Answer:
The final pressure of the whole system is 34.80 atm.
Explanation:
Given that,
Volume = 45.0 ml
Volume of first bulb = 77.0 mL
Pressure = 8.89 atm
Volume of second bulb = 250 mL
Pressure = 2.82 atm
Volume of third bulb = 21.0 mL
Pressure = 8.42 atm
We need to calculate the final pressure of the whole system
Using formula of pressure
[tex]P_{1}V_{1}+P_{2}V_{2}+P_{3}V_{3}+P_{t}V_{t}=P_{f}V_{f}[/tex]
Where, [tex]P_{1}[/tex]= pressure of first bulb
[tex]P_{2}[/tex]= pressure of second bulb
[tex]P_{3}[/tex]= pressure of third bulb
[tex]P_{4}[/tex]= initial pressure of tube
[tex]V_{1}[/tex]= Volume of first bulb
[tex]V_{2}[/tex]=Volume of second bulb
[tex]V_{3}[/tex]= Volume of third bulb
[tex]V_{4}[/tex]= Initial volume of tube
Put the value into the formula
[tex]8.89\times77.0+250\times2.82+21.0\times8.42+0=P_{f}\times45[/tex]
[tex]P_{f}=\dfrac{1566.35}{45}[/tex]
[tex]P_{f}=34.80\ atm[/tex]
Hence, The final pressure of the whole system is 34.80 atm.
Determine the number of revolutions through which a typical automobile tire turns in 1 yr. Suppose the automobile travels 13500 miles each year on tires with radius 0.220 m.
Answer:
Number of revolution made by tire is 1.57 x 10⁷
Explanation:
Radius of tire, r = 0.220 m
Circumference of tire, C = 2πr
Substitute the value of r in the above equation.
C = 2 x π x 0.220 m = 1.38 m
Total distance covered by tire in a year, D = 13500 miles
But 1 mile = 1609.34 m
So, D = 13500 x 1609.34 m
Number of revolutions take by tire, N = [tex]\frac{D}{C}[/tex]
[tex]N=\frac{13500\times1609.34}{1.38}[/tex]
N = 15743543
Total peripheral resistance is related to all of the following except the
Options to the question :
A- blood viscosity
B- osmolarity of interstitial fluid
C- turbulence
D-length of a blood vessel
E- blood vessel diameter
Answer:
Total peripheral resistance is NOT related to B ( osmolarity of interstitial fluid).
Explanation:
Total peripheral resistance( also called systemic vascular resistance) is defined as the total opposition to the flow of blood in systemic circulation. Increase in total peripheral resistance leads to high blood pressure while it's decrease leads to low blood pressure. Factors that contributes to total peripheral resistance in systemic circulation includes:
- blood vessel diameter
- blood viscosity,
- lengthy of a blood vessel and
- turbulence.
Two masses exert a force of 1,161 N on each other when they are 20 km apart. How much force will these two masses exert on each other when they are 40 km apart?
Explanation:
Below is an attachment containing the solution.
A 1 kg rock is suspended by a massless string from one end of a meter stick at the 0 cm mark. What is the mass m suspended from the meter stick at the 75 cm mark if the system is balanced?
Answer:
2 kg
Explanation:
Note: For the meter stick to be balanced,
Sum of clock wise moment must be equal to sum of anti clock wise moment
Wd = W'd' ................ Equation 1
Where W = weight of the rock, d = distance of the meter stick from the point of support, W' = weight of the that must be suspended for the meter stick to be balanced, d' = distance of the mass to the point of support.
make W' the subject of the equation
W' = Wd/d'............... Equation 2
Taking our moment about the support,
Given: W = mg = 1 ×9.8 = 9.8 N, d = 50 cm, d' = (75-50) = 25 cm
Substitute into equation 2
W' = 9.8(50)/25
W' = 19.6 N.
But,
m = W'/g
m = 19.6/9.8
m = 2 kg.
To find the mass M suspended from the meter stick, we can use the principle of rotational equilibrium. By setting the torques on either side of the fulcrum equal to each other, we can calculate the mass M at the 75 cm mark to be 3 kg.
Explanation:To determine the mass M suspended from the meter stick at the 75 cm mark, we can use the principle of rotational equilibrium. Since the system is balanced, the torques on both sides of the meter stick must be equal.
The torque of the rock can be calculated using its weight, which is equal to its mass multiplied by the acceleration due to gravity. The torque of the mass M can be calculated by multiplying its mass by the distance from the fulcrum (75 cm) to its center of mass. By setting the torques equal to each other, we can solve for M.
The torque of the rock: TR = (1 kg)(9.8 m/s²)(0.75 m).
The torque of the mass M: TM = M(9.8 m/s²)(0.25 m).
Setting TR equal to TM and solving for M, we get M = (1 kg)(0.75 m)/(0.25 m) = 3 kg.
