Answer:
1999.46 mm
45.59 s
Explanation:
given that
cylindrical water tank with diameter, D = 3 m
Height of the tank above the ground, h = 2 m
Depth of the water in the tank, d = 2 m
Diameter of hole, d = 0.420 cm
We start by calculating the volume of water in the tank, which is given as
Volume = πr²h
V = (πD²)/4 * h
V = (3.142 * 3²)/4 * 2
V = 28.278/4 * 2
V = 7.07 * 2
V = 14.14 m³
If 1.0 gal of water is equal to 0.0038m³, then
1 gal is 0.0038 = A * h
the area of the tank is 7.07 m²
therefore, 0.0038 = 7.07 * h
h₁ =0.00054 m = 0.54 mm is the height of water that flow out
the change in height of water in the tank = h - h₁ = 2 - 0.00054 = 1.99946 m
b)
Like we stated earlier, 1.0 gal of water is 0.0038m³
to solve this we use the formula
Q = Cd * A * √2gH
where Cd is a discharge coefficient, and is given by 0.9 for water
A is the area of the small hole
A = (πD²)/4
A = (π * 0.0042²)/4
A = 5.54*10^-5 / 4
A = 1.39*10^-5 m²
H= height of the hole from the tank water level = 2m - 0.0042 = 1.9958 m
g = 9.8 m/s²
Q = 0.9 * 1.39*10^-5 m² * √2 * 9.8 * 1.9958
Q = 1.251*10^-5 * 6.25
Q = 7.82*10^-5 m³/s
Q = V/t
t = V/Q = 0.0038m³ / 7.82*10^-5 m³/s
t = 45.59 s
13.4. Young's modulus for iron is 1.9 x 10' Pa. When an iron wire 1.0 m long with a cross-sectional area of 4.0 mm
supports a 100 kg load, the wire stretches by
(
Answer:
0.0129 m
Explanation:
ΔL = FL / (EA)
where ΔL is the deflection,
F is the force,
L is the initial length,
E is Young's modulus,
and A is the cross sectional area.
F = mg = 100 kg × 9.8 m/s² = 9800 N
A = 4.0 mm² × (1 m / 1000 mm)² = 4×10⁻⁶ m²
ΔL = (9800 N) (1.0 m) / ((1.9×10¹¹ Pa) (4×10⁻⁶ m²))
ΔL = 0.0129 m
A parallel plate capacitor consists of two square parallel plates separated by a distance d. the potential across the plates, while keeping everything else constant, what happens to the energy stored in the capacitor?
A) There will be % of the energy stored
B) There will be % of the energy stored
C) The energy stored will remain constant
D) The energy stored will double
E) the energy stored will quadruple
Answer:
E) the energy stored will quadrupled
Explanation:
The correct question is
A parallel plate capacitor consists of two square parallel plates separated by a distance d. If i double the potential across the plates, while keeping everything else constant, what happens to the energy stored in the capacitor?
A) There will be 1/4 of the energy stored
B) There will be 1/2 of the energy stored
C) The energy stored will remain constant
D) The energy stored will double
E) the energy stored will quadruple
The initial energy stored in the capacitor [tex]E_{i}[/tex] = [tex]\frac{1}{2}CV^{2}[/tex]
where C is the capacitance
V is the potential difference
If I double this voltage, while holding every other parameters constant, the new energy stored will be
[tex]E_{n}[/tex] = [tex]\frac{1}{2}C(2V)^{2}[/tex] = [tex]\frac{4}{2}CV^{2}[/tex]
[tex]E_{n}[/tex] = [tex]2CV^{2}[/tex]
dividing new energy stored by the initial energy stored, we have
[tex]E_{n}/E_{i}[/tex] = [tex]2CV^{2}[/tex] ÷ [tex]\frac{1}{2}CV^{2}[/tex] = 4
the energy stored will be quadrupled.
A 24.0-kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 225 N. For the first 11.0 m the floor is frictionless, and for the next 10.0 m the coefficient of friction is 0.20. What is the final speed of the crate after being pulled these 21.0 m?
Answer:
18.83m/s
Explanation:
In the first 11meters, we can calculate the Kinectic energy since we know that
the floor is frictionless then work done by the horizontal force= Kinectic energy gain by the block in the first 11m.
