Answer:
A series of body movements in which a person leaps into the air and then rotates one or more times while airborne is referred to as a flip.
When an athlete flips and lands on the beam, to increase his or her balance, he or she must spread out their hands-on either sides.
In doing this, they redistribute their forces (centre of gravity) to gain equilibrium create inertia to dampen the kinetic energy during the flip.
Cheers!
A person with normal vision can focus on objects as close as a few centimeters from the eye up to objects infinitely far away. There exist, however, certain conditions under which the range of vision is not so extended. For example, a nearsighted person cannot focus on objects farther than a certain point (the far point), while a farsighted person cannot focus on objects closer than a certain point (the near point). Note that even though the presence of a near point is common to everyone, a farsighted person has a near point that is much farther from the eye than the near point of a person with normal vision.
Both nearsightedness and farsightedness can be corrected with the use of glasses or contact lenses. In this case, the eye converges the light coming from the image formed by the corrective lens rather than from the object itself.If a nearsighted person has a far point df that is 3.50 m from the eye, what is the focal length f1 of the contact lenses that the person would need to see an object at infinity clearly?
Answer:
Explanation:
For a person suffering from nearsightedness, far point is 3.5 m . That means , an object beyond 3.5 m will not be clearly visible by the person . Ray of light coming from 3.5 m will be focussed on retina . For an object to be visible from infinity , ray of light coming from infinity should appear to be coming from 3.5 m . In other words , the object placed at infinity should form a virtual image at 3.5 m . This can be done by concave lens.
u = ∝ ; v = 3.5
Using lens formula
[tex]\frac{1}{v} -\frac{1}{u} = \frac{1}{f}[/tex]
[tex]\frac{1}{-3.5} -0 = \frac{1}{f}[/tex]
f = - 3.5 m .
A microstate is a state of a physical system described at the finest level of detail. A macrostate is a state of a physical system that is described in terms of the systems overall or average properties at a macroscopic level. A macrostate will generally consist of many different microstates. In defining a macrostate we ignore what is going on at the microscopic (atomic/molecular) level.
Suppose 24 students are enrolled in a class. For simplicity let’s assume that any student who is present is sitting completely still at his or her assigned seat, facing forward (so we don’t have to worry about students being in different locations, having different motions, etc.).
1. Which of the following would constitute a macrostate description of the class attendance and which would constitute a microstate description.
a. A list of the names of each student present today.
b. The number of students in attendance.
Answer:
a. A list of the names of each student present today. (microstate)
b. The number of students in attendance. (macrostate)
Explanation:
You can fins the answer to this question by comparing the situation of the problem with a system of molecules with discrete energy.
Without importance of which molecules have a specific energy, but rather, what is the total amount of energy, you can get for different configurations of energy the same amount of the total energy. If different configurations of the energies of the molecules give you the same total energy of the system, you say that the macrostate is the same. In the case of the classroom, it does not matter how are distributed the students in the class, the total number of students is always the same. The macrostate is the same for what ever organization of the students in the class.
If you would interested in the energy of each molecules, you will obtain different configurations. In the case of the classroom. The names of the student will define a microstate because in this case there are many configurations.
a. A list of the names of each student present today. (microstate)
b. The number of students in attendance. (macrostate)
As the captain of the scientific team sent to Planet Physics, one of your tasks is to measure g. You have a long, thin wire labeled 1.73 g/m and a 1.30 kg weight. You have your accurate space cadet chronometer but, unfortunately, you seem to have forgotten a meter stick. Undeterred, you first find the midpoint of the wire by folding it in half. You then attach one end of the wire to the wall of your laboratory, stretch it horizontally to pass over a pulley at the midpoint of the wire, then tie the 1.30 kg weight to the end hanging over the pulley. By vibrating the wire, and measuring time with your chronometer, you find that the wire's second harmonic frequency is 200 Hz . Next, with the 1.30 kg weight still tied to one end of the wire, you attach the other end to the ceiling to make a pendulum. You find that the pendulum requires 313 s to complete 200 oscillations. Pulling out your trusty calculator, you get to work.