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Two metal balls are the same size but one weighs twice as much as the other. The balls are dropped from the roof of a single story building at the same instant of time. The time it takes the balls to reach the ground below will be:
a.about half as long for the heavier ball as for the lighter one.
b.about half as long for the lighter ball as for the heavier one.
c,about the same for both balls.
d.considerably less for the heavier ball, but not necessarily half as long.
e..considerably less for the lighter ball, but not necessarily half as long.
Answer:
c, About the same for both balls.
Explanation:
If no air resistance is present, the rate of descent depends only on how far the object has fallen, no matter how heavy the object is. This means that two objects will reach the ground at the same time if they are dropped simultaneously from the same height.
In a manufacturing process, a large, cylindrical roller is used to flatten material fed beneath it. The diameter of the roller is 1.00 m and while being driven into rotation around a fixed axis, its angular position is expressed as
θ = 2.50t2 - 0.600t3
where θ is in radians and t is in seconds.
a) Find the maximum angular speed of the roller
b) what is the maximum tangential speed of the point an the rim of the roller?
c) at what time t should the driven force be removed from the roller so that the roller does not reverse its direction of rotation?
d) Through how many rotations has the roller turned between t=0 and the time found in part c?
Explanation:
Below is an attachment containing the solution.
A man stands on a scale and holds a heavy object in his hands. What happens to the scale reading if the man quickly lifts the object upward and then stops lifting it?
Answer:
Explanation:
When he accelerates the heavy object up , the reading increases because an extra downward normal force acts on it, then scale reading returns to the same reading as when standing stationary, and then decreases as although he is lifting the heavy object , the acceleration is decreasing ,so the extra upward normal force acts.
while driving his sports car at 20.0 m/s down a four lane highway, eddie comes up behind a slow moving dump truck and decides to pass it in the left hand lane. of eddie can accelerate at 5.00 m/s^2, how long will it take for him to reach a speed of 30.0 m/s
Explanation:
Given:
u = 20 m/s
a = 5 m/s^2
v = 30 m/s
t = ?
Use the first kinematic equation of motion:
v = u + at
t = (v - u)/a = 10/5 = 2 seconds
Eddie will take 2 sec to reach a speed of 30 m/s
What is velocity ?
velocity is defined as rate of change of displacement of the object with respect to rate of change in time. In mathematics It is written as :
[tex]\begin{aligned}v&=\frac{\Delta d}{\Delta t}\end{aligned}[/tex]
Here it is given that :
initial speed of car (u) = 20 m/s
acceleration of car (a) = 5 m/s²
final speed of car (v) = 30 m/s²
it is to find the time t to achieve this speed which is calculated using the first equation of motion:
[tex]\begin{aligned}v&=u+at\\30&=20+5t\\&t=\frac{10}{5}\\&=2\text{\:sec}\end{aligned}[/tex]
Therefore, Eddie will take 2 sec to reach a speed of 30 m/s
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Awning windows can be 100% openable and are best used where extreme weather conditions require a tight seal when the window is closed, although these windows are in common use everywhere.1. True2. False
Answer: True
Explanation: Awning windows
are Weather-tight, good choice in damp climates and can protect your home against moisture, even when they’re open during a rainstorm. Because of the way awning windows are constructed, they allow for nearly 100 percent of viable ventilation, without the threat of water seeping into the home. They also offer a superior seal against air pass through
These windows form a slant (of about 45 degree) after opening hence do not open 100%; this slant opening is advantageous in preventing rain from entering the building.
The windows can also be tightly sealed (from the inside) during extreme/cold weather conditions.
True, awning windows can be fully opened and are beneficial for creating a tight seal in extreme weather conditions. They also allow for maximum ventilation while preventing unwanted energy transfer, making them widely used regardless of climate. Design elements like overhangs and window orientation further enhance energy efficiency.
True, awning windows can be 100% openable and are indeed best used in areas that experience extreme weather conditions, thanks to their capability to form a tight seal when closed. The design of these windows allows them to prevent unwanted energy transfers by keeping the harsh weather out when they are closed, while still allowing for maximum ventilation when opened. This makes them highly versatile and thus they are in common use in various locations, not just those with extreme weather. Factors such as the orientation of the windows, the use of overhangs, and the placement of deciduous trees also play important roles in maximizing energy efficiency by modulating the amount of sunlight and heat entering a building.
For instance, overhangs above south-facing windows help keep a house cool in summer by casting a shadow over the window when the sun is high, and allowing sunlight to warm the rooms in winter when the sun stays closer to the horizon. Moreover, in climates where preventing heat gain is crucial, the largest windows may face north to avoid the sun, with south-facing windows being smaller and well insulated to allow for cross-ventilation without admitting much sunlight. Additionally, windows equipped with weatherstripping, and double-paned, low-emissivity glass can significantly reduce energy losses, contributing to a building's overall energy efficiency.