Then Kinetic Energy = wordone
But work done= Force × distance
Kinetic Energy=(225×11)= 2475J
In the next 10.0 m the coefficient of friction is 0.20, the Kinectic energy has equal value to the difference in workdone by both the horizontal and frictional force
K.E= (Force× distance) - ( mass × gravity× coefficient of friction × distance at 10 m)
K.E= [(225×10)-(.20× 9.8×10× 24)]
K.E= 2250-470.4
= 1779.6J
The addition of the Kinectic energy above give us the total Kinectic energy experience by the crate.
Total Kinectic energy= 1779.6+2475
= 4254.6J
the final speed of the crate after being pulled these 21.0 m?
Total distance= 11m + 10m = 21 m
Then the final speed can be calculated from the total Kinectic energy, since we know that
K.E= 0.5mv^2
V= √(2K.E/m)
= √(2×4254.6)/24
Final speed v = √354.55
Final speed v= 18.83m/s
Therefore, the final speed of the crate after being pulled these 21.0 m is 18.83m/s
A fire engine approaches a wall at 5 m/s while the siren emits a tone of 500 Hz frequency. At the time, the speed of sound in air is 340 m/s. How many beats per second do the people on the fire engine hear
Answer:
The values is [tex]f_b =14.9 \ beats/s[/tex]
Explanation:
From the question we are told that
The speed of the fire engine is [tex]v = 5\ m/s[/tex]
The frequency of the tone is [tex]f = 500 \ Hz[/tex]
The speed of sound in air is [tex]v_s = 340 \ m/s[/tex]
The beat frequency is mathematically represented as
[tex]f_b = f_a - f[/tex]
Where [tex]f_a[/tex] is the frequency of sound heard by the people in the fire engine and is is mathematically evaluated as
[tex]f_a = [\frac{v_s + v }{v_s -v} ]* f[/tex]
substituting values
[tex]f_a = [\frac{340 + 5 }{340 -5} ]* 500[/tex]
[tex]f_a = 514.9 \ Hz[/tex]
Thus
[tex]f_b =514.9 - 500[/tex]
[tex]f_b =14.9 \ beats/s[/tex]
A hydraulic press has one piston of diameter 4.0 cm and the other piston of diameter 8.0 cm. What force must be applied to the smaller piston to obtain a force of 1600 N at the larger piston
Answer:
The force is [tex]F_1 = 400.8 \ N[/tex]
Explanation:
From the question we are told that
The first diameter is [tex]d_1 = 4.0 \ cm = 0.04 \ m[/tex]
The second diameter is [tex]d_2 = 8.0 \ cm = 0.08 \ m[/tex]
Generally the first area is
[tex]A_1 = \pi * \frac{d^2_1 }{4}[/tex]
=> [tex]A_1 = 3.142 * \frac{0.04^2}{4}[/tex]
=> [tex]A_1 = 0.00126 \ m^2[/tex]
The second area is
[tex]A_2 = \pi * \frac{d^2_2 }{4}[/tex]
[tex]A_2 = 3.142 * \frac{0.08^2}{4}[/tex]
[tex]A_2 = 0.00503 \ m^2[/tex]
For a hydraulic press the pressure at both end must be equal .
Generally pressure is mathematically represented as
[tex]P = \frac{F}{A}[/tex]
=>
[tex]\frac{F_1}{A_1 } = \frac{F_2}{A_2 }[/tex]
=> [tex]F_1 = \frac{1600}{0.00503} * 0.00126[/tex]
=> [tex]F_1 = 400.8 \ N[/tex]
A sprinter is running a 100m sprint race. The race begins, and stopwatch is started. The sprinter passes the 12m [N] mark at 1.92 s and passes the 59m [N] mark at 7.98s, What was the sprinter's average velocity between the two time marks? Show all your work.