What value of g will you report back to headquarters?
Answer:
The value of g is [tex]g =76.2 m/s^2[/tex]
Explanation:
From the question we are told that
The mass of the weight is [tex]m = 1.30 kg[/tex]
The spring constant [tex]k = 1.73 g/m = 1.73 *10^{-3} \ kg/m[/tex]
The second harmonic frequency is [tex]f = 100 \ Hz[/tex]
The number of oscillation is [tex]N = 200[/tex]
The time taken is [tex]t = 315 \ s[/tex]
Generally the frequency is mathematically represented as
[tex]f = \frac{v}{\lambda}[/tex]
At second harmonic frequency the length of the string vibrating is equal to the wavelength of the wave generated
[tex]l = \lambda[/tex]
Noe from the question the vibrating string is just half of the length of the main string so
Let assume the length of the main string is [tex]L[/tex]
So [tex]l = \frac{L}{2}[/tex]
The velocity of the vibrating string is mathematically represented as
[tex]v = \sqrt{\frac{T}{\mu} }[/tex]
Where T is the tension on the string which can be mathematically represented as
[tex]T = mg[/tex]
So
[tex]v = \sqrt{\frac{mg}{k} }[/tex]
Then
[tex]f = \frac{v}{\frac{L}{2} }[/tex]
=> [tex]v = \frac{fL }{2}[/tex]
=> [tex]\sqrt{\frac{mg}{k} } = \frac{fL}{2}[/tex]
=> [tex]g = \frac{f^2 L^2 \mu}{4m}[/tex]
substituting values
[tex]g = \frac{(100) * (1.73 *10^{-3} )}{(4 * 1.30)} L^2[/tex]
[tex]g = 3.326 m^{-1} s^{-2} L^2[/tex]
Generally the period of oscillation is mathematically represented as
[tex]T_p = 2 \pi \sqrt{\frac{L}{g} }[/tex]
=> [tex]L = \frac{T^2 g}{4 \pi ^2}[/tex]
The period can be mathematically evaluated as
[tex]T_p = \frac{t}{N}[/tex]
substituting values
[tex]T_p = \frac{315}{200}[/tex]
[tex]T_p = 1.575 \ s[/tex]
Therefore
[tex]L = \frac{1.575^2 * g }{4 \pi ^2}[/tex]
[tex]L = 0.0628 ^2 g[/tex]
so
[tex]g = 3.326 m^{-1} s^{-2} L^2[/tex]
substituting for L
[tex]g = 3.326 ((0.0628) g)^2[/tex]
=> [tex]g = \frac{1}{(3.326)* (0.0628)^2}[/tex]
[tex]g =76.2 m/s^2[/tex]
Which forms of energy are this a blender?
Answer:
Electrical energy.
Explanation:
Electrical energy comes into the blender, which turns into magnetic energy in the electric motor. This magnetism repels permanent magnets. The motor spins as well as the blade. So the magnetism is converted to kinetic energy.
A ball thrown downward by initial speed of 3m/s. It hits the ground after 5 second.
How high is throwing point?
Which type of power plant uses the movement of air in nature to generate
electricity?
A. Radiant
B. Wind
C. Coal
D. Hydroelectric
Answer:
Wind
Explanation:
Three long wires are connected to a meter stick and hang down freely. Wire 1 hangs from the 50-cm mark at the center of the meter stick and carries 1.50 A of current upward. Wire 2 hangs from the 70-cm mark and carries 4.00 A of current downward. Wire 3 is to be attached to the meterstick and to carry a specific current, and we want to attach it at a location that results ineach wire experiencing no net force.
(a) Determine the position of wire 3.
Answer:
Three long wires are connected to a meter stick and hang down freely. Wire 1 hangs from the 50-cm mark at the center of the meter stick and carries 1.50 A of current upward. Wire 2 hangs from the 70-cm mark and carries 4.00 A of current downward. Wire 3 is to be attached to the meterstick and to carry a specific current, and we want to attach it at a location that results ineach wire experiencing no net force.
