The energy stored in the capacitor dus to 10C charge and 40V potential difference is 200 Joules.
Firstly we will need to calculate the capacitance of the capacitor. It will further help in calculation of stored energy. The formula to be used for each of the following is -
Q = CV and E = 1/2CV², where Q represents charge, E represents energy, C is capacitance and V is voltage.
Keep the values in formula -
C = 10/40
Performing division
C = 1/4
E = 1/2×1/4×40²
Taking square
E = 1/2×1/4×1600
Performing multiplication and division
E = 200 J
Hence, the stored energy is 200 Joules.
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Two very small +3.00-μC charges are at the ends of a meter stick. Find the electric potential (relative to infinity) at the center of the meter stick.
Answer:
The electric potential at the center of the meter stick is 54 KV.
Explanation:
Electric potential (V) is given as:
i.e V = [tex]\frac{kq}{r}[/tex]
Where: k is the Coulomb constant, q is the charge and r is the distance.
Given: q = 3.0 μC = 3.0 x [tex]10^{-6}[/tex] C, r = 0.5 m
So that,
V = [tex]\frac{9*10^{9}*3.0*10^{-6} }{0.5}[/tex]
= [tex]\frac{2.7*10^{4} }{0.5}[/tex]
V = 54000
= 54 000 volts
The electric potential at the center of the meter stick is 54 KV.
Need help on another homework question
Rick places the blue lens of one pair of 3D glasses over the red lens of another pair. He then looks through both lenses at the same time. What color will he see?
A. blue
B. black
C. red
D. white
Answer:
a
Explanation:
Which list of reaction types are all redox reactions?
A.
Synthesis, decomposition, single-replacement, combustion
B.
Synthesis, double-replacement, combustion, decomposition
C.
Acid-base, single-replacement, double-replacement, synthesis
D.
Decomposition, double-replacement, acid-base, synthesis
A. Synthesis, decomposition, single-replacement, combustion
List of reaction types are redox reactions:
Synthesis, decomposition, single-replacement, combustion.What are redox reaction ?"Redox reactions are oxidation-reduction chemical reactions in which the reactants undergo a change in their oxidation states." The term 'redox' is a short form of reduction-oxidation. All the redox reactions can be broken down into two different processes – a reduction process and an oxidation process.
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For PN junctions, determine if each statement below is True or False:
a. There can be large net current from p-side to n-side under forward bias.
b. There can be large net current from n-side to p-side under reverse bias if reverse bias is sufficiently high.
c. Electron diffusion current flows from n-side to p-side.
d. Electric field magnitude is higher under reverse bias
e. Electrons in the transition region drift from p-side to n-side.
Answer:
a) True
b) True
c) false
d) True
e) True
Explanation:
a) True
In forward bias, the resistance of the p–n junction reduces and hence the electric charges can flow easily.
b) True
In reverse bias condition, the electric current is due to the minority charge carriers (negative).
