Answer:
I = E/R e^{-t/RC}
Explanation:
In a capacitor charging circuit you must have a DC power source, the capacitor, a resistor, and a switch. When closing the circuit,
E -q / c-IR = 0
we replace the current by its expression and divide by the resistance
I = dq / dt
dq / dt = E / R -q / RC
dq / dt = (CE -q) / RC
we solve the equation
dq / (Ce-q) = -dt / RC
we integrate and evaluate for the charge between 0 and q and for the time 0 and t
ln (q-CE / -CE) = -1 /RC (t -0)
eliminate the logarithm
q - CE = CE [tex]e^{-t/RC}[/tex]
q = CE (1 + 1/RC e^{-t/RC} )
In general the teams measure the current therefore we take the derivative to find the current
i = CE (e^{-t/RC} / RC)
I = E/R e^{-t/RC}
This expression is the one that describes the charge of a condensate in a DC circuit
A meter stick balances horizontally on a knife-edge at the 50.0 cm mark. With two 3.62 g coins stacked over the 20.3 cm mark, the stick is found to balance at the 22.5 cm mark. What is the mass of the meter stick
Answer:
0.5792g
Explanation:
The computation of the mass of the meter stick is shown below:
Let us assume the following items
x1 = 50 cm;
m2 = m3 = 3.62 g;
x2 = x3 = 20.3 cm;
xcm = 22.5 cm
Based on the above assumption, now we need to apply the equation of center mass which is given below:
[tex]Xcm = \frac{m1x1 + m2x2 + m3x3}{m1 + m2 + m3} \\\\ 22.5 = \frac{m1\times 50 + 3.62 \times 20.3 + 3.62 \times 20.3}{m1 + 3.62 + 3.62}\\\\ 22.5m1 + 162.9 = 50m1 + 73.486 + 73.486[/tex]
27.5 m1 = 15.928
So, the m1 = 0.5792g
with a speed of 75 m sl. Determine
1) A vehicle of mass 1600 kg moves
the magnitude of its momentum.
Answer:
120000 kgxm/s
Explanation:
momentum is mass times velocity so just multiply 1600 kg times 75 m/s and you get 120000 kgxm/s
3) If a ball launched at an angle of 10.0 degrees above horizontal from an initial height of 1.50 meters has a final horizontal displacement of 3.00 meters, what is its launch velocity
Answer:
35.6 m
Explanation:
The given ball possesses a projectile motion from its initial height. So, the required launch velocity of the ball is 6.55 m/s.
What is launch velocity?The horizontal component of velocity during the projection of an object is known as launch velocity. It is obtained when the horizontal range is known.
Given data -
The angle of projection is, [tex]\theta = 10.0 {^\circ}[/tex].
The initial height of the projection is, h = 1.50 m.
The horizontal displacement is, R = 3.00 m.
The mathematical expression for the horizontal displacement (Range) of the projectile is given as,
[tex]R = \dfrac{u^{2} \times sin2 \theta}{g}[/tex]
here,
u is the launch velocity.
g is the gravitational acceleration.
Solving as,
[tex]u =\sqrt{\dfrac{R \times g}{sin2 \theta}}\\\\\\u =\sqrt{\dfrac{1.50 \times 9.8}{sin(2 \times 10)}}\\\\\\u =\sqrt{\dfrac{1.50 \times 9.8}{sin20^\circ}}\\\\\\u=6.55 \;\rm m/s[/tex]
Thus, we can conclude that the required launch velocity of the ball is 6.55 m/s.
Learn more about the projectile motion here:
https://brainly.com/question/11049671
A proton is released from rest at the positive plate of a parallel-plate capacitor. It crosses the capacitor and reaches the negative plate with a speed of 45000 m/s.
What will be the final speed of an electron released from rest at the negative plate?