Two carts are involved in an elastic collision. Cart A with mass 0.550 kg is moving towards Cart B with mass 0.550 kg, which is initially at rest. Cart A stops after the collision and Cart B begins to move.A)If cart A has an initial velocity of 0.8 m/s , what is the velocity of Cart B after the collision?B)What is the initial kinetic energy of Cart A?C)What is the initial kinetic energy of Cart B?D)What is the final kinetic energy of Cart A?E)What is the final kinetic energy of Cart B?F)Is the kinetic energy conserved for elastic collisions?G)Is the momentum conserved for elastic collisions?
Answer:
A) [tex]v_b=0.8\ m.s^{-1}[/tex] is the final velocity of the cart B after collision.
B) [tex]KE_A=0.176\ J[/tex]
C) [tex]KE_B=0\ J[/tex]
D) [tex]ke_a=0\ J[/tex]
E) [tex]ke_B=0.176\ J[/tex]
F) Yes, here the kinetic energy is conserved because the mass of both the bodies involved in the collision is same.
G) Yes, momentum is always conserved for an elastic collision.
Explanation:
Given:
mass of car A, [tex]m_a=0.55\ kg[/tex]mass of car B, [tex]m_b=0.55\ kg[/tex]initial velocity of car A, [tex]u_a=0.8\ m.s^{-1}[/tex]final velocity of the car A, [tex]v_a=0\ m.s^{-1}[/tex]A)
As given in the question that the cars undergo an elastic collision:
According to the conservation of momentum:
[tex]m_a.u_a+m_b.u_b=m_a.v_a+m_b.v_b[/tex]
[tex]0.55\times 0.8+0.55\times 0=0.55\times 0+0.55\times v_b[/tex]
[tex]v_b=0.8\ m.s^{-1}[/tex] is the final velocity of the cart B after collision.
B)
Initial kinetic energy of cart A:
[tex]KE_A=\frac{1}{2} m_a.u_a^2[/tex]
[tex]KE_A=0.5\times 0.55\times 0.8^2[/tex]
[tex]KE_A=0.176\ J[/tex]
C)
Initial kinetic energy of cart A:
[tex]KE_B=\frac{1}{2} \times m_b.u_b^2[/tex]
[tex]KE_B=0.5\times 0.55\times 0^2[/tex]
[tex]KE_B=0\ J[/tex]
D)
The final kinetic energy of cart A:
[tex]ke_A=\frac{1}{2} m_a.v_a^2[/tex]
[tex]ke_a=0.5\times 0.55\times 0^2[/tex]
[tex]ke_a=0\ J[/tex]
E)
The final kinetic energy of cart B:
[tex]ke_B=\frac{1}{2} m_b.v_b^2[/tex]
[tex]ke_B=0.5\times 0.55\times 0.8^2[/tex]
[tex]ke_B=0.176\ J[/tex]
F)
Yes, here the kinetic energy is conserved because the mass of both the bodies involved in the collision is same.
G)
Yes, momentum is always conserved for an elastic collision.
(a) The velocity of cart B is [tex]0.8m/s[/tex]
(b) Initial kinetic energy of cart A is [tex]0.176J[/tex]
(c) Initial kinetic energy of cart B is 0
(d) The final kinetic energy of cart A is 0
(e) The final kinetic energy of cart B is [tex]0.176J[/tex]
(f) The kinetic energy is conserved
(g) The momentum is conserved
Elastic collision:(a) The momentum of the system must be conserved, which means momentum before the collision must be equal to momentum after the collision.
Momentum before collision = [tex]m_Au_A+m_bu_B=0.55\times0.8+0.55\times0=0.44kgm/s[/tex]
Momentum after collision:
[tex]m_A+m_A+m_Bv_B=0.55\times0+0.55v_B=0.44\\\\v_B=0.8m/s[/tex]
(b) Initial kinetic energy of cart A:
[tex]K_{iA}=\frac{1}{2}m_Au_A^2=0.5\times0.55\times0.64=0.176J[/tex]
(c) Initial kinetic energy of cart B:
[tex]K_{iB}=\frac{1}{2}m_Bu_B^2=0.5\times0.55\times0=0J[/tex]
(d) The final kinetic energy of cart A:
[tex]K_{fA}=\frac{1}{2}m_Av_A^2=0.5\times0.55\times0=0J[/tex]
(e) Initial kinetic energy of cart B:
[tex]K_{iB}=\frac{1}{2}m_B_v_B^2=0.5\times0.55\times0.64=0.176J[/tex]
(f) yes the kinetic energy of the system is conserved for elastic collision
(g) The momentum is conserved in any case including the elastic collision
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(Atwood’s Machine): Two masses, 9 kg and 12 kg, are attached by a lightweight cord and suspended over a frictionless pulley. When released, find the acceleration of the system and the tension in the cord.
