Answer:
D₂ = 2.738 cm
Explanation:
Continuity equation
The continuity equation is nothing more than a particular case of the principle of conservation of mass. It is based on the flow rate (Q) of the fluid must remain constant throughout the entire pipeline.
Since the flow rate is the product of the surface of a section of the duct because of the speed with which the fluid flows, we will have to comply with two points of the same pipeline:
Q = v*A : Flow Equation
where:
Q = Flow in (m³/s)
A is the surface of the cross sections of points 1 and 2 of the duct.
v is the flow velocity at points 1 and 2 of the pipe.
It can be concluded that since the flow rate must be kept constant throughout the entire duct, when the section decreases, the flow rate increases in the same proportion and vice versa.
Data
D₁= 6.0 cm : faucet diameter
v₁ = 5 m/s : speed of fluid in the faucet
v₂ = 20 m/s : speed of fluid in the nozzle
Area calculation
A = (π*D²)/4
A₁ = (π*D₁²)/4
A₂ = (π*D₂²)/4
Continuity equation
Q₁ = Q₂
v₁A₁ = v₂A₂
v₁(π*D₁²)/4 = v₂(π*D₂²)/4 : We divide by (π/4) both sides of the equation
v₁ (D₁)² = v₂(D₂)²
We replace data
6 *(5)² = 20*(D₂)²
150 = 20*(D₂)²
(150 /20) = (D₂)²
7.5 = (D₂)²
[tex]D_{2} = \sqrt{7.5}[/tex]
D₂ = 2.738 cm : nozzle diameter
In 1994 the performer Rod Stewart drew over 3 million people to a concert in Rio de Janeiro, Brazil.
(a) If the people in the group had an average mass of 80.0 kg, what collective gravitational force would the group have on a 4.50-kg eagle soaring 3.00 Ã 10 2 m above the throng? If you treat the group as a point object, you will get an upper limit for the gravitational force.
(b) What is the ratio of that force of attraction to the force between Earth and the eagle
Answers:
a) [tex]8.009(10)^{-7} N[/tex]
b) [tex]1.8(10)^{-8} [/tex]
Explanation:
a) Accoding to the Universal Law of Gravitation we have:
[tex]F_{g}=G\frac{Mm}{d^2}[/tex] (1)
Where:
[tex]F_{g}[/tex] is the gravitational force between the eagle and the throng
[tex]G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}[/tex] is the Universal Gravitational constant
[tex]M=4.5 kg[/tex] is the mass of the eagle
[tex]m=(80 kg)(3(10)^{6} people/kg)=240(10)^{6} kg[/tex] is the mass of the throng
[tex]d=300 m[/tex] is the distance between the throng and the eagle
[tex]F_{g}=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}} \frac{(4.5 kg)(240(10)^{6} kg)}{(300 m)^{2}}[/tex] (2)
[tex]F_{g}=8.009 (10)^{-7} N[/tex] (3) As we can see the gravitational force between the eagle and the throng is quite small.
b) The attraction force between the eagle and Earth is the weight [tex]W[/tex] of the eagle, which is given by:
[tex]W=Mg[/tex] (4)
Where [tex]g=9.8 m/s^{2}[/tex] is the acceleration due gravity on Earth
[tex]W=(4.5 kg)(9.8 m/s^{2})[/tex] (5)
[tex]W=44.1 N[/tex] (6)
Now we can find the ratio between [tex]F_{g}[/tex] and [tex]W[/tex]:
[tex]\frac{F_{g}}{W}=\frac{8.009 (10)^{-7} N}{44.1N}[/tex]
[tex]\frac{F_{g}}{W}=1.8(^{-8})[/tex] As we can see this ratio is also quite small
Two horizontal curves on a bobsled run are banked at the same angle, but one has twice the radius of the other. The safe speed (no friction needed to stay on the run) for the smaller radius curve is v. What is the safe speed on the larger radius curve?
Answer:
safe speed for the larger radius track u= √2 v
Explanation:
The sum of the forces on either side is the same, the only difference is the radius of curvature and speed.
Also given that r_1= smaller radius
r_2= larger radius curve
r_2= 2r_1..............i
let u be the speed of larger radius curve
now, [tex]\sum F = \frac{mv^2}{r_1} =\frac{mu^2}{r_2}[/tex]................ii
form i and ii we can write
[tex]v^2= \frac{1}{2} u^2[/tex]
⇒u= √2 v
therefore, safe speed for the larger radius track u= √2 v
Planets are not uniform inside. Normally, they are densest at the center and have decreasing density outward toward the surface. Model a spherically symmetric planet, with the same radius as the earth, as having a density that decreases linearly with distance from the center. Let the density be 1.30×104 kg/m3 at the center and 2100 kg/m3 at the surface. Part A What is the acceleration due to gravity at the surface of this planet?
The acceleration due to gravity at the surface of a planet depends on its mass and radius, and assumes a uniform density. Since your model has a density that decreases linearly from the center to the surface, the exact value for gravity would require integration over the volume of the planet to account for mass distribution. This arrangement involves advanced calculus.
Explanation:The acceleration due to gravity at the surface of any planet, including Earth, is determined by a constant (G), the mass of the planet (M), and the radius of the planet (R). The formula is g = GM/R². However, this calculation assumes a uniform density throughout the planet, which is often not the case. In reality, like in your model where the density decreases linearly from the center to the surface, finding the precise acceleration due to gravity at the surface becomes more complicated and involves integration over the entire volume of the planet to account for how the mass is distributed.
Given that you provided the densities at the center and surface of the modeled planet, and these densities decrease linearly, one can utilize the formula for the linear density ρ(r) = ρ_center - r(ρ_center - ρ_surface)/R, where R is the radius of the planet, r is the distance from the center, and ρ_center and ρ_surface are the density at the center and surface, respectively. Then, integrate over the volume of the planet to find the total mass.
Once you have the mass, you can use the formula g = GM/R² again to find the acceleration due to gravity at the surface. However, this calculation goes beyond a basic understanding of gravity and requires knowledge of calculus. Without specific numbers for the mass and the integration result, I cannot provide the exact value for surface gravity in this case.
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The acceleration due to gravity at a planet's surface depends on the planet's radius, mass and the linear decrease of density from center to surface. The formula of this acceleration is G×M/r², considering that M is the planet's mass obtained by the product of volume and average density. However, as the density changes linearly, the force of gravity also decreases linearly from the center to the surface.