c) False
direction of diffusion current is in the direction of movement of positive charge i,e towards n side
d) True
Because the breakdown of charge carriers occur due to which the current increases rapidly
e) True
It has been suggested that rotating cylinders several miles in length and several miles in diameter be placed in space and used as colonies. Inhabitants of the space colonies would live on the inside surface of the cylinder. Inertial effects would resemble gravity's influence and keep them 'plastered to the surface.' Suppose that you are an inhabitant of a space colony which is 1070 miles in length and 4.86 miles in diameter. How many revolutions per hour must the cylinder have in order for the occupants to experience a centripetal acceleration equal to the acceleration of gravity
Answer:
the required revolution per hour is 28.6849
Explanation:
Given the data in the question;
we know that the expression for the linear acceleration in terms of angular velocity is;
[tex]a_{c}[/tex] = rω²
ω² = [tex]a_{c}[/tex] / r
ω = √( [tex]a_{c}[/tex] / r )
where r is the radius of the cylinder
ω is the angular velocity
given that; the centripetal acceleration equal to the acceleration of gravity a [tex]a_{c}[/tex] = g = 9.8 m/s²
so, given that, diameter = 4.86 miles = 4.86 × 1609 = 7819.74 m
Radius r = Diameter / 2 = 7819.74 m / 2 = 3909.87 m
so we substitute
ω = √( 9.8 m/s² / 3909.87 m )
ω = √0.002506477 s²
ω = 0.0500647 ≈ 0.05 rad/s
we know that; 1 rad/s = 9.5493 revolution per minute
ω = 0.05 × 9.5493 RPM
ω = 0.478082 RPM
1 rpm = 60 rph
so
ω = 0.478082 × 60
ω = 28.6849 revolutions per hour
Therefore, the required revolution per hour is 28.6849
The required revolution per hour is 28.6849
Calculation of revolution per hour:
The expression for the linear acceleration with respect to the angular velocity is
= rω²
So,
ω² = / r
ω = √( / r )
Here r represent the radius of the cylinder
ω represent the angular velocity
Now
diameter = 4.86 miles = 4.86 × 1609 = 7819.74 m
And,
Radius r = Diameter / 2 = 7819.74 m / 2 = 3909.87 m
So,
ω = √( 9.8 m/s² / 3909.87 m )
ω = √0.002506477 s²
ω = 0.0500647 ≈ 0.05 rad/s
Now
1 rad/s = 9.5493 revolution per minute
So,
ω = 0.05 × 9.5493 RPM
ω = 0.478082 RPM
Now
1 rpm = 60 rph
so
ω = 0.478082 × 60
ω = 28.6849 revolutions per hour
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An ideal gas in a 50.0 L tank has a
pressure of 2.45 atm at 22.5°C.
How many moles of gas are in
the tank?
Answer:
5.05225 moles
Explanation:
The computation of the number of moles of gas in the tank is shown below:
Given that
Volume = V = 50 L = 50.0 × 10^-3m^3
Pressure = P = 2.45 atm = 2.45 × 101325
Temperature = T = 22.5°C = (22.5 + 273)k = 295.5 K
As we know thta the value of gas constant R is 8.314 J/mol.K
Now
PV = nRT
n = PV ÷ RT
= ((2.45 × 101325) (50.0 × 10^-3)) ÷ ((8.314) (295.5))
= 5.05225 moles
m
A 3.0 kg model train going right at 2.8 bumps into another 2.0 kg model train car moving in the same
S
m
direction at 1.6 . The heavier train car has a final speed of 2.2 to the right.
S
S
What is the final speed of the lighter 2.0 kg train car?
Answer:
it’s 2.5 m/s
Explanation:
i’m too lazy but trust
This question can be solved by using the law of conservation of momentum.
The final speed of the lighter 2 kg train is " 2.5 ".
When two moving objects collide with each other, the law of conservation of momentum can be applied to them as follows:
[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2[/tex]
where,
m₁ = mass of heavier train = 3 kg
m₂ = mass of lighter train = 2 kg
u₁ = initial speed of heavier train = 2.8
u₂ = initial speed of lighter train = 1.6
v₁ = final speed of heavier train = 2.2
v₂ = final speed of lighter train = ?
Therefore,
(3 kg)(2.8) + (2 kg)(1.6) = (3 kg)(2.2) + (2 kg)(v₂)
[tex]v_2 = \frac{5 kg}{2 kg}[/tex]
v₂ = 2.5
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The attached picture illustrates the law of conservation of momentum.
. You have two carts, one which is empty and has mass m. The second cart is of the same mass but loaded with twice the mass of the empty cart i.e. it has mass 3m. You push each of them (one at a time) with the same constant force, over the same distance, starting from rest. After you have pushed them through this distance, you remove the force. How will the kinetic energy of the loaded and empty carts compare to each other
Answer:
Their kinetic energies would be the same
Explanation:
This is because, since the force, F acting on them moves the same distance, d, the work done by the force is W = Fd.
Now, from work-kinetic energy principles,
W = ΔK where ΔK = change in kinetic energy of the carts.
Since the work-done is the same for both carts, their change in kinetic energies would also be the same.
Since they start from rest, ΔK = K' - K = K' - 0 = K'
So, the kinetic energies of the carts would be the same
If all pairs of adjacent sides of a quadrilateral are congruent then it is called _________.
