Point on Cartesian plane,
x-coordinate = 4.53
y - coordinate = 0.8
What are Polar and Cartesian coordinates?In the Cartesian system the coordinates are perpendicular to one another with the same unit length on both axes. A Polar coordinate system is determined by a fixed point, a origin or pole, and a zero direction or axis. Each point is determined by an angle and a distance relative to the zero axis and the origin.
Given,
Polar coordinates of point (r , θ)
r = 4.6
θ = 10°
Cartesian coordinates of point (x , y)
x = r cosθ
⇒ x = 4.6 × cos 10°
⇒ x = 4.6 × 0.9848
⇒ x = 4.53
y = r sinθ
⇒ y = 4.6 × sin 10°
⇒ y = 4.6 × 0.174
⇒ y = 0.8
Hence, Cartesian coordinates of the point (4.53 , 0.8)
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Find the distance traveled by a boy if he walks 3 km north and then turns and walks 1 km to the west.
Include both magnitude and direction.
Answer:
Explanation:
Distance = total length traveled = 3 + 1 = 4 km
Displacement² = 3² + 1² = 10
Displacement = √10 = 3.16
direction = tan⁻¹(3/-1) = 71.6⁰ north of west
a block hangs in equilibrium from a vertical spring. the equilibrium position sags by 7.00 cm when a second identical block is added. what is the oscillation frequency of the two-block system?
The oscillation frequency of the two-block system is 1.6 Hz.
What is equilibrium position?When a system is in equilibrium, neither its internal energy state nor its state of motion tend to change over time. A simple mechanical body is said to be in equilibrium if it neither experiences linear acceleration nor angular acceleration; unless it is disturbed by an external force, it will remain in that state indefinitely. If all of the forces acting on a single particle are vector summated to zero, equilibrium results. In addition to the states described above for the particle, a rigid body is said to be in equilibrium if the vector sum of all torques acting on it equals zero, keeping its state of rotational motion constant.
[tex]$$The oscillation frequency of the two-block is system is then found from Equation (3):$$\begin{aligned}f & =\frac{1}{2 \pi} \sqrt{\frac{k}{2 m}} \end{aligned}[/tex]
[tex]\begin{aligned}& =\frac{1}{2 \pi} \sqrt{\frac{196 \mathrm{~s}^{-2}}{2}} \\& =1.6 \mathrm{~Hz}\end{aligned}$$\\\\\text{where $2 m$ is the mass of the two blocks}[/tex]
Thus, the oscillation frequency of the two-block system is 1.6 Hz.
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1-m^3 volume of water is contained in a rigid container. Estimate the change in the volume of the water when a piston applies a pressure of 35 MPa.
The estimated change in water volume is -0.00056 [tex]m^3[/tex], which represents the volume reduction due to the applied pressure.
Where can I find volume?The formula for volume change is:
ΔV = -V * ΔP/B
where:
ΔV = volume change
V = initial volume
ΔP = pressure change
B = bulk modulus of water
After plugging in the values it looks like this:
ΔV = -1 [tex]m^3[/tex] * (35 x 10^6 Pa) / (2.2 x [tex]10^9[/tex] Pa)
= -0.00056[tex]m^3[/tex]
Therefore, the estimated change in volume of water is -0.00056 [tex]m^3[/tex], which represents the decrease in volume due to applied pressure.
How do you manually compute volume?The volume is determined by multiplying the dimensions of length, width, and height. Good news for cubes comes from the fact that each of these dimensions weighs exactly the same. As a result, we can add three to the length of any side. The formula that results is as follows: Front * Back Volume
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Determine the number of cubic feet of air per minute required to cook a room having a sensible heat gain of 4500 btu per hour to a temperature of 78 F dry bulb, if the air enters the room at a temperature of 63 F and the outside temperature is 93F
It is given that, the heat gained is 4500 btu per hour. The temperature difference here is 30 F and the specific heat of air is 0.24 btu/lb°F. Then the cubic feet of air per minute is 138.8 CFM.