(a) Determine the position of wire 3.
b) Determine the magnitude and direction of current in wire 3
Explanation:
a) [tex]F_{net} \text {on wire }3=0[/tex]
[tex]\frac{\mu_0 I_1 I_3}{2 \pi x} = \frac{\mu I_2 I_3}{2 \pi (0.2+x)} \\\\\frac{1.5}{x} =\frac{4}{0.2+x} \\\\0.03+1.5x=4x\\\\x=0.012m\\\\=1.2cm[/tex]
position of wire = 50 - 1.2
= 48.8cm
b) [tex]F_{net} \text {on wire }1=0[/tex]
[tex]\frac{\mu _0 I_1 I_3}{2 \pi (1.2)} = \frac{\mu _0 I_1 I_2}{2 \pi (20)} \\\\\frac{I_3}{1.2} =\frac{4}{20} \\\\I_3=0.24A[/tex]
Direction ⇒ downward
What was the revolution in atomic theory produced by the discovery of the electron?
Explanation:
the atom are comprised of particle. this atomic theory also proves that atoms cannot be destroyed nor created and it composed of many particles. it is also said that atoms of the same element can be identical.
Tidal forces are gravitational forces exerted on different parts of a body by a second body. Their effects are particularly visible on the earth's surface in the form of tides. To understand the origin of tidal forces, consider the earth-moon system to consist of two spherical bodies, each with a spherical mass distribution. Let re be the radius of the earth, m be the mass of the moon, and G be the gravitational constant.
(a) Let r denote the distance between the center of the earth and the center of the moon. What is the magnitude of the acceleration ae of the earth due to the gravitational pull of the moon? Express your answer in terms of G, m, and r.
(b) Since the gravitational force between two bodies decreases with distance, the acceleration a_near experienced by a unit mass located at the point on the earth's surface closest to the moon is slightly different from the acceleration a_far experienced by a unit mass located at the point on the earth's surface farthest from the moon. Give a general expression for the quantity a_near - a_far. Express your answer in terms of G, m, r, and re.
Answer:
Explanation:
radius of earth = re
mass of the noon = m
mass of the earth = E
distance between earth and moon = r
acceleration of earth ae
force on earth = GMm / r²
acceleration of the earth
ae = force / mass
= GMm / (r² x M )
= Gm / r²
b ) The point on the earth nearest to moon will be at a distance of r - re
a_near = Gm / ( r - re)²
The point farthest on the earth to moon will be at a distance of r + re
a_ far = Gm / ( r + re )²
1. What is the potential energy of a 5.0-kg
object located 2.0 m above the ground?
A. 2.5 J
C. 98 J
B. 10 J
D. 196 J
Answer:
C. 98 J
Explanation:
The appropriate formula is ...
PE = mgh . . . . . m is mass; below, m is meters
PE = (5 kg)(9.8 m/s^2)(2 m) = 98 kg·m^2/s^2
PE = 98 J
how much external energy is required to bring three identical point charges (20uc) from infinity and place them at the corners of an equilateral triangle with side of 2 meter length:
Answer:
U = 269.4 kJ
Explanation:
The energy required to place the three charges from infinity is given by:
[tex]U=k\frac{q_1q_2}{r_{1,2}}+k\frac{q_1q_3}{r_{1,3}}+k\frac{q_2q_3}{r_{2,3}}[/tex]
In this case, you have that
q1 = q2 = q3 = q = 20uC
r12 = r13 = r23 = r = 2m
k: Coulomb constant = 8.98*10^9 NM^2/C^2
Then, you replace the values of q, r and k in the equation for the energy U:
[tex]U=3k\frac{q^2}{r}\\\\U=3(8.98*10^9Nm^2/C^2)\frac{(20*10^{-6}C)^2}{2m}=269400\ J=269.4\ kJ[/tex]
hence, the required energy is 269.4 kJ
When two forces on the same object or equal and opposite these forces are called
Answer:
These two forces are called action and reaction forces
Explanation:
____is the distance traveled during a specific unit of time
In a certain part of the North America Nebula, the amount of interstellar extinction in the visual wavelength band is 1.1 magnitudes. The thickness of the nebula is estimated to be 20 pc and it is located 700 pc from Earth. Suppose that a B spectral class main-sequence star is observed in the direction of the nebula and that the absolute visual magnitude of the star is known to be M(V) = -1.1 from spectroscopic data. Neglect any other sources of extinction between the observer and the nebula. Show all your work, assumptions, equations, and units.1. Find the apparent visual magnitude of the star if it is lying just in front of the nebula.2. Find the apparent visual magnitude of the star if it is lying just behind the nebula.
