Answer:
The acceleration is [tex]a = 5 \ m/s^2[/tex]
Explanation:
From the question we are told that
The coefficient of kinetic friction is [tex]\mu_k = 0.24[/tex]
The coefficient of static friction is [tex]\mu_s = 0.75[/tex]
The horizontal force is [tex]F_h = 93 \ N[/tex]
Generally the static frictional force is mathematically represented as
[tex]F_F = \mu_s * (m * g )[/tex]
The static frictional force is the equivalent to the maximum possible force for which the crate does not begin to slide So
[tex]F_h = F_F = \mu_s * (m * g )[/tex]
=> [tex]93 = \mu_s * (m * g )[/tex]
=> [tex]m = \frac{93}{\mu_s * g }[/tex]
substituting values
[tex]m = \frac{93}{0.75 * 9.8 }[/tex]
[tex]m = 12.65 \ kg[/tex]
When the crate is already sliding the frictional force is
[tex]F_s = \mu_k *(m * g )[/tex]
substituting values
[tex]F_s = 0.24 * 12.65 * 9.8[/tex]
[tex]F_s = 29.82 \ N[/tex]
Now the net force when the horizontal force is applied during sliding is
[tex]F_{net} = F_h - F_s[/tex]
substituting values
[tex]F_{net} = 93 - 29.8[/tex]
[tex]F_{net} = 63.2 \ N[/tex]
This net force is mathematically represented as
[tex]F_{net } = m * a[/tex]
Where a is the acceleration of the crate
So
[tex]a = \frac{F_{net}}{m }[/tex]
[tex]a = \frac{ 63.2}{12.65 }[/tex]
[tex]a = 5 \ m/s^2[/tex]
A load of 223,000 N is placed on an aluminum column 10.2 cm in diameter. If the column was originally 1.22 m high find the amount that the column has shrunk.
Answer:
0.4757 mm
Explanation:
Given that:
Load P = 223,000 N
the length of the height of the aluminium column = 1.22 m
the diameter of the aluminum column = 10.2 cm = 0.102 m
The amount that the column has shrunk ΔL can be determined by using the formula:
[tex]\Delta L = \dfrac{PL}{AE_{Al}}[/tex]
where;
A = πr²
2r = D
r = D/2
r = 0.102/2
r = 0.051
A = π(0.051)²
A = 0.00817
Also; the young modulus of aluminium [tex]E_{Al}[/tex] is:
[tex]E_{Al}= 7*10^{10} \Nm^{-2}[/tex]
[tex]\Delta L = \dfrac{PL}{AE_{Al}}[/tex]
[tex]\Delta L = \dfrac{223000* 1.22}{0.00817* 7*10^{10}}[/tex]
ΔL = 4.757 × 10⁻⁴ m
ΔL = 0.4757 mm
Hence; the amount that the column has shrunk is 0.4757 mm
Explain the purpose of hot gravity filtration. Why is it good to use the stemless funnel for this experiment
Answer: It is done to prevent the necessary compound from solidifying along with the debasements. It expels any insoluble pollutions from the appropriate response (as opposed to separating the predetermined item). With since quite a while ago stemmed channels, the gems kick off inside the progression because the arrangement cools, obstructing the pipe. utilizing a stemless channel keeps this from occurring.
Explanation:
it is good to use the stemless funnel for hot gravity filtration experiment, to prevent the necessary compound from solidifying, expels any insoluble pollutions from the appropriate response.
what is hot gravity filtration ?Recrystallization is the process of getting pure crystals from an impure compound in a solvent and Hot gravity filtration remove the impurities from a solution prior to recrystallization.
In this technique the filtration equipment and the sample are heated and the filtration is needed for recrystallization which requires a hot solution as it need to be supersaturated for crystals to form on cooling.
Hot solutions hold more solute in a suspension than a cold solution as the solubility of solids increases with a increase in temperature, that means saturated solution contain more dissolved solute.