(A) rectangle (B) parallelogram (C) trapezium, (D) rhombus
Answer:
D
Explanation:
If you need an explanation feel free to ask.
A parallel-plate capacitor is connected to a battery until it is fully charged. Then, the capacitor is disconnected from the battery and connected to two uncharged parallel plates that make up a capacitor. The potential between the plates of the initial capacitor will
Answer:
The potential between the plates will decrease.
Explanation:
An insulator is usually placed between the parallel plates and is also called a dielectric because it makes the amount of charge a capacitor can accommodate to increase at a particular potential difference.
Furthermore, the dielectric effect will make the electric field of the charged capacitor which is not connected to a source of supply to decrease.
Now, when the battery is removed, the charge Q remains constant and Capacity C will increase.
Formula for the potential difference is here;
V = Q/C
Since the numerator Q is constant and the denominator C increases, it means the potential difference V will decrease
A car hits a tree with a force of 45 N, the mass of the tree is 65g. What is the resulting acceleration?
a. 0.69 m/s2
b. 692 m/s2
c. 2,925 m/s2
d. 2.93 m/s2
Answer:
i think 692m/s2 is the correct answer
A tank initially holds 100 gallons of salt solution in which 50 lbs of salt has been dissolved. A pipe fills the tank with brine at the rate of 3 gpm, containing 2 lbs of dissolved salt per gallon. Assuming that the mixture is kept uniform by stirring, a drain pipe draws out of the tank the mixture at 2 gpm. Find the amount of salt in the tank at the end of 30 minutes.
A. 171.24 lbs
B. 124.11 lbs
C. 143.25 lbs
D. 105.12 lbs
Answer:
A. 171.24 Ibs
Explanation:
To find the amount of salt in the tank,
Let Q = Amount of salt in the mixture
And let 100 + (3-2)t = 100 + t be the volume of mixture at anytime t.
Rate of gain - Rate of loss = dQ / dt
Concentration of salt = Q / (100+t)
For the linear differential equation,
dQ / dt = 3(2) - 2 [Q/ (100 + t)]
dQ /dt + Q [2 / (100 + t)] = 6
The general solution of the linear differential equation is:
Q (i.f) = ∫ A(t) (i.f) dt + C
Therefore,
i.f = e ^ ∫ P(t) dt
And P(t) = 2 / (100 + t)
i.f = e ^ ∫ 2 / (100 + t)
= e ^ 2㏑ (100 + t)
= e ^ ㏑ (100 + t) ^2 = (100 + t) ^2
Q(100 + t) ^ 2 = ∫6 (100 + t) ^2 dt + C
Q(100 + t) ^2 = 2(100 + t) ^ 3 + C
When t = 0, Q = 50
Therefore,
50( 100) ^2 = 2(100) ^3 + C
C = -1.5 * 10 ^6
therefore, when t = 30,
Q (100 + 30) ^2 = 2(100 + 30) ^3 - 1.5 * 10 ^6
Q (400) ^2 = 2(130) ^3 - 1.5 * 10 ^6
Q = 171.24 Ibs
The amount of salt in the tank at the end of 30 minutes is 171.24 lbs.