Explanation:To calculate the acceleration due to gravity at the surface of the planet, we have to consider the planet's radius, mass and density. Given the density at the center and surface, we can calculate the average density which is the total mass of the planet divided by the total volume. In this spherically symmetric planet model, we can use the formula for the volume of a sphere, which is 4/3πr³, with r being the Earth's radius. We consider that mass (M) equals density (ρ) times volume (V), and the force of gravity (F) is G×(M1×M2)/r², where G is the gravitational constant. In this case, M1 is the mass of the planet and M2 is the mass of the object where we want to know the acceleration, and r is the distance between the centers of the two masses, or in this case the radius of the planet. As force is also mass times acceleration, we can replace F in the formula with M2 times a (acceleration), and find that acceleration is G×M1/r². However, as the density changes linearly from the center to the surface, the force of gravity will also decrease linearly, affecting the acceleration.
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A student pushes a 21-kg box initially at rest, horizontally along a frictionless surface for 10.0 m and then releases the box to continue sliding. If the student pushes with a constant 10 N force, what is the box's speed when it is released?
Answer:v=3.08 m/s
Explanation:
Given
mass of student [tex]m=21 kg[/tex]
distance moved [tex]d=10 m[/tex]
Force applied [tex]F=10 N[/tex]
acceleration of system during application of force is a
[tex]a=\frac{F}{m}=\frac{10}{21}=0.476 m/s^2[/tex]
using [tex]v^2-u^2=2 as[/tex]
where v=final velocity
u=initial velocity
a=acceleration
s=displacement
[tex]v^2-0=2\times 0.476\times 10[/tex]
[tex]v=\sqrt{9.52}[/tex]
[tex]v=3.08 m/s[/tex]
A 60-kg woman stands on the very end of a uniform board, which is supported one-quarter of the way from one end and is balanced. What is the mass of the board?
a. 60 kg
b. 30 kg
c. 20 kg
d. 15 kg
e. 120 kg
The correct option can be seen in Option A.
The diagrammatic expression of the question can be seen in the image attached below.
From the given question, we are being informed that the uniform board is balanced. As a result, the torque(i.e. a measurement about how significantly a force acts on a body for it to spin about an axis) acting on the right-hand side of the balance point should be equal to that of the left-hand side.
Mathematically;
[tex]\mathbf{\tau_{_{right}}= \tau_{_{left}}}[/tex]
Given that the mass of the woman = 60 kg
[tex]\mathbf{\tau =\dfrac{m\times g \times l}{\mu}}[/tex]
[tex]\mathbf{\tau_{left} =\dfrac{m\times g \times l}{\mu}}---(1)[/tex]
[tex]\mathbf{\tau_{_{right}} =\dfrac{60 \times g \times l}{\mu}}---(2)[/tex]
Equating both (1) and (2) together, we have:
[tex]\mathbf{\dfrac{m\times g \times l}{\mu} =\dfrac{60 \times g \times l}{\mu} }[/tex]
Dividing like terms on both side
mass (m) = 60 kg
As such, the correct option can be seen in Option A.
Thus, we can conclude that from the 60-kg woman who stands on the very end of a uniform board, the mass of the board on the other end is also 60 kg.
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A 40 kg slab rests on a frictionless floor. A 10 kg block rests on top of the slab (Fig. 6-58). The coefficient of static friction µstat beltween the block and the slab is 0.70, whereas their kinetic friction coefficient µkin is 0.40. The 10 kg block is pulled by a horizontal force with a magnitude of 100 N.
Answer:
[tex]\text { The "resulting action" on the slab is } 0.98 \mathrm{m} / \mathrm{s}^{2}[/tex]
Explanation:
Normal reaction from 40 kg slab on 10 kg block
M × g = 10 × 9.8 = 98 N
Static frictional force = 98 × 0.7 N
Static frictional force = 68.6 N is less than 100 N applied
10 kg block will slide on 40 kg slab and net force on it
= 100 N - kinetic friction
[tex]=100-(98 \times 0.4)\left(\mu_{\text {kinetic }}=0.4\right)[/tex]
= 100 - 39.2
= 60.8 N
[tex]10 \mathrm{kg} \text { block will slide on } 40 \mathrm{kg} \text { slab with, } \frac{\mathrm{Net} \text { force }}{\text { mass }}[/tex]
[tex]10 \mathrm{kg} \text { block will slide on } 40 \mathrm{kg} \text { slab with }=\frac{60.8}{10}[/tex]
[tex]10 \mathrm{kg} \text { block will slide on } 40 \mathrm{kg} \text { slab with }=6.08 \mathrm{m} / \mathrm{s}^{2}[/tex]
[tex]\text { Frictional force on 40 kg slab by 10 kg block, normal reaction \times \mu_{kinetic } }[/tex]
Frictional force on 40 kg slab by 10 kg block = 98 × 0.4
Frictional force on 40 kg slab by 10 kg block = 39.2 N
[tex]40 \mathrm{kg} \text { slab will move with } \frac{\text { frictional force }}{\text { mass }}[/tex]
[tex]40 \mathrm{kg} \text { slab will move with }=\frac{39.2}{40}[/tex]
40 kg slab will move with = [tex]0.98 \mathrm{m} / \mathrm{s}^{2}[/tex]
[tex]\text { The "resulting action" on the slab is } 0.98 \mathrm{m} / \mathrm{s}^{2}[/tex]
An aircraft component is fabricated from an aluminum alloy that has a plane strain fracture toughness of 39 . It has been determined that fracture results at a stress of 208 MPa when the maximum (or critical) internal crack length is 2.82 mm. a) Determine the value of for this same component and alloy at a stress level of 270 MPa when the maximum internal crack length is 1.41 mm.
Answer:
[tex]27.57713\ MPa\sqrt{m}[/tex]
Explanation:
Y = Fracture parameter
a = Crack length
[tex]\sigma[/tex] = Stress in part
Plane strain fracture toughness is given by
[tex]K_I=Y\sigma\sqrt{\pi a}\\\Rightarrow Y=\frac{K_I}{\sigma\sqrt{\pi a}}\\\Rightarrow Y=\frac{39}{270\times \sqrt{\pi 0.00282}}\\\Rightarrow Y=1.53462[/tex]
When a = 1.41 mm
[tex]K_I=Y\sigma\sqrt{\pi a}\\\Rightarrow K_i=1.53462\times 270\sqrt{\pi 0.00141}\\\Rightarrow K_I=27.57713\ MPa\sqrt{m}[/tex]
The value of plane strain fracture toughness is [tex]27.57713\ MPa\sqrt{m}[/tex]
Given: G = 6.67259 × 10−11 N m2 /kg2 . A 438 kg geosynchronous satellite orbits a planet similar to Earth at a radius 1.94 × 105 km from the planet’s center. Its angular speed at this radius is the same as the rotational speed of the Earth, and so they appear stationary in the sky. That is, the period of the satellite is 24 h . What is the force acting on this satellite? Answer in units of N.