Answer:
Acceleration = 1.428m/s2
Tension = 102.85N
Explanation:
The detailed solution is attached
Answer:
The acceleration of the system is 1.401 m/s² and
The tension in the cord is 100.902 N
Explanation:
Let the 9 kg mass be m
Let the 12 kg mass be M
By Newton's second law of motion we have
For the 9 kg mass, T - mg = ma and for the 12 kg mass we have T - Mg = -Ma
Here we took the upward acceleration as positive a of the 9 kg mass and the downward acceleration of the 12 kg mass as -a
Solving for T for the 9 kg mass we have
T = mg + ma
Substituting the value of T in to the 12 kg mass equation, we have
mg + ma - Mg = -Ma or a = [tex](\frac{M-m}{M+m} )g[/tex] therefore the acceleration is
1.401 m/s²
and the tension is T = mg + ma = 9×(9.81+1.401) = 100.902 N
The relationship between heat (q), work (w), and internal energy (U) can be described with which of the following?a. q = ΔU × wb. ΔU = qwc. w = ΔUqd. ΔU = q + w
Answer:
d. ΔU = q + w
Explanation:
Internal energy is the sum of kinetic energy (which comes from motion of molecules) and potential energy ( which comes from chemical bonds between atoms and other intermolecular forces that maybe present).
First law of thermodynamics states that: "the change in internal energy of a closed system is equal to the energy added to it in form of heat(q) and work (w) done on the system by the surroundings.
Mathematically written as:
[tex]E_{internal}[/tex] = q + w
Conventionally, the first law is based on the system doing work and the surrounding doing work.
[tex]E_{internal}[/tex] can also be written as ΔU.
Therefore [tex]E_{internal}[/tex] = ΔU = q + w
ΔU = q + w
A box contains 9 new light bulbs and 6 used light bulbs. Each light bulb is the same size and shape. Meredith will randomly select 2 light bulbs from the box without replacement. What is the probability Meredith will select a new light bulb and then a used light bulb
Answer:
(9/35) = 0.257
Explanation:
Box contains 9 new light bulbs and 6 used light bulbs, total number of bulbs = 15.
Probability of selecting two bulbs; a new light bulb and then, a used light bulb in that order = [(probability of selecting a new bulb) × (probability of selecting a used bulb from the rest)] = [(9/15) × (6/14)] = (9/35) = 0.257
The probability that Meredith will select a new bulb and then a used bulb in sequential order without replacement is around 26%.
Explanation:The question asks about the probability of selecting a new light bulb and then a used light bulb from a box containing 9 new light bulbs and 6 used light bulbs. Probability events like these are solved using multiplication rules of probability that each event is independent.
First, the probability of picking a new bulb is 9/15 (total bulbs are 15). If you pick one out, you don't replace it, so there are only 14 bulbs left. Thus, the probability of picking a used bulb now is 6/14. Hence the probability of both events happening in order is the multiplication of both, i.e., (9/15)*(6/14) = 54/210 = 0.257 or around 26%.
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Approximately how fast is Jupiter orbiting the Sun? Approximately how fast is Jupiter orbiting the Sun? 10 km/skm/s a little less than 15 km/skm/s 20 km/skm/s cannot be determined from the information provided
Answer:
The correct answer is
a little less than 15 km/s.
Explanation:
The distance between the sun and Jupiter varies by about 75 million km between the perihelion and the aphelion with an average distance of 778 million km from the sun for which it takes Jupiter about 12 years to complete its orbit round the sun giving it an orbital speed of about 13.07 km/s
The size of Jupiter is more than the twice the combined size of all the other planets, which is about 1.300 times the size of earth.
Jupiter is orbiting the Sun from b. 15 km/s.
Jupiter orbits the Sun at an average distance (semi-major axis) of about 5.2 Astronomical Units (AU), where 1 AU is the average distance from the Earth to the Sun, approximately 149.6 million kilometers.
Using Kepler's third law of planetary motion and the fact that Jupiter takes approximately 11.86 Earth years to complete one orbit around the Sun, we can estimate its average orbital speed. The orbital speed (v) of a planet can be roughly calculated by the formula:
[tex]v = \frac{2 \pi r}{T}[/tex]
where:
[tex]r[/tex] is the average orbital radius (semi-major axis in meters), and[tex]T[/tex] is the orbital period in seconds.Converting the semi-major axis from AU to kilometers:-
[tex]r = 5.2 \times 149.6 \times 10^6 \text{ km} \\= 777.92 \times 10^6 \text{ km}[/tex]
Converting the orbital period from years to seconds:
[tex]T = 11.86 \text{ years} \times 3.154 \times 10^7 \text{ seconds/year} \\= 3.74 \times 10^8 \text{ seconds}[/tex]
Now, plugging in these values to our formula:
[tex]v = \frac{2 \pi \times 777.92 \times 10^6}{3.74 \times 10^8} \text{ km/s} \\\approx 13.07 \text{ km/s} \ {or} 15 km[/tex]
Sphere 1 with radius R_1 has positive charge q, Sphere 2 with radius 4.50 R_1 is far from sphere 1 and initially uncharged. The separated spheres are then connected with a wire then uncouth to retain only negligible charge. (a) What is the ratio V_1/V_2 of the final potentials of the spheres? (b) what fraction of q ends up on sphere? (c) What fraction of q ends up on sphere 2? (d) What is the ratio q_1/q_2 of the surface charge densities of the spheres?