When the hot solution cool down, it will be supersaturated and hold more dissolved solute than its cold. The main objective to choose a solvent is that it dissolves the compound when heated, but that doesn’t dissolve the impurity at high temperatures.
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HELP ILL MARK BRAINLIEST PLS!!!!
A patch of mud has stuck to the surface of a bicycle tire as shown. The stickiness of
the mud is the centripetal or tension force that keeps the mud on the tire as it spins.
Has work been done on the mud as the tire makes one revolution, if the mud stays
on the tire? Explain.
Answer:
Yes, work has been done on the mud.
Explanation:
Work is done on a body, when a force is applied on the body to move it through a certain distance. In the case of the mud, the tire exerts a centripetal force on the mud. The centripetal force moves the mud along a path that follows the circle formed by the tire in one revolution of the tire. The total distance traveled is the circumference of the circle formed. The work done on the mud is therefore the product of the centripetal force on the mud from the tire, and the circumference of the circle formed by the tire, usually expressed in radian.
The fastest pitched baseball was clocked at 47 m/s. Assume that the pitcher exerted his force (assumed to be horizontal and constant) over a distance of 1.0 m, and a baseball has a mass of 145 g.(a) What force did he produce on the ball during this record-setting pitch? (b) Draw free-body diagrams of the ball during the pitch and just after it left the pitcherâs hand.
Answer:
Explanation:
F ×1 = 0.5×0.145×47×47
F = 160.15 N
what is the largest star in our night sky
A 32-cm-long solenoid, 1.8 cm in diameter, is to produce a 0.30-T magnetic field at its center. If the maximum current is 4.5 A, how many turns must the solenoid have?
Answer:
16,931 turnsExplanation:
The magnetic field produced is expressed using the formula
[tex]B = \frac{\mu_0NI}{L}[/tex]
B is the magnetic field = 0.30T
I is the current produced in the coil = 4.5A
[tex]\mu_0[/tex] is the magnetic permittivity in vacuum = 1.26*10^-6Tm/A
L is the length of the solenoid = 32 cm = 0.32 m
N is the number of turns in the solenoid.
Making N the subject of the formula from the equation above;
[tex]B = \frac{\mu_0NI}{L}\\\\BL = \mu_0NI\\\\Dividing\ both\ sides \ by \ \mu_0I\\\\\frac{BL}{\mu_0I} =\frac{\mu_oNI}{\mu_0I} \\\\[/tex]
[tex]N = \frac{BL}{\mu_0I}[/tex]
Substituting the give values to get N;
[tex]N = \frac{0.3*0.32}{1.26*10^{-6} * 4.5}\\\\N = \frac{0.096}{0.00000567} \\\\N = 16,931.21[/tex]
The number of turns the solenoid must have is approximately 16,931 turns
During a move, Jonas and Matías carry a 115kg safe to the third floor of a building, covering a height of 6.6m.
1) = what work do they do?
2) = what power do they develop if the work is done in 5.5 minutes?
Answer:
work is =7590joules
power = 23watts
Answer:
1) 7590 Joules
2) 23 Watts
Explanation:
1) Work = force × distance
W = mgh
W = (115 kg) (10 m/s²) (6.6 m)
W = 7590 J
2) Power = work / time
P = W / t
P = (7590 J) / (330 s)
P = 23 W
A conventional current of 8 A runs clockwise in a circular loop of wire in the plane, with center at the origin and with radius 0.078 m. Another circular loop of wire lies in the same plane, with its center at the origin and with radius 0.03 m. How much conventional current must run counterclockwise in this smaller loop in order for the magnetic field at the origin to be zero
Answer:
I2 = 3.076 A
Explanation:
In order to calculate the current in the second loop, you take into account that the magnitude of the magnetic field at the center of the ring is given by the following formula:
[tex]B=\frac{\mu_oI}{2R}[/tex] (1)
I: current in the wire
R: radius of the wire
μo: magnetic permeability of vacuum = 4π*10^-7 T/A
In the case of the two wires with opposite currents and different radius, but in the same plane, you have that the magnitude of the magnetic field at the center of the rings is:
[tex]B_T=\frac{\mu_oI_1}{2R_1}-\frac{\mu_oI_2}{2R_2}[/tex] (2)
I1: current of the first ring = 8A
R1: radius of the first ring = 0.078m
I2: current of the second ring = ?