Answer:
449.37412 N
Explanation:
G = Gravitational constant = 6.67259 × 10⁻¹¹ m³/kgs²
m = Mass of satellite = 438 kg
M = Mass of planet
T = Time period of the satellite = 24 h
r = Radius of planet = [tex]1.94\times 10^8\ m[/tex]
The time period of the satellite is given by
[tex]T=2\pi\sqrt{\frac{r^3}{GM}}\\\Rightarrow M=4\pi^2\frac{r^3}{T^2G}\\\Rightarrow M=4\pi^2\times \frac{(1.94\times 10^8)^3}{(24\times 3600)^2\times 6.67259\times 10^{-11}}\\\Rightarrow M=5.78686\times 10^{26}\ kg[/tex]
The gravitational force is given by
[tex]F=G\frac{Mm}{r^2}\\\Rightarrow F=6.67259\times 10^{-11}\times \frac{5.78686\times 10^{26}\times 438}{(1.94\times 10^8)^2}\\\Rightarrow F=449.37412\ N[/tex]
The force acting on this satellite is 449.37412 N
Gravity, is often known as gravitation. The gravitational force between the planet and the satellite can be written as 0.4332 N.
What is gravitational force?Gravity, often known as gravitation, is the universal force of attraction that acts between all matter in mechanics. It is the weakest known force in nature, and so has no bearing on the interior properties of ordinary matter.
[tex]F =G\dfrac{m_1m_2}{r^2}[/tex]
As it is given that the value of the gravitational constant is G = 6.67259 × 10−11 N m² /kg², while the radius of the planet is 1.94×10⁵. And the time taken by the satellite to revolve around the planet is 24 hours. therefore, the Mass of the planet can be written as,
The Time period of the satellite is given as:
[tex]T = 2\pi \sqrt{\dfrac{r^3}{GM}}[/tex]
Substitute the values in the formula,
[tex]24 = 2\pi \sqrt{\dfrac{(1.94 \times 10^5)^3}{6.67259 \times 10^{-11}M}}\\\\M = 5.78686 \times 10^{17}\rm\ kg[/tex]
Thus, the mass of the planet can be written as 5.5786×10¹⁷ kg.
Now, the gravitational force can be written as,
[tex]F = G\dfrac{Mm}{r^2}\\\\F = 6.67259\times 10^{-11} \times \dfrac{5.5786 \times 10^{17} \times 438}{1.94 \times 10^5}\\\\ F = 0.4332\rm\ N[/tex]
Hence, the gravitational force between the planet and the satellite can be written as 0.4332 N.
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A 44.0 kg uniform rod 4.90 m long is attached to a wall with a hinge at one end. The rod is held in a horizontal position by a wire attached to its other end. The wire makes an angle of 30.0° with the horizontal, and is bolted to the wall directly above the hinge. If the wire can withstand a maximum tension of 1450N before breaking, how far from the wall can a 69.0kg person sit without breaking the wire?
Answer:
x ≤ 3.6913 m
Explanation:
Given
Mrod = 44.0 kg
L = 4.90 m
Tmax = 1450 N
Mman = 69 kg
A: left end of the rod
B: right end of the rod
x = distance from the left end to the man
If we take torques around the left end as follows
∑τ = 0 ⇒ - Wrod*(L/2) - Wman*x + T*Sin 30º*L = 0
⇒ - (Mrod*g)*(L/2) - (Mman*g)*x + Tmax*Sin 30º*L = 0
⇒ - (44*9.8)*(4.9/2) - (69*9.8)*x + (1450)*(0.5)*(4.9) = 0
⇒ x ≤ 3.6913 m
As a civil engineer for your city, you have been assigned to evaluate the purchase of spring-loaded guard rails to prevent cars from leaving the road. In response to a request for proposals*, one company states their guard rails are perfect for the job. Each section of their guard rails consists of two springs, each having a force constant 3.13400 105 N/m with a maximimum distance of compression of 0.614 m. (According to the manufacturer, beyond this compression the spring loses most of its ability to absorb an impact elastically.) The largest vehicle the guardrails are expected to stop are trucks of mass 4550.000 kg. What is the maximum speed at which these guard rails alone can be expected to bring such vehicles to a halt within the stated maximum compression distance? (Assume the vehicles can strike the guard rail head on and that the springs are perfectly elastic.)
____ m/s
Given your result which section of road often features such a speed
School Zone
Large Road
Highway
Guard rails are pointless if the acceleration they create seriously injures passengers. One important safety factor is the acceleration experienced by passengers during a collision. Calculate the maximum acceleration of the vehicle during the time in which it is in contact with the guard rail.
_____ m/s^2
If the highway department considers 20 g\'s the maximum safe acceleration, is this guard rail safe in regards to acceleration?
Answer:
a) v = 7,207 m / s
, b) a = 42.3 m / s²
Explanation:
We will solve this exercise using the concept of mechanical energy, We will write it in two points before the car touches the springs and in point of maximum compression
Initial
Em₀ = K = ½ m v²
Final
[tex]Em_{f}[/tex] = 2 Ke = ½ k x²
The two is placed because each barred has two springs and each does not exert the same force
Emo = [tex]Em_{f}[/tex]
½ m v² = 2 ½ k x²
v = √(2k/m) x
v = √ (2 3,134 10⁵/4550) 0.614
v = 7,207 m / s
Let's take this speed to km / h
v = 5,096 m / s (1km / 1000m) (3600s / 1h)
v = 25.9 km / h
This speed is common in school zones
Let's use kinematics to calculate the average acceleration
vf² = v₀² - 2 a x
0 = v₀² - 2 a x
a = v₀² / 2 x
a = 7,207²/2 0.614
a = 42.3 m / s²
We buy this acceleration with the acceleration of gravity
a / g = 42.3 / 9.8
a / g = 4.3
This acceleration is well below the maximum allowed
A liquid of density 1290 kg/m 3 1290 kg/m3 flows steadily through a pipe of varying diameter and height. At Location 1 along the pipe, the flow speed is 9.83 m/s 9.83 m/s and the pipe diameter d 1 d1 is 12.1 cm 12.1 cm . At Location 2, the pipe diameter d 2 d2 is 17.7 cm 17.7 cm . At Location 1, the pipe is 8.35 m higher than it is at location 2. Ignoring viscosity, calculate the difference between fluid pressure at location 2 and the fluid pressure at location 1.
