to you
☆☆☆☆☆☆☆☆☆☆☆
Help me with my physics, please
A 1.64 kg mass on a spring oscillates horizontal frictionless surface. The motion of the mass is described by the equation: X = 0.33cos(3.17t). In the equation, x is measured in meters and t in seconds. What is the maximum energy stored in the spring during an oscillation?
Answer:
[tex]K.E_{max}=0.8973J[/tex]
Explanation:
From the question we are told that:
Mass [tex]m=1.64kg[/tex]
Equation of Mass
[tex]X=0.33cos(3.17t)[/tex]...1
Generally equation for distance X is
[tex]X=Acos(\omega t)[/tex]...2
Therefore comparing equation
Angular Velocity [tex]\omega=3.17rad/s[/tex]
Amplitude A=0.33
Generally the equation for Max speed is mathematically given by
[tex]V_{max}=A\omega[/tex]
[tex]V_{max}=0.33*3.17[/tex]
[tex]V_{max}=1.0461m/s[/tex]
Therefore
[tex]K.E_{max}=0.5mv^2[/tex]
[tex]K.E_{max}=0.5*1.64*(1.0461)^2[/tex]
[tex]K.E_{max}=0.8973J[/tex]
PLEASE HELP ME WITH THIS ONE QUESTION
What is the rest energy of a proton? (c = 2.9979 x 10^9 m/s, mp = 1.6726 x 10^-27)
[tex]E_0=1.5033×10^{-10}\:\text{J}[/tex]
Explanation:
The rest energy [tex]E_0[/tex] of a proton of mass [tex]m_p[/tex] is given by
[tex]E_0 = m_pc^2[/tex]
[tex]\:\:\:\:\:\:\:=(1.6726×10^{-27}\:\text{kg})(2.9979×10^8\:\text{m/s})^2[/tex]
[tex]\:\:\:\:\:\:\:=1.5033×10^{-10}\:\text{J}[/tex]
Part of your electrical load is a 60-W light that is on continuously. By what percentage can your energy consumption be reduced by turning this light off
Answer:
Following are the solution to the given question:
Explanation:
Please find the complete question in the attached file.
The cost after 30 days is 60 dollars. As energy remains constant, the cost per hour over 30 days will be decreased.
[tex]\to \frac{\$60}{\frac{30 \ days}{24\ hours}} = \$0.08 / kwh.[/tex]
Thus, [tex]\frac{\$0.08}{\$0.12} = 0.694 \ kW \times 0.694 \ kW \times 1000 = 694 \ W.[/tex]
The electricity used is continuously 694W over 30 days.
If just resistor loads (no reagents) were assumed,
[tex]\to I = \frac{P}{V}= \frac{694\ W}{120\ V} = 5.78\ A[/tex]
Energy usage reduction percentage = [tex](\frac{60\ W}{694\ W} \times 100\%)[/tex]
This bulb accounts for [tex]8.64\%[/tex] of the energy used, hence it saves when you switch it off.
An electrostatic paint sprayer has a 0.100 m diameter metal sphere at a potential of 30.0 kV that repels paint droplets onto a grounded object. (a) What charge (in C) is on the sphere?(b) What charge must a 0.100-mg drop of paint have to arrive at the object with a speed of 10.0 m/s?
Answer:
A) q = 1.67 × 10^(-7) C
B) q = 1.67 × 10^(-10) C
Explanation:
We are given;
Potential; V = 30 KV = 30000 V
Radius of sphere; r = diameter/2 = 0.1/2 = 0.05 m
A) To find the charge of the sphere, we will use the formula;
V = kq/r
Where;
q is the charge
k is electric force constant = 9 × 10^(9) N.m²/C²
Thus;
q = Vr/k
q = (30000 × 0.05)/(9 × 10^(9))
q = 1.67 × 10^(-7) C
B) Now, potential energy here is a formula; U = qV
However, for the drop of paint to move, the potential energy will be equal to the kinetic energy. Thus;
qV = ½mv²
q = mv²/2V
Where;
v is speed = 10 m/s
V = 30000 V
m = mass = 0.100 mg = 0.1 × 10^(-6) Thus;
q = (0.1 × 10^(-6) × 10²)/(2 × 30000)
q = 1.67 × 10^(-10) C
. Set the applied force to Force necessary to Keep the box Moving without accelerating. Restart the animation. Just before the box hits the wall, stop the animation. What can you tell me about relative magnitudes of the frictional force and the applied force
Answer:
elative magnitude of the two forces is the same and they are applied in a constant direction.