What is sensible heat transfer ?The sensible heat transfer in a system can be calculated using the equation below:
q = CFM × 1.08 ×ΔT
q = CFM x 0.075 lb/ft3 x 60 min/hour x 0.24 btu/lb°F x ∆T
where, 0.24 btu/lb°F is the specific heat of the dry air.
Given that q = 4500 btu/hour.
temperature difference = 93 F - 63 F.
Then 4500 btu/hr = CFM × 1.08 × 30 F
CFM of air = 4500 btu/hr /(1.08 × 30 F ) = 138.8 CFM.
There for the number of cubic feat of air per minute is 138.8.
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A cylindrical soda can has a diameter of 6 cm and a height of 12 cm when upright. When the can is placed on its side, the soda reaches a height of 5 cm. What percentage of the volume of the can does the soda fill? (you may use a calculator for the final calculation. ).
The soda fills approximately 41.67% of the volume of the can.
How to find the percentage of the can's volume that is filled with soda?The volume of the cylindrical soda can is given by the formula V = πr^2h,
where
r is the radius of the can h is the height of the canWe are given that the diameter of the can is 6 cm, so the radius is 3 cm. Also, the height of the can is 12 cm when upright. Therefore, the volume of the can is:
V = π(3 cm)^2(12 cm) = 108π cubic centimeters.
When the can is placed on its side and the soda fills to a height of 5 cm, the volume of soda in the can is:
V_soda = π(3 cm)^2(5 cm) = 45π cubic centimeters.
To find the percentage of the can's volume that is filled with soda, we can divide the volume of the soda by the total volume of the can and multiply by 100:
Percentage = (V_soda / V) x 100%
= (45π / 108π) x 100%
= 41.67%
Therefore, the soda fills approximately 41.67% of the volume of the can.
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Events in the environment trigger physiological arousal, which in turn, give the rise to the experience of emotion.
If the stars Betelgeuse and Rigel each have the same luminosity, but the temperature of Betelgeuse is lower than Rigel, which star has the greater surface area? A. Betelgeuse B. Rigel C. They are the same size. D. There is insufficient information to answer this question.
Option A is Correct. Betelgeuse star has a larger surface area if Betelgeuse and Rigel both have the same brightness but Rigel's temperature is lower.
Rigel is usually always brighter than Orionis, despite the fact that the second-brightest star in each constellation is commonly referred to as "beta" (Betelgeuse). Therefore, red-glow objects are cooler than white-glow or blue-glow objects, despite what your water taps may claim.
Therefore, blue-white stars like Rigel and red stars like Betelgeuse have substantially lower surface temperatures. At 3,500 Kelvin, Betelgeuse's surface is warm. The brightness of Betelgeuse is extremely intense (40,000 times that of our Sun), but its surface is cold (less than 4000 K).
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. consider fictitious elements z and m. element z pulls on electrons a lot. element m does not pull on electrons much. determine which reaction will happen.
When considering fictitious elements Z and M, it is likely that element Z will have a higher ionization energy, meaning that more energy will be needed to remove an electron from it.
This means that when the two elements react, element Z will be more likely to gain electrons, while element M will be more likely to lose electrons. This means that element Z will be oxidized, while element M will be reduced.
Ionization energy, also known as ionization potential, is the amount of energy required to remove an electron from an atom or a molecule.
This process is referred to as ionization, and it involves the transfer of an electron from a neutral atom or molecule to form an ion with a positive charge.
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when stress is applied to an object and the object changes shape, we call this distortion, or strain. label the four images below with the type of strain shown.
Stretching, shear, shortening, and unstrained types of strain A c D 2. Displacement, rotation, and distortion are the three ways that deformation manifests.
What is an object in physics?In physics, an object is a visible grouping of matter that may move together in three dimensions via translation or rotation and may be contained by an obvious border. Everything has a unique identity that exists independently of any other features.
Is it a force or an object?A force is a push or pull that an object feels as a result of contact with another thing. Each time two things come into contact, a force is applied to each one of them.