Express your answer to two significant figures and include the appropriate units
V=_________
Answer:
2.1x10^6m/s
Explanation:
One electron has a charge of –1.602e-19 C
mass of electron is 9.1e-31 kg
mass of proton is 1.6726e−27 kg
mass ratio is 1.6726e−27 / 9.1e-31 = 1838
The force is constant, F
distance is constant, d
a = F/m
a increases by a factor 1838, as m decreases by that factor
a = a₀1838
v₀² = 2a₀d
v² = 2a₀d1838
v²/v₀² = 2a₀d1838 / 2a₀d = 1838
v² = 1838v₀² = 1838(45000)²
v = 45000√1838 = 2.1e6 m/s
Unpolarized light passes through a vertical polarizing filter, emerging with an intensity I0. The light then passes through a horizontal filter, which blocks all of the light; the intensity transmitted through the pair of filters is zero. Suppose a third polarizer with axis 45 ? from vertical is inserted between the first two.
What is the transmitted intensity now?
Express your answer in terms of I0. I got I0/8. But this is not right. I guess they want a number?
Answer:
I₂ = 0.25 I₀
Explanation:
To know the light transmitted by a filter we must use the law of Malus
I = I₀ cos² θ
In this case, the intensity of the light that passes through the first polarizer is I₀, it reaches the second polarized, which is at 45⁰, therefore the intensity I1 comes out of it.
I₁ = I₀ cos² 45
I₁ = I₀ 0.5
this is the light that reaches the third polarizer, which is at 45⁰ with respect to the second, from this comes the intensity I₂
I₂ = I₁ cos² 45
I₂ = (I₀ 0.5) 0.5
I₂ = 0.25 I₀
this is the intensity of the light transmitted by the set of polarizers
Blood is pumped from the heart at a rate of 5.0 L/min into the aorta (of radius 1.0 cm). Determine the speed of blood through the aorta in units of m/s
Answer:
The speed of blood through the aorta is 0.265 m/s
Explanation:
Given;
volumetric flow rate, Q = 5.0L/min = 0.005 m³/min x 1min/60s = 8.333 x 10⁻⁵ m³/s
radius of the aorta, r = 1.0 cm = 0.01 m
Area of the aorta = πr²
Area of the aorta = π(0.01)² = 3.142 x 10⁻⁴ m²
Volumetric flow rate is given by;
Q = Av
where;
v is the speed of blood through the aorta
v = Q /A
v = (8.333 x 10⁻⁵ ) / (3.142 x 10⁻⁴)
v = 0.265 m/s
Therefore, the speed of blood through the aorta is 0.265 m/s
The blood's speed through the aorta will be "0.265 m/s". To understand the calculation, check below.
Blood and AortaAccording to the question,
Volumetric flow rate, Q = 5.0 L/min or,
= 0.005 × [tex]\frac{1 \ min}{60 \ s}[/tex]
= 8.333 × 10⁻⁵ m³/s
Aorta's radius, r = 1.0 cm
We know the formula,
→ Q = A × v
or,
Speed, v = [tex]\frac{Q}{A}[/tex]
By substituting the values,
= [tex]\frac{8.333\times 10^{-5}}{3.142\times 10^{-4}}[/tex]
= 0.265 m/s
Thus the above answer is correct.
Find out more information about Blood here:
https://brainly.com/question/9797618
A 4g bullet, travelling at 589m/s embeds itself in a 2.3kg block of wood that is initially at rest, and together they travel at the same velocity. Calculate the percentage of the kinetic energy that is left in the system after collision to that before.
Answer:
The percentage of the kinetic energy that is left in the system after collision to that before is 0.174 %
Explanation:
Given;
mass of bullet, m₁ = 4g = 0.004kg
initial velocity of bullet, u₁ = 589 m/s
mass of block of wood, m₂ = 2.3 kg
initial velocity of the block of wood, u₂ = 0
let the final velocity of the system after collision = v
Apply the principle of conservation of linear momentum
m₁u₁ + m₂u₂ = v(m₁+m₂)
0.004(589) + 2.3(0) = v(0.004 + 2.3)
2.356 = 2.304v
v = 2.356 / 2.304
v = 1.0226 m/s
Initial kinetic energy of the system
K.E₁ = ¹/₂m₁u₁² + ¹/₂m₂u₂²
K.E₁ = ¹/₂(0.004)(589)² = 693.842 J
Final kinetic energy of the system
K.E₂ = ¹/₂v²(m₁ + m₂)
K.E₂ = ¹/₂ x 1.0226² x (0.004 + 2.3)
K.E₂ = 1.209 J
The kinetic energy left in the system = final kinetic energy of the system
The percentage of the kinetic energy that is left in the system after collision to that before = (K.E₂ / K.E₁) x 100%
= (1.209 / 693.842) x 100%
= 0.174 %
Therefore, the percentage of the kinetic energy that is left in the system after collision to that before is 0.174 %
Two capacitors, CA and CB, are such that CA > CB. These are connected with a battery in various ways: each individually, series, and parallel. Rank these four cases according to the total amount of charge, greatest first.