Explanation:
Newton's second law states that the sum of the forces is equal to the mass times the acceleration
∑ F = m a
in this case there are two forces on the x axis
F_applied - fr = 0
since they indicate that the velocity is constant, consequently
F_applied = fr
the relative magnitude of the two forces is the same and they are applied in a constant direction.
A solenoid 10.0 cm in diameter and 85.1 cm long is made from copper wire of diameter 0.100 cm, with very thin insulation. The wire is wound onto a cardboard tube in a single layer, with adjacent turns touching each other. What power must be delivered to the solenoid if it is to produce a field of 8.90 mT at its center
Answer:
P = 29.3 W
Explanation:
The magnetic field in a solenoid is
B = μ₀ n i
i = B /μ₀ n
where n is the density of turns
We can use a direct rule of proportions or rule of three to find the number of turns, 1 a turn has a diameter of 0.100 cm = 10⁻³ m, in the length of
L= 85.1 cm = 0.851 m how many turns there are
#_threads = 0.851 / 10⁻³
#_threads = 8.50 10³ turns
the density of turns is
n = # _threads / L
n = 8.51 103 / 0.851
n = 104 turn / m
the current that must pass through the solenoid is
i = 8.90 10-3 / 4pi 10-7 104
i = 0.70823 A
now let's find the resistance of the copper wire
R = ρ L / A
the resistivity of copper is ρ = 1.72 10⁻⁸ Ω m
wire area
A = π r²
A = π (5 10⁻⁴)
A = 7,854 10⁻⁷ m²
let's find the length of wire to build the coil, the length of a turn is
Lo = 2π r = ππ d
Lo = π 0.100
Lo = 0.314159 m / turn
With a direct proportion rule we find the length of the wire to construct the 8.5 103 turns
L = Lo #_threads
L = 0.314159 8.50 10³
L = 2.67 10³ m
resistance is
R = 1.72 10⁻⁸ 2.67 10₃ / 7.854 10⁻⁷
R = 5,847 10¹
R = 58.47 ohm
The power to be supplied to the coil is
P = VI = R i²
P = 58.47 0.70823²
P = 29.3 W
A uniform metre rule of mass 10g is balanced on a knife edge placed at 45cm mark. Calculate the distance of a mass 25g from the pivot
Answer:
The distance of a mass 25g from the pivot is 18cm
Explanation:
Given
[tex]m_1 = 10g[/tex]
[tex]d_1 = 45cm[/tex]
[tex]m_2 = 25g[/tex]
Required
Distance of m2 from the pivot
To do this, we make use of:
[tex]m_1 * d_1 = m_2 * d_2[/tex] --- moments of the masses
So, we have:
[tex]10 * 45= 25* d_2[/tex]
[tex]450= 25* d_2[/tex]
Divide both sides by 25
[tex]18= d_2[/tex]
Hence:
[tex]d_2 = 18[/tex]
The conductivity of intrinsic semiconductors, such as silicon and germanium, can be increased when small amounts of group 3A or group 5A elements are added by a process called doping.
a. True
b. False
Answer:
a. True
Explanation:
A semiconductor can be defined as a crystalline solid substance that has its conductivity lying between that of a metal and an insulator, due to the effects of temperature or an addition of an impurity. Semiconductors are classified into two main categories;
1. Extrinsic semiconductor.
2. Intrinsic semiconductor.
An intrinsic semiconductor is a crystalline solid substance that is in its purest form and having no impurities added to it. Examples of intrinsic semiconductor are Germanium and Silicon.