R2: radius of the first second = 0.03m
To find the values of the current of the second ring, which makes the magnitude of the magnetic field equal to zero, you solve the equation (2) for I2:
[tex]\frac{\mu_oI_2}{2R_2}=\frac{\mu_oI_1}{2R_1}\\\\I_2=I_1\frac{R_2}{R_1}=(8A)\frac{0.03m}{0.078m}=3.076A[/tex]
The current of the second ring is 3.076A and makes that the magntiude of the total magnetic field generated for both rings is equal to zero.
The combustion of propane (C3H8) in the presence of excess oxygen yields CO2 and H2O: C3H8 (g) + 5O2 (g) → 3CO2 (g) + 4H2O (g) When 2.0 mol of O2 are consumed in this reaction, ________ mol of CO2 are produced.
Answer:
1.2
Explanation:
2.0 mol O₂ × (3 mol CO₂ / 5 mol O₂) = 1.2 mol CO₂
"A plane has an airspeed of 142 m/s. A 16.0 m/s wind is blowing southward at the same time as the plane is flying. If the velocity of the plane relative to Earth is due east, what is the magnitude of that velocity
Answer:
vr = 142.90 m/s
the magnitude of its relative velocity is 142.90 m/s
Explanation:
Given;
A plane has an airspeed of 142 m/s (eastward)
vi = 142 m/s
16.0 m/s wind is blowing southward at the same time as the plane is flying
vb = 16.0m/s
Writing the relative velocity vector, we have;
Taking north and south as positive and negative y axis respectively, east and west as positive and negative x axis respectively.
v = 142i - 16j
The magnitude of the velocity is;
vr = √(vi^2 + vb^2)
vr = √(142^2 + 16^2)
vr = √(20420)
vr = 142.8985654231 m/s
vr = 142.90 m/s
the magnitude of its relative velocity is 142.90 m/s
Applying Gaussâs Law
When a charge is placed on a metal sphere, it ends up in equilibrium at the outer surface. Use this information to determine the electric field of +3.0 μC charge put on a 5.0-cm aluminum spherical ball at the following two points in space: (a) a point 1.0 cm from the center of the ball (an inside point) and (b) a point 10 cm from the center of the ball (an outside point).
Answer:
a) E = 0
b) E = 2.697 MN/C
Explanation:
Solution:-
- The Gauss Law makes life simpler by allowing us to determine the Electric Field strength ( E ) of symmetrically charged objects. By choosing an appropriate Gaussian surface and determine the flux ( Φ ) that passes through an imaginary closed surface.
- The Law states that the net flux ( Φ ) that passes through a Gaussian surface is proportional to the net charged ( Q ) stored within that surface. We can mathematically express the flux ( Φ ) as follows:
Φ = Q / εo
Where, 1 / εo : The proportionality constant
εo: The permittivity of free space = 8.85*10^-12
- The flux produced by a charged object is also given in form of a surface integral of Electric Field ( E ) over the entire surface area ( A ) of the Gaussian surface as follows:
Φ = [tex]_S\int\int [ E ] . dA[/tex]
- We can combine the two relations as follows:
[tex]_S\int\int [ E ] . dA[/tex] = Q / εo
- Now we will consider a charged metal sphere. The important part to note is that the charge on a conducting sphere ( Q ) uniformly distributed on the outside surface of the charged sphere.