a. (CA) > (C) > (CA and CB in series) > (CA and Co in parallel)
b. (CA)>(Cb)> (CA and CB in parallel) > (CA and CB in series)
c. (CA and CB in series) > (CA) > (CB) > (CA and CB in parallel)
d. (CA and Cg in parallel) > (CA) > (CB) > (CA and Cg in series)
Answer:
b. (CA)>(Cb)> (CA and CB in parallel) > (CA and CB in series)
Explanation:
This is because capacitors in series is the sum of the reciprocal of the capacitances while that of parallel is the sum of the individual capacitances
Two point charges attract each other with an electric force of magnitude F. If one charge is reduced to one-third its original value and the distance between the charges is doubled, what is the resulting magnitude of the electric force between them
Answer:
F' = F/12
Therefore, the electrostatic force is reduced to one-twelve of its original value.
Explanation:
The electrostatic force of attraction or repulsion between to charges is given by Coulomb's Law:
F = kq₁q₂/r² ---------- equation 1
where,
F = Electrostatic Force
k = Coulomb's Constant
q₁ = magnitude of first charge
q₂ = magnitude of 2nd charge
r = distance between charges
Now, if we double the distance between charges and reduce one charge to one-third value, then the force will become:
F' = kq₁'q₂'/r'²
where,
q₁' = (1/3)q₁
q₂' = q₂
r' = 2r
Therefore,
F' = k(1/3 q₁)(q₂)/(2r)²
F' = (1/12)kq₁q₂/r²
using equation 1:
F' = F/12
Therefore, the electrostatic force is reduced to one-twelve of its original value.
Consider the following three objects, each of the same mass and radius:
(1) a solid sphere
(2) a solid disk
(3) a hoop
All three are released from rest at the top of an inclined plane. The three objects proceed down the incline undergoing rolling motion without slipping. Use work-kinetic energy theorem to determine which object will reach the bottom of the incline first.
a) 1, 2, 3
b) 2, 3, 1
c) 3, 1, 2
d) 3, 2, 1
e) All three reach the bottom at the same time.
Answer:
Explanation:a 1
On a certain planet a body is thrown vertically upwards with an initial speed of 40 m / s. If the maximum height was 100 m, the acceleration due to gravity is
a) 15 m / s 2
b) 12.5 m / s 2
c) 8 m / s 2
d) 10 m / s 2
Answer:
C) 8 m/s²
Explanation:
Given:
v₀ = 40 m/s
v = 0 m/s
Δy = 100 m
Find: a
v² = v₀² + 2aΔy
(0 m/s)² = (40 m/s)² + 2a (100 m)
a = -8 m/s²
A car travels at 100 km / h, collides head-on against a pole. Assuming the vehicle stopped at 2.2 seconds after impact, calculate the magnitude of the deceleration suffered by the driver.
Answer:
12.6 m/s²
Explanation:
First, convert to m/s.
100 km/h × (1000 m/km) × (1 hr / 3600 s) = 27.8 m/s
a = Δv / Δt
a = (0 m/s − 27.8 m/s) / 2.2 s
a = -12.6 m/s²
A magnetic field is entering into a coil of wire with radius of 2(mm) and 200 turns. The direction of magnetic field makes an angle 25° in respect to normal to surface of coil. The magnetic field entering coil varies 0.02 (T) in every 2 seconds. The two ends of coil are connected to a resistor of 20 (Ω).