The given parameters:
Initial volume of the tank, i = 100 gallonsRate of gain of salt = 3 gpmRate of loss of salt = 2 gpmThe linear differential equation of the salt solution is calculated as follows;
[tex]\frac{dx}{dt} = Gain - loss[/tex]
where;
x is the salt concentrationThe salt concentration at time t, is calculated as follows;
[tex]\frac{dx}{dt} = 2(3) - 2(\frac{X}{100 + t} )\\\\\frac{dx}{dt} = 6 - 2(\frac{X}{100 + t} )\\\\\frac{dx}{dt} +2(\frac{X}{100 + t} ) =6[/tex]
Apply the general solution of linear differential equation as follows;
[tex]X(f) = \int\limits {At} \, dt \ + C\\\\f = e^{\int\limits {At} \, dt}\\\\ f = e^{\int\limits {\frac{2}{100 + t} } \, dt}\\\\f = e^{2 ln(100 + t)}\\\\f = (100 + t)^2[/tex]
[tex]X(100 + t)^2 = \int\limits {6(100 + t)^2} \, dt \ + \ C\\\\ X(100 + t)^2 = 2(100 + t)^3 + C[/tex]
When t = 0 and X = 50
[tex]50(100 + 0)^2 = 2(100+ 0)^3 + C\\\\C = -1.5 \times 10^6[/tex]
When t = 30 min, the concentration is calculated as;
[tex]X (100 + 30)^2 = 2(100 + 30)^3- 1.5 \times 10^6\\\\X(130)^2 = 2(130)^3 - 1.5\times 10^6\\\\X(130)^2 = 2894000\\\\X = \frac{2894000}{130^2} \\\\X = 171.24 \ lbs[/tex]
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A point charge, Q1 = -4.2 μC, is located at the origin. A rod of length L = 0.35 m is located along the x-axis with the near side a distance d = 0.45 m from the origin. A charge Q2 = 10.4 μC is uniformly spread over the length of the rod.Part (a) Consider a thin slice of the rod, of thickness dx, located a distance x away from the origin. What is the direction of the force on the charge located at the origin due to the charge on this thin slice of the rod? Part (b) Write an expression for the magnitude of the force on the point charge, |dF|, due to the thin slice of the rod. Give your answer in terms of the variables Q1, Q2, L, x, dx, and the Coulomb constant, k. Part (c) Integrate the force from each slice over the length of the rod, and write an expression for the magnitude of the electric force on the charge at the origin. Part (d) Calculate the magnitude of the force |F|, in newtons, that the rod exerts on the point charge at the origin.
Answer:
a) attractiva, b) dF = [tex]k \frac{Q_1 \ dQ_2}{dx}[/tex], c) F = [tex]k Q_1 \frac{Q_2}{d \ (d+L)}[/tex], d) F = -1.09 N
Explanation:
a) q1 is negative and the charge of the bar is positive therefore the force is attractive
b) For this exercise we use Coulomb's law, where we assume a card dQ₂ at a distance x
dF = [tex]k \frac{Q_1 \ dQ_2}{dx}[/tex]
where k is a constant, Q₁ the charge at the origin, x the distance
c) To find the total force we must integrate from the beginning of the bar at x = d to the end point of the bar x = d + L
∫ dF = [tex]k \ Q_1 \int\limits^{d+L}_d {\frac{1}{x^2} } \, dQ_2[/tex]
as they indicate that the load on the bar is uniformly distributed, we use the concept of linear density
λ = dQ₂ / dx
DQ₂ = λ dx
we substitute
F = [tex]k \ Q_1 \lambda \int\limits^{d+L}_d \, \frac{dx}{x^2}[/tex]
F = k Q1 λ ([tex]-\frac{1}{x}[/tex])
we evaluate the integral
F = k Q₁ λ [tex](- \frac{1}{d+L} + \frac{1}{d} )[/tex]
F = k Q₁ λ [tex]( \frac{L}{d \ (d+L)})[/tex]
we change the linear density by its value
λ = Q2 / L
F = [tex]k Q_1 \frac{Q_2}{d \ (d+L)}[/tex]
d) we calculate the magnitude of F
F =9 10⁹ (-4.2 10⁻⁶) [tex]\frac{10.4 10x^{-6} }{0.45 ( 0.45 +0.35)}[/tex]
F = -1.09 N
the sign indicates that the force is attractive
Answer:
a)Toward the rod
b)|dF| = k|Q1|Q2(dx/L)/x^2
c)|F| = k|Q1|Q2/(d(d+L))
d)Plug in for answer c and solve
Explanation:
A)
Q1 is negative and Q2 is positive so it is an attractive force to where the rod is located.
B)
The formula for Force due to electric charges is F=kQ1Q2/r^2
In this case, Q2 is distrusted through the length of the rod as opposed to a single point charge. As such Q2 is actually Q2*dx/L as dx is a small portion of the full length, L.
The radius between Q1 and Q2 depends on the section of the rod taken so r will be the variable x distance from Q1.
The force is only from a small portion of the rod so more accurately, we are finding |dF| as opposed to the full force, F, caused by the whole rod.