Answer:
velocity = 7.7558 m/s
Explanation:
s = u × t
s - distance u - velocity t - time
59 - 12 = (7.98 - 1.92) *u
47 = ( 6.06) u
u = 7.7558 m/s
When a sprinter is running a 100m sprint race. The race begins, and the stopwatch is started. The sprinter passes the 12m [N] mark at 1.92 s and passes the 59m [N] mark at 7.98s, then the sprinter's average velocity between the two-time marks would be
What is Velocity?The total displacement covered by any object per unit of time is known as velocity. The velocity of an object depends on the magnitude as well as the direction of the object.
the mathematical expression for velocity is given by
velocity = total displacement /time
As given in the problem a sprinter is running a 100m sprint race. The race begins, and the stopwatch is started. The sprinter passes the 12m [N] mark at 1.92 s and passes the 59m [N] mark at 7.98s
The total displacement covered by the sprinter
S = 59-12
S= 47 m
The total time is taken by the sprinter
T= 7.98 -1.92
= 6.06 seconds
Velocity = S/T
=47/6.06
=7.75 m/s
Thus. the average velocity of the sprinter would be 7.75 m/s
Learn more about Velocity from here
brainly.com/question/18084516
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Calculate the force that must be applied to an object weighing 25 pounds for it to be in equilibrium, if it is on a plane inclined at 45 °
Answer:
17.7 lb
Explanation:
Assuming the force is parallel to the incline, draw a free body diagram of the object. There are 3 forces:
Normal force N pushing perpendicular to the incline,
Weight force mg pulling down,
and applied force F pushing parallel to the incline.
Sum the forces in the parallel direction:
∑F = ma
F − mg sin 45° = 0
F = mg sin 45°
F = (25 lb) sin 45°
F = 17.7 lb
a vehicle moving with a uniform a acceleration of 2m/s and has a velocity of 4m/s at a certain time .
what is will its velocity be
a: 1s later
b: 5s later.
Answer:
1 second later the vehicle's velocity will be:
[tex]v(1)= 6\,\,\frac{m}{s} \\[/tex]
5 seconds later the vehicle's velocity will be:
[tex]v(5)=14\,\,\frac{m}{s}[/tex]
Explanation:
Recall the formula for the velocity of an object under constant accelerated motion (with acceleration "[tex]a[/tex]"):
[tex]v(t)=v_0+a\,t[/tex]
Therefore, in this case [tex]v_0=4\,\,\frac{m}{s}[/tex] and [tex]a=2\,\,\frac{m}{s^2}[/tex]
so we can estimate the velocity of the vehicle at different times just by replacing the requested "t" in the expression:
[tex]v(t)=v_0+a\,t\\v(t)=4+2\,\,t\\v(1)=4+2\,(1) = 6\,\,\frac{m}{s} \\v(5)=4+2\,(5)=14\,\,\frac{m}{s}[/tex]
What is the mass of erath
What is the root mean square speed of an As4 particle as it is sublimed? (Assume at high temperatures arsenic acts like an ideal gas; the Boltzmann constant, kB, can be approximated as 1.4 x 10-23 J∙K-1) and the kinetic energy of the gas is equal to 3/2kB T
Answer: [tex]v_{rms} =[/tex] 273m/s
Explanation: Root Mean Square Speed of an atom or molecule is the speed of a particle in a gas. It is the average speed a particle in a gas can have.
It can be calculated:
[tex]\frac{1}{2}mv^{2} = \frac{3}{2} k_{B}T[/tex]
[tex]v^{2} = \frac{3k_{B}T}{m}[/tex]
[tex]v = \sqrt{\frac{3k_{B}T}{m}}[/tex]
m is mass of one atom or molecule in kg.
An atom of Arsenic sublimes at 614°C. Converting to Kelvin:
T = 614 + 273 = 887K
Molecular mass of As4 is approximately 0.3kg.
[tex]m = \frac{0.3}{6.10^{23}}[/tex]
[tex]m=5.10^{-25}[/tex]kg
Calculating Root mean square speed :
[tex]v = \sqrt{\frac{3*1.4.10^{-23}*887}{5.10^{-25}}[/tex]
[tex]v = \sqrt{745.08.10^{2}[/tex]
v = 273m/s
The root mean square speed of As4 is approximately 273m/s.
A block that weighs 200 N is initially at rest on a horizontal surface where the coefficient of static friction is 0.500 and the coefficient of kinetic friction is not yet known.
A) What is the gravitational force on this block?
B) What is the mass of the block?
C) What is the normal force on this block?
D) What is the smallest force needed to start moving the block?
E) If you were pushing the block with the force you found in C, what force would the block exert on you?
F) If you were pushing the block with the force you found in C, and the block began to accelerate at 1.96 meters per second squared, what is the net force on the block?