Answer:
[tex]\Delta P=1060184.8946\ Pa[/tex]
[tex]P_1=124651.2383\ Pa[/tex]
Explanation:
Given:
density of liquid, [tex]\rho=1290\ kg.m^{-3}[/tex]speed of flow at location 1, [tex]v_1=9.83\ m.s^{-1}[/tex]diameter of pipe at location 1, [tex]d_1=0.121\ m[/tex]diameter of pipe at location 2, [tex]d_2=0.177\ m[/tex]height of pipe at location 1, [tex]z_1=8.35\ m[/tex]We know the Bernoulli's equation of in-compressible flow:
[tex]\frac{P}{\rho.g} +\frac{v^2}{2g} + z=constant[/tex] ........................(1)
Cross sectional area of pipe at location 2:
[tex]A_2=\pi \frac{d_2^2}{4}[/tex]
[tex]A_2=\pi\times \frac{0.177^2}{4}[/tex]
[tex]A_2=0.0246\ m^2[/tex]
Cross sectional area of pipe at location 1:
[tex]A_1=\pi \frac{d_1^2}{4}[/tex]
[tex]A_1=\pi\times \frac{0.121^2}{4}[/tex]
[tex]A_1=0.0115\ m^2[/tex]
Using continuity equation:
[tex]A_1.v_1=A_2.v_2[/tex]
[tex]0.0115\times 9.83=0.0246\times v_2[/tex]
[tex]v_2=4.5953\ m.s^{-1}[/tex]
Now apply continuity eq. on both the locations:
[tex]\frac{P_1}{\rho.g} +\frac{v_1^2}{2g} + z_1= \frac{P_2}{\rho.g} +\frac{v_2^2}{2g} + z_2[/tex]
[tex](P_2-P_1) = \rho.g [\frac{v_1^2}{2g} + z_1-\frac{v_2^2}{2g} ][/tex]
[tex]\Delta P=1290\times 9.8 [\frac{9.83^2}{19.6} + 8.35-\frac{4.5953^2}{19.6} ][/tex]
[tex]\Delta P=154266.016\ Pa[/tex]...................................Ans (a)
Now the mass flow rate at location 1:
[tex]\dot{m_1}=\rho\times \dot{V}[/tex]
[tex]\dot{m_1}=1290\times (0.0115\times 9.83)[/tex]
[tex]\dot{m_1}=145.828\ kg.s^{-1}[/tex]
Now pressure at location 1:
[tex]P_1=\frac{\dot{m_1}\times v_1}{A_1}[/tex]
[tex]P_1=\frac{145.828\times 9.83}{0.0115}[/tex]
[tex]P_1=124651.2383\ Pa[/tex] ...................................Ans (b)
The difference between fluid pressure at location 2 and fluid pressure at location 1 is mathematically given as
dP = 114 kPa
What is the difference between fluid pressure at location 2 and fluid pressure at location 1.?Question Parameter(s):
Generally, the Bernoulli's equation is mathematically given as
P + ρ*g*y + v² =pipe constant
Where
A1*v1 = A2*v2
π*(0.105/2)²*9.91 = π*(0.167/2)²*v2
v2 = 3.9 m/s
Therefore
P1 + ρ*g*y1 + v1² = P2 + ρ*g*y2 + v2²
dP = 1290*9.8*9.01 + 9.91² - 3.9²
dP = 114 kPa
In conclusion, difference between fluid pressure is
dP = 114 kPa
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A fairground ride spins its occupants inside a flying saucer-shaped container. If the horizontal circular path the riders follow has a radius of 5.17 m, at what angular velocity will the riders be subjected to a centripetal acceleration whose magnitude is equal to 1.50 times the acceleration due to gravity?
Answer:
Angular velocity will be 2.843 rad/sec
Explanation:
We have given the radius r = 5.17 m
Centripetal acceleration [tex]a_c=1.5g=1.5\times 9.8=14.7m/sec^2[/tex]
We know that centripetal acceleration is given by
[tex]a_c=\frac{v^2}{r}[/tex]
And linear velocity is given by [tex]v=\omega r[/tex]
[tex]a_c=\frac{(\omega r)^2}{r}=\omega ^2r[/tex]
[tex]14.7=\omega ^2\times 5.17[/tex]
[tex]\omega =2.843rad/sec[/tex]
A parallel beam of light in air makes an angle of 43.5 ∘ with the surface of a glass plate having a refractive index of 1.68. You may want to review (Pages 1080 - 1086) . For related problemsolving tips and strategies, you may want to view a Video Tutor Solution of Reflection and refraction.
a. What is the angle between the reflected part of the beam and the surface of the glass? θθ = nothing ∘
b. What is the angle between the refracted beam and the surface of the glass? θθ = nothing ∘
a. The angle between the reflected part of the beam and the surface of the glass is [tex]\(43.5^\circ\).[/tex]
b. The angle between the refracted beam and the surface of the glass is [tex]\(64.5^\circ\).[/tex]
To solve this problem, we need to apply the laws of reflection and refraction. Let's address each part separately.
Part (a): Angle Between the Reflected Beam and the Surface of the Glass
The law of reflection states that the angle of incidence is equal to the angle of reflection. The angle of incidence is given as 43.5° with respect to the surface of the glass. However, angles in optics are typically measured with respect to the normal (a line perpendicular to the surface).
So, the angle of incidence with respect to the normal (which we'll call [tex]\(\theta_i\)[/tex] ) is:
[tex]\[ \theta_i = 90^\circ - 43.5^\circ = 46.5^\circ \][/tex]
Since the angle of incidence equals the angle of reflection:
[tex]\[ \theta_r = \theta_i = 46.5^\circ \][/tex]
Therefore, the angle between the reflected part of the beam and the surface of the glass is:
[tex]\[ 90^\circ - \theta_r = 90^\circ - 46.5^\circ = 43.5^\circ \][/tex]
So, the angle between the reflected beam and the surface of the glass is:
[tex]\[ 43.5^\circ \][/tex]
Part (b): Angle Between the Refracted Beam and the Surface of the Glass
For the refracted beam, we need to apply Snell's Law, which is:
[tex]\[ n_1 \sin(\theta_i) = n_2 \sin(\theta_t) \][/tex]
Where:
- [tex]\( n_1 \)[/tex] is the refractive index of the first medium (air), [tex]\( n_1 = 1.00 \)[/tex],
- [tex]\( \theta_i \)[/tex] is the angle of incidence with respect to the normal, [tex]\( \theta_i = 46.5^\circ \),[/tex]
- [tex]\( n_2 \)[/tex] is the refractive index of the second medium (glass), [tex]\( n_2 = 1.68 \)[/tex],
- [tex]\( \theta_t \)[/tex] is the angle of refraction with respect to the normal.
