What is the total surface charge qint on the interior surface of the conductor (i.e., on the wall of the cavity)
Answer: hello your question is incomplete below is the missing part
A spherical cavity is hollowed out of the interior of a neutral conducting sphere. At the center of the cavity is a point charge, of positive charge q.
answer:
- q
Explanation:
Since the spherical cavity was carved out of a neutral conducting sphere hence the electric field inside this conductor = zero
given that there is a point charge +q at the center of the spherical cavity hence for the electric field inside the conductor to be = zero the total surface charge qint on the wall of the cavity will be -q
suppose you have a block resting on a horizontal smooth surface. th block with a mass m is attached to a horizontal spring which is fixed at one end. the spring can be compressed and stretched. the mass is pulled to one side then released what is the formula required
The time period of the spring is 2[tex]\pi[/tex][√(m/k)].
What is meant by spring constant ?The spring constant of a spring is defined as the measurement of ratio of the force that is exerted on the spring to the displacement caused by it.
Here,
The mass of the block = m
Let F be the applied force on the spring and k be the spring constant.
When the mass attached to the spring is pulled to one side then released, it executes SHM.
Therefore we can write that, the applied force,
F = kx
Restoring force = -kx
According to Newton's law, we know that,
F = ma
So,
ma = -kx
Therefore, the acceleration,
a = (-k/m) x
For an SHM, the acceleration is given as,
a = -ω²x
Therefore, we can write that,
-ω²x = (-k/m) x
ω² = k/m
So, the time period of the spring,
T = 2[tex]\pi[/tex]/ω
T = 2[tex]\pi[/tex][√(m/k)]
Hence,
The time period of the spring is 2[tex]\pi[/tex][√(m/k)].
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Strategies for good health management involve:
A Avoiding stressful situations that may cause depression or moodiness insomnia, or lack motivation.
B) Denying, ignoring, or repressing feelings or problems, so that you don't have to face them.
Eating your favorite foods, imagining yourself working out (mind is power), sleeping a few hours a day, so as to make
the most of party time.
D Eating healthy, maintaining and ideal weight, resting, exercising, and establishing healthy relationships.
Answer:
D
Explanation:
This is a great way to manage health.
A would be avoiding everything which isnt good.
B. would be emotionally draining and damaging to bottle feelings and ignore them.
C. is unhealthy to not exercise and eat food while doing nothing.
The velocity of an object increases at a constant rate from 20 m/s to 50 m/s in 10 s.Find the acceleation
Answer:
[tex]{ \bf{v = u + at}} \\ 50 = 20 + (a \times 10) \\ 30 = 10a \\ { \tt{acceleration = 3 \: {ms}^{ - 2} }}[/tex]
Write one advantage of MKS system over CGS system.
How are Newton’s 1 and 2 law related?
At an airport, luggage is unloaded from a plane into the three cars of a luggage carrier, as the drawing shows. The acceleration of the carrier is 0.12 mls2, and friction is negligible. The coupling bars have negligible mass. By how much would the tension in each of the coupling bars A, B, and C change if 39 kg of luggage were removed from car 2 and placed in (a) car I and (b) car 3
Answer:
a) ΔT₁ = -4.68 N, ΔT₂ = 4.68 N, b) ΔT₂ = 4.68 N, ΔT₁ = 4.68 N
Explanation:
In this exercise we will use Newton's second law.
∑F = m a
Let's start with the set of three cars
F_total = M a
F_total = M 0.12
where the total mass is the sum of the mass of each charge
M = m₁ + m₂ + m₃
This is the force with which the three cars are pulled.
