Answer:
a) The distance between the ineas doubles, intensity decreases with distance
b) The distance between the ineas doubles
Explanation:
The diffraction pattern of a grid is given as a percentage
d sin θ = m λ
where d give the distance between two consecutive lines, θ it is at an angle, λ the wavelength and m is an integer that determines the order of diffraction, let's not forget that the entire spectrum is at a value of m and then it is repeated.
Let's apply this to our case
a) distance from grid to observation screen doubles
Here we have two effect:
* the energy of the source is constant, it must be distributed over a surface, therefore the intensity decreases with distance
* The other factor can be found using trigonometry
tan θ= y / L
where y is the distance from the central maximum to the line under study and L is the distance to the screen
In general, diffraction experiments cover very small angles
tan θ = sin θ/ cos θ = sin θ
we substitute
sinθ = y / L
we subtitle into the diffraction equation
d y / L = m λ
y = L / d m λ
L = 2 L₀
y = 2 L₀ m λ / d
we see that by doubling the distance to the screen the lines we are seeing are separated by double
b) When the density of lines doubles, it means that in the same distance I have twice as many lines, therefore the distance between two consecutive lines is reduced by half
d = d₀o / 2
y = (L m λ) / d
y = (L m λ/ d₀) 2
we see that The distance between the ineas doubles
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 charged particle is moving perpendicular to a magnetic field in a circle with a radius r. An identical charge particle enters the field, with v prependicular to B, but with a higher speed than the first particle. Compared with the radius of the circle for the first particle, the radius of the circular path of the second particle is
Answer:
The second particle will move through the field with a radius greater that the radius of the first particle
Explanation:
For a charged particle, the force on the particle is given as
[tex]F = \frac{mv^{2} }{r}[/tex]
also recall that work is force times the distance traveled
work = F x d
so, the work on the particle = F x d,
where the distance traveled by the particle in one revolution = [tex]2\pi r[/tex]
Work on a particle = 2πrF = [tex]2\pi mv^{2}[/tex]
This work is proportional to the energy of the particle.
And the work is also proportional to the radius of travel of the particles.
Since the second particle has a bigger speed v, when compared to the speed of the first particle, then, the the second particle has more energy, and thus will move through the field with a radius greater that the radius of the first particle.
A positive test charge of 5.00 E-5 C is places in an electric field. The force on it is 0.751 N. The magnitude of the electric field at the location of the test charge is
1.50 E4 N/C
1.52 E5 N/C
3.75 E4 N/C
3.75 E5 N/C
6.75 E4 N/C
Answer:
[tex]\boxed{\mathrm{1.50 \: E^4 \: N/C}}[/tex]
Explanation:
[tex]\displaystyle \mathrm{E=\frac{F_e}{q} }[/tex]
[tex]\displaystyle \mathrm{Electric \: field \: strength \: (N/C)=\frac{Electric \: force \: (N)}{Charge \: (C)} }[/tex]
[tex]\displaystyle \mathrm{E=\frac{0.751}{5.00\: E^{-5}} }[/tex]
[tex]\displaystyle \mathrm{E=\frac{0.751}{0.00005} }[/tex]
[tex]\displaystyle \mathrm{E=15020}[/tex]
The magnitude of the electric field at the location of the test charge is [tex]1.50E^{4}N/C[/tex]
Given that,
A positive test charge of 5.00 E-5 C is places in an electric field. The force on it is 0.751 N.Based on the above information, the calculation is as follows:
[tex]= 0.751 \div 5.00E^{-5}\\\\= 0.751 \div 0.00005[/tex]
= 15020
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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
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
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]
A solid non-conducting sphere of radius R carries a charge Q1 distributed uniformly. The sphere is surrounded by a concentric spherical shell of inner radius Ra and outer radius Rb . The shell carries a total charge Q2 distributed uniformly in its volume. What is the net electric field at a radial distance r such that R < r < Ra
Answer:
E = k Q₁ / r²
Explanation:
For this exercise that asks us for the electric field between the sphere and the spherical shell, we can use Gauss's law
Ф = ∫ E .dA = [tex]q_{int}[/tex] / ε₀
where Ф the electric flow, qint is the charge inside the surface
To solve these problems we must create a Gaussian surface that takes advantage of the symmetry of the problem, in this almost our surface is a sphere of radius r, that this is the sphere of and the shell, bone
R <r <R_a
for this surface the electric field lines are radial and the radius of the sphere are also, therefore the two are parallel, which reduces the scalar product to the algebraic product.