Using Snell's Law, we can solve for [tex]\(\theta_t\):[/tex]
[tex]\[ 1.00 \sin(46.5^\circ) = 1.68 \sin(\theta_t) \][/tex]
[tex]\[ \sin(\theta_t) = \frac{\sin(46.5^\circ)}{1.68} \][/tex]
Calculating [tex]\(\sin(46.5^\circ)\):[/tex]
[tex]\[ \sin(46.5^\circ) \approx 0.723 \][/tex]
So,
[tex]\[ \sin(\theta_t) = \frac{0.723}{1.68} \approx 0.430 \][/tex]
Now we find [tex]\(\theta_t\):[/tex]
[tex]\[ \theta_t = \sin^{-1}(0.430) \approx 25.5^\circ \][/tex]
The angle between the refracted beam and the surface of the glass is:
[tex]\[ 90^\circ - \theta_t = 90^\circ - 25.5^\circ = 64.5^\circ \][/tex]
So, the angle between the refracted beam and the surface of the glass is:
[tex]\[ 64.5^\circ \][/tex]
A beam of x-rays with wavelength λ = 0.300 nm is directed toward a sample in which the x-rays scatter off of electrons that are effectively free. The wavelength of the outgoing electrons is measured as a function of scattering angle, where a scattering angle of 0 means the direction of the x-rays was unchanged when passing through the sample. When looking at all possible scattering angles, what are the longest and shortest wavelengths that the scattered x-rays can have?
Answer:
Explanation:
The problem relates to Compton Effect in which electrons are scattered due to external radiation . The electron is scattered out and photons relating to radiation also undergo scattering at angle θ .
The formula relating to Compton Effect is as follows
[tex]\lambda_f-\lambda_i=\frac{h}{m_0c} (1-cos\theta)[/tex]
Here [tex]\lambda_i[/tex] = 3 0 x 10⁻¹¹
For longest [tex]\lambda_f[/tex] θ =180°
[tex]\lambda_f[/tex] = [tex]\lambda_i + \frac{2\times h}{m_0c}[/tex]
= .3 x 10⁻⁹ + [tex]\frac{2\times6.6\times 11^{-34}}{9\times10^{-31}\times3\times10^8}[/tex]
= .348 nm
For shortest wavelength θ = 0
Putting this value in the given formula
[tex]\lambda_f=\lambda_i[/tex]
[tex]\lambda_f[/tex] = .3 nm
Which of the following is a TRUE statement?
a. It is possible for heat to flow spontaneously from a hot body to a cold one or from a cold one to a hot one, depending on whether or not the process is reversible or irreversible.
b. It is not possible to convert work entirely into heat.
c. The second law of thermodynamics is a consequence of the first law of thermodynamics.
d. It is impossible to transfer heat from a cooler to a hotter body.
e. All of these statements are false.
Answer:
e. All of these statements are false.
Explanation:
As we know that heat transfer take place from high temperature to low temperature.
It is possible to convert all work into heat but it is not possible to convert all heat in to work some heat will be reject to the surrounding.
The first law of thermodynamics is the energy conservation law.
Second law of thermodynamics states that it is impossible to construct a device which convert all energy into work without rejecting the heat to the surrounding.
By using heat pump ,heat can transfer from cooler body to the hotter body.
Therefore all the answer is False.
The true statement among the given options is that the entropy of a system can be reduced by cooling it.
The correct answer to the student's question regarding true statements about thermodynamics is option (c) It is always possible to reduce the entropy of a system, for instance, by cooling it. This statement aligns with the principles of thermodynamics, which affirm that entropy, a measure of disorder or randomness, can decrease in a system if energy is removed from the system, such as by lowering its temperature. However, this does not violate the second law of thermodynamics because entropy may still increase in the overall process when considering the surroundings.
In contrast, options (a) and (b) are false because they wrongly imply irreversibility in scenarios where reversibility is possible. Specifically, it is possible to reverse an entropy increase by cooling a system and it is also possible to convert some amount of thermal energy back into mechanical energy, although not with 100% efficiency due to inherent thermodynamic losses.
The second law also articulates that heat transfer occurs spontaneously from a higher to a lower temperature body and not in the reverse direction without external work, implying that a spontaneous flow of heat from a colder to a warmer body is impossible, as is complete conversion of heat to work in a cyclical process.
Why is fusion an appealing energy source?
Fusion products are generally not radioactive.
Extremely high temperatures are required.
The reaction can be confined by available structural materials.
Extremely high pressures are required.
To take place the process of nuclear fusion basically seeks to reach heavy nuclei through light nuclei. Reaching this process implies a release of energy that is what makes this process attractive because it is possible to obtain significant volumes of energy. The procedure to arrive at this process also implies a high cost concerning high temperatures and exorbitant pressures as it is necessary to be able to overcome the barrier of electrostatic repulsion.
This process does not generate any type of radioactive waste like other processes, therefore it is not as dangerous as nuclear fission. For this reason the correct answer is A. Fusion products are generally not radioactive.
A parallel plate capacitor is connected to a battery that maintains a constant potential difference between the plates. If the plates are pulled away from each other, increasing their separation, what happens to the amount of charge on the plates?
a. The amount of the charge decreases, because the capacitance increases.
b. Nothing happens; the amount of charge stays the same.
c. The amount of the charge increases, because the capacitance increases.
d. The amount of the charge increases, because the capacitance decreases.
e. The amount of the charge decreases, because the capacitance decreases.
When the separation between the plates of a parallel plate capacitor is increased, the amount of charge on the plates decreases due to the decrease in capacitance (option e), with the voltage remaining constant.
When parallel plate capacitor plates are pulled away from each other while connected to a battery maintaining a constant potential difference, the capacitance decreases. This is because the capacitance is inversely proportional to the distance between the plates. As the capacitance decreases, the charge on the plates also decreases since the voltage (V) remains constant, and the relation between charge (Q), capacitance (C), and voltage (V) is given by Q = CV. Therefore, the amount of charge on the plates decreases because the capacitance decreases (option e).