- Lets consider a case, where we set up our Gaussian surface ( spherical ) with radius ( r ) < radius of the charged metal surface ( a ). We will use the combined relation and determine the Electric Field ( E ) within a charged metal sphere as follows:
[tex]E. ( 4\pi*r^2 ) = \frac{Q_e_n_c}{e_o} \\\\E = \frac{Q_e_n_c}{e_o4\pi*r^2}[/tex]
- However, the amount of charge enclosed in our Gaussian surface is null or zero. As all the charge is on the surface r = a. Hence (Q_enc = 0 ),
[tex]E = 0[/tex] ..... ( r < a )
- For the case when we set up our gaussian surface with radius ( r ) > radius of the charged metal surface ( a ). We placed a charge of Q = +3.0uC on the surface of the metal sphere. Therefore, the electric field strength at a distance ( r ) from the center of metal sphere is:
[tex]E = \frac{Q_e_n_c}{e_o*4*\pi*r^2 } = k\frac{Q_e_n_c}{r^2 }[/tex] .... ( r > a )
- The above relation turns out to be the Electric Field strength ( E ) produced by a point charge at distance ( r ) from the center. Where, k = 8.99*10^9 is the Coulomb's constant.
a) The radius of the charged metal sphere is given to be a = 5.0 cm. The first point r = 1.0 cm lies within the metal sphere. We looked at the first case where, ( r < a ) the enclosed charge is zero. Hence, the magnitudue of Electric Field Strength ( E ) is zero. ( E = 0 )
b) The second point lies at 10 cm from the center. For this we will use the second case where, ( r > a ). The Electric Field Strength due to a point charge with an enclosed charge of Q = +3.0 uC is:
[tex]E = ( 8.99*10^9 ) * \frac{3.0*10^-^6}{0.1^2} \\\\E = 2697000 N / C[/tex]
Answer: The electric field strength at point 10 cm away from the center is 2.697 MN/C
Which scientist's work led to our understanding of how planets move around
the Sun?
A. Albert Einstein
B. Lord Kelvin
C. Johannes Kepler
D. Edwin Hubble
Answer:
Johannes KeplerExplanation:
He made rules about planetary motion.The scientist Johannes Kepler was a German astronomer.He found out that the planets evolved around the Sun.He also made the laws of planetary motion.Hope this helped,
Kavitha
the time required for one cycle, a complete motion that returns to its starting point, it called the_____. period medium frequency periodic motion
Answer:
The time required for one cycle, a complete motion that returns to its starting point,it is called periodic motion
Explanation:
I hope this will help you:)
Suppose that 300 keV X-ray photons are aimed at a zinc cube (Zinc, Z = 30). According to the chart below, what effect will predominate when the X-rays hit the metal?
a) Photoelectric Effect 3
b) Compton Effect 3
c) Pair Production
Answer:
the answer is option A = photoelectric effect
Explanation:
If the threshold frequency of a metal is lower than the energy of X-rays, then photoelectric effect will happen.
How much work is needed to move an object from one position to another when both positions are located the same distance from the center of the earth
Answer:
The product of the object's weight and the horizontal distance between the two positions.
Explanation:
Work is the product of force and the distance through which this force is moved. The distance moved can be vertical, or horizontal. For two bodies located the same distance from the center of the earth, the work done will be the product of the weight of the product and the horizontal distance between the two positions. If the vertical work is needed, then the work is zero, since there is no height gradient between them.
B. Write short notes on:
1. Horticulture
2. Pisciculture
3. Aviculture
4. Veterinary science
5. Intensive farming.
1. Horticulture is the agriculture of plants, mainly for food, materials, comfort and beauty for decoration.
2.Pisciculture also known as fish farming is the rearing of fish for food in enclosures such as fish ponds or tanks.
3.Aviculture is the practice of keeping and breeding birds, especially of wild birds in captivity. Aviculture is generally focused on not only the raising and breeding of birds, but also on preserving avian habitat, and public awareness campaigns.
