Answer:
The correct answer is A. The current moving through the wire made the nail into an electromagnet.
An electromagnet is created when a current is passed through a wire. The flow of electrons creates a magnetic field, which can attract or repel other magnetic objects. In this case, when the student wrapped the copper wire around the nail and connected it to a battery, the flow of electrons created a magnetic field around the nail. This magnetic field allowed the student to pick up the staples because they were attracted to the nail, which was now acting as an electromagnet.
Four capacitors are of the same capacity. If three of them are connected in
parallel and the remaining one is connected in series with this combination, the
resultant capacity is 16.5 μF. Find the capacity of each capacito
Answer:
bbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb
Explanation:
A certain asteroid is approximately 50 AU from the Sun. According to Kepler’s third law, what is the orbital period of this asteroid in terms of Earth years?
The orbital period of this asteroid is approximately 353.55 Earth years.
What is the orbital period of this asteroid?
Kepler's third law states that the square of the orbital period (in years) of a planet or asteroid is proportional to the cube of its semi-major axis (in AU).
Therefore, we can use the following formula to calculate the orbital period:
(T / 1 year)^2 = (a / 1 AU)^3
where;
T is the orbital period in years, and a is the semi-major axis in AU.Plugging in the values for the asteroid at 50 AU, we get:
(T / 1 year)^2 = (50 / 1 AU)^3
(T / 1 year)^2 = 125000
T / 1 year = √(125000)
T = 353.55 Earth years (rounded to two decimal places)
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State any five (5) clinical uses of heat enery .
The therapeutic effects of heat include increasing the extensibility of collagen tissues; decreasing joint stiffness; reducing pain; relieving muscle spasms; reducing inflammation, edema, and aids in the post acute phase of healing; and increasing blood flow.
What is the role of heat energy in the body?In the process of ATP production by cells throughout the body, approximately 60 percent of the energy produced is in the form of heat used to maintain body temperature. Thermoregulation is an example of negative feedback.
Why is heat used as a treatment during health care?
Heat treatments should be used for chronic conditions to help relax and loosen tissues, and to stimulate blood flow to the area. Heat is a great treatment option for chronic conditions caused from old injuries or even arthritis. Heat can be applied before participating in activities to help limber up.
So we can conclude that heat is one of the most important form of energy for our daily life.
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A 65 kg person sits on a 3.6 kg chair. Each leg of the chair makes contact with the floor in a circle that is 1.1 cm in diameter.
Find the pressure exerted on the floor by each leg of the chair, assuming the weight is evenly distributed.!!!!!!!!!!!!!!!!!!!!URGENT!!!!!!!!!!!!!!!!1
The pressure exerted on the floor by each leg of the chair is [tex]7.080685*10^6M Pa[/tex]
Given :
Mass of person = 65 kg
Mass of chair = 3.6 kg
Diameter of contact point = 1.1 cm = D
Radius of contact point = 1.1/2 = 0.55 cm
The total mass of chair and person = 65 + 3.6 = 68.6 kg = M
Acceleration due to gravity = 9.81 m/s²
Force acting on the floor F
F=mg
F= 68.6 * 9.81
F = 672 .9 N
Area of the contact point
A = [tex]\pi r^2[/tex]
A = [tex]\pi (0.55)^2[/tex]
A = [tex]0.00003025\pi[/tex]
Pressure = force/Area
Pressure = [tex]672.9/0.00003025\pi[/tex]
Pressure =[tex]7.080685*10^6[/tex]
What is pressure in short answer?
Pressure is an expression of force exerted on a surface per unit area. if a force F is applied on area A, then pressure P=F/A.
What is the formula for pressure, force, and area?
Pressure is defined as force per unit area. It is expressed as P = F/A, where P is pressure in pascals, F is the force in newtons, and A is the area in square meters.
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A motorist runs out of gas on a level road 210 m from a gas station. The driver pushes the 1,180 kg car to the gas station. If a 140 N force is required to keep the car moving, how much work (in J) does the driver do?
Answer:
the driver does 29,400 J of work in pushing the car to the gas station.
Explanation:
To find the amount of work done by the driver, we need to know the force applied by the driver and the distance the car is moved.