A) Calculate Emf induced in coil
B) Calculate the current in resistor
C) Calculate the power delivered to resistor by Emf
Answer:
a) 2.278 x 10^-5 volts
b) 1.139 x 10^-6 Ampere
c) 2.59 x 10^-11 W
Explanation:
The radius of the wire r = 2 mm = 0.002 m
the number of turns N = 200 turns
direction of the magnetic field ∅ = 25°
magnetic field strength B = 0.02 T
varying time = 2 sec
The cross sectional area of the wire = [tex]\pi r^{2}[/tex]
==> A = 3.142 x [tex]0.002^{2}[/tex] = 1.257 x 10^-5 m^2
Field flux Φ = BA cos ∅ = 0.02 x 1.257 x 10^-5 x cos 25°
==> Φ = 2.278 x 10^-7 Wb
The induced EMF is given as
E = NdΦ/dt
where dΦ/dt = (2.278 x 10^-7)/2 = 1.139 x 10^-7
E = 200 x 1.139 x 10^-7 = 2.278 x 10^-5 volts
b) If the two ends are connected to a resistor of 20 Ω, the current through the resistor is given as
[tex]I[/tex] = E/R
where R is the resistor
[tex]I[/tex] = (2.278 x 10^-5)/20 = 1.139 x 10^-6 Ampere
c) power delivered to the resistor is given as
P = [tex]I[/tex]E
P = (1.139 x 10^-6) x (2.278 x 10^-5) = 2.59 x 10^-11 W
If an X-ray beam of wavelength 1.4 × 10-10 m makes an angle of 30° with a set of planes in a crystal causing first order constructive interference, what is the plane spacing?
Answer:
Plane spacing, [tex]d=1.4\times 10^{-10}\ m[/tex]
Explanation:
It is given that,
Wavelength of x-ray, [tex]\lambda=1.4\times 10^{-10}\ m[/tex]
Angle the x-ray made with a set of planes in a crystal causing first order constructive interference is 30 degrees
We need to find the plane spacing. It is based on Bragg's law such that,
[tex]2d\sin\theta=n\lambda[/tex]
d is plane spacing
n = 1 here
[tex]d=\dfrac{\lambda}{2\sin\theta}\\\\d=\dfrac{1.4\times 10^{-10}}{2\times \sin (30)}\\\\d=1.4\times 10^{-10}\ m[/tex]
So, the plane spacing is [tex]1.4\times 10^{-10}\ m[/tex].
A 1200 kg aircraft going 30 m/s collides with a 2000 kg aircraft that is parked and they stick together after the collision and are going 11.3 m/s after the collision. If they skid for 14.7 seconds before stopping, how far did they skid
Answer:
83.055 m
Explanation:
According to the given scenario, the calculation of skid distance is shown below:-
[tex]S = \frac{1}{2} \times (u + v) \times t[/tex]
Where
u = 11.3
v = 0
t = 14.7
Now placing these values to the above formula,
So,
[tex]S = \frac{1}{2} \times (11.3 + 0) \times 14.7[/tex]
= 83.055 m
Therefore for computing the skid distance we simply applied the above formula i.e by considering the all items given in the question
Diamagnetic materialsA) have small negative values of magnetic susceptibility.B) are those in which the magnetic moments of all electrons in each atom cancel.C) experience a small induced magnetic moment when placed in an external magnetic field.D) exhibit the property of diamagnetism independently of temperature.E)are described by all
Answer:
C) experience a small induced magnetic moment when placed in an external magnetic field.
Explanation:
Diamagnetics materials are those that experience a small induced magnetic moment when placed in an external magnetic field. These materials, such as bismuth, copper, silver and lead, have elementary magnets in their compositions. When they are exposed to an external magnetic cap, these elemental magnets tend to follow an orientation contrary to the external magnetic field. As a result, a magnetic field is created in the opposite direction to the external magnetic field.
The cost of buying shirts is partly constant and partly varies with the number of shirts bought. When the number of shirts is 5 the cost is #240, also, 10 shirts costs #400. find the cost when 300 shirts were bought
Answer:
The cost of the buying the shirts is #9680
Explanation:
let the cost of buying shirt = C
let the number of shirt bought = N
The following equation can be generated based on the statement above;
C = k + Nb
When the cost, C = #240, the number of shirt = 5
240 = k + 5b ------ equation (1)
where;
k and b are constants
When the cost, C = #400, the number of shirt = 10
400 = k + 10b ------ equation (2)
From equation (1), make k the subject of the formula;
k = 240 - 5b ---- equation (3)
Substitute in the value of k into equation (2)
400 = k + 10b
400 = (240 - 5b) + 10b
400 = 240 - 5b + 10b
400 - 240 = -5b + 10b
160 = 5b
b = 160 / 5
b = 32
From equation (3), calculate k
k = 240 - 5b
k = 240 -5(32)
k = 240 - 160
k = 80
When the number of shirts bought = 300, the cost of the buying the shirts =
C = k + Nb
C = 80 +32N
Where;
N is the number of shirts
C = 80 + 32(300)
C = 80 + 9600
C = #9680
Therefore, the cost of the buying the shirts is #9680
g A tube open at both ends, resonated at it's fundamental frequency, to a sound wave traveling at 330m/s. If the length of the tube is 4cm, find the frequency of the sound wave.