Now let's write this law for each vehicle
car 1
F_total - T₁ = m₁ a
T₁ = F_total - m₁ a
car 2
T₁ - T₂ = m₂ a
T₂ = T₁ - m₂ a
car 3
T₂ = m₃ a
note that tensions are forces of action and reaction
a) They tell us that 39 kg is removed from car 2 and placed on car 1
m₂’= m₂ - 39
m₁'= m₁ + 39
m₃ ’= m₃
they ask how much each tension varies, let's rewrite Newton's equations
The total force does not change since the mass of the set is the same F_total ’= F_total
car 1
F_total ’- T₁ ’= m₁’ a
T₁ ’= F_total - m₁’ a
T₁ ’= (F_total - m₁ a) - 39 a
T₁ '= T₁ - 39 0.12
ΔT₁ = -4.68 N
car 2
T₁’- T₂ ’= m₂’ a
T₂ ’= T₁’- m₂’ a
T₂ '= (T₁'- m₂ a) + 39 a
T₂ '= T₂ + 39 0.12
ΔT₂ = 4.68 N
b) in this case the masses remain
m₁ '= m₁
m₂ ’= m₂ - 39
m₃ ’= m₃ + 39
we write Newton's equations
car 3
T₂ '= m₃' a
T₂ ’= (m₃ + 39) a
T₂ '= m₃ a + 39 a
T₂ '= T₂ + 39 0.12
ΔT₂ = 4.68 N
car 1
F_total - T₁ ’= m₁’ a
T₁ ’= F_total - m₁ a
car 2
T₁' -T₂ '= m₂' a
T₁ ’= T₂’- m₂’ a
T₁ '= (T₂'- m₂ a) + 39 a
T₁ '= T₁ + 39 0.12
ΔT₁ = 4.68 N
The tension in each of the coupling bars A, B, and C of the luggage carrier changes as,
When luggage were removed from car 2 and placed in car 1, the tension is A and C does not change and the tension in B is decreased by 4.68 N.When luggage were removed from car 2 and placed in car 3, the tension is A and B does not change and the tension in C is increased by 4.68 N.What is tension force?Tension is the pulling force carried by the flexible mediums like ropes, cables and string.
Tension in a body due to the weight of the hanging body is the net force acting on the body.
At an airport, luggage is unloaded from a plane into the three cars of a luggage carrier, as the drawing shows. The acceleration of the carrier is 0.12 m/s², and friction is negligible.
The acceleration is the same, Tension due to the horizontal component of the forces for car 1, 2 and 3 can be given as,
[tex]\sum F_{1h}=T_A-T_B=m_1a\\\sum F_{2h}=T_B-T_C=m_2a\\\sum F_{3h}=T_C=m_3a[/tex]
On solving the above 3 equation, we get the values of tension in each bar as,
[tex]T_A=(m_1+m_2+m_3)a\\T_B=(m_3+m_2)a\\T_C=m_3a[/tex]
Case 1- When 39 kg of luggage were removed from car 2 and placed in car IThe tension is A and C does not change for this case. The acceleration of the carrier is 0.12 m/s². Thus, the change in tension is B is,
[tex]\Delta T_B=39\times0.12\\\Delta T_B=4.68\rm \;N[/tex]
Case 2- When 39 kg of luggage were removed from car 2 and placed in car IIIThe tension is A and B does not change for this case. The acceleration of the carrier is 0.12 m/s². Thus, the change in tension is B is,
[tex]\Delta T_C=39\times0.12\\\Delta T_C=4.68\rm \;N[/tex]
Hence, the tension in each of the coupling bars A, B, and C of the luggage carrier changes as,
When luggage were removed from car 2 and placed in car 1, the tension is A and C does not change and the tension in B is decreased by 4.68 N.When luggage were removed from car 2 and placed in car 3, the tension is A and B does not change and the tension in C is increased by 4.68 N.
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A child is outside his home playing with a metal hoop and stick. He uses the stick to keep the hoop of radius 45.0 cm rotating along the road surface. At one point the hoop coasts downhill and picks up speed. (a) If the hoop starts from rest at the top of the hill and reaches a linear speed of 6.35 m/s in 11.0 s, what is the angular acceleration, in rad/s2, of the hoop? rad/s2 (b) If the radius of the hoop were smaller, how would this affect the angular acceleration of the hoop? i. The angular acceleration would decrease. ii. The angular acceleration would increase. iii. There would be no change to the angular acceleration.