Answer:
Explanation:
capacitance of sphere 2 will be 4.5 times sphere 1
a ) when spheres are in contact they will have same potential finally . So
V_1 / V_2 = 1
b )
Charge will be distributed in the ratio of their capacity
charge on sphere1 = q x 1 / ( 1 + 4.5 )
= q / 5.5
fraction = 1 / 5.5
c ) charge on sphere 2
= q x 4.5 / 5.5
fraction = 4.5 / 5.5
d ) surface charge density of sphere 1
= q /( 5.5 x A ) where A is surface area
surface charge density of sphere 2
= q x 4.5 /( 5.5 x 4.5² A ) where A is surface area
= q /( 5.5 x 4.5 A )
q_1/q_2 = 4.5
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A numerical scale of earthquake magnitude that takes into account the size of the fault rupture is the ____.
A) Richter scale
B) modifies Mercalli scale
C) moment magnitude scale
D) epicentral distance scale
Answer:
d. epicentral distance scale
Explanation:
The depth of focus from the epicenter, called as Focal Depth, is an important parameter in determining the damaging potential of an earthquake. Most of the damaging earthquakes have shallow focus with focal depths less than about 70km. Distance from epicenter to any point of interest is called epicentral distance
9. Over a certain region of space, the electric potential is V = . (a) Find the expressions for the x, y, and z components of the electric field over this region. (b) What is the magnitude of the field at the point P that has coordinates (1.00, 0, 22.00) m?
Answer:
a)
The expression for the electric potential in this problem is:
[tex]V=5x-3x^2y+2yz[/tex]
where
x, y, z are the three spatial coordinates
The relationship between components of the electric field and electric potential is:
[tex]E_x=-\frac{dV}{dx}\\E_y=-\frac{dV}{dy}\\E_z=-\frac{dV}{dz}[/tex]
Therefore, we have to calculate the derivatives of the potential over the three variables.
Doing so, we find:
[tex]E_x=-\frac{d}{dx}(5x-3x^2y+2yz)=-(5-6xy)=6xy-5[/tex]
[tex]E_y=-\frac{d}{dy}(5x-3x^2y+2yz)=-(-3x^2+2z)=3x^2-2z[/tex]
[tex]E_z=-\frac{d}{dz}(5x-3x^2y+2yz)=-(2y)=-2y[/tex]
b)
Here we want to find the magnitude of the electric field at the point P that has coordinates
P (1.00, 0, 22.00) m
First of all, we find the components of the electric field at that point by substituting
x = 1.00
y = 0
z = 22.0
We find:
[tex]E_x=6xy-5=6(1)(0)-5=-5 N/C\\E_y=3x^2-2z=3(1)^2-2(22)=-41 N/C\\E_z=-2y=-2(0)=0[/tex]
Now, the magnitude of the electric field is given by
[tex]E=\sqrt{E_x^2+E_y^2+E_z^2}[/tex]
And by substituting,
[tex]E=\sqrt{(-5)^2+(-41)^2+0}=41.3 N/C[/tex]
Consider the following tasks: A. Draw a closed curve around the system. B. Identify "the system" and "the environment." C. Draw a picture of the situation. Which of these are steps used to identify the forces acting on an object? 1. A only 2. B only 3. C only 4. A, B, and C 5. None of them
Answer:
Explanation:
To identify forces acting on an object
1. Draw a closed curve around the system.
2.Identify "the system" and "the environment."
3. Draw a picture of the situation. Which of these are steps used to identify the forces acting on an object?
All of these are correct.
A close loop doesn't allow external forces to act on the system and this only allows the the forces on the object to act alone.
Drawing the free body diagram will show the forces acting on the body.
Also, isolating the system from the environment so that it won't be affected by external forces or resistance
1) Calculate the torque required to accelerate the Earth in 5 days from rest to its present angular speed about its axis. 2) Calculate the energy required. 3) Calculate the average power required.
Final answer:
To find the torque to accelerate Earth in 5 days to its angular speed, calculate the moment of inertia and use τ = Iα. The energy required is given by E_k = (1/2)Iω^2, and average power is P = E_k / t.