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A tank for scuba diving is designed to contain 50 standard cubic feet (SCF) of air when filled to a pressure of 3,000 pounds per square inch (gage) at an ambient temperature of 80 F. Calculate the interior volume of the tank and its length if the inside diameter is 6 in. A standard cubic foot of gas occupies one cubic foot at standard temperature and pressure (T=15C and p=101.3kPa absolute ).
The interior volume of the tank is 2.70 ft³ and its length is 34.0 inches.
What is pressure?
The force delivered perpendicularly to an object's surface per unit area across which this force is dispersed is known as pressure.
To calculate the interior volume of the tank, we can use the ideal gas law:
PV = nRT
At standard temperature and pressure (STP), one mole of an ideal gas occupies 22.4 liters (0.79 cubic feet). Therefore, 50 SCF of air is equivalent to:
n = (50 SCF) / (0.79 SCF/mol) = 63.3 mol
The temperature and pressure of the tank are not at STP, so we need to use the ideal gas law to calculate the volume. Rearranging the equation to solve for V, we get:
V = nRT/P
We can assume that the temperature remains constant at 80 F (which is equivalent to 26.7 C), and the pressure is 3000 psi. The gas constant R is 8.314 J/(mol K), or 10.73 psi·ft³/(mol K).
[tex]$\mathrm{V = \frac{ (63.3 mol) \times (10.73 psi\cdot ft^3 / (mol K)) \times (26.7 C + 273.15)}{(3000 psi)} }[/tex]
= 2.70 cubic feet
To find the length of the tank, we can use the formula for the volume of a cylinder:
[tex]\mathbf{V = \pi r^2h}[/tex]
where r is the radius (which is half the diameter) and h is the height (which is the length of the tank). We know that the diameter is 6 inches, so the radius is 3 inches (0.25 feet).
Substituting the known values into the equation, we get:
2.70 cubic feet = π (0.25 ft)² h
Solving for h, we get:
h = 34.0 inches
Therefore, the interior volume of the tank is 2.70 cubic feet and its length is 34.0 inches
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four cars undergo acceleration as described by the data in the following table. car initial velocity (m/s) final velocity (m/s) time period (s) a 2.0 11.0 3.0 b -5.0 3.0 2.0 c 1.0 -5.0 2.0 d 0.0 25.0 10.0
According to the given data the acceleration of each car is a) 3.0 m/^s b) 4.0 m/s^2 c) -3.0 m/s^2 d) 2.5 m/s^2
What is an acceleration?The rate about which velocity fluctuates is titled acceleration. Acceleration typically signals a shift in speed, but not necessarily. An item that continues a circular course while maintaining a constant speed is still moving forward because the orientation of its motion is rotating.
The following equation can be used to determine each car's acceleration: acceleration = (final velocity - beginning velocity) / time period. For each car, the computation is as follows:
a) acceleration = (11.0 - 2.0) / 3.0 = 3.0 m/s^
b) acceleration = (3.0 - (-5.0)) / 2.0 = 4.0 m/s^2
c) acceleration = (-5.0 - 1.0) / 2.0 = -3.0 m/s^2
d) acceleration = (25.0 - 0.0) / 10.0 = 2.5 m/s^2
Because the vehicle is slowing down rather than accelerating, the acceleration of automobile "c" is negative.
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How much work is required to assemble eight identical point charges, each of magnitude q, at the corners of a cube of side s? Note: Assume a reference level of potential V = 0 at r = . (Use k_e for ke and q and s as necessary.)
U =
Work required to assemble eight identical point charges, each of magnitude q, at the corners of a cube of side 's' is [tex]\frac{Kq^{2} }{s} [12+6\sqrt{2} +\frac{4}{\sqrt{3}} ][/tex] where K is [tex]\frac{1}{4\pi e }[/tex].
Here, we have a cube of side 's' unit with point charges on all the vertices of the cube. So, we get 12 pairs of charges having separation of length s. 12 pairs of charges on each faces (on the diagonals) having separation of √2s and 4 pairs of charges on the body diagonals of the cube having separation of √3s.