Answer:
frequency =4125Hz
Explanation:
L = 4cm = 0.04m
f =v/2L
f = 330/2 x 0.04
f = 4125Hz
Light with a frequency of 5.70×10^14 Hz travels in a block of glass that has an index of refraction of 1.56. What is the wavelength of the light in the glass?
✔First you have to calculate the light's speed in the glass,
You know that in the air and in the void (where the refraction index n is zero) the light's speed C corresponds to 3,0 x 10^8 m/s
So We have :
V = C/n
V = 3,0 x 10^8/1,56 V ≈ 1,92 x 10^8 m/s✔ Now, you know the light's speed in glass, and you know that : the wavelength λ is the quotient of light's speed V on its frequency ν, so :
λ = V/ ν
λ = 1,82 x 10^8/5,70 x 10^14 λ ≈ 3.40 x 10^-7 m λ ≈ 340 nmwhich of the following statements is not true Negatively charged objects attract other negatively charged objects. Positively charged objects attract negatively charged objects. Positively charged objects attract neutral objects. Negatively chargers objects attract neutral objects.
Answer:
negatively charged object attract other negatively objects
Explanation:
opposites attract
Answer:
negativelycharged objects attract other negatively charged objects
Explanation:
unlike charges attract like charges repel
A 2 m tall, 0.5 m inside diameter tank is filled with water. A 10 cm hole is opened 0.75 m from the bottom of the tank. What is the velocity of the exiting water? Ignore all orificelosses.
Answer:
4.75 m/s
Explanation:
The computation of the velocity of the existing water is shown below:
Data provided in the question
Tall = 2 m
Inside diameter tank = 2m
Hole opened = 10 cm
Bottom of the tank = 0.75 m
Based on the above information, first we have to determine the height which is
= 2 - 0.75 - 0.10
= 2 - 0.85
= 1.15 m
We assume the following things
1. Compressible flow
2. Stream line followed
Now applied the Bernoulli equation to section 1 and 2
So we get
[tex]\frac{P_1}{p_g} + \frac{v_1^2}{2g} + z_1 = \frac{P_2}{p_g} + \frac{v_2^2}{2g} + z_2[/tex]
where,
P_1 = P_2 = hydrostatic
z_1 = 0
z_2 = h
Now
[tex]\frac{v_1^2}{2g} + 0 = \frac{v_2^2}{2g} + h\\\\V_2 < < V_1 or V_2 = 0\\\\Therefore\ \frac{v_1^2}{2g} = h\\\\v_1^2 = 2gh\\\\ v_1 = \sqrt{2gh} \\\\v_1 = \sqrt{2\times 9.8\times 1.15}[/tex]
= 4.7476 m/sec
= 4.75 m/s
Unlike a roller coaster, the seats in a Ferris wheel swivel so that the rider is always seated upright. An 80-ft-diameter Ferris wheel rotates once every 24 s.What is the apparent weight of a 40 kg passenger at the lowest point of the circle?What is the apparent weight of a 40 kg passenger at the highest point of the circle?