E A = q_{int} /ε₀
The charge inside the surface is Q₁, since the other charge Q₂ is outside the Gaussian surface, therefore it does not contribute to the electric field
q_{int} = Q₁
The surface area is
A = 4π r²
we substitute
E 4π r² = Q₁ /ε₀
E = 1 / 4πε₀ Q₁ / r²
k = 1/4πε₀
E = k Q₁ / r²
g Suppose that you charge a 3 F capacitor in a circuit containing eight 3.0 V batteries, so the final potential difference across the plates is 24.0 V. How much charge is on each plate
The complex, highly technical formula for capacitors is
Q = C V
Charge = (capacitance) (voltage)
Charge = (3 F) (24 V)
Charge = 72 Coulombs
The positive plate of the capacitor is missing 72 coulombs worth of electrons. They were sucked into positive terminal of the battery stack.
The negative plate of the capacitor has 72 coulombs worth of extra electrons. They came from the negative terminal of the battery stack.
You should be aware that this is a humongous amount of charge ! An average lightning bolt, where electrons flow between a cloud and the ground for a short time, is estimated to transfer around 15 coulombs of charge !
The scenario in the question involves a "supercapacitor". 3 F is is no ordinary component ... One distributor I checked lists one of these that's able to stand 24 volts on it, but that product costs $35 apiece, you have to order at least 100 of them at a time, and they take 2 weeks to get.
Also, IF you can charge this animal to 24 volts, it will hold 864J of energy. You'd probably have a hard time accomplishing this task with a bag of leftover AA batteries.
A circuit element maintains a constant resistance. If the current through the circuit element is doubled, what is the effect on the power dissipated by the circuit element
Answer:
This shows that the power dissipated by the circuit element is four times its original power if the current is doubled.Explanation:
The formula for calculating the power expended in a circuit is P = I²R... 1
i is the current (in amperes)
R is the resistance (in ohms)
If a circuit element maintains a constant resistance and the current through the circuit element is doubled, then new current I₂ = 2I
New power dissipated P₂ = (I₂)²R
P₂ = (2I)²R
P₂ = 4I²R ... 2
Dividing equation 2 by 1 will give;
P₂/P = 4I²R/I²R
P₂/P = 4
P₂ = 4P
This shows that the power dissipated by the circuit element is four times its original power if the current is doubled.
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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A rigid massless rod is rotated about one end in a horizontal circle. There is a particle of mass m1 attached to the center of the rod and a particle of mass m2 attached to the outer end of the rod. The inner section of the rod sustains a tension that is 5 times as great as the tension that the outer section sustains. Find the ratio m1/m2
Answer:
m₁/m₂ =8
Explanation:
Writing the centripetal force equation: where T₁ is inner tension and T₂ is outer tension
T₂ = m₂ × ω²R....equation 1
T₁-T₂ = m₁ × ω²R/2
Also T₁=5T₂
5T₂-T₂ = m₁ × ω²R/2
4 × 2T₂ = m₁ × ω²R
8T₂ = m₁ × ω²R
dividing it by equation 1
8 = m₁/m₂
Hence m₁/m₂ =8
how is mirage formed
Answer:
Mirages are formed when the ground is really hot and the air is really cold. The hot ground will warm a layer of the air closest to the ground. When the light moves through the cold air and hits the warm air it bends creating the U shaped bend
Explanation:
Hope this helps!
Three rocks of equal mass are thrown with identical speeds from the top of the same building (identical height). Rock X is thrown vertically downward, rock Y is thrown vertically upward, and rock Z is thrown horizontally.Required:Which rock has the greatest speed just before it hits the ground? Assume air resistance is negligible.
Answer:Three rocks of equal mass are thrown
Explanation:
The phenomena: Three rocks of equal mass are thrown with identical speeds from the top of the same building (identical height). Rock X is thrown vertically downward, rock Y is thrown vertically upward, and rock Z is thrown horizontally.
The exact ____ to be removed by the patient is determined according to the examination
Answer:
organ
Explanation:
The organ removal is the process that involves deep examination. The vital organs may be removed or transplanted if they suffer some sort of disease. The human body has two lungs and two kidneys, if there is a failure or disease in one of these organ the patient can survive without any transplant. In case of heart disease the transplant is the only option that a patient may have. The exact procedure for removal of these organ is determined after deep examination of patient and identifying his history of diseases.