Zirconium tungstate is an unusual material because its volume shrinks with an increase in temperature for the temperature range 0.3 K to 1050 K (where it decomposes). In fact, the volumetric coefficient of thermal expansion is –26.4 × 10–6/K. Determine the ratio ΔV/V0 for the above mentioned temperature range. Express your answer in percent.
Answer:
2.771208%
Explanation:
[tex]\Delta V[/tex] = Change of volume
[tex]V_0[/tex] = Initial volume
[tex]\Delta T[/tex] = Change in temperature = (0.3-1050)
[tex]\beta[/tex] = Volumetric coefficient of thermal expansion = [tex]-26.4\times 10^{-6}\ /K[/tex]
Volumetric expansion of heat is given by
[tex]\frac{\Delta V}{V_0}=\beta \Delta T\\\Rightarrow \frac{\Delta V}{V_0}=-26.4\times 10^{-6}\times (0.3-1050)\\\Rightarrow \frac{\Delta V}{V_0}=0.02771208[/tex]
Finding percentage
[tex]\frac{\Delta V}{V_0}=0.02771208\times 100=2.771208\%[/tex]
The ratio of change of volume to initial volume is 2.771208%
To calculate the ratio ΔV/V0 for zirconium tungstate, the volume change can be determined using the volumetric coefficient of thermal expansion. The formula for the ratio is ΔV/V0 = (Volume change/Initial volume) × 100.
Explanation:The ratio ΔV/V0 can be calculated using the formula:
ΔV/V0 = (Volume change/Initial volume) × 100
Given that the volumetric coefficient of thermal expansion for zirconium tungstate is -26.4 × 10^(-6)/K, we can use this value to calculate the volume change. The volume change can be found by multiplying the coefficient of thermal expansion by the change in temperature:
Volume change = (-26.4 × 10^(-6)/K) × (1050 K - 0.3 K)
Using this value, we can calculate the ratio ΔV/V0:
ΔV/V0 = (Volume change/Initial volume) × 100
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A 45.0-kg girl is standing on a 166-kg plank. The plank, originally at rest, is free to slide on a frozen lake, which is a flat, frictionless surface. The girl begins to walk along the plank at a constant velocity of 1.48 m/s to the right relative to the plank. (Let the direction the girl is moving in be positive. Indicate the direction with the sign of your answer.)
1. What is her velocity relative to the surface of the ice?
2. What is the velocity of the plank relative to the surface of ice?
Answer:
-0.31563 m/s
1.16437 m/s
Explanation:
[tex]m_1[/tex] = Mass of girl = 45 kg
[tex]m_2[/tex] = Mass of plank = 166 kg
[tex]v_1[/tex] = Velocity of girl relative to plank = 1.48 m/s
[tex]v_2[/tex] = Velocity of the plank relative to ice surface
In this system the linear momentum is conserved
[tex](m_1+m_2)v_2+m_1v_1=0\\\Rightarrow v_2=-\frac{m_1v_1}{m_1+m_2}\\\Rightarrow v_2=-\frac{45\times 1.48}{45+166}\\\Rightarrow v_2=-0.31563\ m/s[/tex]
Velocity of the plank relative to ice surface is -0.31563 m/s
Velocity of the girl relative to the ice surface is
[tex]v_1+v_2=1.48-0.31563=1.16437\ m/s[/tex]
Unpolarized light is passed through an optical filter that is oriented in the vertical direction.
If the incident intensity of the light is 46 W/m2 , what is the intensity of the light that emerges from the filter? (Express your answer to two significant figures.)
In order to solve this problem it is necessary to apply the concepts related to intensity and specifically described in Malus's law.
Malus's law warns that
[tex]I = I_0 cos^2\theta[/tex]
Where,
[tex]\theta=[/tex] Angle between the analyzer axis and the polarization axis
[tex]I_0 =[/tex]Intensity of the light before passing through the polarizer
The intensity of the beam from the first polarizer is equal to the half of the initial intensity
[tex]I = \frac{I_0}{2}[/tex]
Replacing with our the numerical values we get
[tex]I = \frac{46}{2}[/tex]
[tex]I = 23W/m^2[/tex]
Therefore the intensity of the light that emerges from the filter is [tex]23W/m^2[/tex]
A noninverting op-amp circuit with a gain of 96 V/V is found to have a 3-dB frequency of 8 kHz. For a particular system application, a bandwidth of 32 kHz is required. What is the highest gain available under these conditions?
Final answer:
To find the highest available gain for an op-amp circuit with a required bandwidth, we use the Gain-Bandwidth Product (GBP), which is a constant. For a GBP of 768 kHz and a bandwidth of 32 kHz, the maximum available gain is 24 V/V.
Explanation:
The question concerns determining the highest available gain for an op-amp circuit given a required bandwidth.
The Gain-Bandwidth Product (GBP) is a constant for an op-amp and is found by multiplying the current gain with its corresponding 3-dB frequency.
With an initial gain of 96 V/V and a 3-dB frequency of 8 kHz, the GBP can be calculated as 96 V/V * 8 kHz = 768 kHz.
To meet the requirement of a 32 kHz bandwidth with the same GBP (because GBP is constant), we can calculate the maximum gain as follows:
GBP = Gain * Bandwidth, which gives us
Gain = GBP / Bandwidth.
Plugging in the numbers, we get
Gain = 768 kHz / 32 kHz, resulting in a maximum gain of 24 V/V under the conditions of a 32 kHz bandwidth.
Model the concrete slab as being surrounded on both sides (contact area 24 m2) with a 2.1-m-thick layer of air in contact with a surface that is 5.0 ∘C cooler than the concrete. At sunset, what is the rate at which the concrete loses thermal energy by conduction through the air layer?
Final answer:
The rate at which the concrete loses thermal energy by conduction through the air layer can be calculated using Fourier's Law of Heat Conduction. The formula involves the thermal conductivity, area, temperature difference, and thickness of the air layer. However, without the thermal conductivity value for air, the calculation cannot be completed.
Explanation:
To calculate the rate at which the concrete slab loses thermal energy by conduction through the surrounding air layer at sunset, we can apply Fourier's Law of Heat Conduction. This law states that the heat transfer rate (Q) through a material is directly proportional to the temperature difference across the material (ΔT), the area through which heat is being transferred (A), and the thermal conductivity (k), and inversely proportional to the thickness of the material (L).
The formula to calculate the rate of heat loss is given by Q = k*A*(ΔT/L), where ΔT is the temperature difference between the two sides of the material, A is the contact area, k is the thermal conductivity of the material, and L is the thickness of the material.