Answer:
a
[tex]F_A =425.42 \ N[/tex]
b
[tex]F_A_H = 358.58 \ N[/tex]
Explanation:
From the question we are told that
The diameter of the Ferris wheel is [tex]d = 80 \ ft = \frac{80}{3.281} = 24.383[/tex]
The period of the Ferris wheel is [tex]T = 24 \ s[/tex]
The mass of the passenger is [tex]m_g = 40 \ kg[/tex]
The apparent weight of the passenger at the lowest point is mathematically represented as
[tex]F_A_L = F_c + W[/tex]
Where [tex]F_c[/tex] is the centripetal force on the passenger, which is mathematically represented as
[tex]F_c =m * r * w^2[/tex]
Where [tex]w[/tex] is the angular velocity which is mathematically represented as
[tex]w = \frac{2* \pi }{T}[/tex]
substituting values
[tex]w = \frac{2* 3.142 }{24}[/tex]
[tex]w = 0.2618 \ rad/s[/tex]
and r is the radius which is evaluated as [tex]r = \frac{d}{2}[/tex]
substituting values
[tex]r = \frac{24.383}{2}[/tex]
[tex]r = 12.19 \ ft[/tex]
So
[tex]F_c = 40 * 12.19* (0.2618)^2[/tex]
[tex]F_c = 33.42 \ N[/tex]
W is the weight which is mathematically represented as
[tex]W = 40 * 9.8[/tex]
[tex]W = 392 \ N[/tex]
So
[tex]F_A = 33.42 + 392[/tex]
[tex]F_A =425.42 \ N[/tex]
The apparent weight of the passenger at the highest point is mathematically represented as
[tex]F_A_H = W- F_c[/tex]
substituting values
[tex]F_A_H = 392 - 33.42[/tex]
[tex]F_A_H = 358.58 \ N[/tex]
A hoop, a solid disk, and a solid sphere, all with the same mass and the same radius, are set rolling without slipping up an incline, all with the same initial kinetic energy.
Which goes furthest the incline?
a. The hoop
b. The disk
c. The sphere
d. They all roll to the same height
Answer:
The sphere
Explanation:
Because it has a smaller inertia (I) value in the explanation in the attached file
A man is at a car dealership, looking for a car to buy. He looks at the sticker on the driver’s window of a car and sees that the price of the car is $32,540. Why is the long shadow of scarcity visible at the car dealership? Check all that apply.
Answer:
he will see the sticker because its behind a window bruh and thats a big daddy stack of greens
Explanation:
Assume that a lightning bolt can be represented by a long straight line of current. If 15.0 C of charge passes by in a time of 1.5·10-3s, what is the magnitude of the magnetic field at a distance of 24.0 m from the bolt?
Answer:
The magnitude of the magnetic field is 8.333 x 10⁻⁷ T
Explanation:
Given;
charge on the lightening bolt, C = 15.0 C
time the charge passes by, t = 1.5 x 10⁻³ s
Current, I is calculated as;
I = q / t
I = 15 / 1.5 x 10⁻³
I = 10,000 A
Magnetic field at a distance from the bolt is calculated as;
[tex]B = \frac{\mu_o I}{2\pi r}[/tex]
where;
μ₀ is permeability of free space = 4π x 10⁻⁷
I is the current in the bolt
r is the distance of the magnetic field from the bolt
[tex]B = \frac{\mu_o I}{2\pi r} \\\\B = \frac{4\pi *10^{-7} 10000}{2\pi *24} \\\\B = 8.333 *10^{-5} \ T[/tex]
Therefore, the magnitude of the magnetic field is 8.333 x 10⁻⁷ T
If a system has 4.50×102 kcal of work done to it, and releases 5.00×102 kJ of heat into its surroundings, what is the change in internal energy (ΔE or Δ????) of the system?
The change in internal energy (ΔE) of the system is equal to -18823 Kilojoules.
Given the following data:
Quantity of heat = [tex]5.00 \times 10^2 \;kJ[/tex]Work done = [tex]4.50 \times 10^2 \;kcal[/tex]Conversion:
1 kcal = 4.184 kJ
[tex]4.50 \times 10^2 \;kcal[/tex] = [tex]4.50 \times 10^2 \times 4.184 = 18828 \; kJ[/tex]
To determine the change in internal energy (ΔE) of the system, we would apply the first law of thermodynamics.
Mathematically, the first law of thermodynamics is given by the formula:
[tex]\Delta E = Q - W[/tex]
Where;
[tex]\Delta E[/tex] is the change in internal energy.Q is the quantity of heat released.W is the work done.Substituting the given parameters into the formula, we have;
[tex]\Delta E = 5 - 18828\\\\\Delta E = -18823[/tex]
Change in internal energy, E = -18823 Kilojoules
Read more: https://brainly.com/question/20599052
In a shipping yard, a crane operator attaches a cable to a 1,390 kg shipping container and then uses the crane to lift the container vertically at a constant velocity for a distance of 33 m. Determine the amount of work done (in J) by each of the following.
a) the tension in the cable.
b) the force of gravity.