The force applied by the driver is given as 140 N, and the distance the car is moved is given as 210 m.
Work done can be calculated using the formula:
W = F * d * cos(θ)
where W is the work done, F is the force applied, d is the distance moved, and θ is the angle between the force and the displacement. Since the force and displacement are in the same direction, θ = 0, and the cosine of 0 is 1.
So,
W = F * d * cos(θ) = 140 N * 210 m * 1 = 29,400 J
A positively charged solid conducting sphere is contained within a negatively charged conducting spherical shell as shown. The magnitude of the total charge on each sphere is the same.
1)Which of the following statements best describes the electric field in the region between the spheres?
A.The field points radially outward
B.The field points radially inward
C.The field is zero
2.Which of the following statements best describes the electric field in the region outside the red sphere?
A.The field points radially outward
B.The field points radially inward
C.The field is zero
CAN u give me a explanation.
1) Statements best describes the electric field in the region between the spheres field points radially outward.
2) Statements best describes the electric field in the region outside the red sphere The field is zero.
For positive charge, E field lines are projecting outward of positive charge.
For negative charge, E field lines are projecting inward of negative charge.
Then, between the spheres,
because of q, E field lines are projecting outward of q and inward of- q;
and because of- q, E filed lines are projecting outward of q and inward of- q
So, The field points radially outward
2) C still, net charge inside the gaussian face is zero, If we draw a gaussian face outside the spheres.
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what is life? can you explain in please
A startled deer runs 520 m at 20.0° north of east for half a minute, then turns and runs 380 m at 55.0° north of west for 15.0 seconds and stops. What is the average velocity of the deer during this time?
Answerbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb
Explanation:
A startled deer runs 520 m at 20.0° north of east for half a minute, then turns and runs 380 m at 55.0° north of west for 15.0 seconds and stops. Therefore, 60m/s is the average velocity of the deer during this time.
What is average velocity?Motion may be defined using physical quantity concepts such as speed, velocity, duration, displacement, as well as acceleration. Sir Isaac Newton provided the correct explanation of motion.
All of these quantities are explained in terms of a single quantity, time. In this section, we will look at average velocity, its mathematical representation, and its graphical depiction.
average velocity = 520 m+ 380 m / 15=60m/s
Therefore, 60m/s is the average velocity of the deer during this time.
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Which sample of water will contain the smallest amount of heat energy?
10 g at 10°C
20 g at 10°C
20 g at 20°C
20 g at 40°C
Answer:
10 g at 10°C
Explanation:
Answer:
10 g at 10°C
Explanation:
Heat energy is directly proportional to the mass and the temperature of a substance. When comparing two samples of the same substance, the one with higher mass contains more heat energy. Similarly, a substance with higher temperature contains more heat energy compared to the same substance with a lower temperature.
In this case, the smallest amount of heat energy would be in the sample of 10g at 10°C. This is because it has the lowest mass and temperature compared to the other three options.
ALLEN
They include cash,
accounts receivable, notes receivable,
inventory and prepaid expenses. Step 8
explains each one in detail.
1.
You are the owner of an appliance store.
Choose four categories from above and
explain how that asset may fit into your
business. For example, if I owned a
company that installed furnaces, my
inventory would include different types of
furnaces and air conditioners as well as
controls to operate them. Accounts
receivable would include the installations
we did but have not been paid for yet.
As the owner of an appliance store, the following four categories of assets could apply to my business:
What are the categories of assets that could apply to my business?Cash: Cash is the most liquid asset and represents the money that the business has on hand or in the bank. As an appliance store owner, cash can be used to pay for business expenses such as rent, utilities, employee wages, and inventory purchases.
Accounts receivable: Accounts receivable represents the money that customers owe to the business for goods or services that have been sold but not yet paid for. In the context of an appliance store, accounts receivable would include sales made on credit, where customers are allowed to take the appliances and pay for them at a later date.
Inventory: Inventory represents the goods that the business has on hand to sell to customers. For an appliance store, inventory would include appliances such as refrigerators, washing machines, dryers, and ovens. The value of the inventory can fluctuate depending on market demand, changes in technology, and other factors.