Basically, the number of free electrons in an intrinsic semiconductor is equal to the number of holes. Also, the number of holes and free electrons in an intrinsic semiconductor is directly proportional to the temperature; as the temperature increases, the number of holes and free electrons increases and vice-versa.
In an intrinsic semiconductor, each free electrons (valence electrons) produces a covalent bond.
Generally, a process referred to as doping can be used to increase the conductivity of an intrinsic semiconductor such as silicon or germanium, by adding small amounts of impurities found in group 3A or group 5A elements.
A car hurtles off a cliff and crashes on the canyon floor below. Identify the system in which the net momentum is zero during the crash.
Solution :
It is given that a car ran off from a cliff and it crashes on canyon floor. Now the system of a car as well as the earth together have a [tex]\text{ net momentum of zero}[/tex] when the car crashes on the canyon floor, thus reducing the momentum of the car to zero. The earth also stops its upward motion and it also reduces the momentum to zero.
NEED HELP ASAP. Please show all work.
A point on a rotating wheel (thin hoop) having a constant angular velocity of 200 rev/min, the wheel has a radius of 1.2 m and a mass of 30 kg. ( I = mr2 ).
(a) (5 points) Determine the linear acceleration.
(b) (4 points) At this given angular velocity, what is the rotational kinetic energy?
Answer:
Look at work
Explanation:
a) I am not sure if you want tangential or centripetal but I will give both
Centripetal acceleration = r*α
Since ω is constant, α is 0 so centripetal acceleration is 0m/s^2
Tangential acceleration = ω^2*r
convert 200rev/min into rev/s
200/60= 10/3 rev/s
a= 100/9*1.2= 120/9= 40/3 m/s^2
b) Rotational Kinetic Energy = 1/2Iω^2
I= mr^2
Plug in givens
I= 43.2kgm^2
K= 1/2*43.2*100/9=2160/9=240J
Based on the information in the table, what
is the acceleration of this object?
t(s) v(m/s)
0.0
9.0
1.0
4.0
2.0
-1.0
3.0
-6.0
A. -5.0 m/s2
B. -2.0 m/s2
C. 4.0 m/s2
D. 0.0 m/s2
Answer:
Option A. –5 m/s²
Explanation:
From the question given above, the following data were obtained:
Initial velocity (v₁) = 9 m/s
Initial time (t₁) = 0 s
Final velocity (v₂) = –6 m/s
Final time (t₂) = 3 s
Acceleration (a) =?
Next, we shall determine the change in the velocity and time. This can be obtained as follow:
For velocity:
Initial velocity (v₁) = 9 m/s
Final velocity (v₂) = –6 m/s
Change in velocity (Δv) =?
ΔV = v₂ – v₁
ΔV = –6 – 9
ΔV = –15 m/s
For time:
Initial time (t₁) = 0 s
Final time (t₂) = 3 s
Change in time (Δt) =?
Δt = t₂ – t₁
Δt = 3 – 0
Δt = 3 s
Finally, we shall determine the acceleration of the object. This can be obtained as follow:
Change in velocity (Δv) = –15 m/s
Change in time (Δt) = 3 s
Acceleration (a) =?
a = Δv / Δt
a = –15 / 3
a = –5 m/s²
Thus, the acceleration of the object is
–5 m/s².
Solids diffuse because the particles cannot move.
A. Can
B. Not enough info
C. Cannot
D. Sometimes will
Solids cannot diffuse.
A car is moving at a speed of 60 mi/hr (88 ft/sec) on a straight road when the driver steps on the brake pedal and begins decelerating at a constant rate of 10ft/s2 for 3 seconds. How far did the car go during this 3 second interval?
Answer:
219 ft
Explanation:
Here we can define the value t = 0s as the moment when the car starts decelerating.