Answer:
a) A = 449526 J, b) 449526 J
Explanation:
In this exercise they do not ask for the work of different elements.
Note that as the box rises at constant speed, the sum of forces is chorus, therefore
T-W = 0
T = W
T = m g
T = 1,390 9.8
T = 13622 N
Now that we have the strength we can use the definition of work
W = F .d
W = f d cos tea
a) In this case the tension is vertical and the movement is vertical, so the tension and displacement are parallel
A = A x
A = 13622 33
A = 449526 J
b) The work of the force of gravity, as the force acts in the opposite direction, the angle tea = 180
W = T x cos 180
W = - 13622 33
W = - 449526 J
Consider the three dip1acement vectors A = (3i - 3j) m, B = (i-4j) m, and C = (-2i + 5j) m. Use the component method to determine:
(a) the magnitude and direction of the vector D=A+B+C and
(b) the magnitude and direction of E=-A - B + C.
Answer:
(a) [tex]\vec D = 2\,i - 2\,j[/tex], (b) [tex]\vec E = -6\,i + 12\,j[/tex]
Explanation:
Let be [tex]\vec A = 3\,i - 3\,j\,[m][/tex], [tex]\vec B = i - 4\,j\,[m][/tex] and [tex]\vec C = -2\,i + 5\,j \,[m][/tex], each resultant is found by using the component method:
(a) [tex]\vec D = \vec A + \vec B + \vec C[/tex]
[tex]\vec D = (3\,i - 3\,j) + (i-4\,j) + (-2\,i+5\,j)\,[m][/tex]
[tex]\vec D = (3\,i + i -2\,i)+(-3\,j-4\,j+5\,j)\,[m][/tex]
[tex]\vec D = (3 + 1 -2)\,i + (-3-4+5)\,j\,[m][/tex]
[tex]\vec D = 2\,i - 2\,j[/tex]
(b) [tex]\vec E = -\vec A - \vec B + \vec C[/tex]
[tex]\vec E = -(3\,i-3\,j)-(i - 4\,j)+(-2\,i+5\,j)[/tex]
[tex]\vec E = (-3\,i + 3\,j) +(-i+4\,j) + (-2\,i + 5\,j)[/tex]
[tex]\vec E = (-3\,i-i-2\,i) + (3\,j+4\,j+5\,j)[/tex]
[tex]\vec E = (-3-1-2)\,i + (3+4+5)\,j[/tex]
[tex]\vec E = -6\,i + 12\,j[/tex]
Which of the following represents a concave mirror? +f,-f,-di,+di
Answer:
fully describes the concave mirror is + f
Explanation:
A concave mirror is a mirror where light rays are reflected reaching a point where the image is formed, therefore this mirror has a positive focal length, the amount that fully describes the concave mirror is + f
This allows defining a sign convention, for concave mirror + f, the distance to the object is + d0 and the distance to the image is + di
Answer:
+f
Explanation:
because you have to be really dumb to get an -f
The inner and outer surface temperature of a glass window 10 mm thick are 25 and 5 degree-C, respectively. What is the heat loss through a 1 m x 3 m window
Answer:
The heat loss is [tex]H = 8400\ W[/tex]
Explanation:
From the question we are told that
The thickness is [tex]t = 10 \ mm = 0.01 \ m[/tex]
The inner temperature is [tex]T_i = 25 ^oC[/tex]
The outer temperature is [tex]T_o = 5 ^oC[/tex]
The length of the window is L = 1 m
The width of the window is w = 3 m
Generally the heat loss is mathematically represented as
[tex]H = \frac{k * A * \Delta T}{t}[/tex]
Where k is the thermal conductivity of glass with value [tex]k = 1.4\ W/m \cdot K[/tex]
and A is the area of the window with value
[tex]A = 1 * 3[/tex]
[tex]A = 3 \ m^2[/tex]
substituting values
[tex]H = \frac{1.4 * 3 * (23-5)}{0.01}[/tex]
[tex]H = 8400\ W[/tex]