Answer:
Explanation:
The apparent magnitude of a star is related to the distance modulus as follows
[tex]m_{\lambda}= M_{\lambda}+5log_{10}d-5+A_{\lambda}[/tex]
[tex]m_{\lambda}= \text {absolute visual magnitude}[/tex]
d = distance in parsec
[tex]A_{\lambda}=\text {interstellar extinction}[/tex]
Substitute
absolute visual magnitude = -1.1
distance =700pc
interstellar extinction = 0
to determine the apparent visual magnitude of the star lying in front of nebula
[tex]m_{\lambda}= M_{\lambda}+5log_{10}d-5+A_{\lambda}[/tex]
[tex]=-1.1+5\log_{10}(700)-5+0\\\\=8.12[/tex]
Thus, the apparent visual magnitude of the star lying in front of nebula is 8.12
b) Substitute
absolute visual magnitude = -1.1
distance =700pc
interstellar extinction = 1.1
to determine the apparent visual magnitude of the star lying behind nebula
[tex]m_{\lambda}= M_{\lambda}+5log_{10}d-5+A_{\lambda}[/tex]
[tex]=-1.1+5\log_{10}(700)-5+1-1\\\\=9.22[/tex]
the apparent visual magnitude of the star lying behind nebula is 9.22
c)
without taking extinction i.e 0, the distance of the star lying just behind nebula is calculated as follows
[tex]m_{\lambda}= M_{\lambda}+5log_{10}d-5[/tex]
[tex]d=10^{(m_\lambda-M_{\lambda_5)/5}[/tex]
[tex]d=10^{(9.22+1.1+5)/5}\\\\=158.79pc[/tex]
Thus, without taking extinction , the distance of the star lying just behind nebula is 158.79pc
Compare the distance of nebula measured from earth with consideration of extinction to the distance of nebula without consideration of extinction
[tex]\frac{d_e}{d} =\frac{700pc}{1158.8pc}[/tex]
= 60.4%
thus, the percentage error in determining the distance if the interstellar extinction neglected is 60.4%
A typical machine tests the tensile strength of a sheet of material cut into a standard size of 5.00 centimeters wide by 10.0 centimeters long. The machine consists of one clamp that holds the entire width (5.00 centimeters) so that it hangs vertically. A second clamp is placed on the lower end of the object, to which a variable downward force is applied. The force is slowly increased until the object ruptures, and the breaking force is recorded.
A strip of aluminum foil with a thickness of 15.0 micrometers and matching the size recommendations of the machine is placed in the machine and tested. The force needed to rupture the foil is found to be 233 newtons. What is the tensile strength of the aluminum foil sample?
Answer:
Explanation:
tensile strength is stress that is needed to break the wire made of the material .
Here force required to break the sheet of material = 233 N
cross sectional area of the foil = breadth x thickness
= 5 x 10⁻² x 15 x 10⁻⁶ m²
= 75 x 10⁻⁸ m²
breaking stress = force / cross sectional area
= 233 / 75 x 10⁻⁸
= 3.1 x 10⁸ Pa .
Tensile strength = 3.1 x 10⁸ Pa .
An electricity-generating station needs to deliver 20 MW of power to a city 1.0 km away. A common voltage for commercial power generators is 22 kV, but a step-up transformer is used to boost the voltage to 230 kV before transmission.
(a) If the resistance of the wires is 2.0 ohms and the electricity costs about 10 cents/kWh, estimate what it costs the utility company to send the power to the city for one day?