Prepaid expenses: Prepaid expenses represent the costs that the business has paid in advance for goods or services that will be used in the future. In the context of an appliance store, prepaid expenses may include things like rent payments, insurance premiums, and maintenance contracts. By prepaying for these expenses, the business can ensure that it has access to the resources it needs to operate efficiently.
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HELPPPP
i need this done quickly
Answer:
F(Net) = F(a) + F(g) + F(f) + FN.
Drop Tower NASA operates a 2.2-second drop tower at the Glenn Research Center in Cleveland, Ohio. At this facility, experimental packages are dropped from the top of the tower, on the 8th floor of the building. During their 2.2 seconds of free fall, experiments experience a microgravity environment similar to that of a spacecraft in orbit.
What is the drop distance of a 2.2- s tower?
(part a)
How fast are the experiments traveling when they hit the air bags at the bottom of the tower?
(part b)
If the experimental package comes to rest over a distance of 0.80 m upon hitting the air bags, what is the average stopping acceleration?
(part c)
The drop distance is 23.7402 m
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
g = Acceleration due to gravity = 9.81 m/s² = a
Equation of motion
S = ut + ½ at²
S = 0 × t × ½ × 9.81 × 2.2²
S= 23.7402 m
The drop distance is23.7402 m
v = u + at
v = 0 + 9.81 × 2.2
v= 21. 582 m/s
When they hit the airbags at the bottom of the tower the speed of the experiments is 21.582 m/s
The final speed of the fall will be the initial velocity of stopping
v² - u² = 2as
a = v² - u²/ 2s
a= 310.521816 m/s²
The average stopping acceleration is 310.521816 m/s²
What is displacement?
The temporal rate of change of the position vector can also be used to describe velocity. The velocity of P (for example, a point representing the position of a passenger walking on the train) may be referred to as a relative velocity as opposed to an absolute velocity if one considers a moving initial position, or equivalently a moving origin (e.g., an initial position or origin which is fixed to a train waggon which in turn moves on its rail track).To know more about displacement, click the link given below:
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For each of the following file extensions, select the correct file format from the drop-down menu. . Avi. Rtf. Pdf. Txt.
The following file extensions are
.avi: Audio Video Interleave file format
.rtf: Rich Text Format file format
.pdf: Portable Document Format file format
.txt: Text File format
File extensions are used by computer operating systems and applications to determine which program to use to open a file and how to handle it.
A file extension is a series of characters that follow the last period in a filename and indicates the format of the file. It is a way of identifying the type of data stored in a file, such as a text document, image, audio or video file, spreadsheet, or executable program.
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who suggested mutation theory
Answer:
Hugo de VriesThe mutation Theory of Evolution was proposed by Hugo de Vries.Answer:
Hugo de Vries
Explanation:
In the history of evolutionary biology, Hugo dr Vries is known as a proponent of the mutation theory of evolution, in which new species are believed to arise by single mutational events.
The following table shows the distance from the sun of some unknown planets of equal mass.
Distance from Sun
Planet Distance from Sun
W 1.5 AU
X 0.723 AU
Which of the following best explains which planet revolves at a faster speed?
Planet X, because the sun pulls it with a greater force
Planet W, because the sun pulls it with a greater force
Planet W, because the gravitational force is weakened by distance
Planet X, because the gravitational force is strengthened by distance
(D was wrong, I'm taking the test again)
The correct answer is : Planet X, because the sun pulls it with a greater force
The speed at which a planet revolves around the sun is determined by its distance from the sun and the gravitational force between the planet and the sun. The closer a planet is to the sun, the stronger the gravitational force, and the faster it will orbit. Therefore, Planet X, which is closer to the sun than Planet W, will revolve around the sun at a faster speed, assuming that both planets have equal masses. Thus, the correct answer is "Planet X, because the gravitational force is strengthened by distance." The farther a planet is from the sun, the weaker the gravitational force, but this would result in a slower orbital speed, not a faster one.To know more about planets visit:
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Answer: A A A !!!!
Explanation: just did the test, got it right thx thx!!!