G) If the block accelerates at 1.96 meters per second squared, what is the coefficient of kinetic friction?
Answer:
A) W= 200N, B) m = 20.4 kg , C) N = 200 N , D) F = 100 N ,
E) F_reaction = 200 N , F) F _net = 40 N
Explanation:
This exercise can be solved using Newton's second law
A) The gravitational force is the weight of the block
Fg = W = m g
W = 200 N
B) m = W / g
m = 200 / 9.8
m = 20.4 kg
C) Normal is the reaction of the floor to the weight of the block
N-W = 0
N = W
N = 200 N
D) we write Newton's second law on the x-axis
F - fr = 0
F = fr
the friction force equation is
fr = μ_s N
fr = μ_s W
subtitute
F = 0.50 200
F = 100 N
E) As the forces in the natural are in pairs, by Newton's third law or law of action and reaction, the block responds with a force of equal magnitude, but opposite direction
F_reaction = 200 N
F and G) We write Newton's second law
F - fr = m a
fr = N = μ_k mg
F - μ_k m g = m a
μ_k = (F - ma) / mg
μ_k = (200 - 20.4 1.96) / 200
μ_k = 0.8
In general, the coefficient of kinetic friction is lower than the static one
the net force is
F_net = F -fr = F - μ_k W
F_net = 200 - 0.8 200
F _net = 40 N
What is the opposite of 4/4
Answer: -1 or -4/4
Explanation:
4/4 simplifies to 1 and the opposite of 1 is -1.
Average speed is calculated by dividing distance traveled by
time. How is average velocity calculated?
Answer:
The average velocity of an object is its total displacement divided by the total time taken. In other words, it is the rate at which an object changes its position from one place to another. Average velocity is a vector quantity. The SI unit is meters per second.
hope this helps you
Answer:
displacement divided by time
Explanation:
2.18) The calorie is a unit of energy defined as the amount of energy needed to raise 1 g of water by 1oC. a) How many calories are required to bring a pot of water at 1oC to a boil
Answer:
Q = 80000 cal
Explanation:
La expresin para el calor es
Q = m [tex]c_{e}[/tex]ce ΔT
el calor especifico del agua es ce= 1 cal/gr ºC
la temperatura de ebullición del agua es Ef = 100ºC y partimos de la temperatura ambiente To= 20 OC
Q = m 1 ( 100 – 20)
q= 80 m
en el ejercicio no se da la masa de agua, pro podemos suponer, pote lleva 1 litros = 100 G
Q = 80 1000
Q = 80000 cal
The rate at which energy is emitted from an accelerating charge that has charge q and acceleration a is given by dE/dt= q^2 a^2/ 6πεc^3, where c is the speed of light.
Required:
a. If a proton with a kinetic energy of 5.0 MeV is travelling in a particle accelerator in a circular orbit with a radius of 0.540 m, what fraction of its energy does it radiate per second?
b. Consider an electron orbiting with the same speed and radius. What fraction of its energy does it radiate per second?
Answer:
The value of fraction of energy is [tex]2.23\times10^{-11}[/tex]
The value of fraction of energy is [tex]7.53\times10^{-5}[/tex]
Explanation:
Given that,
Charge = q
Acceleration = a
The rate at which energy is emitted from an accelerating charge
[tex]\dfrac{dE}{dt}=\dfrac{q^2a^2}{6\pi\epsilon c^3}[/tex]....(I)
We know that,
Acceleration for circular motion is
[tex]a=\dfrac{v^2}{r}[/tex]....(II)
The kinetic energy is
[tex]K.E=\dfrac{1}{2}mv^2[/tex]
[tex]v^2=\dfrac{2(K.E)}{m}[/tex]
Put the value of v in equation (II)
[tex]a=\dfrac{2(K.E)}{mr}[/tex]
Put the value of a in equation (I)
[tex]\dfrac{dE}{dt}=\dfrac{q^2(\dfrac{2(K.E)}{mr})^2}{6\pi\epsilon c^3}[/tex]
[tex]\dfrac{dE}{dt}=\dfrac{q^24(K.E)^2}{6\pi\epsilon c^3\times m^2 r^2}[/tex]
[tex]\dfrac{\dfrac{dE}{dt}}{K.E}=\dfrac{q^24(K.E)}{6\pi\epsilon c^3\times m^2 r^2}[/tex]
Suppose that,