Answer:
Explanation:
Power to be transferred = 20 x 10⁶ W .
Voltage at which power is transferred V = 230 x 10³ V .
If current in the carrying wire is I
V x I = Power
230 x 10³ x I = 20 x 10⁶
I = 86.9565 A
Power loss in the transmission line
I² R , I is current and R is resistance
= 86.9565² x 2
= 15122.86 W
= 15.123 KW
In one day power loss
= 15.123 x 24 kWH .
= 363 kWH .
Cost = 363 x 10
= 3630 cent
= 36.30 dollar .
Electromagnetic radiation is emitted when a charged particle moves through a medium faster than the local speed of light. This radiation is known as Cherenkov radiation. Cherenkov radiation is found in many interesting places such as particle detectors and nuclear reactors and can even be seen by astronauts when cosmic rays traverse their eyes. It should be stressed that the particle is never going faster than the speed of light in vacuum (or ccc), just faster than the speed of light in the material (which is always less than ccc). The creation of Cherenkov radiation occurs in much the same way that a sonic boom is created when a plane is moving faster than the speed of sound in the air. The various wavefronts that propagate in the material add coherently to create an effective shock wave. In this problem, you will become familiar with this type of radiation and learn how to use its properties to get information about the particles that created it. Part A What is the threshold velocity vthreshold(water)vthreshold(water)v_threshold (water) (i.e., the minimum velocity) for creating Cherenkov light from a charged particle as it travels through water (which has an index of refraction of n
Answer:
to create the particle the speed must be greater than 2.25 10⁸ m / s
Explanation:
In this exercise we must use the relation of the index of refraction with the speed of light in a vacuum and a material medium
n = c / v
where c is the speed of light in the vacuum, v the speed of light in the material medium and n the ratio of rafraccio
in this case they give us that the medium matter water them that has a refractive index of
n = 1,333
we clear
v = c / n
let's calculate
v = 3 10⁸ / 1,333
v = 2.25 10⁸ m / s
to create the particle the speed must be greater than 2.25 10⁸ m / s
You are assigned the design of a cylindrical, pressurized water tank for a future colony on Mars, where the acceleration due to gravity is 3.71 m/s^2. The pressure at the surface of the water will be 150 kPa , and the depth of the water will be 13.6 m . The pressure of the air outside the tank, which is elevated above the ground, will be 93.0 kPa .
A) Find the net downward force on the tank's flat bottom, of area 2.15 m^2 , exerted by the water and air inside the tank and the air outside the tank.
Answer:
630.93 kN of force.
Explanation:
Pressure inside the tank is 150 kPa
The acceleration due to gravity on Mars g is 3.71 m/s^2.
The depth of water h is 13.6 m.
Pressure due to air outside tank is 93 kPa
The density of water p is 1000 kg/m^3
Pressure of the water on the tank bottom will be equal to pgh
Pressure of water = pgh
= 1000 x 3.71 x 13.6 = 50456 Pa
= 50.456 kPa.
Total pressure at the bottom of the tank will be pressure within tank and pressure due to water and pressure outside tank.
Pt = (150 + 50.456 + 93) = 293.456 kPa
Force at the bottom of the tank will be pressure times area of tank bottom.
F = Pt x A
F = 293.456 x 2.15 m^2 = 630.93 kN
How much time is needed to push a 5,000 N car 50 meters if you are using a machine with a power of 4,500 W?
Answer:
55.56
Explanation:
5000N * 50 = 250000/4500= 55.55555555 or 55.56
If the frequency of a wave is tripled, what happens to the period of the wave?
Answer:
if the frequency of the wave if tripled then period of wave gets tripled
When the wave frequency is tripled, the period of the wave becomes one-third of its original.
Relationship between frequency and period of a waveThe frequency (f) of a wave is inversely proportional to the period (T) of the wave.
Mathematically,
[tex]$f \propto \frac{1}{T}$[/tex]
Thus, when frequency increases, the period decrease and when frequency decreases, the period of the wave increases.
So when frequency of a wave increases by three time, the period of the wave decreases by three times.