Explanation:
To calculate the torque required to accelerate the Earth from rest to its present angular speed about its axis in 5 days, we first need to determine the Earth's moment of inertia (I) and its angular acceleration (α). The Earth's moment of inertia can be estimated using the formula I = (2/5)MR^2, where M is the mass of the Earth and R is its radius. The angular speed (ω) is 2π rad/day since there are 2π radians in a full rotation and the Earth completes one rotation per day. The angular acceleration, α, is then ω divided by the time in seconds to accelerate, which is 5 days or 5 x 24 x 3600 seconds. The formula τ = Iα is used to calculate torque.
To find the energy required, we can use the rotational kinetic energy formula E_k = (1/2)Iω^2, where ω is the final angular velocity. Subsequently, the average power required is the energy divided by the time over which it is expended, P = E_k / t, where t is the time in seconds.
A block in the shape of a rectangular solid has a crosssectional area of 3.50 cm2 across its width, a front-to-rear length of 15.8 cm, and a resistance of 935 0. The block’s material contains 5.33 $ 1022 conduction electrons/m3. A potential difference of 35.8 V is maintained between its front and rear faces. (a) What is the current in the block? (b) If the current density is uniform, what is its magnitude? What are (c) the drift velocity of the conduction electrons and (d) the magnitude of the electric field in the block?
Answer:
a) 38.3mA
b) 109.396A/m^2
c) 1.283cm/s
d) 226.582V/m
Explanation:
The equation for current is given by:
I = V/R = 35.6/ 935 = 0.03829 A Approximately 38.3mA
b) The equation to find the magnitude is given by:
J = I/A = 0.03829/ 0.000350m^2
J = 109.396A/m^2
c) The equation to calculate drift velocity of the electron is given by: Vd = J/ ne = 109.396/( 5.33×10^23 )(1.60×10^-19)
Vd = 0.01283 approximately 1.283cm/s
d) The magnitude of electric field in the block, E = V/L = 35.8/ 0.158m = 226.582V/m
Chromatic aberration in lenses is a result of which wave property of light?
Answer:
Dispersion
Explanation:
Chromatic aberration refers to the failure of a lenses to focus all the wavelengths or colours in white light at the same point. This is due to dispersion of light in glass. The refractive index of the lenses elements varies with different wavelengths of light. The different colours of white light will travel at different speeds through the lens hence they cannot all be focused at the same point. Dispersion results from the refraction of the various wavelengths of light to varying degrees.
Chromatic aberration in lenses is a result of the property of light, wavelength. It leads to dispersion of different colors of light due to the dependence of lens's refractive index on color or wavelength, causing different colors to have different focal points. This can be partially corrected in optical systems using multiple lenses of different materials.
Explanation:Chromatic aberration in lenses is caused by the property of light known as wavelength. This concept is related to the wave nature of light, where the wavelength is the distance between successive peaks or troughs of a light wave. Chromatic aberration occurs because the index of refraction in a lens is dependent on the light's wavelength, which is related to its color. Because the index of refraction varies with wavelength, different colors (wavelengths) of light are bent by different amounts when they pass through a lens and hence have different focal points.
For example, in a single convex lens, violet light, which has a shorter wavelength, is bent more than red light, which has a longer wavelength. Therefore, violet light converges to a focal point closer to the lens compared to red light, which results in the dispersion of colors, as depicted in Figure 26.28. However, this chromatic aberration can be partially corrected by employing a two-lens system, often consisting of a converging and a diverging lens made of different materials with different dispersions, referred to as an achromatic doublet. However, completely rectifying chromatic aberration is typically not possible, which is why systems such as telescopes often prefer mirrors as they do not depend on wavelength for reflection.
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You have a fixed length of wire and need to design a generator that will produce the greatest peak emf for a given frequency and magnetic field strength. You should use (a) a one-turn square coil, (b) a two-turn square coil, (c) either a one- or a two-turn square coil because both give the same peak emf for a given frequency and magnetic field strength.
Final answer:
The peak emf from a generator can be maximized by increasing the number of turns, increasing the area of the wired loop, and increasing the frequency. Both the one-turn and two-turn square coils can produce the same peak emf for a given frequency and magnetic field strength.
Explanation:
The peak emf from a generator can be maximized by maximizing the number of turns, maximizing the area of the wired loop, or maximizing the frequency. The one-turn and two-turn square coils are both options that can be used, as they both give the same peak emf for a given frequency and magnetic field strength. However, the other factors, such as the number of turns and area of the loop, may differ between the two options and can affect the overall performance of the generator.
A first-order reaction has a rate constant of 0.241/min. If the initial concentration of A is 0.859 M, what is the concentration of A after 10.0 minutes? 0.0772 M 0.00334 M 0.736 M 0.280 M
Answer:
option A.