Two balls are dropped to the ground from different heights. One is dropped 1.5 s after the other, but they
both strike the ground at the same time, 5.0 s after the first was dropped. (a) What is the difference in the
heights from which they were dropped? (b) From what height was the first ball dropped? Ans. (a) 63 m;
(b) 0.12 km
Answer:
Height difference: approximately [tex]63\; {\rm m}[/tex].
The first ball was dropped from a height of approximately [tex]123\; {\rm m}[/tex].
(Assumptions: both balls were released from rest, air friction is negligible, and that [tex]g = 9.81\; {\rm m\cdot s^{-2}}[/tex].)
Explanation:
Under the assumptions, both ball would accelerate at a constant [tex]a = (-g) = (-9.81)\; {\rm m\cdot s^{-2}}[/tex].
Let [tex]t[/tex] denote the time since the first ball was released.
Height of the first ball at time [tex]t[/tex] can be modelled with the SUVAT equation [tex]h(t) = (1/2)\, a\, t^{2} + u\, t + h_{0}[/tex], where [tex]u[/tex] is the initial velocity. However, since [tex]u = 0\; {\rm m\cdot s^{-1}}[/tex] by assumption, this equation simplifies to [tex]h(t) = (1/2)\, a\, t^{2} + h_{0}[/tex].
Since this ball reached the ground after [tex]t = 5.0\; {\rm s}[/tex], [tex]h(5.0) = 0\; {\rm m}[/tex]. In other words:
[tex]\begin{aligned}\frac{1}{2}\, (-9.81)\, (5.0)^{2} + h_{0} = 0\end{aligned}[/tex].
Simplify and solve for the initial height of this ball, [tex]h_{0}[/tex]:
[tex]\begin{aligned}h_{0} &= -\frac{1}{2}\, (-9.81)\, (5.0)^{2} \\ &\approx 123\; {\rm m}\end{aligned}[/tex].
In other words, the first ball was dropped from a height of approximately [tex]123\; {\rm m}[/tex].
Similarly, the height of the second ball may be modelled as [tex]h(t) = (1/2)\, a\, t^{2} + h_{0}[/tex].
Since this ball reached the ground [tex]t = (5.0 - 1.5)\; {\rm s} = 3.5\; {\rm s}[/tex] after being released, [tex]h(3.5) = 0\; {\rm m}[/tex]. The initial height of this ball would be:
[tex]\begin{aligned}h_{0} &= -\frac{1}{2}\, (-9.81)\, (3.5)^{2} \\ &\approx (-60)\; {\rm m}\end{aligned}[/tex].
Subtract the initial height of the second ball from that of the first ball to find the difference in initial height:
[tex](123 - 60) \; {\rm m} \approx 63\; {\rm m}[/tex].
A rock is thrown upward with a velocity of 14
meters per second from the top of a 50
meter high cliff, and it misses the cliff on the way back down. When will the rock be 2
meters from ground level? Round your answer to two decimal places.
Answer:
The rock will be 2 meters from the ground after 1.29 seconds.
Explanation:
The motion of the rock can be described by the following kinematic equation:
h = vi * t + 0.5 * a * t^2
where h is the height of the rock above the ground, vi is the initial velocity (14 m/s), t is time, and a is the acceleration due to gravity (-9.8 m/s^2).
Setting h = 2 m and solving for t, we get:
2 = 14 * t + 0.5 * (-9.8) * t^2
Expanding and solving for t, we get:
t = 1.29 seconds
So, the rock will be 2 meters from the ground after approximately 1.29 seconds.
What is the mass density of a liquid that exerts a pressure of 180 kPa at a depth of 41 m?
Answer: no answer
Explanation:
The bottom cube has mass m2 = 4.8kg and sits on a surface without friction. The top cube has mass m1 = 2.2kg. The friction between the two cubes is μ = 0.5. What is the maximum force that can be applied on the upper cube so that the two cubes move together?
The maximum force that can be applied on the upper cube is 26.29N.
Solving for the the maximum force that can be applied on the upper cube, we have:The maximum force that can be applied on the upper cube so that the two cubes move together is given by the equation:
F_max = μ(m1 + m2)g
where,
μ = the coefficient of friction
m1 and m2 = the masses of the top and bottom cube
g = the acceleration due to gravity.