[tex]\dfrac{\dfrac{dE}{dt}}{K.E}=R[/tex]
So,
[tex]R=\dfrac{q^2\times4(K.E)}{6\pi\epsilon c^3\times m^2 r^2}[/tex]....(III)
(a). For proton,
We need to calculate the fraction of its energy does it radiate per second
Using equation (III)
[tex]R=\dfrac{4\times(1.6\times10^{-19})^2\times5.0\times1.6\times10^{-19}\times10^{6}\times6\times10^{9}}{(3\times10^{8})^3\times(1.67\times10^{-27})^2\times(0.540)^2}[/tex]
[tex]R=2.23\times10^{-11}[/tex]
(b). For electron,
We need to calculate the fraction of its energy does it radiate per second
Using equation (III)
[tex]R=\dfrac{4\times(1.6\times10^{-19})^2\times5.0\times1.6\times10^{-19}\times10^{6}\times6\times10^{9}}{(3\times10^{8})^3\times(9.1\times10^{-31})^2\times(0.540)^2}[/tex]
[tex]R=0.0000753[/tex]
Hence, The value of fraction of energy is [tex]2.23\times10^{-11}[/tex]
The value of fraction of energy is [tex]7.53\times10^{-5}[/tex]
Using the Bohr model, find the ionization energy of the ground He ion.
Answer:
54.4ev
Explanation:
Using
E = - me⁴Z²/8Eo²h²n²
Where n= 1
Z= 1
h=6.6E-31Js
me= 9.1x 10-31kg
Eo=8.85E-12m-3 kg-1 s4 A2
So by substituting
E= -(9.1x 10-31kg)⁴x 1²/8(8.85E-12m-3 kg-1 s4 A2)² x (6.6E-31Js)² x 1²
So E= 54.4ev
Show that (a)KE=1/2mv2
Answer:
[tex] \boxed{ \bold{ \huge{ \boxed{ \sf{see \: below}}}}}[/tex]
Explanation:
[tex] \underline{ \bold{ \sf{To \: prove \: that \: kinetic \: energy = \frac{1}{2} m {v}^{2} }}}[/tex]
Let us consider, a body of mass ' m ' is lying at rest ( initial velocity = 0 ) on a smooth surface. Let a constant force F displaces this body in its own direction by a displacement ' d '. Let 'v' be it's final velocity. The work done ' W ' by the force is given by :
[tex] \sf{W = FD}[/tex]
⇒[tex] \sf{W = m \: \times a \: \times s} \: \: \: \: \: \: \: \: \: ( \: ∴ \: f \: = \: ma \: ; \: s \: = d)[/tex]
⇒[tex] \sf{W = m \: \times \frac{v - u}{t} \times \frac{u + v}{2} \times t \: \: \: \: \: \: \: \: \: (∴ \: a = \frac{v - u}{t} and \: s = \frac{u + v}{2} \times t}[/tex]
⇒[tex] \sf{W = m \times \frac{ {v}^{2} - {u}^{2} }{2} }[/tex]
⇒[tex] \sf{W = \frac{1}{2} m {v}^{2} \: \: \: \: \: \: \: \: \: \: \: \: (since, \: initial \: velocity(u) = 0)}[/tex]
The work done becomes the kinetic energy of the body. Thus, the kinetic energy of a body of mass ' m : moving with the velocity equal to 'v ' is 1 / 2 mv²
∴ [tex] \sf{KE= \frac{1}{2} m {v}^{2} }[/tex]
[tex] \sf{ \underline{ \bold{ {proved}}}}[/tex]
Hope I helped!
Best regards!!
During which stage of the water cycle does water from the ocean form clouds?
Answer:
Condensation
Explanation:
Condensation is the process by which water vapor in the air is changed into liquid water. Condensation is crucial to the water cycle because it is responsible for the formation of clouds.
Answer:
This process is called condensation.
Explanation:
When a cloud becomes full of liquid water, it falls from the sky as rain or snow.
masses of 3kg on a smooth horizontal table.It is connected by a light string passing at the edge of the table to another mass of 2kg hanging vertically.When to the system is released from rest with what acceleration do the mass move
Answer:
aaawwwwwwwsssaaaasasss
Leticia timed how fast five apple slices turned brown (oxidate) after being being dipped in different preservatives such as lemon juice, fruit freshener, salt water and lime soda. another part of the experiment had the apple slices simply set out without any chemical on them. all parts of the experiment had the apple slices in the same indoor conditions such as humidity temperature and lighting; also only one variety of apple-red delicious was used.