4. Veterinary medicine is the branch of medicine that deals with the prevention, control, diagnosis, and treatment of disease, disorder, and injury in animals. Along with this, it also deals with animal rearing, husbandry, breeding, research on nutrition and product development.
5. Intensive agriculture, also known as intensive farming and industrial agriculture, is a type of agriculture, both of crop plants and of animals, with higher levels of input and output per cubic unit of agricultural land area.
Hope this helps.
A 60 kg person is in a head-on collision. The car's speed at impact is 15 m/s . Estimate the net force on the person if he or she is wearing a seat belt and if the air bag deploys.
Complete question:
Seat belts and air bags save lives by reducing the forces exerted on the driver and passengers in an automobile collision. Cars are designed with a "crumple zone" in the front of the car. In the event of an impact, the passenger compartment decelerates over a distance of about 1 m as the front of the car crumples. An occupant restrained by seat belts and air bags decelerates with the car. In contrast, a passenger not wearing a seat belt or using an air bag decelerates over a distance of 5mm.
(a) A 60 kg person is in a head-on collision. The car's speed at impact is 15 m/s . Estimate the net force on the person if he or she is wearing a seat belt and if the air bag deploys.
Answer:
The net force on the person as the air bad deploys is -6750 N backwards
Explanation:
Given;
mass of the passenger, m = 60 kg
velocity of the car at impact, u = 15 m/s
final velocity of the car after impact, v = 0
distance moved as the front of the car crumples, s = 1 m
First, calculate the acceleration of the car at impact;
v² = u² + 2as
0² = 15² + (2 x 1)a
0 = 225 + 2a
2a = -225
a = -225 / 2
a = -112.5 m/s²
The net force on the person;
F = ma
F = 60 (-112.5)
F = -6750 N backwards
Therefore, the net force on the person as the air bad deploys is -6750 N backwards
Unpolarized light enters a polarizer with vertical polarization axis. The light that passes through passes another polarizer with transmission axis at 40 degrees to the horizontal. What is the intensity of the light after the second polarizer expressed as a fraction of the original intensity
Answer:
I = 0.2934 I₀
Explanation:
The expression that governs the transmission of polarization is
I = I₀ cos² θ
Let's apply this to our case, when the unpolarized light enters the first polarized, the polarized light that comes out has the intensity of
I₁ = I₀ / 2
this is the light that enters the second polarizer
I = I₁ cos² θ
we substitute
I = I₀ / 2 cos² 40
I = I₀ 0.2934
I = 0.2934 I₀
What is the equivalent resistance between the points A and B of the network?
Explanation:
First, simplify the circuit. Then calculate the parallel and consecutive resistances to find the answer.
Suppose that the voltage of the battery in the circuit is 3.9 V, the magnitude of the magnetic field (directed perpendicularly into the plane of the paper) is 1.1 T, and the length of the rod between the rails is 0.22 m. Assuming that the rails are very long and have negligible resistance, find the maximum speed attained by the rod after the switch is closed.
Answer:
v = 16.11 m/s
Explanation:
In order to calculate the maximum speed of the rod, you use the following formula:
[tex]\epsilon=vBLsin\theta[/tex] (1)
ε = voltage of the circuit = 3.9V
v: maximum speed of the rod = ?
B: magnitude of the magnetic field = 1.1T
L: length of the rod = 0.22m
θ: angle between the direction of motion of the rod and the direction of the magnetic field = 90°
You solve the equation (1) for v and replace the values of the other parameters:
[tex]v=\frac{\epsilon}{BLsin\theta}=\frac{3.9V}{(1.1T)(0.22m)sin90\°}\\\\v=16.11\frac{m}{s}[/tex]
The maximum speed of the rod is 16.11 m/s
The upward velocity of a 2540kg rocket is v(t)=At + Bt2. At t=0 a=1.50m/s2. The rocket takes off and one second afterwards v=2.00m/s. Determine the constants A and B with units.