Substituting in the values, we get:
F_max = 0.5(2.2kg + 4.8kg)(9.81m/s²)
F_max = 26.29N
Therefore, the maximum force that can be applied on the upper cube is 26.29N.
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Find the moment of inertia for the following objects:
A. A uniform disc of mass M and radius R rotating about its axis. Replace the sum in 3.31 by the appropriate integral and do the integral in polar coordinates.
B. A uniform solid sphere rotating about a diameter. Replace the sum in 3.31 by the appropriate integral and do the integral in spherical coordinates.
The moment of inertia of a uniform disc of mass M and radius R rotating about its axis is [tex](1/2) MR^2[/tex].
The moment of inertia of a uniform solid sphere of mass M and radius R rotating about a diameter is [tex](8/5) MR^2[/tex].
The moment of inertia of a uniform disc of mass M and radius R rotating about its axis can be found by integrating the moment of inertia of small elements of mass dm located at a distance r from the axis of rotation.
Using polar coordinates, we can write dm = (M/πR^2)rdrdθ, where r ranges from 0 to R and θ ranges from 0 to 2π.
The moment of inertia of each element is given by dI = dm r^2. Therefore, we have:
I = ∫dI
= ∫[tex]r^2 dm[/tex]
= ∫₀²π ∫₀ᴿ (M/πR^2)r³drdθ
= (M/πR^2) ∫₀²π [∫₀ᴿ r³dr] dθ
= (M/πR^2) ∫₀²π [(1/4)R^4] dθ
= (M/πR^2) (1/4)R^4 (2π)
= [tex](1/2) MR^2[/tex]
The moment of inertia of a uniform solid sphere of mass M and radius R rotating about a diameter can be found by integrating the moment of inertia of small elements of mass dm located at a distance r from the diameter. Using spherical coordinates, we can write dm = (M/4πR^3)r^2sinθdrdθdφ, where r ranges from 0 to R, θ ranges from 0 to π, and φ ranges from 0 to 2π. The moment of inertia of each element is given by dI = dm r^2sin^2θ. Therefore, we have:
I = ∫dI = ∫r^2sin^2θ dm = ∫₀²π ∫₀ᴾ ∫₀ᴿ (M/4πR^3)r^4sin^3θdrdθdφ
= (M/4πR^3) ∫₀²π ∫₀ᴾ [∫₀ᴿ r^4sin^3θdr] dθdφ
= (M/4πR^3) ∫₀²π ∫₀ᴾ [(2/5)R^5sin^3θ] dθdφ
= (2/5) MR^2 ∫₀²π [∫₀ᴾ sin^3θ dθ] dφ
= (2/5) MR^2 ∫₀²π [(-cosθ + (3/2)cos^3θ/3)|₀ᴾ] dφ
= (8/15) MR^2 ∫₀²π dφ
= (8/15) MR^2 (2π)
= (8/5) MR^2
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A car driving at a constant speed of 20 m/s (meters per
second) turns right, down a street.
1. Is the car accelerating during the turn? Why or why not?
2. Would you feel any of the movement during this event?
Why or why not?
EXTRA HELP
What is changing as the car turns? What changes qualify as
an acceleration? How can you feel acceleration?
Answer:
it is accelerating because it is at constant speed.
A 0.0272 kg arrow is fired with a velocity of 87.3 m/s from a 0.0245 kg bow. Determine the equivalent impulse value imparted on the arrow. (hint: you can also use newton's third law of motion)* A.2.37 Ns B.3.11 Ns C.113.5 Ns D.214.2 Ns
Answer:
Explanation:
C. 113.5 Ns
A pendulum bob of mass 5 kg is displaced from the vertical until it is 0.25 m above its lowest point. Upon release, it rises to a height of 0.24 m on the other side. How much energy was lost due to friction? Assuming half the total energy loss takes place during the downward swing, with what speed does the bob pass though its lowest point?
The energy lost due to friction is 0.49 J. And the speed of the bob passing through its lowest point will be 2.19 m/s.