Answer:
a) [tex] \alpha = 1.28 rad/s^{2} [/tex]
b) Option ii. The angular acceleration would increase
Explanation:
a) The angular acceleration is given by:
[tex] \omega_{f} = \omega_{0} + \alpha t [/tex]
Where:
[tex] \omega_{f} [/tex]: is the final angular speed = v/r
v: is the tangential speed = 6.35 m/s
r: is the radius = 45.0 cm = 0.45 m
[tex]\omega_{0}[/tex]: is the initial angular speed = 0 (the hoop starts from rest)
t: is the time = 11.0 s
α: is the angular acceleration
Hence, the angular acceleration is:
[tex] \alpha = \frac{\omega}{t} = \frac{v}{r*t} = \frac{6.35 m/s}{0.45 m*11.0 s} = 1.28 rad/s^{2} [/tex]
b) If the radius were smaller, the angular acceleration would increase since we can see in the equation that the radius is in the denominator ([tex] \alpha = \frac{v}{r*t} [/tex]).
Therefore, the correct option is ii. The angular acceleration would increase.
I hope it helps you!
A resident of a lunar colony needs to have her blood pressure checked in one of her legs. Assume that we express the systemic blood pressure as we do on earth and that the density of blood does not change. Suppose also that normal blood pressure on the moon is still 120/80 (which may not actually be true).
Required:
If a lunar colonizer has her blood pressure taken at a point on her ankle that is 1.5 m below her heart, what will be her systemic blood-pressure reading, expressed in the standard way, if she has normal blood pressure? The acceleration due to gravity on the moon is 1.67 m/s^2
Answer:
The pressure is 2505 Pa.
Explanation:
Height, h = 1.5 m
density of blood, d = 1000 kg/cubic meter
Gravity, g = 1.67 m/s^2
let the pressure is P.
The pressure due to the fluid is given by
P = h d g
P = 1.5 x 1000 x 1.67
P = 2505 Pa
PLEASE HELP ME WITH THIS ONE QUESTION
The half-life of Barium-139 is 4.96 x 10^3 seconds. A sample contains 3.21 x 10^17 nuclei. What is the decay constant for this decay?
Explanation:
hope this will help u
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A car of mass 500 kg increases its velocity from 40 metre per second to 60 metre per second in 10 second find the distance travelled and amount of force applied
Answer:
it is answer of u are question
Mary and her younger brother Alex decide to ride the carousel at the State Fair. Mary sits on one of the horses in the outer section at a distance of 2.0 m from the center. Alex decides to play it safe and chooses to sit in the inner section at a distance of 1.1 m from the center. The carousel takes 5.8 s to make each complete revolution.
Required:
a. What is Mary's angular speed %u03C9M and tangential speed vM?
b. What is Alex's angular speed %u03C9A and tangential speed vA?
Answer:
you can measure by scale beacause we dont no sorry i cant help u but u can ask me some other Q
What will be the potential difference measured by an ideal voltmeter in the circuit of the figure?
Answer:
The voltage across 150 ohm resistor is 6 volts.
Explanation:
Given that,
Resistors having resistances 150 ohms and 300 ohms are in series. Their equivalent is :
R = 150 + 300
R = 450 ohms
Let I is the current in the circuit. Using Ohm's law,
V = IR
[tex]I=\dfrac{V}{R}\\\\I=\dfrac{18}{450}\\\\I=0.04\ A[/tex]
The current in series remains the same while potential divides. So,
[tex]V_1=IR_1\\\\V_1=0.04\times 150\\\\=6\ V[/tex]
So, the voltage across 150 ohm resistor is 6 volts.
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
~~~~NEED HELP ASAP~~~~
Block A slides into block B along a frictionless surface. They are moving in the direction from left o the right.
Block A= 3kg
Block B= 4kg
Block A velocity before collision =30m/s.
Block B velocity before collision = 15 m/s
The velocity of block B after the collision is 20m/s.
a.) What is the velocity of block A after collision?
b.) Is the collision elastic? Show work to explain answer why or why not.
Answer:
Block A velocity is 23.33 m/s and the collission is not elastic.