Answer:
The value of A is 1.5m/s^2 and B is 0.5m/s^³
Explanation:
The mass of the rocket = 2540 kg.
Given velocity, v(t)=At + Bt^2
Given t =0
a= 1.50 m/s^2
Now, velocity V(t) = A*t + B*t²
If, V(0) = 0, V(1) = 2
a(t) = dV/dt = A+2B × t
a(0) = 1.5m/s^²
1.5m/s^² = A + 2B × 0
A = 1.5m/s^2
now,
V(1) = 2 = A× 1 + B× 1^²
1.5× 1 +B× 1 = 2m/s
B = 2-1.5
B = 0.5m/s^³
Now Check V(t) = A× t + B × t^²
So, V(1) = A× (1s) + B× (1s)^² = 1.5m/s^² × 1s + 0.5m/s^³ × (1s)^² = 1.5m/s + 0.5m/s = 2m/s
Therefore, B is having a unit of m/s^³ so B× (1s)^² has units of velocity (m/s)
An alarm clock is plugged into a 120 volt outlet and has a resistance of 15,000 ohms. How much power does it use?
Answer:
The power used is 0.96 watts.
Explanation:
Recall the formula for electric power (P) as the product of the voltage applied times the circulating current:
[tex]P=V\,\,I[/tex]
and recall as well that the circulating current can be obtained via Ohm's Law as the quotient of the voltage applied divided the resistance:
[tex]V=I\,\,R\\I=\frac{V}{R}[/tex]
Then we can re-write the power expression as:
[tex]P=V\,\,I=V\,\,\frac{V}{R} =\frac{V^2}{R}[/tex]
which in our case becomes:
[tex]P=\frac{V^2}{R}=\frac{120^2}{15000} =0.96\,\,watts[/tex]
The magnetic field strength at the north pole of a 2.0-cm-diameter, 8-cm-long Alnico magnet is 0.10 T. To produce the same field with a solenoid of the same size, carrying a current of 1.8 A , how many turns of wire would you need
Answer:
The number of turns of the solenoid is 3536 turns
Explanation:
Given;
magnetic field of the solenoid, B = 0.1 T
current in the solenoid, I = 1.8 A
length of the solenoid, L = 8cm = 0.08m
The magnetic field near the center of the solenoid is given by;
B = μ₀nI
Where;
μ₀ is permeability of free space = 4π x 10⁻⁷ m/A
n is number of turns per length
I is the current in the coil
The number of turns per length is calculated as;
n = B / μ₀I
n = (0.1 ) / (4π x 10⁻⁷ x 1.8)
n = 44203.95 turns/m
The number of turns is calculated as;
N = nL
N = (44203.95)(0.08)
N = 3536 turns
Therefore, the number of turns of the solenoid is 3536 turns
A dinner plate falls vertically to the floor and breaks up into three pieces, which slide horizontally along the floor. Immediately after the impact, a 320-g piece moves along the x-axis with a speed of 2.00 m/s and a 355-g piece moves along the y-axis with a speed of 1.50 m/s. The third piece has a mass of 100 g. In what direction relative to the +x-axis does the third piece move?
(a) 39.8º from the +x-axis
(b) 36.9° from the +x-axis
(c) 39.9° from the +x-axis
(d) 216.9° from the +x-axis
(e) 219.8° from the +X-axis
Answer:
M1 Vx1 + M2 Vx2 + M3 Vx3 = 0 conservation of momentum in x direction
Vx3 = -(M1 Vx1 + M2 Vx2 ) / M3
Vx3 = - 320 * 2 / 100 = -6.4 m/s M2 has no x-component of momentum
Likewise:
Vy3 = -(M1 Vy1 + M2 Vy2 ) / M3
Vy3 = - 355 * 1.5 / 100 = -5.33 m/s
tan theta = -5.33 / -6.4 = .833 where theta is in the third quadrant and measured from the negative x-axis
theta = 39.8 deg
180 + 39.8 = 219.8 from the positive x-axis
A wheel has a rotational inertia of 16 kgm2. Over an interval of 2.0 s its angular velocity increases from 7.0 rad/s to 9.0 rad/s. What is the average power done by the torque
Answer:
128.61 WattsExplanation:
Average power done by the torque is expressed as the ratio of the workdone by the toque to time.