The final formula is |dF| = k|Q1|Q2(dx/L)/x^2
C)
Integrating with respect to the only changing variable, x, which spans the length of the rod, from radius = d to d+L we get this:
F = integral from d to d+L of k|Q1|Q2(dx/L)/x^2
factor out constants
F = kQ1Q2/L * integral d to d+L(1/x^2)dx
F = kQ1Q2/L * (-1/x)| from d to d+L
F = kQ1Q2/L * (-1/d+L - -1/d)
F = kQ1Q2/L * (-d/(d(d+L)) + (d+L)/(d(d+L))
F = kQ1Q2/L * (L)/(d(d+L))
F = kQ1Q2/(d(d+L))
D)
Plug in the given values into c and you have your answer.
An aluminum wire having a cross-sectional area equal to 2.20 10-6 m2 carries a current of 4.50 A. The density of aluminum is 2.70 g/cm3. Assume each aluminum atom supplies one conduction electron per atom. Find the drift speed of the electrons in the wire.
Answer:
The drift speed of the electrons in the wire is 2.12x10⁻⁴ m/s.
Explanation:
We can find the drift speed by using the following equation:
[tex] v = \frac{I}{nqA} [/tex]
Where:
I: is the current = 4.50 A
n: is the number of electrons
q: is the modulus of the electron's charge = 1.6x10⁻¹⁹ C
A: is the cross-sectional area = 2.20x10⁻⁶ m²
We need to find the number of electrons:
[tex] n = \frac{6.022\cdot 10^{23} atoms}{1 mol}*\frac{1 mol}{26.982 g}*\frac{2.70 g}{1 cm^{3}}*\frac{(100 cm)^{3}}{1 m^{3}} = 6.03 \cdot 10^{28} atom/m^{3} [/tex]
Now, we can find the drift speed:
[tex]v = \frac{I}{nqA} = \frac{4.50 A}{6.03 \cdot 10^{28} atom/m^{3}*1.6 \cdot 10^{-19} C*2.20 \cdot 10^{-6} m^{2}} = 2.12 \cdot 10^{-4} m/s[/tex]
Therefore, the drift speed of the electrons in the wire is 2.12x10⁻⁴ m/s.
I hope it helps you!
Two people each do 100 joules of work by pushing a crate to the right. During this process, 50 joules of heat is generated from the
friction between the floor and the crate. How much energy is gained by the crate during this process?
A car travels 150 kilometers west in 3 hours. What is its average velocity?
Your answer:
150 km/hr
50 km/hr
50 km/hr west
150 km/hr west
Answer:
C= 50km/hr west
Explanation:
150/3= 50
Because it asks for velocity, make sure to include the direction as well.
A loaded wagon of mass 10,000 kg moving with a speed of 15 m/s strikes a stationary wagon of the same mass making a perfect inelastic collision. What will be the speed of coupled wagons after collision?
Answer:
7.5 m/s
Explanation:
Unfortunately, I don't have an explanation but I guessed the correct answer.
A 14.0-g wad of sticky clay is hurled horizontally at a 90-g wooden block initially at rest on a horizontal surface. The clay sticks to the block. After impact, the block slides 7.50 m before coming to rest. If the coefficient of friction between block and surface is 0.650, what was the speed of the clay immediately before impact
Answer:the speed of the clay immediately before impact =72.58m/s
Explanation:
Given that
mass of the stick clay, M₁= 14.0 g = 0.014 kg
mass of the block ,M₂= 90 g = 0.09 kg
Therefore the total mass= (M₁+M₂) = 104g = 0.104 kg
Also, distance, s = 7.50 m
coefficient of friction μ= 0.650
Acceleration due to gravity ,g = 9.8 m/s²
Using the Work- Energy theorem,
change in kinetic energy = work done
final kinetic energy(K₂) - initial kinetic energy(K₁) = force, F x coefficient of friction, μ x distance,s
The final kinetic energy is zero because after the impact, the block with the clay comes to a stop after 7.50m
kinetic energy =Work done
0.5 x m x v²=coefficient of friction, μ x force(F) x distance,s(Since force = m g )
0.5 x m x v²= μ x m x g x s
0.5 x 0.104 x v² = 0.650 x 0.104x 9.8 x 7.5
v²= 0.650 x 0.104x 9.8 x 7.5 / 0.5 x 0.104
v²==95.55
V = 9.77 m/s
Using the conservation of momentum formulae where
M₁ V₁ + M₂ V₂ = (M₁ + M₂ ) V
Since V₂ which is the velocity of block is zero as the block is initially at rest, We now have that
M₁ V₁ = (M₁ + M₂ ) V
0.014 kg x V₁ = 0.104 x 9.77
V₁=0.104 x 9.77 / 0.014
V=72.58m/s
Bill is walking to the store and he walks the first 500m in 60s. He then runs 1000m in 90s. After stopping for 45s, he was the remaining 450m to the store in 50s. What is the average velocity for Bills entire
trip?