Explanation:
a) m1v1 + m2v2 = m1v1' + m2v2'
Plug in givens
90+60=3v1'+80
solve for v1'= 23.33m/s
b) Find the initial and final kinetic energy of Block B
Ki= 1/2(4)(15)^2 + 1/2(3)(30)^2 = 1800 J
Kf= 1/2(4)(20)^2 + 1/2(3)*(23.33)^2= 1616.433J
Since Ki does not equal Kf the collision is not elastic
When Peter tosses an egg against a sagging sheet, the egg doesn't break due to
A) reduced impulse.
B) reduced momentum.
C) both of these
D) neither of these
It has to do with impulse or force. Just how the sheet has no volume. There is no sufficient impulse to crack the shell.
What is force?A force is an effect that can alter an object's motion according to physics. An object with mass can change its velocity, or accelerate, as a result of a force. An obvious way to describe force is as a push or a pull. A force is a vector quantity since it has both magnitude and direction.
The sagging sheet gives the impact with the egg additional time, which prevents the egg from breaking when it is hurled against it. This lessens the force the egg would have applied to the wall had it been flung at it.
It has to do with impulse or force. Just how the sheet has no volume. There is no sufficient impulse to crack the shell.
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What is the pH of a solution with a hydrogen ion concentration of 2.0x10^3.(Use 3 digits)
Answer:
2.70
Explanation:
pH = -log[H+]
pH = -log[2.0x10^-3]
pH = 2.70
A rope, under a tension of 221 N and fixed at both ends, oscillates in a second-harmonic standing wave pattern. The displacement of the rope is given by y = (0.10 m)(sin πx/2) sin 12πt, where x = 0 at one end of the rope, x is in meters, and t is in seconds.
What are:
a. the length of the rope.
b. the speed of the waves on the rope
c. the mass of the rope
d. If the rope oscillates in a third-harmonic standing wave pattern, what will be the period of oscillation.
Answer:
sup qwertyasdfghjk
Explanation:
Gradual shifting or movement of a time series to relatively higher or lower values over a longer period of time is called _____.
Answer:
Gradual shifting of a time series to relatively higher or lower values over a long period of time is called a Trend.
A transverse sine wave with an amplitude of 2.50 mm and a wavelength of 1.80 m travels, from left to right along a long, horizontal stretched string with a speed of 36.0 m s. I Take the origin at the left end of the undisturbed string. At time t = 0 the left end of the string has its maximum upward displacement,
(a) What is the frequency of the wave?
(b) What is the angular frequency of the wave?
(c) What is the wave number of the wave?
(d) What is the function y(x,t) that describes the wave?
(e) What is y(t) for a particle at the left end of the string?
(f) What is y(t) for a particle 1.35 m to the right of the origin?
(g) What is the maximum magnitude of transverse velocity of any particle of the string?
(h) Find the transverse displacement of a particle 1.35 m to the right of the origin at time t = 0.0625 s.
(i) Find the transverse velocity of a particle 1.35 m to the right of the origin at time t = 0.0625 s.
Explanation:
Given that,
Amplitude, A = 2.5 nm
Wavelength,[tex]\lambda=1.8\ m[/tex]
The speed of the wave, v = 36 m/s
At time t = 0 the left end of the string has its maximum upward displacement.
(a) Let f is the frequency. So,
[tex]f=\dfrac{v}{\lambda}\\\\f=\dfrac{36}{1.8}\\\\f=20\ Hz[/tex]
(b) Angular frequency of the wave,
[tex]\omega=2\pi f\\\\=2\pi \times 20\\\\=125.7\ rad/s[/tex]
(c) The wave number of the wave[tex]=\dfrac{1}{\lambda}[/tex]
[tex]=\dfrac{1}{1.8}\\\\=0.56\ m^{-1}[/tex]
A wheel has a diameter of 10m and weight 360N what minimum horizontal force is necessary to pull the wheel over a brick 0.1m when a force is applied at the wheel
A baseball of mass 0.145 kg is thrown at a speed of 40.0 m/s. The batter strikes the ball with a force of 15,000 N; the bat and ball are in contact for 0.500 ms. The force is exactly opposite to the original direction of the ball. Determine the final speed of the ball.
The final speed of the ball is 91.72 m/s.