What is the conservation of energy?The mass is 5 kg and the heights are h₁ 0.25 m and h₂ = 0.24. And the difference in the height is given as,
h₁ - h₂ = 0.25 - 0.24
h₁ - h₂ = 0.01
The energy loss due to friction is given as,
Energy loss = mgh₁ - mgh₂
Energy loss = mg (h₁ - h₂)
Energy loss = 5 x 9.8 x 0.01
Energy loss = 0.49 J
The energy lost due to friction is 0.49 J.
The velocity is given as,
KE = mgh₁ - E / 2
(1/2) x 5 x v² = 5 x 9.8 x 0.25 - 0.49 / 2
2.5v² = 12.25 - 0.245
v² = 4.802
v = 2.19 meters per second
The speed of the bob passing through its lowest point will be 2.19 m/s.
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An object starts from rest at the origin and moves along the x axis subject to the force shown in the figure below. If the mass of
the object is 38.0 kg, determine the speed of the object at the following positions. Each unit along the position axis is 2.00 m and
each unit along the force axis is 3.00 N.
The speeds of the object at the various positions are :
At x = A; speed = 2.06 m/sAt x = B; speed = 2.38 m/sAt x = speed = 0 m/sWhat is the formula to determine the velocity or speed?The formula to determine the velocity or speed is given below:
velocity or speed = √(2s * F/m)
where:
s the position or displacementF is the forcem is the massThe speeds of the object at the various points are given below:
At x = A: force = 9 N; s = 6 m
speed = √(2 * 6 * 9 / 38)
speed = 2.06 m/s
At x = B; Force = 9 N; s = 12 m
speed = √(2 * 12 * 9 / 38)
speed = 2.38 m/s
At x = C; Force = 0 N; s = 18 m
speed = √(2 * 12 * 0 / 38)
speed = 0 m/s
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The speed of light in vacuum is 3* 10^8. If the refractive index of a transparent liquid is 4/3, then what is the speed of light in the liquid?
The speed of light in vacuum is 3 × 10⁸ m/s. If the refractive index of a transparent liquid is 4/3, then the speed of light in water will be 2.25 × 10⁸ m/s.
What is Refractive index?Refractive Index is a value calculated from the ratio of the speed of light in a vacuum to the speed of light in the second medium of greater density. The refractive index variable is most commonly symbolized by the letter n or n'.
R.I. of water = Speed of light in vacuum / speed of light in water
n' = 4/3
Speed of light in vacuum = 3 × 10⁸
n' = Speed of light in vacuum / speed of light in water
Speed of light in water = Speed of light in vacuum/ n'
Speed of light in water = 3 × 10⁸/ 4/3
Speed of light in water = 2.25 × 10⁸
Therefore, the speed of light in water medium will be 2.25 × 10⁸ m/s.
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Blood flows through a major artery at 1.2 m/s over a distance of 0.50m then at 0.70 m/s for another 0.50 m through a smaller artery.
Is the average speed of the blood 0.95 m/s greater than 0.95 m/s or less than 0.95 m/s (part a)
Verify your answer to Part A by calculating the average speed.
(part b)
The average speed of the blood is less than 0.95 m/s and the average speed is 0.884 m/s.
What is average speed?
Average speed is a measure of the rate of change in position of an object over a certain period of time. It is calculated by dividing the total distance traveled by the total time taken. In other words, it is the rate at which an object covers a given distance. Average speed is useful in determining the efficiency of a vehicle, or of a person, in covering a particular distance. For example, if a person drives for 2 hours and travels a total distance of 200 kilometers, the average speed of the body is 100 km/hr (200km / 2 hrs).
The average speed is,
[tex]V_{avg} = \frac{d_{1}+ d_{2} }{\frac{d_{1} }{V_{1} }+ \frac{d_{2} }{V_{2} } }[/tex]
[tex]= \frac{0.50m + 0.50m}{\frac{0.50m}{1.2m/s}+\frac{0.50m}{0.70m/s} }[/tex]
= 0.884 m/s
Due to the average speed is 0.884 m/s the average speed of the blood becomes less than 0.95 m/s.