Unfortunately, without the thermal conductivity value for air in the provided data, we cannot calculate the exact rate of heat loss for this specific scenario. Thermal conductivity is required to solve this problem, and it's typically obtained from tables in textbooks or scientific references.
A truck horn emits a sound with a frequency of 238 Hz. The truck is moving on a straight road with a constant speed. If a person standing on the side of the road hears the horn at a frequency of 220 Hz, then what is the speed of the truck? Use 340 m/s for the speed of the sound.
Answer:
[tex]v_s=27.8m/s[/tex]
Explanation:
If the person hearing the sound is at rest, then the equation for the frequency heard [tex]f[/tex] given the emitted frequency [tex]f_0[/tex], the speed of the truck [tex]v_s[/tex] and the speed of sound [tex]c[/tex] will be:
[tex]f=f_0\frac{c}{c+v_s}[/tex]
Where [tex]v_s[/tex] will be positive if the truck is moving away from the person, and negative otherwise. We then do:
[tex]\frac{f}{f_0}=\frac{c}{c+v_s}[/tex]
[tex]\frac{f_0}{f}=\frac{c+v_s}{c}=1+\frac{v_s}{c}[/tex]
[tex]v_s=c(\frac{f_0}{f}-1)=(340m/s)(\frac{238Hz}{220Hz}-1)=27.8m/s[/tex]
A plane wave with a wavelength of 500 nm is incident normally ona single slit with a width of 5.0 × 10–6 m.Consider waves that reach a point on a far-away screen such thatrays from the slit make an angle of 1.0° with the normal. Thedifference in phase for waves from the top and bottom of the slitis:
A) 0
B) 0.55 rad
C) 1.1 rad
D) 1.6 rad
E) 2.2 rad
To solve this exercise it is necessary to use the concepts related to Difference in Phase.
The Difference in phase is given by
[tex]\Phi = \frac{2\pi \delta}{\lambda}[/tex]
Where
[tex]\delta =[/tex] Horizontal distance between two points
[tex]\lambda =[/tex] Wavelength
From our values we have,
[tex]\lambda = 500nm = 5*10^{-6}m[/tex]
[tex]\theta = 1\°[/tex]
The horizontal distance between this two points would be given for
[tex]\delta = dsin\theta[/tex]
Therefore using the equation we have
[tex]\Phi = \frac{2\pi \delta}{\lambda}[/tex]
[tex]\Phi = \frac{2\pi(dsin\theta)}{\lambda}[/tex]
[tex]\Phi = \frac{2\pi(5*!0^{-6}sin(1))}{500*10^{-9}}[/tex]
[tex]\Phi= 1.096 rad \approx = 1.1 rad[/tex]
Therefore the correct answer is C.
The phase difference for waves from the top and bottom of the slit can be calculated by using the formula for calculating phase difference. With the provided values for wavelength, angle and slit width, the calculated phase difference is 0.55 rad.
Explanation:The phase difference of the waves is directly related to the path difference between them and can be calculated by using the formula:
Φ = 2 π × (d / λ) × sin(θ).
Where φ is the phase difference, π is Pi, d is the slit width, λ is the wavelength and θ is the incident angle.
Let's plug the provided numbers into our formula:
Φ = 2 π × (5.0x10-6 m / 500x10-9 m) × sin(1.0°).
Φ = 2 π × 10 × sin(1.0°)= 0.55 rad.
So, the correct answer is (B) 0.55 rad.
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Water flows through a horiztonal pipe at a rate of 94 ft3/min. A pressure gauge placed on a 3.3 inch diameter section of the pipe reads 15 psi.
What is the gauge pressure in a section of pipe where the diameter is 5.2 inches?
Answer:
The gauge pressure is 1511.11 psi.
Explanation:
Given that,
Flow rate = 94 ft³/min
Diameter d₁=3.3 inch
Diameter d₂ = 5.2 inch
Pressure P₁= 15 psi
We need to calculate the pressure on other side
Using Bernoulli equation
[tex]P_{1}+\dfrac{1}{2}\rho v_{1}^2=P_{2}+\dfrac{1}{2}\rho v_{2}^2[/tex]
We know that,
[tex]V=Av[/tex]
[tex]v=\dfrac{V}{A}[/tex]
Where, V = volume
v = velocity
A = area
Put the value of v into the formula
[tex]P_{1}+\dfrac{1}{2}\rho (\dfrac{V}{A_{1}})^2=P_{2}+\dfrac{1}{2}\rho (\dfrac{V}{A_{2}})^2[/tex]
Put the value into the formula
[tex]15+\dfrac{1}{2}\times0.36\times(\dfrac{2707.2\times4}{\pi\times(3.3)^2})^2=P_{2}+\dfrac{1}{2}\times0.36\times(\dfrac{2707.2\times4}{\pi\times(5.2)^2})^2[/tex]
[tex]P_{2}=15+\dfrac{1}{2}\times0.036\times(\dfrac{2707.2\times4}{\pi\times(3.3)^2})^2-\dfrac{1}{2}\times0.036\times(\dfrac{2707.2\times4}{\pi\times(5.2)^2})^2[/tex]
[tex]P_{2}=1525.8\ psi[/tex]
We need to calculate the gauge pressure
Using formula of gauge pressure
[tex]P_{g}=P_{ab}-P_{atm}[/tex]
Put the value into the formula
[tex]P_{g}=1525.8-14.69[/tex]
[tex]P_{g}=1511.11\ psi[/tex]
Hence, The gauge pressure is 1511.11 psi.
A sinusoidal electromagnetic wave is propagating in a vacuum in the +z-direction.
Part A
If at a particular instant and at a certain point in space the electric field is in the +x-direction and has a magnitude of 3.40V/m , what is the magnitude of the magnetic field of the wave at this same point in space and instant in time?
Part B
What is the direction of the magnetic field?
Answer:
a) 1.13 10-8 T. b) +y direction
Explanation:
a)
For an electromagnetic wave propagating in a vacuum, the wave speed is c = 3. 108 m/s.
At a long distance from the source, the components of the wave (electric and magnetic fields) can be considered as plane waves, so the equations for them can be written as follows:
E(z,t) = Emax cos (kz-ωt-φ) +x
B(z,t) = Bmax cos (kz-ωt-φ) +y
In an electromagnetic wave, the magnetic field and the electric field, at any time, and at any point in space, as the perturbation is propagating at a speed equal to c (light speed in vacuum), are related by this expression:
Bmax = Emax/c
So, solving for Bmax:
Bmax = 3.4 V/m / 3 108 m/s = 1.13 10-8 T.
b) As we have already said, in an electromagnetic wave, the electric field and the magnetic field are perpendicular each other and to the propagation direction, so in this case, the magnetic field propagates in the +y direction.