Power = Workdone by torque/time
Workdone by the torque = [tex]\tau \theta[/tex] = [tex]I\alpha * \theta[/tex]
I is the rotational inertia = 16kgm²
[tex]\theta = angular\ displacement[/tex]
[tex]\theta = 2 rev = 12.56 rad[/tex]
[tex]\alpha \ is \ the\ angular\ acceleration[/tex]
To get the angular acceleration, we will use the formula;
[tex]\alpha = \frac{\omega_f^2- \omega_i^2}{2\theta}[/tex]
[tex]\alpha = \frac{9.0^2- 7.0^2}{2(12.54)}\\\alpha = 1.28\ rad/s^{2}[/tex]
Workdone by the torque = 16 * 1.28 * 12.56
Workdone by the torque = 257.23 Joules
Average power done by the torque = Workdone by torque/time
= 257.23/2.0
= 128.61 Watts
Air is cooled in a process with constant pressure of 150 kPa. Before the process begins, air has a specific volume of 0.062 m^3/kg. The final specific volume is 0.027 m^3/kg. Find the specific work in the process.
Answer:
The pressure is constant, and it is P = 150kpa.
the specific volumes are:
initial = 0.062 m^3/kg
final = 0.027 m^3/kg.
Then, the specific work can be written as:
[tex]W = \int\limits^{vf}_{vi} {Pdv} \, = P(vf - vi) = 150kPa*(0.0027 - 0.062)m^3/kg = -5.25 kPa*m^3/kg.[/tex]
The fact that the work is negative, means that we need to apply work to the air in order to compress it.
Now, to write it in more common units we have that:
1 kPa*m^3 = 1000J.
-5.25 kPa*m^3/kg = -5250 J/kg.
A meat baster consists of a squeeze bulb attached to a plastic tube. When the bulb is squeezed and released, with the open end of the tube under the surface of the basting sauce, the sauce rises in the tube to a distance h, as the drawing shows. Using 1.013 105 Pa for the atmospheric pressure and 1200 kg/m3 for the density of the sauce, find the absolute pressure in the bulb when the distance h is (a) 0.15 m and (b) 0.10 m.
Answer:
(a) P = 103064 Pa = 103.064 KPa
(b) P = 102476 Pa = 102.476 KPa
Explanation:
(a)
First we need to find the gauge pressure:
Gauge Pressure = Pg = (density)(g)(h)
Pg = (1200 kg/m³)(9.8 m/s²)(0.15 m)
Pg = 1764 Pa
So, the absolute Pressure is:
Absolute Pressure = P = Atmospheric Pressure + Pg
P = 1.013 x 10⁵ Pa + 1764 Pa
P = 103064 Pa = 103.064 KPa
(b)
First we need to find the gauge pressure:
Gauge Pressure = Pg = (density)(g)(h)
Pg = (1200 kg/m³)(9.8 m/s²)(0.1 m)
Pg = 1176 Pa
So, the absolute Pressure is:
Absolute Pressure = P = Atmospheric Pressure + Pg
P = 1.013 x 10⁵ Pa + 1176 Pa
P = 102476 Pa = 102.476 KPa
The absolute pressure in the bulb is approximately 1.031 x 10⁵ Pa when h = 0.15 m and 1.025 x 10⁵ Pa when h = 0.10 m.
Absolute pressure is the total pressure exerted by a fluid, including both the pressure from the fluid itself and the atmospheric pressure. It is the sum of the gauge pressure, which is the pressure above atmospheric pressure, and the atmospheric pressure. Absolute pressure is measured relative to a complete vacuum, where the pressure is zero.