We know that the Work done required to assemble 2 charges = (K x q1 x q2)/ r where q1 and q2 are the amount of charges and r is the separation.
Therefore, the total work done U :
⇒12 x [Kq²/s] + 12 x [Kq²/√2s] + 4 x [Kq²/√3s]
⇒[tex]\frac{Kq^{2} }{s} [12+6\sqrt{2} +\frac{4}{\sqrt{3}} ][/tex] units
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A ball is dropped, from
rest, from the top of a building. There are two motion sensors positioned
outside of two windows in the building to record the velocities of the ball
as it passes them. Sensor A is located above sensor B. Sensor A records
the ball’s velocity to be -15.5 m
s
and sensor B records the ball’s velocity to
be -17.2 m
s
. a) how far apart are the sensors?
b) how far from sensor a is the top of the building where the ball was released
c) if the time from when the ball passes sensor b until it hits the ground is 3.5s how tall is the building
Answer:
The sensors are approximately [tex]2.83\; {\rm m}[/tex] apart.
Sensor [tex]\texttt{a}[/tex] is approximately [tex]12.2\; {\rm m}[/tex] from the top of the building.
Height of the building is approximately [tex]135\; {\rm m}[/tex].
(Assumptions: air resistance on the ball is negligible; [tex]g = 9.81\; {\rm m\cdot s^{-2}}[/tex].)
Explanation:
Make use of the SUVAT equation [tex]x = (v^{2} - u^{2}) / (2\, a)[/tex], where [tex]x[/tex] represents displacement, [tex]v[/tex] and [tex]u[/tex] are the final and initial velocity, and [tex]a[/tex] is the acceleration.
In other words, as the velocity of the object changes from [tex]u\![/tex] to [tex]v\![/tex] at a rate of [tex]a[/tex], position of the object would have changed by [tex]x = (v^{2} - u^{2}) / (2\, a)[/tex].
a)When the ball is at sensor [tex]\texttt{a}[/tex], velocity of the ball was [tex]u = (-15.5)\; {\rm m\cdot s^{-1}}[/tex].
Shortly after, when the ball is at sensor [tex]\texttt{b}[/tex], velocity of the ball was [tex]v = (-17.2)\; {\rm m\cdot s^{-2}}[/tex].
Under the assumptions, [tex]a = (-g) = (-9.81)\; {\rm m\cdot s^{-2}}[/tex]. Apply the SUVAT equation [tex]x = (v^{2} - u^{2}) / (2\, a)[/tex] to find the change in the position of the ball between sensor [tex]\texttt{a}[/tex] ([tex]u = (-15.5)\; {\rm m\cdot s^{-1}}[/tex]) and [tex]\texttt{b}[/tex] ([tex]v = (-17.2)\; {\rm m\cdot s^{-2}}[/tex]):
[tex]\begin{aligned}x &= \frac{(-17.2)^{2} - (-15.5)^{2}}{2\, (-9.81)} \; {\rm m} \approx (-2.83)\; {\rm m}\end{aligned}[/tex].
Hence, the distance between sensor [tex]\texttt{a}[/tex] and [tex]\texttt{b}[/tex] is approximately [tex]2.83\; {\rm m}[/tex].
b)Since the ball was released from rest, the initial velocity of the ball would be [tex]u = 0\; {\rm m\cdot s^{-1}}[/tex].
Let [tex]v[/tex] denote the velocity of the ball at sensor [tex]\texttt{a}[/tex]: [tex]v = (-15.5)\; {\rm m\cdot s^{-1}}[/tex].
Apply the SUVAT equation [tex]x = (v^{2} - u^{2}) / (2\, a)[/tex] to find the change in the position of the ball between the top of the roof ([tex]u = 0\; {\rm m\cdot s^{-1}}[/tex]) and sensor [tex]\texttt{a}[/tex] ([tex]v = (-15.5)\; {\rm m\cdot s^{-1}}[/tex]):
[tex]\begin{aligned}x &= \frac{(-15.5)^{2} - (0)^{2}}{2\, (-9.81)} \; {\rm m} \approx (-12.2)\; {\rm m}\end{aligned}[/tex].