Given the following data:
Mass of baseball = 0.145 kgInitial speed = 40.0 m/sForce = 15,000 NewtonTime = 0.500 milliseconds (ms) to seconds = 0.0005 seconds.To find the final speed of the ball, we would use the following formula:
[tex]F = \frac{M(V - U)}{t}[/tex]
Where:
F is the force applied. u is the initial speed. v is the final speed. t is the time measured in seconds.Substituting the parameters into the formula, we have;
[tex]15000 = \frac{0.145(V \;- \;40)}{0.0005}\\\\15000(0.0005) = 0.145(V \;- \;40)\\\\7.5 = 0.145V - 5.8\\\\0.145V = 7.5 + 5.8\\\\0.145V = 13.3\\\\V = \frac{13.3}{0.145}[/tex]
Final speed, V = 91.72 m/s
Therefore, the final speed of the ball is 91.72 m/s.
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A 0.060 kg ball hits the ground with a speed of –32 m/s. The ball is in contact with the ground for 45 milliseconds and the ground exerts a +55 N force on the ball.
What is the magnitude of the velocity after it hits the ground?
Answer:
9.25 m/s
Explanation:
a. What do you mean by chromatic aberration in lenses?
A tire is filled with air at 22oC to a gauge pressure of 240 kPa. After driving for some time, if the temperature of air inside the tire is 45oC, what fraction of the original volume of air must be removed to maintain the pressure at 240 kPa?
Answer:
7.8% of the original volume.
Explanation:
From the given information:
Temperature [tex]T_1[/tex] = 22° C = 273 + 22 = 295° C
Pressure [tex]P_1[/tex] = 240 kPa
Temperature [tex]T_2[/tex] = 45° C
At initial temperature and pressure:
Using the ideal gas equation:
[tex]P_1V_1 =nRT_1[/tex]
making V_1 (initial volume) the subject:
[tex]V_1 = \dfrac{nRT_1}{P_1}[/tex]
[tex]V_1 = \dfrac{nR*295}{240}[/tex]
Provided the pressure maintained its rate at 240 kPa, when the temperature reached 45° C, then:
the final volume [tex]V_2[/tex] can be computed as:
[tex]V_2 = \dfrac{nR*318}{240}[/tex]
Now, the change in the volume ΔV = V₂ - V₁
[tex]\Delta V = \dfrac{nR*318}{240}- \dfrac{nR*295}{240}[/tex]
[tex]\Delta V = \dfrac{23nR}{240}[/tex]
∴
The required fraction of the volume of air to keep up the pressure at (240) kPa can be computed as:
[tex]= \dfrac{\dfrac{23nR}{240}}{ \dfrac{295nR}{240}}[/tex]
[tex]= {\dfrac{23nR}{240}} \times { \dfrac{240}{295nR}}[/tex]
[tex]= 0.078[/tex]
= 7.8% of the original volume.
А pressure gauge with a measurement range of 0-10 bar has a quoted inaccuracy of £1.0% f.s. (+1% of full-scale reading). (a) What is the maximum measurement error expected for this instrument? (b) What is the likely measurement error expressed as a percentage of the or reading if this pressure gauge is measuring a pressure of 1 bar?
Answer:
I am not able to answer this question please don't mind...Explanation:
please marks me as brainliests...Infrared radiation from young stars can pass through the heavy dust clouds surrounding them, allowing astronomers here on Earth to study the earliest stages of star formation, before a star begins to emit visible light. Suppose an infrared telescope is tuned to detect infrared radiation with a frequency of 4.39 THz. Calculate the wavelength of the infrared radiation.
Answer:
[tex]\lambda=6.83\times 10^{-5}\ m[/tex]
Explanation:
Given that,
An infrared telescope is tuned to detect infrared radiation with a frequency of 4.39 THz.
We know that,
1 THz = 10¹² Hz
So,
f = 4.39 × 10¹² Hz
We need to find the wavelength of the infrared radiation.
We know that,
[tex]\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{4.39\times 10^{12}}\\\\=6.83\times 10^{-5}\ m[/tex]
So, the wavelength of the infrared radiation is [tex]6.83\times 10^{-5}\ m[/tex].