Answer:
letra A segundo o couculo a divisão e completa
Why are carbon atoms able to form many organic compounds?
A. Carbon atoms have strong attraction to other elements.
B. Carbon atoms attract electrons from other atoms.
C. Carbon atoms can form many types of bonds with other carbon.
D. All of the above
Answer:
yo imma so I dunno find out yourself
Explanation:
dhdhdhdnndsisijjsksskskekekekkekssisisieieieiiwiwiwieiwieidjdjddi?
The force of gravity on a person or object on the surface of a planet is called
A. gravity
ОВ.
B. free fall
OC
c. terminal velocity
D. weight
Answer:
D. Weight
Explanation:
Hope that helps:)
A crate rests on a flatbed truck which is initially traveling at 17.9 m/s on a level road. The driver applies the brakes and the truck is brought to a halt in a distance of 46.1 m. If the deceleration of the truck is constant, what is the minimum coefficient of friction between the crate and the truck that is required to keep the crate from sliding
Answer:
The minimum coefficient of friction required is 0.35.
Explanation:
The minimum coefficient of friction required to keep the crate from sliding can be found as follows:
[tex] -F_{f} + F = 0 [/tex]
[tex] -F_{f} + ma = 0 [/tex]
[tex] \mu mg = ma [/tex]
[tex] \mu = \frac{a}{g} [/tex]
Where:
μ: is the coefficient of friction
m: is the mass of the crate
g: is the gravity
a: is the acceleration of the truck
The acceleration of the truck can be found by using the following equation:
[tex] v_{f}^{2} = v_{0}^{2} + 2ad [/tex]
[tex] a = \frac{v_{f}^{2} - v_{0}^{2}}{2d} [/tex]
Where:
d: is the distance traveled = 46.1 m
[tex] v_{f}[/tex]: is the final speed of the truck = 0 (it stops)
[tex]v_{0}[/tex]: is the initial speed of the truck = 17.9 m/s
[tex] a = \frac{-(17.9 m/s)^{2}}{2*46.1 m} = -3.48 m/s^{2} [/tex]
If we take the reference system on the crate, the force will be positive since the crate will feel the movement in the positive direction.
[tex] \mu = \frac{a}{g} [/tex]
[tex] \mu = \frac{3.48 m/s^{2}}{9.81 m/s^{2}} [/tex]
[tex] \mu = 0.35 [/tex]
Therefore, the minimum coefficient of friction required is 0.35.
I hope it helps you!
125 cm of gas are collected at 15 °C and
755 mmHg pressure. Calculate the volume of
the gas at s.t.p.
two faer coin and unbayers
dice are thrown together list the
Sample space
determine the probabilities that
A head and even number
A prime number and atleast a tail
A hedgehog lives in a backyard in England. Every night, the house owner puts out a bowl of canned cat food and hard-boiled egg. The hungry hedgehog eats some of the food, then stops when it is no longer hungry. This pattern helps the hedgehog to maintain a steady energy level and weight.
What is the name for keeping a stable internal environment despite changes in the outside environment?