In other words, the distance between sensor [tex]\texttt{a}[/tex] and the top of the roof would be approximately [tex]12.2\; {\rm m}[/tex].
c)Another SUVAT equation, [tex]v = u + a\, t[/tex], gives the velocity of an object after accelerating for a duration of [tex]t[/tex] (at a rate of [tex]a[/tex], starting from an initial velocity of [tex]u[/tex].)
Make use of this SUVAT equation to find the velocity of the ball right before hitting the ground. Specifically, velocity of the ball was [tex]u = (-17.2)\; {\rm m\cdot s^{-1}}[/tex] at sensor [tex]\texttt{b}[/tex]. After accelerating at [tex]a = (-g) = (-9.81)\; {\rm m\cdot s^{-2}}[/tex] for [tex]t = 3.5\; {\rm s}[/tex], velocity of the ball would be:
[tex]\begin{aligned}v &= u + a\, t \\ &= (-17.2)\; {\rm m\cdot s^{-1}} + (-9.81)\, (3.5)\; {\rm m\cdot s^{-1}} \\ &= (-51.535)\; {\rm m\cdot s^{-1}} \end{aligned}[/tex].
Again, the velocity of the ball was [tex]u = 0\; {\rm m\cdot s^{-1}}[/tex] at the top of the building. Apply the SUVAT equation [tex]x = (v^{2} - u^{2}) / (2\, a)[/tex] to find the change in the position of the ball between the top of the building ([tex]u = 0\; {\rm m\cdot s^{-1}}[/tex]) and right before hitting the ground ([tex]v = (-51.535)\; {\rm m\cdot s^{-1}}[/tex]):
[tex]\begin{aligned}x &= \frac{(-51.535)^{2} - (0)^{2}}{2\, (-9.81)} \; {\rm m} \approx (-135)\; {\rm m}\end{aligned}[/tex].
Water (H2O) is critical to sustain life. One of its major roles is creating and breaking bonds in chemical reactions. Other significant roles of water include: a. composing up to 30% of the human body. b. acting as a solvent in the transport of dissolved substances around the body. c. the protection and support of organs.
Other significant roles of water are:
Acting as a solvent in the transport of dissolved substances around the body (option B).As protection and support of organs (option C).Water is a critical component to sustain life, including in the human body. In general, about 60% of the human body is water, but the percentage can vary from 45% to 75% based on age, sex, and hydration level. That's why option A is not correct, since there is way more than 30% of the human body is water.
Besides that, water is also important in keeping our organs healthy. It helps kidneys to remove waste, keep the blood vessels open, prevent constipation, and many more.
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in order to take quality notes, a student should have read the assigned chapter before listening to the lecture.
Answer:
it is true
Explanation:
A machinist with normal vision has a near point at 25 cm. This machinist wears +4.25-diopter eyeglasses in order to do very close work. With these eyeglasses, what is the near point of the machinist? Assume that he wears the glasses extremely close to his eyes. 12 cm 10 cm 7 cm 15 cm 17 cm
The near point of the machinist with the eyeglasses is 7 cm. This is because the power of the eyeglasses is +4.25 diopters. The near point is inversely related to the power of the lenses.
What is the power ?The power is the capacity or ability to do something or act in a particular way. It can refer to physical strength, mental strength, the ability to influence people or events, or the capacity to achieve goals. Power can also refer to knowledge, control or authority over something or someone. It can be used to accomplish great things and to gain advantage over others, but it can also be abused or misused. It is important to use power responsibly and ethically, as it has the potential to cause great harm.
Therefore, when the power of the lenses increases, the near point decreases. Therefore, the near point of the machinist with the +4.25-diopter eyeglasses would be 7 cm.
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Each of the following figures shows a person (not to scale) located on Earth at either 40°N or 40°S latitude. Rank the figures based on how much time the person spends in daylight during each 24-hour period, from most to least. To rank items as equivalent, overlap them.