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]
What is the equivalent temperature in kelvin if you have a metal at 50°F?
Answer:
The required temperature is 283 K.
Explanation:
[tex]T\:=\:\left(50-32\right)\times \frac{5}{9}+273\\\\T=283\:K[/tex]
Best Regards!
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)
According to the law of conservation of energy, if the ocean water cools, then something else should warm. What is it that warms?
Answer:
The answer is air
Explanation:
The ocean is salty because it contains many dissolved ions. As these charged particles move with the water in strong ocean currents, they feel a force from the earthâs magnetic field. Positive and negative charges are separated until an electric field develops that balances this magnetic force. This field produces measurable potential differences that can be monitored by ocean researchers. The Gulf Stream moves northward off the east coast of the United States at a speed of up to 3.5 m/s . Assume that the current flows at this maximum speed and that the earthâs field is 50 μT tipped 60 degrees below horizontal.
Part A
What is the direction of the magnetic force on a singly ionized negative chlorine ion moving in this ocean current? Choose best answer
(a) east
(b) west
(c) up
(d) down
Part BWhat magnitude electric field is necessary to exactly balance this magnetic force?
Choose best answer.
(a) 1.8 x 10-4 N/C
(b) 1.5 x 10-4 N/C
(c) 1.0 x 10-4 N/C
Given that,
Speed = 3.5 m/s
Magnetic field = 50μT
Angle = 60°
(A). We need to find the direction of magnetic force
Using formula of magnetic force
[tex]\vec{F}=q(\vec{v}\times\vec{B})[/tex]
Here, [tex](\vec{v}\times\vec{B})[/tex]= down
But , charge is negative.
So, the direction of magnetic force will be up.
(B). We need to calculate the magnetic electric field
Using formula of magnetic force
[tex]F=qvB\sin\theta[/tex]
[tex]qE=qvB\sin\theta[/tex]
[tex]E=vB\sin\theta[/tex]
Where, v = speed
B = magnetic field
Put the value into the formula
[tex]E=3.5\times50\times10^{-6}\sin60[/tex]
[tex]E=0.000151\ N/C[/tex]
[tex]E=1.5\times10^{-4}\ N/C[/tex]
Hence, (A). The direction of magnetic force is UP
(c) is correct option
(B). The magnetic electric field is [tex]1.5\times10^{-4}\ N/C[/tex]
(b) is correct option
Having aced your introductory physics course, you are hired as a summer intern at NASA. You are sent as part of a team to explore the possibility of mining iron from a small, airless spherical asteroid with a radius of 499 km and a surface acceleration of 2.6 m/sec2.
1) What is the mass of the asteroid?
2) What is the minimum vertical velocity you could give a 4 kg rock from the surface of this asteroid so that it continues to move upwards forever (i.e. it would never return to the surface)?
3) What is the minimum vertical velocity you could give an 8 kg rock from the surface of this asteroid so that it continues to move upwards forever (i.e. it would never return to the surface)?
4) How far will the 4 kg rock go if it leaves the asteroid's surface with a vertical velocity of 1000 m/s?
5) What will the speed of the 8 kg rock be at the surface of the asteroid if it is dropped from a point 1000 km above the surface of the asteroid?
Answer:
1) M = 9.7*10^21 kg
2, 3) 1610.81 m/s
4) 312.911 km
5) 1315.65 m/s
Explanation:
Given:-
- The radius of the asteroid, R = 499 km
- The surface acceleration, g = 2.6 m/s^2
Solution:-
- To determine the mass of the asteroid ( M ) we will use the relation for the gravitational acceleration produced by a spherical object of mass ( M ) as follows:
[tex]g = G\frac{M}{R^2}[/tex]
Where,
G: the universal gravitational constant = 6.674*10^−11
- Use the radius of asteroid ( R ) and the given surface acceleration ( g ) and solve for the mass ( M ) of the asteroid using the relation given above:
[tex]M = \frac{gR^2}{G} \\\\M = \frac{2.6*(499,000)^2}{6.674*10^-^1^1} \\\\M = 9.7*10^2^1 kg[/tex]
- To determine the escape velocity of the mass of rock ( m = 4kg ) from the gravitational pull of the asteroid. We will use the conservation of energy principle.