Identify each of the following independent variable, dependent variable, constants, control experiment and repeated trials.
Answer:
Independent Variable: Choice of preservative
Dependent Variable: Time it took to brown
Controlled Variables/Constants: Climate, Size of apple slices, and amount of preservative on each slice
Control: Apple Slices without preservatives on them
Repeated Trials: One?
Explanation:
PLEASE SEND GIVE THE ANSWER AS FAST AS POSSIBLE
Q1) State similarities & difference between the laboratory thermometers & the clinical thermometers
Q2) Give two examples for conductors & insulators.
Q3) Give reason: Wearing more layers of clothing during winter keeps
us warm
Answer:
(1): Similarities Both thermometers are used to measure temperature and both of them use mercury,Differences Clinical thermometer is used to measure human body temperature whereas laboratory thermometer is used to measure temperature of other object which has higher temperature than human body temperature
(2)Examples of conductors include metals, aqueous solutions of salts, Examples of insulators include plastics, Styrofoam, paper, rubber, glass and dry air.
(3)Air acts as insulator of heat. This layer prevents our body heat to escape in the surroundings. More layers of thin clothes will allow more air to get trapped and as a result we will not feel cold. So wearing more layers of clothing during winter keeps us warmer than wearing just one thick piece of clothing
Explanation:
Describe two homeostasis mechanisms
Answer:
The interpretation including its given subject is listed in the subsection below on explanatory.
Explanation:
Homeostasis should be any mechanism of self-regulation from which an individual seeks to preserve equilibrium when adapting to requirements better suited towards its existence.
Two mechanisms are given below:
Blood Pressure: Pressure is regulated by either a homeostatic system through which the blood becomes circulated throughout the individual's body.Body Temperature: The body requires multiple forms of heat regulatory oversight such as:Endothermic is something a living creature can hold one's own internal body temperature. Ectothermic is really where the external atmosphere absorbs temperature.A small, dense ball is launched from ground level at an angle of 50° above the horizontal. The ballâs initial speed is 22 m/s, and it lands on a hard, level surface at the same height from which it was launched. The ball then bounces and reaches a height of 75% its peak height it achieved at launch. Assume air resistance is negligible.(a) Find the maximum height reached by the ball during its first parabolic arc. (b) Find the distance between the launch point to where the ball lands the first time. (c) Find the distance between the launch point to where the ball lands the second time?
Answer:
a) Hmax = 10.86 m
b) Rx = 48.64 m
c) Rx¹ = 36.48m
Explanation:
Given that Ф = 50°
v₀ = 22 m/s
a)
hmax = v₀²sin²Ф / 2g
hmax = 22² × sin²50 / 2 × 9.8
hmax = 14.49 m
Hmax = 75% of hmax
Hmax = 0.75 × 14.49
Hmax = 10.86 m
the maximum height reached by the ball during its first parabolic arc is 10.86 m
b)
Rx = v₀²sin2Ф / g
Rx = 22² × sin 100 / 9.8
Rx = 48.64 m
the distance between the launch point to where the ball lands the first time is 48.64 m
c)
Rx¹ = 75% 0f Rx
Rx¹ = 0.75 × 48.64
Rx¹ = 36.48m
the distance between the launch point to where the ball lands the second time is 36.48m
The piston of a hydraulic elevator used for lifting trucks has a 0.3m radius. What pressure is required to lift a truck of 2500 kg mass? What force was applied to the small piston if it has a radius of 3cm?
Answer:
1. 88370.45 N/[tex]m^{2}[/tex]
2. 2500 N.
Explanation:
Pressure, P, is define as the force, F, per unit area, A, applied on/ by an object.
i.e P = [tex]\frac{F}{A}[/tex]
1. The area, A, of the piston of the hydraulic elevator can be determined by;
A = [tex]\pi[/tex][tex]r^{2}[/tex]
where r is the radius of the piston.