Explanation:
given,
rate constant. k = 0.241/min
[A_0] = 0.859 M
[A_t] = ?
t = 10 minutes.
using first order reaction formula
[tex]k=\dfrac{2.303}{t}log(\dfrac{[A_0]}{[A_t]})[/tex]
[tex]0.241=\dfrac{2.303}{10}log(\dfrac{[0.859]}{[A_t]})[/tex]
[tex]log(\dfrac{[0.859]}{[A_t]}) = 1.0464[/tex]
[tex]\dfrac{[0.859]}{[A_t]} = 11.129[/tex]
[tex][A_t]=\dfrac{0.859}{11.129}[/tex]
[tex][A_t]=0.0772\ M[/tex]
the concentration of A after 10 minutes is equal to 0.0772 M.
Hence, the correct answer is option A.
A straight wire segment 2 m long makes an angle of 30degrees with a uniform magnetic field of 0.37 T. Find the magnitude of the force on the wire if it carries a current of 2.2 A.
Answer : 0.814 newton
Explanation:
force (magnetic) acting on the wire is given by
F= ? , I=2.2amp , B = 0.37 T
F = B i l sin (theta) = 0.37 x 2.2 x 2x 0.5 = 0.814N
A straight wire segment 2 m long makes an angle of 30degrees with a uniform magnetic field of 0.37 T. The magnitude of the force is
F= 0.814N
What is the magnitude of the force?Generally, the equation for the magnitude of the force is mathematically given as
[tex]F = B i l sin (\theta)[/tex]
Therefore
F= 0.37 x 2.2 x 2x 0.5
F= 0.814N
In conclusion, the magnitude of the force
F= 0.814N
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A 48.0-turn circular coil of radius 5.50 cm can be oriented in any direction in a uniform magnetic field having a magnitude of 0.480 T. If the coil carries a current of 23.3 mA, find the magnitude of the maximum possible torque exerted on the coil.
Answer:
Explanation:
Given that,
Number of turn is 48
N=48
Radius is 4.8cm
r=0.048m
Magnetic Field
B=0.48T
Current in coil
i=23.3mA
i=0.233A
Maximum Torque?
Maximum torque occur at angle 90°
Torque is given as
τ = N•I•A•B•sinθ
Where N is number of turn =48
I is current in coil =0.233A
A is area of circular coil form
Area of a circle is given as
A=πr²
A=π×0.048²
A=0.007238m²
B is magnetic field =0.48T
Maximum torque occurs at 90°
τ = N•I•A•B•sinθ
τ=48×0.233×0.007238×0.48×Sin90
τ = 0.0389Nm
This torque is large enough to exert the coil
Given Information:
Magnetic field = B = 0.480 T
Current = I = 23.3 mA = 0.0233 A
Number of turns = N = 48 turns
Radius = r = 5.50 cm = 0.055 m
Required Information:
Maximum possible torque = τ = ?
Answer:
Maximum possible torque = 0.0051 N.m
Explanation:
We know that toque τ is given by
τ = NIABsin(θ)
Where N is the number of turns of the circular coil, I is the current flowing through the circular coil, A is the area of circular coil, B is the magnetic field induced in the circular coil.
The area of the circular coil is
A = πr²
A = π(0.055)²
A = 0.009503 m²
The maximum torque is possible when θ = 90°
τ = 48*0.0233*0.009503*0.480*sin(90°)
τ = 0.0051 N.m
In the absence of an electric field, a radioactive beam strikes a fluorescent screen at a single point. When an electric field is applied, the radioactive beam is separated into three different components. One of the components is deflected toward the positive electrode because it is negatively charged, one of the components is deflected toward the negative electrode because it is positively charged, and one component is not deflected in any direction; instead, it moves along a straight path. Identify the charges possessed by the different components of the radioactive beam by observing their behavior under the influence of an electric field.
Answer: Beta, alpha and gamma ray
Explanation: The component being deflected to the positive side is the beta radiation because it is negatively charged and thus is attracted by the positive terminal of the Electric Field.
The component being deflected to the negative side is the alpha radiation, it's is positively charged and thus being attracted by the negative part of the Electric Field.
The component that went straight down without deflection is the gamma radiation. It is neutral and possess no charge and thus is not deflected.
Answer:
Beta rays (symbol β) are negatively charged because they are electrons.
Alpha rays (symbol α) are positively charged because they are helium nuclei.
Gamma rays (symbol γ or. ) are neutral, since they are photons only. Comprises of the shortest electromagnetic wavelength and thus provides the highest photon energy.
A soccer ball is rolling with a constant acceleration of -1.5 m/s2 . At ti = 0 s, the ball has an instantaneous velocity of 14 m/s. After 3s of rolling, what would be the velocity of the ball (in m/s)?