In fluid mechanics, absolute pressure is important for determining the forces and behaviors of fluids in various systems. It is commonly expressed in units such as pascals (Pa), atmospheres (atm), pounds per square inch (psi), or torr.
The absolute pressure in the bulb can be calculated using the following formula:
P = P₀ + ρgh
where:
P is the absolute pressure in the bulb,
P₀ is the atmospheric pressure (1.013 x 10⁵ Pa),
ρ is the density of the sauce (1200 kg/m³),
g is the acceleration due to gravity (9.8 m/s²), and
h is the height of the sauce in the tube.
(a) When h = 0.15 m:
P = 1.013 x 10⁵ Pa + (1200 kg/m³) x (9.8 m/s²) x (0.15 m)
P ≈ 1.013 x 10⁵ Pa + 1764 Pa
P ≈ 1.031 x 10⁵ Pa
(b) When h = 0.10 m:
P = 1.013 x 10⁵ Pa + (1200 kg/m³) x (9.8 m/s²) x (0.10 m)
P ≈ 1.013 x 10⁵ Pa + 1176 Pa
P ≈ 1.025 x 10⁵ Pa
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5. Two men, Joel and Jerry, push against a car that has stalled, trying unsuccessfully to get it moving. Jerry stops after 10 min, while Joel is able to push for 5.0 min longer. Compare the work they do on the car
Answer:
The work done by both Joel and Jerry is equal to 0 J.
Explanation:
The work done on a body by an external agency is the product of the force applied on the body and the distance through which the body moves. Therefore,
W = F.d
where,
W = Work Done on the Body
F = Force Applied on the Body
d = displacement covered by the body
In the given case of both Joel and Jerry, they are unable to move the car. It means that the displacement covered by the car is zero. Hence,
W = F(0)
W = 0 J (For both Joel and Jerry)
In an undergraduate physics lab, a simple pendulum is observed to swing through 71 complete oscillations in a time period of 1.80 min. What are the period and length of the pendulum
Explanation:
We have
A simple pendulum is observed to swing through 71 complete oscillations in a time period of 1.80 min.
The frequency of a pendulum is equal to the no of oscillation per unit time. so,
[tex]f=\dfrac{N}{t}\\\\f=\dfrac{71}{1.8\times 60}\\\\f=0.65\ Hz[/tex]
Tim period is reciprocal of frequency. So,
[tex]T=\dfrac{1}{0.65}\\\\T=1.53\ s[/tex]
The time period of a pendulum is given by :
[tex]T=2\pi \sqrt{\dfrac{l}{g}}[/tex]
l is length of pendulum
[tex]l=\dfrac{T^2g}{4\pi ^2}\\\\l=\dfrac{T^2g}{4\pi ^2}\\\\l=\dfrac{(1.53)^2\times 9.8}{4\pi ^2}\\\\l=0.58\ m[/tex]
So, the period and length of the pendulum are 1.53 s and 0.58 m respectively.
An 75-kg hiker climbs to the summit of Mount Mitchell in western North Carolina. During one 2.0-h period, the climber's vertical elevation increases 540 m. Determine the change in gravitational potential energy of the climber-Earth system.
Answer:
The change in gravitational potential energy of the climber-Earth system is [tex]\Delta PE = 396900 \ J[/tex]
Explanation:
From the question we are told that
The mass of the hiker is [tex]m = 75 \ kg[/tex]
The time taken is [tex]T = 2 \ hr = 2 * 3600 = 7200 \ s[/tex]
The vertical elevation after time T is [tex]H = 540 \ m[/tex]
The change in gravitational potential is mathematically represented as
[tex]\Delta PE = mgH[/tex]
here g is the acceleration due to gravity with value [tex]g = 9.8 \ m/s^2[/tex]
substituting values
[tex]\Delta PE = 75 * 9.8 * 540[/tex]
[tex]\Delta PE = 396900 \ J[/tex]