A wheel rotates about a fixed axis with a constant angular acceleration of 3.3 rad/s2. The diameter of the wheel is 21 cm. What is the linear speed (in m/s) of a point on the rim of this wheel at an instant when that point has a total linear acceleration with a magnitude of 1.7 m/s2
Answer:
The the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s
Explanation:
We are given that
Angular acceleration, [tex]\alpha=3.3 rad/s^2[/tex]
Diameter of the wheel, d=21 cm
Radius of wheel, [tex]r=\frac{d}{2}=\frac{21}{2}[/tex] cm
Radius of wheel, [tex]r=\frac{21\times 10^{-2}}{2} m[/tex]
1m=100 cm
Magnitude of total linear acceleration, a=[tex]1.7 m/s^2[/tex]
We have to find the linear speed of a at an instant when that point has a total linear acceleration with a magnitude of 1.7 m/s2.
Tangential acceleration,[tex]a_t=\alpha r[/tex]
[tex]a_t=3.3\times \frac{21\times 10^{-2}}{2}[/tex]
[tex]a_t=34.65\times 10^{-2}m/s^2[/tex]
Radial acceleration,[tex]a_r=\frac{v^2}{r}[/tex]
We know that
[tex]a=\sqrt{a^2_t+a^2_r}[/tex]
Using the formula
[tex]1.7=\sqrt{(34.65\times 10^{-2})^2+(\frac{v^2}{r})^2}[/tex]
Squaring on both sides
we get
[tex]2.89=1200.6225\times 10^{-4}+\frac{v^4}{r^2}[/tex]
[tex]\frac{v^4}{r^2}=2.89-1200.6225\times 10^{-4}[/tex]
[tex]v^4=r^2\times 2.7699[/tex]
[tex]v^4=(10.5\times 10^{-2})^2\times 2.7699[/tex]
[tex]v=((10.5\times 10^{-2})^2\times 2.7699)^{\frac{1}{4}}[/tex]
[tex]v=0.418 m/s[/tex]
Hence, the the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s
The block in the drawing has dimensions L0×2L0×3L0,where L0 =0.2 m. The block has a thermal conductivity of 150 J/(s·m·C˚). In drawings A, B, and C, heat is conducted through the block in three different directions; in each case the temperature of the warmer surface is 35 ˚C and that of the cooler surface is 16 ˚C Determine the heat that flows in 6 s for each case.
Answer:
1140 J, 6840 J, 10260 J
Explanation:
Lo x 2 Lo x 3 Lo, Lo = 0.2 m, K = 150 J/(s · m · C˚) , T = 35 ˚C, T' = 16 ˚C,
time, t = 6 s
The heat conducted is
[tex]H = \frac{K A (T - T') t}{d}\\\\H = \frac{150\times 3\times 0.2\times 0.2\times (35-16) \times 6}{3\times 0.2}\\\\H = 1140 J[/tex]
The heat conducted is
[tex]H = \frac{K A (T - T') t}{d}\\\\H = \frac{150\times 3\times 0.2\times 2\times0.2\times (35-16) \times 6}{3\times 0.2}\\\\H = 6840 J[/tex]
The heat conducted is
[tex]H = \frac{K A (T - T') t}{d}\\\\H = \frac{150\times 3\times 0.2\times 2\times0.2\times (35-16) \times 6}{2\times 0.2}\\\\H = 10260 J[/tex]
Two pistons are connected to a fluid-filled reservoir. The first piston has an area of 3.002 cm2, and the second has an area of 315 cm2. If the first cylinder is pressed inward with a force of 50.0 N, what is the force that the fluid in the reservoir exerts on the second cylinder?
Answer:
The force on the second piston is 5246.5 N .
Explanation:
Area of first piston, a = 3.002 cm^2
Area of second piston, A = 315 cm^2
Force on first piston, f = 50 N
let the force of the second piston is F.
According to the Pascal's law
[tex]\frac{f}{a} = \frac{F}{A}\\\\\frac{50}{3.002}=\frac{F}{315}\\\\F = 5246.5 N[/tex]