At this point, the acceleration of the car is given by the equation:
A(t) = -10 ft/s^2
Where the negative sign is because the car is decelerating.
To get the velocity equation of the car, we integrate over time, to get:
V(t) = (-10 ft/s^2)*t + V0
Where V0 is the initial velocity of the car, we know that this is 88 ft/s
Then the velocity equation is:
V(t) = (-10 ft/s^2)*t + 88ft/s
To get the position equation we need to integrate again, this time we get:
P(t) = (1/2)*(-10 ft/s^2)*t^2 + (88ft/s)*t + P0
Where P0 is the initial position of the car, we do not know this, but it does not matter for now.
We want to find the total distance that the car traveled in a 3 seconds interval.
This will be equal to the difference in the position at t = 3s and the position at t = 0s
distance = P(3s) - P(0s)
= ( (1/2)*(-10 ft/s^2)*(3s)^2+ (88ft/s)*3s + P0) - ( (1/2)*(-10 ft/s^2)*(0s)^2 + (88ft/s)*0s + P0)
= ( (1/2)*(-10 ft/s^2)*9s^2+ (88ft/s)*3s + P0) - ( P0)
= (1/2)*(-10 ft/s^2)*9s^2+ (88ft/s)*3s = 219ft
The car advanced a distance of 219 ft in the 3 seconds interval.
write down the following units in the ascending of their value A) mm nm cm um B) 1m 1cm 1km 1mm. convert the following units into SI without changing their values? A)3500g B)2.5km C)2h
Answer:
A) nm, um, mm, cm
B) 1mm, 1cm, 1m, 1km
A) 3500g, B) 2500m, C) 7200 seconds
To calculate the gravitational potential energy of a statue on a 10-meters-tall platform, you would have to know the statue's ______________
Answer:
mass
Explanation:
In a double-slit experiment, the slit spacing is 0.120 mm and the screen is 2.00 m from the slits. Find the wavelength (in nm) if the distance between the central bright region and the third bright fringe on a screen is 2.75 cm.
Answer:
[tex]\lambda=550nm[/tex]
Explanation:
From the question we are told that:
The slit spacing [tex]d_s=0.120mm=>0.120*10^{-3}[/tex]
Screen distance [tex]d_{sc}=2.0m[/tex]
Third Distance [tex]X=2.75cm=>2.75*10^{-2}[/tex]
Generally the equation for Wavelength is mathematically given by
[tex]\lambda=\frac{Xd_s}{n*d_{sc}}[/tex]
Where
n=number of screens
[tex]n=3[/tex]
Therefore
[tex]\lambda=\frac{2.75*10^{-2}*0.120*10^{-3}}{3*2}[/tex]
[tex]\lambda=0.055*10^{-5}[/tex]
[tex]\lambda=550nm[/tex]
How do the magnitude and direction of the electric field on the left side of the dipole compare to the right side for the same distance
Answer:
The magnitude of the electric field is same while the direction at the left and at the right is opposite to each other.
Explanation:
The direction of the electric field due to the dipole on the axial line is same as the direction of dipole moment.
The magnitude of the electric field due to an electric dipole on its axial line is
[tex]E=\frac{2kp}{r^3}[/tex]
where, k is the constant, p is the electric dipole moment and r is the distance from the center of dipole.
The magnitude of the electric field is same while the direction at the left and at the right is opposite to each other.
I need help with this physics question.
The acceleration will increase by 61.3%.
Explanation:
The centripetal acceleration [tex]a_c[/tex] is given by
[tex]a_c = \dfrac{v^2}{r}[/tex]
If the velocity of the object increases by 27.0%, then its new velocity v' becomes
[tex]v' = 1.270v[/tex]
The new centripetal acceleration [tex]a'_c[/tex] becomes
[tex]a'_c = \dfrac{(1.270v)^2}{r} = 1.613 \left(\dfrac{v^2}{r} \right)[/tex]
[tex]\:\:\:\:\:\:\:\:\:= 1.613a_c[/tex]