- The conservation of energy principle states:
[tex]K.E_i + P.E_i = K.E_f + P.E_f[/tex]
Where,
K.Ei: The initial kinetic energy of the rock of mass ( m )
P.Ei: The potential energy of the system at the surface
K.Ef: The final kinetic energy of the rock of mass ( m )
P.Ef: The final potential energy of the system at infinite
- The gravitational potential energy of the system of an object of mass (m ) at any distance ( r ) from the center of the more massive body ( M ) is given as:
[tex]P.E = - G\frac{M*m}{r}[/tex]
- The escape velocity is just enough initial velocity ( ve ) that allows an object to cross the gravitational effect of the massive body. Once the effect of the gravity is insignificantly small ( infinite ). Almost all of the kinetic energy has been lost by doing work against the gravitational pull.
- Therefore, K.Ef = P.Ef = 0 ( at infinity ).
[tex]0.5mv_e^2 - G\frac{Mm}{r} = 0\\\\v_e = \sqrt{2G\frac{M}{r} } = \sqrt{2G\frac{M}{R} } \\\\v_e = \sqrt{2(6.674*10^-^1^1)\frac{9.7*10^2^1}{499,000} }\\\\v_e = 1610.81 \frac{m}{s}[/tex]
- From the above relationship derived for an object at the surface of an asteroid body to escape the grasp of the gravitational pull is independent of the mass of the object ( m ). Hence, whatever the mass of the object is it does not affect the required escape velocity.
Answer: The rock of mass m = 8 kg and m = 4 kg require minimum vertical velocity of 1610.81 m/s to escape the asteroid gravitational pull.
- For the final part we will again apply the principle of conservation of energy for the system of asteroid of mass ( M ) and an object of mass ( m ). We have:
[tex]K.E_i + P.E_i = K.E_f + P.E_f[/tex]
Where,
K.Ei: The initial kinetic energy of the rock of mass ( m ) = 0 ( dropped)
P.Ei: The potential energy of the system at altitude ( h )
K.Ef: The final kinetic energy of the rock of mass ( m ) at surface
P.Ef: The final potential energy of the system at surface
[tex]0 - G\frac{Mm}{( R + h )} = 0.5mv^2 - G\frac{Mm}{ R } \\\\v^2 = 2GM* [ \frac{1}{R} - \frac{1}{R+h} ] \\\\v = \sqrt{2(6.674*10^-^1^1)*(9.70*10^2^1)* [ \frac{1}{499,000} - \frac{1}{1499000}]}\\\\v = 1315.65 \frac{m}{s}[/tex]
Answer: The speed of the 8kg rock upon hitting the surface of asteroid would be 1315.65 m/s
- Similarly, how far away a rock of mass ( m ) can go to from the surface of asteroid if it leaves the surface with initial velocity vi = 1000 m/s. We will use the energy conservation expression derived in the previous part. We have:
[tex]0.5mv^2 - G\frac{Mm}{R} = G\frac{Mm}{R + h} \\\\\frac{0.5v^2}{GM} - \frac{1}{R} = -\frac{1}{R + h}\\\\R + h = \frac{1}{-\frac{0.5v^2}{GM} + \frac{1}{R}} \\\\h = \frac{1}{-\frac{0.5v^2}{GM} + \frac{1}{R}} - R\\\\h = \frac{1}{-\frac{0.5(1000)^2}{(6.674*10^-^1^1)*(9.7*10^2^1)} + \frac{1}{(499000)}} - 499000\\\\h = 312.911 km[/tex]
Answer: The 4 kg rock would be able to travel 312.911 km above the asteroid surface if it had an initial velocity of 1000 m/s.
Which describes any compound that has at least one element from group 17? Halide;noble gas; metalliod; transition metal
Answer:
Halide
Explanation:
It has at least one element from the halogen group (17)
Halide describes any compound that has at least one element from group 17, therefore the correct option is option A.
What are halides?When the elements belonging to group 17 of the periodic table form ionic compounds with other electropositive elements, then these compounds are known as halides.
These elements from group 17 are also known as halogens. Generally, these halides have very high electronegativity as they reside on the right side of the periodic table.
Generally, the valency of the halogens element involved in the halide compound is one and they form ionic compounds with the alkali and alkaline earth metals.
Thus, halides are compounds that have at least one element from group 17.