Therefore, the average speed of the blood is less than 0.95 m/s and the average speed is 0.884 m/s.
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aInfinite Sheets of Charge | 1|2 In both cases shown below, the colored lines represent positive (blue) and negative (red) charged planes. The magnitudes of the charge per unit area on each plane is the same. Case A Case B 1) In which case is the magnitude of the electric field at point P bigger? Case A Case B They are the same Submit (Survey Question) 2) Explain your reasoning Submit
Because the charge densities per unit area on every plane are equal. Hence, the strength of the electric field around point P is the same in situations A and B.
Explain the electric filed due to infinite line of charge?Gauss' law can be used to determine the electric field of such an infinite line charge with such a uniform linear charge density.
The electric field is of equal strength at each point of a cylinder and thus is pointed outward when a Gaussian surface with radius r is modeled as a cylinder.
We discovered that electric field is radially oriented away as from line charge and that its strength decreases in inverse proportion to its distance.
The colored lines in the two instances below denote the positive (blue) and negative (red) charged planes, respectively.Because the charge densities per unit area on every plane are equal.Hence, the strength of the electric field around point P is the same in situations A and B.
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Identify the correct sequence of events from this passage.
A. Martin Cooper invented the first cell phone.
B. AT&T discussed the concept of a wireless phone.
C. The FDA ruled that the health risks of cell phones were small.
D. Cell phone became the most common electronic device.
Answer:
The correct sequence of events is:
A. Martin Cooper invented the first cell phone.
B. AT&T discussed the concept of a wireless phone.
C. Cell phone became the most common electronic device.
D. The FDA ruled that the health risks of cell phones were small.
Continue where the red line would would go
Refraction would cause the red light to bend at the air/glass boundary.
What is refraction?While the question is unclear, I assume that you are trying to know something about the refraction of the red light across the air/glass boundary. I will tell you something about refraction!
Refraction is a phenomenon of light where it changes direction when it enters a medium with a different refractive index. Refractive index is a measure of how much a medium bends light that passes through it. When light passes from a medium with a lower refractive index to a medium with a higher refractive index, it bends towards the normal (an imaginary line perpendicular to the surface at the point of incidence). On the other hand, when light passes from a medium with a higher refractive index to a medium with a lower refractive index, it bends away from the normal.
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Part A- Variation in the speed of sound Sound is a longitudinal wave, and its speed depends on the medium through which it propagates. In air, sound travels at 343 m/s. In an experiment, you observo a sound with a frequency of 500 Hz and a wavelength of 6.4 m. What is the speed of sound in this differont medium? Express your answer with the appropriate units. HÅ ? Value Units Submit Request Answer Sounds vary in intensity and loudness across a very wide range, and we use the logarithmic decibel scale to quantify sound. Sort and match the following sounds with their decibel level. Place each sound in the appropriate in for its decibel level. Only one sound may be placed in each bin. Reset Help Operating a lawnmower Your clothing rusting Telephone hold music A whisper across a football field Abrary with people studying -20 dB 10 dB 30 dB 60 dB 90 dB Part D- Calculating the limits of hearing In a large perfectly quiet room, at what maximum distance could you hear a bee with a power output of 1W? Recall that the threshold of hearing is 10-12 W/m² Express your answer with the appropriate units. HA ? T Value Units Submit Request Answer
The Velocity is 4270 m/s it is equal to product of frequency and wavelength.
Given: -
Frequency(f) = 700 Hz
Wavelength(λ) = 6.1m
velocity = f x λ
velocity = 700 x 6.1
velocity = 4270 m/s
Wavelength is a fundamental concept in physics, particularly in the study of waves, which includes electromagnetic radiation, light, and sound waves. It is defined as the distance between two successive points of the same phase, such as the crest or trough of a wave. This distance is usually measured in units of meters, and it is inversely proportional to the frequency of the wave.
In electromagnetic radiation, the wavelength is an important property that determines the type of radiation, such as radio waves, microwaves, infrared radiation, visible light, ultraviolet radiation, X-rays, and gamma rays. Each type of radiation has a unique range of wavelengths that can be measured using instruments like a spectroscope or a radio antenna.
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