A = [tex]\frac{22}{7}[/tex] × [tex](0.3)^{2}[/tex]
= [tex]\frac{22}{7}[/tex] × 0.09
= 0.2829 [tex]m^{2}[/tex]
Pressure required to lift a truck of 2500 kg mass can be determined as;
P = [tex]\frac{F}{A}[/tex]
F = W = mg
= 2500 × 10
= 25 000 N
So that,
P = [tex]\frac{25000}{0.2829}[/tex]
= 88370.45 N/[tex]m^{2}[/tex]
The pressure required is 88370.45 N/[tex]m^{2}[/tex].
2. [tex]\frac{F_{1} }{A_{1} }[/tex] = [tex]\frac{F_{2} }{A_{2} }[/tex]
Area of small piston = [tex]\pi[/tex][tex]r^{2}[/tex]
= [tex]\frac{22}{7}[/tex] × [tex](0.03)^{2}[/tex]
= 0.02829 [tex]m^{2}[/tex]
So that,
[tex]\frac{25000}{0.2829}[/tex] = [tex]\frac{F_{2} }{0.02829}[/tex]
[tex]F_{2}[/tex] = 2500 N
The force applied to the small piston is 2500 N.
Calculate the following expression: 2.36 + 3.38 + 0.355 + 1.06 =
the answer wil be 2.36+3.38+0.355+1.06=7.155
Two sources emit beams of light of wavelength 550 nm. The light from source A has an intensity of 10 μW/m2, and the light from source B has an intensity of 20 μW/m2. This is all we know about the two beams. Which of the following statements about these beams are correct? A) Beam B carries twice as many photons per second as beam A. B) A photon in beam B has twice the energy of a photon in beam A. C) The frequency of the light in beam B is twice as great as the frequency of the light in beam A. D) A photon in beam B has the same energy as a photon in beam A. E) None of the above statements are true.
Answer:
A) Beam B carries twice as many photons per second as beam A.
Explanation:
If we have two waves with the same wavelength, then their intensity is proportional to their power, or the energy per unit time.
We also know that the amount of photon present in an electromagnetic beam is proportional to the energy of the beam, hence the amount of beam per second is proportional to the power.
With these two facts, we can say that the intensity is a measure of the amount of photon per second in an electromagnetic beam. So we can say that beam B carries twice as more power than beam A, or Beam B carries twice as many photons per second as beam A.
A 150 W driveway light produces 9.3 × 104 J of light energy over a 4.0 h period. If the light uses 2.06 × 106 J of energy in the 4.0 h, then what is its efficiency? Show all your work.
Answer:
4.5%
Explanation:
efficiency = energy out / energy in
e = 9.3×10⁴ J / 2.06×10⁶ J
e = 0.045
Estimate the distance (in cm) between the central bright region and the third dark fringe on a screen 5.00 m from two double slits 0.500 mm apart illuminated by 500-nm light.
Answer:
1.25cm
Explanation:
Using
Minimum, as dsinစ = (m+1/2) lambda
Third dark fringe m= 2
dsinစ = (2+1/2)lambda
d(y/L)= (5/2) lambda
Y= 5/2* lambda *L/d
So substituting
=[ (500E-9m)(5m)/0.5E-3] 5/2
=0.0125m
= 1.25cm
Explanation:
A teapot with a surface area of 700 cm2 is to be plated with silver. It is attached to the negative electrode of an electrolytic cell containing silver nitrate (Ag+ NO3-). If the cell is powered by a 12.0-V battery and has a resistance of 1.30 , how long does it take for a 0.133-mm layer of silver to build up on the teapot? (The density of silver is 10.5 multiplied by 103 kg/m3.)
Open the Faraday Law simulation and discover what you can about induction. Make a list of ways to cause induction.
Answer:
Current can be induced in a lot of ways. I'll try and list as much of these methods as I can here.
1. Moving a bar magnet relative to a wire coil
2. Moving a wire coil relative to a magnet.
3. Move a wire coil that has electricity flowing though it relative to a wire coil without electricity flowing through it.
4. Moving a current carrying circuit relative to a non-current carrying circuit
5. Rapidly opening and closing the switch of a current carrying circuit beside a non current carrying circuit.
6. Moving a bar magnet through the middle of a wire coil.
7. Moving a wire coil through a magnet
8. Moving a circuit relative to a magnetic field.
The main idea is to cause a change in the magnetic field, or to change the available area of the wire loop, or to change the angle between the field and the loop.