Earth constantly moves. The Earth rotates around its axis once per day. The axis is the fictitious line passing through the middle of the planet, from the North to the South Poles.
What is Earths latitude?
It appears that the sun is moving across the sky as Earth spins, the sun is actually whirling. There are 24 hours in a day because it takes 24 hours to complete one rotation.
In other words, if the sun is visible in the morning beginning around 6:00 AM, the Earth will have spun around by 6:00 AM the following morning, and the sun will be seen in roughly the same location.
The Earth circles the sun and rotates on its axis. The Earth completes one full rotation of the sun in a little over 365 days.
Therefore, Earth constantly moves. The Earth rotates around its axis once per day. The axis is the fictitious line passing through the middle of the planet, from the North to the South Poles.
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a long, straight metal rod has a radius of 4.60 cm and a charge per unit length of 27.4 nc/m. find the electric field at the following distances from the axis of the rod, where distances are measured perpendicular to the rod's axis
The distances from the axis of the rod (a long, straight metal rod has a radius of 4.60 cm and a charge per unit length of 27.4 nc/m.), where distances are measured perpendicular to the rod's axis
(a) 3.20 cm = 15.39 x 10³ N/c
(b) 2.0 cm = 2.43 x 10³ N/c
(c) 200 cm = 2.4 x 10² N/c
Given a long, infinite length and a constant charge per unit, the electric field is given by:
E = λ /2πε₀r
Where,
λ = wavelength (m)
ε₀ = 8.8542 x 10⁻¹²
r = radius (m)
Hence,
The distance are measured perpendicular to the rod's axis:
a. 3.20 cm
E = (27.4 x 10⁻⁹) / 2π (8.8542 x 10⁻¹²) (0.0320)
= 15.39 x 10³ N/c
b. 20 cm
E = (27.4 x 10⁻⁹) / 2π (8.8542 x 10⁻¹²) (0.2)
= 2.43 x 10³ N/c
c. 200 cm
E = (27.4 x 10⁻⁹) / 2π (8.8542 x 10⁻¹²) (2)
= 2.4 x 10² N/c
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____________ occurs when a stress is removed and the material does not go back to its original shape.a. Plastic deformation b. Constant deformation c. Unchangeable deformation d. Permanent deformation
Plastic deformation occurs when stress is removed and the material does not go back to its original shape.
The capacity of a solid material to undergo permanent deformation, also known as plastic deformation, is defined as the ability of a material to change its shape in a way that is irreversible in response to external pressures being applied.
Plasticity can be seen, for instance, in a piece of solid metal that has been bent or pounded into a new shape. This is because permanent changes have occurred inside the material itself. Yielding is a term used in the field of engineering to describe the shift from an elastic behavior to a plastic behavior.
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A proton and an electron of masses 1.67x10^-27kg and 9.1x10^-31kg respectively are separated by a distance of 1m. Determine: a) Force due to gravitatiobal attraction (Fg). [G=6.67x10^-11Nm^2/kg^2] [F=GMm/d^2]
The force of gravitational attraction between the proton and electron is 8.23 x 10^-62 N.
Gravitational Force CalculationThe force of gravitational attraction between two masses can be calculated using the formula F = G * (m1 * m2) / d^2, where F is the force, G is the gravitational constant, m1 and m2 are the masses of the two objects, and d is the distance between them.
For the proton and electron, the force of gravitational attraction can be calculated as follows:
F = G * (1.67 x 10^-27 kg) * (9.1 x 10^-31 kg) / (1 m)^2
G = 6.67 x 10^-11 Nm^2/kg^2
Plugging in the values, we get:
F = 6.67 x 10^-11 Nm^2/kg^2 * (1.67 x 10^-27 kg) * (9.1 x 10^-31 kg) / (1 m)^2
= 8.23 x 10^-62 N
So the force of gravitational attraction between the proton and electron is 8.23 x 10^-62 N.