A. hormonal control
B. stimulus and response
C. balance
D. homeostasis
Answer: I think it C and B but I am really confident in C
Explanation:
The new springs will be identical to the original springs, except the force constant will be 5655.00 N/m smaller. When James removes the original springs, he discovers that the length of each spring expands from 8.55 cm (its length when installed) to 12.00 cm (its length with no load placed on it). If the mass of the car body is 1355.00 kg, by how much will the body be lowered with the new springs installed, compared to its original height
Answer:
Explanation:
For original spring , compression in spring due to a load of 1355 kg is
x = 12 - 8.55 = 3.45 cm = .0345 m
spring constant = W / x
= 1355 x 9.8 / .0345
= 384898.55 N /m
Spring constant of new spring
k = 384898.55 - 5655 = 379243.55 N /m
New compression for new spring
= W / k
= 1355 x 9.8 / 379243.55
= .035 m
= 3.50 cm
Difference of compression = 3.50 - 3.45
= .05 cm .
In later case , car will be more lowered by .05 cm .
Potential energy is energy due to motion.
True or False?
Answer:
true
Explanation:
Answer:
true
Explanation:
please give brainlest need 1
What is net force?
OA. a push or a pull
B. A measure of how fast an object is moving
OC. The amount of energy an object has
D. The combination of all forces acting on an object.
Answer:
D
Explanation:
The net force is the combination of all forces acting on an object.
a passenger on cruise between San Juan, Puerto Rico and Miami, Florida accidentally drops a souvenir metal cube over the side of the boat, into the water. Each side of the metal cube measures 1 meter. The cube promptly sinks to the deepest part of the Puero Rico Trench. Once at the bottom, what pressure does the cube experience? Neglect Atmospheric Pressure. Use wikipedia to see depth of Trench!
Answer:
P = 84.1 MPa
Explanation:
The pressure at the bottom of column of of salt water of height h, is given by the following expression:[tex]P = \rho * g * h (1)[/tex]
where ρ = density of salt water (in Kg/m³),
g = acceleration due to gravity (in m/s²)
h = height of the column of water.
Replacing by their values in (1):[tex]P = \rho * g * h = 1023.6kg/m3*9.8m/s2*8380m = 84.1 MPa (2)[/tex]
Neglecting the atmospheric pressure, the pressure on the cube at the bottom of Puerto Rico Trench is given by (2):P = 84.1 MPa.On Earth, the number flux of solar neutrinos from the p-p chain is:
f_neutrino = 2fo/2.62MeV
Other nuclear reactions in the Sun supplement this neutrino flux with a small additional flux of higher-energy neutrinos. A neutrino detector in Japan, named SuperKamiokande, consists of a tank of 50 kton of water, surrounded by photomultiplier tubes. The tubes detect the flash of Cerenkov radiation emitted by a recoiling electron when a high-energy neutrino scatters on it.
Required:
a. How many electrons are there in the water of the detector?
b. Calculate the detection rate for neutrino scattering, in events per day.
Answer:
Explanation:
The volume of the tank = 50 kton
50 kton = 5 × 10⁷ kg
Since 18 grams of water will contain: 10 electrons × 6.023 × 10²³
Then;
5× 10⁷ kg will contain [tex]( \dfrac{5 \times 10^7 \times 10^3}{18}) \times 10 \times 6.023 \times 10^{23}[/tex]
= 1.67 × 10³⁴ electrons
(b)
Suppose:
[tex]f_{neutrino} = \dfrac{2f_o}{26.2 MeV} = 6.7\times 10^{10} \ s^{-1} cm^{-2}[/tex]
Then;
10⁻⁶ of [tex]f_{neutrino} = 6.7 \times 10^{10} \times 10^{-6} \ s^{-1} cm^{-2}[/tex]
[tex]=6.7 \times 10^{4}\ s^{-1} cm^{-2}[/tex]
Thus, the number of high energy neutrinos which will interact with water is:
= [tex]6.7 \times 10^4 \times \sigma[/tex]
= [tex]6.7 \times 10^4 \times 10^{-43}[/tex]
= [tex]6.7 \times 10^{-39} s^{-1}[/tex]
For 1.67 × 10³⁴ electrons, the detection rate is:
[tex]6.7 \times 10^{-39} \times 1.67 \times 10^{34}[/tex]
[tex]= 11.19 \times 10^{-5} \ s^{-1}[/tex]
= 9.668 per day