Answer:
[tex]v_{f} = 9.5\,\frac{m}{s}[/tex]
Explanation:
The final velocity is:
[tex]v_{f} = v_{o} + a \cdot t[/tex]
[tex]v_{f} = 14\,\frac{m}{s} + (-1.5\,\frac{m}{s^{2}} )\cdot (3\,s)[/tex]
[tex]v_{f} = 9.5\,\frac{m}{s}[/tex]
The velocity of the soccer ball after 3 s of rolling would be 9.5 m/s.
For a soccer ball rolling with a constant acceleration of -1.5 m/s², we can use the equation for velocity with constant acceleration: v = (initial velocity) + (acceleration)(time). At t = 3 s, the ball has an initial velocity of 14 m/s and has been rolling for 3 s, so we can plug these into the equation to find the velocity:
v = 14 m/s + (-1.5 m/s²)(3 s) = 14 m/s - 4.5 m/s = 9.5 m/s
Therefore, after 3 s of rolling, the velocity of the ball would be 9.5 m/s.
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Johnnie's father worked his entire career in the automotive manufacturing industry in Michigan. When Johnnie entered the workforce, the auto industry was in decline, so he instead found a professional career in the booming information technology sector, an industry that did not exist when his father began his career. This is an example of
a. structural mobility.
b. horizontal mobility.
c. exchange mobility.
d. social reproduction.
Final answer:
Johnnie's move from automotive manufacturing to the IT sector due to changes in the economy is an example of structural mobility, which reflects broad societal shifts affecting whole groups within the social class system. Option a
Explanation:
Johnnie's career change from the declining auto industry that his father was a part of, to the burgeoning information technology sector, is an example of structural mobility. This type of mobility occurs due to large-scale economic and societal shifts rather than individual actions. For instance, in the past, industrialization led to the general upliftment of the population, causing widespread upward structural mobility. In more recent times, factors such as recessions and outsourcing have resulted in many experiencing downward structural mobility, as traditional jobs have become obsolete or have moved overseas.
Moreover, technological advancements have altered the kinds of skills employers seek in their workforce, leading to a situation where some individuals may find themselves structurally unemployed if they lack the necessary analytical and communication skills, despite having traits like reliability and physical strength that were formerly highly valued.
Therefore, Johnnie's entrance into a different industry than his father's, brought upon by changes in the economic landscape, is a reflection of the broader trend where entire groups in society move up or down the social class ladder due to changes in the social structure itself.
The velocity of the transverse waves produced by an earthquake is 8.9 km/s, and that of the longitudinal waves is 5.1 km/s. A seismograph records the arrival of the transverse waves 68 s before the arrival of the longitudinal waves. How far away is the earthquake?
Answer: The distance is 723.4km
Explanation:
The velocity of the transverse waves is 8.9km/s
The velocity of the longitudinal wave is 5.1 km/s
The transverse one reaches 68 seconds before the longitudinal.
if the distance is X, we know that:
X/(9.8km/s) = T1
X/(5.1km/s) = T2
T2 = T1 + 68s
Where T1 and T2 are the time that each wave needs to reach the sesmograph.
We replace the third equation into the second and get:
X/(9.8km/s) = T1
X/(5.1km/s) = T1 + 68s
Now, we can replace T1 from the first equation into the second one:
X/(5.1km/s) = X/(9.8km/s) + 68s
Now we can solve it for X and find the distance.
X/(5.1km/s) - X/(9.8km/s) = 68s
X(1/(5.1km/s) - 1/(9.8km/s)) = X*0.094s/km= 68s
X = 68s/0.094s/km = 723.4 km
Two speakers, one directly behind the other, are each generating a 240-Hz sound wave. What is the smallest separation distance between the speakers that will produce destructive interference at a listener standing in front of them?
Answer:
The smallest separation distance between the speakers is 0.71 m.
Explanation:
Given that,
Two speakers, one directly behind the other, are each generating a 240-Hz sound wave, f = 240 Hz
Let the speed of sound is 343 m/s in air. The speed of sound is given by the formula as :
[tex]v=f\lambda\\\\\lambda=\dfrac{v}{f}\\\\\lambda=\dfrac{343\ m/s}{240\ Hz}\\\\\lambda=1.42\ m[/tex]
To produce destructive interference at a listener standing in front of them,
[tex]d=\dfrac{\lambda}{2}\\\\d=\dfrac{1.42}{2}\\\\d=0.71\ m[/tex]
So, the smallest separation distance between the speakers is 0.71 m. Hence, this is the required solution.
According to the question,
Frequency, f = 240 HzSpeed of sound, v = 343 m/sThe wavelength will be:
→ [tex]\lambda = \frac{v}{f}[/tex]
By substituting the values, we get
[tex]= \frac{343}{240}[/tex]
[tex]= 1.43 \ m[/tex]
hence,
The smallest separation distance will be:
= [tex]\frac{\lambda}{2}[/tex]
= [tex]\frac{1.43}{2}[/tex]
= [tex]0.715 \ m[/tex]
Thus the above solution is correct.
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