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Amy, who has a mass of 68 kg, is riding her bicycle in the country. She is traveling at 11m / s when she begins to coast down a hill. She then coasts up another hill that is 15 m high and comes to a stop just as she reaches the top. How high was the first hill, in m? (Round your answer to one decimal place if necessary .)
The first hill was 6.17 meters high.
What is the velocity?Velocity is defined as the displacement of the object in a given amount of time and is referred to as velocity.
The work done on Amy and her bicycle by gravitational force can be calculated as follows:
W = mgh
where m is the mass of Amy and her bicycle, g is the acceleration due to gravity (9.8 m/s²), and h is the change in height.
When Amy is coasting down the first hill, her potential energy is converted to kinetic energy, so the work done by gravity is equal to the initial increase in kinetic energy:
W = (1/2)mv²
where v is the velocity. Setting these two equations equal to each other, we get:
mgh = (1/2)mv²
Solving for h, we get:
h = (1/2)v²/g
Plugging in the known values, we get:
h = (1/2)(11 m/s)²/9.8 m/s² = 6.17 m
Thus, the first hill was 6.17 m high.
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Determine the current at point A in this circuit.
A) 2.6 A
B) 1.6 A
C) 1.8 A
D) 2.4 A
Answer:
2.6 A
Explanation:
Firstly here we need to find out the net resistance. 30Ω and 40Ω are grouped in parallel and to these 10Ω and 15Ω are in series. So we can find the net resistance as ,
[tex]\implies R_{net}=\dfrac{R_1R_2}{R_1+R_2} \\[/tex]
[tex]\implies R_{net}=\dfrac{30\times 40}{30+70}\Omega \\[/tex]
[tex]\implies R_{net}=\dfrac{120}{7}\Omega \\[/tex]
Again,
[tex]\implies R_{net}= R_1 + R_2+R_3 \\[/tex]
[tex]\implies R_{net}=\dfrac{120}{7}+10+15\Omega \\[/tex]
[tex]\implies R_{net}= \dfrac{120+105+70}{7}\Omega\\[/tex]
[tex]\implies R_{net}= \dfrac{295}{7}\Omega \approx 42\Omega \\[/tex]
Now use Ohm's law as ,
[tex]\implies V = iR \\[/tex]
[tex]\implies 110V = i\times 42\Omega\\[/tex]
[tex]\implies i =\dfrac{110}{42} A\\[/tex]
[tex]\implies \underline{\underline{ i = 2.6\ A }} \\[/tex]
and we are done!
the effective mass for electrons (in conduction band) is 5 times that of holes (in valence band), approximately how ef be from ev g
Holes are much lighter and more mobile than electrons in semiconductors. This difference in effective mass is a key factor in determining the electronic properties of materials such as B. Its conductivity, charge carrier mobility, and optical properties.
What do you understand by the the effective mass for electrons?The effective mass (meff) of an electron or hole in a solid is a measure of how a particle behaves as if it had a certain mass in a vacuum, even when interacting with crystal lattices and other particles in the solid. is. In general, the effective mass of an electron or hole depends on its energy and momentum.
In semiconductors with an energy bandgap, the valence band (VB) is separated from the conduction band (CB) by the bandgap energy (Eg). The energy of CB is higher than that of VB, the electrons in CB are free to move and conduct electricity, while the holes in VB are immobile and carry positive charge.
If the effective mass of electrons in CB is five times the effective mass of holes in VB, then the effective mass ratio can be expressed as
meff,electron / meff,hole = 5
Taking the reciprocals of both sides and rearranging them gives:
meff,hole / meff,electron = 1/5
This means that the effective mass of holes in VB is about one-fifth that of electrons in CB. In other words, holes are much lighter and more mobile than electrons in semiconductors. This difference in effective mass is a key factor in determining the electronic properties of materials such as: B. Its conductivity, charge carrier mobility, and optical properties.
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Non skull bones of the axial skeleton
Answer:
humerus and all hand bones
Explanation:
not skull bones
in axial skel.