The return-air ventilation duct in a home has a cross-sectional area of 900 cm^2. The air in a room that has dimensions 5.0 m x 11.0 m ×x 2.4 m is to be completely circulated in a 50-min cycle.
1) What is the speed of the air in the duct? (Express your answer to two significant figures.)
To solve the problem it is necessary to apply the concepts related to the flow rate of a fluid.
The flow rate is defined as
[tex]Q = Av[/tex]
Where,
[tex]Q = Discharge (m^3/s)[/tex]
[tex]A = Area (m^2)[/tex]
v = Average speed (m / s)
And also as
[tex]Q = \frac{V}{t}[/tex]
Where,
V = Volume
t = time
Let's start by finding the total volume according to the given dimensions, that is to say
[tex]V = 5*11*2.4[/tex]
[tex]V = 132m^3[/tex]
The entire cycle must be completed in 50 min = 3000s
In this way we know that the [tex]132m ^ 3[/tex] must be filled in 3000s, that is to say that there should be a flow of
[tex]Q = \frac{V}{t}[/tex]
[tex]Q = \frac{132}{3000}[/tex]
[tex]Q = 0.044m^3/s[/tex]
Using the relationship to find the speed we have to
[tex]Q = Av[/tex]
[tex]v = \frac{Q}{A}[/tex]
Replacing with our values,
[tex]v = \frac{0.044}{900*10^{-4}m^2}[/tex]
[tex]v = 0.488m/s[/tex]
Therefore the air speed in the duct must be 4.88m/s
What is the distance from axis about which a uniform, balsa-wood sphere will have the same moment of inertia as does a thin-walled, hollow, lead sphere of the same mass and radius R, with the axis along a diameter, to the center of the balsa-wood sphere?
Answer:
[tex]D_{s}[/tex] ≈ 2.1 R
Explanation:
The moment of inertia of the bodies can be calculated by the equation
I = ∫ r² dm
For bodies with symmetry this tabulated, the moment of inertia of the center of mass
Sphere [tex]Is_{cm}[/tex] = 2/5 M R²
Spherical shell [tex]Ic_{cm}[/tex] = 2/3 M R²
The parallel axes theorem allows us to calculate the moment of inertia with respect to different axes, without knowing the moment of inertia of the center of mass
I = [tex]I_{cm}[/tex] + M D²
Where M is the mass of the body and D is the distance from the center of mass to the axis of rotation
Let's start with the spherical shell, axis is along a diameter
D = 2R
Ic = [tex]Ic_{cm}[/tex] + M D²
Ic = 2/3 MR² + M (2R)²
Ic = M R² (2/3 + 4)
Ic = 14/3 M R²
The sphere
Is =[tex]Is_{cm}[/tex] + M [[tex]D_{s}[/tex]²
Is = Ic
2/5 MR² + M [tex]D_{s}[/tex]² = 14/3 MR²
[tex]D_{s}[/tex]² = R² (14/3 - 2/5)
[tex]D_{s}[/tex] = √ (R² (64/15)
[tex]D_{s}[/tex] = 2,066 R
A thin flashlight beam traveling in air strikes a glass plate at an angle of 52° with the plane of the surface of the plate. If the index of refraction of the glass is 1.4, what angle will the beam make with the normal in the glass?
To solve this problem it is necessary to apply Snell's law and thus be able to calculate the angle of refraction.
From Snell's law we know that
[tex]n_1sin\theta_1 = n_2 sin\theta_2[/tex]
Where,
n_i = Refractive indices of each material
[tex]\theta_1[/tex] = Angle of incidence
[tex]\theta_2[/tex] = Refraction angle
Our values are given as,
[tex]\theta_1 = 38\°[/tex]
[tex]n_1 = 1[/tex]
[tex]n_2 = 1.4[/tex]
Replacing
[tex]1*sin38 = 1.4*sin\theta_2[/tex]
Re-arrange to find [tex]\theta_2[/tex]
[tex]\theta_2 = sin^{-1} \frac{sin38}{1.4}[/tex]
[tex]\theta_2 = 26.088°[/tex]
Therefore the angle will the beam make with the normal in the glass is 26°
About once every 30 minutes, a geyser known as Old Faceful projects water 11.0 m straight up into the air.
Use g = 9.80 m/s2, and take atmospheric pressure to be 101.3 kPa. The density of water is 1000 kg/m3.
(a) What is the speed of the water when it emerges from the ground? m/s
(b) Assuming the water travels to the surface through a narrow crack that extends 9.00 m below the surface, and that the water comes from a chamber with a large cross-sectional area, what is the pressure in the chamber?
Answers:
a) [tex]8820 m/s[/tex]
b) [tex]189500 Pa[/tex]
Explanation:
We have the following data:
[tex]t=30 min \frac{60 s}{1 min}=1800 s[/tex] is the time
[tex]h=11 m[/tex] is the height the water reaches vertically
[tex]g=9.8 m/s^{2}[/tex] is the acceleration due gravity
[tex]P_{air}=101.3 kPa=101.3(10)^{3} Pa[/tex] is the pressure of air
[tex]\rho_{water}=1000 kg/m^{3}[/tex] is the density of water
Knowing this, let's begin:
a) Initial speed of waterHere we will use the following equation:
[tex]h=h_{o}+V_{o}t-\frac{g}{2}t^{2}[/tex] (1)
Where:
[tex]h_{o}=0 m[/tex] is the initial height of water
[tex]V_{o}[/tex] is the initial speed of water
Isolating [tex]V_{o}[/tex]:
[tex]V_{o}=\frac{1}{t}(h+\frac{g}{2}t^{2})[/tex] (2)
[tex]V_{o}=\frac{1}{1800 s}(11 m+\frac{9.8 m/s^{2}}{2}(1800 s)^{2})[/tex]
[tex]V_{o}=8820.006 m/s \approx 8820 m/s[/tex] (3)
b) Pressure in the chamber
In this part we will use the following equation:
[tex]P=\rho_{water} g d + P_{air}[/tex] (4)
Where:
[tex]P[/tex] is the absolute pressure in the chamber
[tex]d=9 m[/tex] is the depth
[tex]P=(1000 kg/m^{3})(9.8 m/s^{2})(9 m) + 101.3(10)^{3} Pa[/tex]
[tex]P=189500 Pa[/tex] (5)