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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
5- The inside of a Carnot refrigerator is maintained at a temperature of 277 K, while the temperature in the kitchen is 303 K. Using 2159 J of work, how much heat can this refrigerator remove from its inside compartment
Answer:
The heat is [tex]H = 23001.65 \ J[/tex]
Explanation:
From the question we are told that
The inside temperature is [tex]T_i = 277 \ K[/tex]
Te kitchen temperature is [tex]T_k = 303 \ K[/tex]
The workdone is [tex]W = 2159 \ J[/tex]
The coefficient of performance of the refrigerator is mathematically represented as
[tex]COP = \frac{1}{\frac{T_k}{T_i} -1}[/tex]
substituting values
[tex]COP = \frac{1}{\frac{303}{277} -1}[/tex]
[tex]COP = \frac{277}{26}[/tex]
[tex]COP = 10.65[/tex]
Now this coefficient of performance of the refrigerator can also be represented mathematically as
[tex]COP = \frac{H}{W}[/tex]
Where H is the heat which the refrigerator removes from the inside component
So
[tex]H = COP * W[/tex]
substituting values
[tex]H = 10.65 * 2159[/tex]
[tex]H = 23001.65 \ J[/tex]
A tugboat tows a ship at a constant velocity. The tow harness consists of a single tow cable attached to the tugboat at point A that splits at point B and attaches to the ship at points C and D. The two rope segments BC and BD angle away from the center of the ship at angles of ϕ = 25.0 ∘ and θ = 23.0 ∘, respectively. The tugboat pulls with a force of 3700 lb . What are the tensions TBC and TBD in the rope segments BC and BD?
Explanation:
perpendicular to the rope
Tcd×sin23 = Tbd×sin25
Tcd = 1.08160 Tbd
along the rope
Tcd×cos23 + Tbd×cos25 =F
1.08160×Tbd×cos23 + Tbd×cos25 = 3700 lb
Tbd (1.08160×cos23°+cos25°)=3700
Therefore, Tbd = 1945.3965 lb
Tcd = 1945.3965×1.0816 = 2104.14085 lb.
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.
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
At what minimum angle will you get total internal reflection of light traveling in flint glass and reflected from water?
Answer:
θ_c = 53.65°
Explanation:
The point after which the light ray had started reflecting internally will be when the reflecting angle is at 90°. The incident angle at this point is called the critical angle and this can be calculated through Snell's law as;
n1 sin θ_c = n2 sin 90
Where;
n1 is the refractive index of the medium through which the incident rays will pass through.
n2 is the Refractive index of the medium through which the refracted rays will pass through.
θ_c is the critical angle at which the incident ray will reflect totally internally.
Now, since sin 90 = 1
Thus;
n1 sin θ_c = n2
θ_c = sin^(-1) (n2/n1)
Now, we are told that the reflection travels in flint glass and reflected from water.
Thus, the first medium is flint glass and the second medium is water.
So, from tables,
Refractive index of flint glass; n1 = 1.655
Refractive index of water; n2 = 1.333
Thus;
θ_c = sin^(-1) (1.333/1.655)
θ_c = 53.65°
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
1) Using only astronomical data, calculate the speed of the planet Venus in its essentially circular orbit around the sun.
Venus = 4.87x10^24
2) Using only astronomical data, calculate the gravitational force that the sun must be exerting on Venus.
Answer:
1) v_orbit = 3.49*10^4 m/s
2) F = 5.51*10^22 N
Explanation:
1) In order to calculate the speed of Venus in its orbit, you use the following formula:
[tex]v_{orbit}=\sqrt{\frac{GM_s}{R}}[/tex] (1)
v_orbit: speed of Venus = ?
G: Cavendish's constant = 6.674*10^-11.m^3kg^-1s^-2
Ms: mass of the sun = 1.98*10^30 kg
R: distance between the center of Sun and the center of Venus = 1.08*10^11m
You replace the values of the parameters in the equation (1):
[tex]v_{orbit}=\sqrt{\frac{(6.674*10^{-11}m^3kg^{-1}s^{-2})(1.98*10^{30})}{1.08*10^{11}m}}\\\\v_{orbit}=3.49*10^4\frac{m}{s}[/tex]
The speed of Venus in its orbit around the Sun is 3.49*10^4 m/s
2) The force is given by the following formula:
[tex]F=G\frac{M_vM_s}{R^2}[/tex]
Ms: mass of Venus = 4.87*10^24 kg
[tex]F=(6.674*10^{-11}m^3kg^{-1}s-2})\frac{(4.87*10^{24}kg)(1.98*10^{30}kg)}{(1.08*10^{11}m)^2}\\\\F=5.51*10^{22}N[/tex]
The Sun exertes on Venus a force of 5.51*10^22 N
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