If the atmosphere was much thicker than it is now, how would the sun appear? A. The sun would appear the same. B. The sun would appear blue-violet. C. The sun would appear green-blue. D. The sun would appear red-orange. E. The sun would appear yellow-green.
If the atmosphere was much thicker than it is now, the sun would appear red-orange. Thus, the correct option for this question is D.
What is an atmosphere?An atmosphere may be defined as a type of layer that significantly involves the mixture that significantly surrounds any celestial body or planets like earth and mars.
According to the context of this question, if the atmosphere was twice as thick as it is now, the sky when the sun is directly above you would be a stronger blue and the sun would look yellower. The red coloration of the sun at sunrise and sunset would begin earlier at sunset and last longer in the morning. It is due to refraction.
Therefore, if the atmosphere was much thicker than it is now, the sun would appear red-orange. Thus, the correct option for this question is D.
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what is the potential energy of a charged ball?
Answer:
Explanation:
Potential energy means the energy of a material when it is not moving.
Ep= mgh
That's going to depend on the height of the ball above ground, and the amount of charge on the ball.
Sadly, you haven't shared either quantity with us.
All living things release energy from their food in a process called________, which happens inside their cells
Recall from the video the steps for properly focusing on a specimen using a compound microscope. Remember to always use both eyes when looking through binocular lenses! Place the steps for focusing a microscope in the correct order from left to right.
1)Click the scanningobjective into positionabove the specimen.
2)Use the coarse adjustment knob to move the stage all the way up.
3)Use the fine adjustment knob until the specimen is in focus.
4)Adjust the iris diaphragm and/or the light control knob to improve contrast.
5)Use the mechanical stage controls to move a given structure to the center of your field of view.
The steps are already in correct order for focusing on specimen while using a microscope.
What is microscope used for?
The Latin word "microscopium," from which the English word "microscope" is derived, is a combination of the Greek words "mikros," which means "little," and "skopein," which means "to look at."
A microscope is a device that enlarges an image of a small object to expose details that the unassisted eye cannot perceive. The optical microscope, which employs visible light focused through lenses, is the most used type of microscope.
Small rectangles of clear glass or plastic called microscope slides are used to support specimens while they are being studied under a microscope.
Hence, the step are in correct order 1), 2), 3), 4), and 5) for focusing on specimen while using a microscope.
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The water is flowing through a tapping pipe having diameter 400mm and 200mm at section 1 and 2 respectively. The discharge through the pipe is 0.06m³/sec. The section 1 is. 10m above datum line and section 2 is 6m above datum line. Find the intensity of pressure at section 2 to that if that of section 1 is 420 KN/m².
The intensity of pressure at section 2 is approximately 345.64 kN/m^2.
What is pressure?
Pressure is described as the force applied perpendicular to the surface of an object per unit area over which that force is distributed.
To determine the intensity of pressure at section 2, we need to use the equation of continuity and the Bernoulli equation:
we will substitute the values from the continuity equation into the Bernoulli equation which will give us :
P_1 + 1/2 * p * (Q_1 / A_1)^2 + p * g * h_1 = P_2 + 1/2 * p * (Q_2 / A_2)^2 + p * g * h_2
Substitute all the given values into the equation
Solving for P_2 = 420,000 N/m^2 + 47.06 KN/m^2 - 71.42 KN/m^2 = 345.64 KN/m^2
In conclusion, the intensity of pressure at section 2 is approximately 345.64 kN/m^2.
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A space probe traveling from Earth will hit the asteroid belt at what distance? A. 3 AU B. 30 AU C. 50 AU D. 5,000 AU
A space probe traveling from Earth will hit the asteroid belt at 3 AU. Hence, option (A) is correct.
What is the asteroid belt?The Solar System's asteroid belt is a torus-shaped region that is located between the orbits of Jupiter and Mars and is centered on the Sun.
It is home to a large number of asteroids and minor planets, which are solid bodies with erratic shapes. The discovered objects range in size, but they are all considerably smaller than planets.
On average, they are around one million kilometers (or six hundred thousand miles) away. To distinguish it from other asteroid populations in the Solar System.
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