Hence, when the frequency is tripled, the period of the wave becomes one-third of its original value.
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A metal ring 5.00 cm in diameter is placed between the north and south poles of large magnets with the plane of its area perpendicular to the magnetic field. These magnets produce an initial uniform field of 1.12 T between them but are gradually pulled apart, causing this field to remain uniform but decrease steadily at 0.220 T/s.
Required:
What is the magnitude of the electric field induced in the ring?
Answer:
Ein: 2.75*10^-3 N/C
Explanation:
The induced electric field can be calculated by using the following path integral:
[tex]\int E_{in} dl=-\frac{\Phi_B}{dt}[/tex]
Where:
dl: diferencial of circumference of the ring
circumference of the ring = 2πr = 2π(5.00/2)=15.70cm = 0.157 m
ФB: magnetic flux = AB (A: area of the loop = πr^2 = 1.96*10^-3 m^2)
The electric field is always parallel to the dl vector. Then you have:
[tex]E_{in}\int dl=E_{in}(2\pi r)=E_{in}(0.157m)[/tex]
Next, you take into account that the area of the ring is constant and that dB/dt = - 0.220T/s. Thus, you obtain:
[tex]E_{in}(0.157m)=-A\frac{dB}{dt}=-(1.96*10^{-3}m^2)(-0.220T/s)=4.31*10^{-4}m^2T/s\\\\E_{in}=\frac{4.31*10^{-4}m^2T/s}{0.157m}=2.75*10^{-3}\frac{N}{C}[/tex]
hence, the induced electric field is 2.75*10^-3 N/C
A lion and a pig participate in a race over a 2.20 km long course. The lion travels at a speed of 18.0 m/s and the pig can do 2.70 m/s. The lion runs for 1.760 km and then stops to tease the slow-moving pig, which eventually passes by. The lion waits for a while after the pig passes and then runs toward the finish line. Both animals cross the finish line at the exact same instant. Assume both animals, when moving, move steadily at their respective speeds.
(a) How far (in m) is the pig from the finish line when the lion resumes the race? (b) For how long in time (in s) was the lion stationary?
The lion covers a distance of [tex]\left(18.0\frac{\rm m}{\rm s}\right)t[/tex] after [tex]t[/tex] seconds, so it reaches the 1.760 km mark at time
[tex]\left(18.0\dfrac{\rm m}{\rm s}\right)t=1760\,\mathrm m\implies t\approx97.8\,\mathrm s[/tex]
The pig travels a distance of [tex]\left(2.70\frac{\rm m}{\rm s}\right)t[/tex], so that it has moved
[tex]\left(2.70\dfrac{\rm m}{\rm s}\right)(97.8\,\mathrm s)=264\,\mathrm m[/tex]
in the time it takes for the lion to move 1.760 km.
(a) The lion has 0.44 km left in the race, which would take it
[tex]\left(18.0\dfrac{\rm m}{\rm s}\right)t=440\,\mathrm m\implies t\approx24.4\,\mathrm s[/tex]
to finish.
In order for the lion and pig to cross the finish line at the same time, the lion needs to resume running once the pig has 24.4 s remaining to the finish line; this happens when it is
[tex]\left(2.70\dfrac{\rm m}{\rm s}\right)(24.4\,\mathrm s)\approx\boxed{65.9\,\mathrm m}[/tex]
away from the end.
(b) The lion is stationary for as long as it takes the pig to cover the distance between [65.9 m away from the finish line] and [264 m from the starting line], or (2.20 km - 65.9 m) - 264 m = 1.87 km, which takes it
[tex]\left(2.70\dfrac{\rm m}{\rm s}\right)t=1870\,\mathrm m\implies t=\boxed{935\,\mathrm s}[/tex]
(a) The distance of the pig from the finish line is 428.3 m
(b) The lion was stationary for 4.32 s
The given parameters;
total distance covered by both animals, d = 2.20 km = 2,200 mspeed of the lion, [tex]v_l[/tex] = 18 m/sspeed of the pig, [tex]v_p[/tex] = 2.7 m/sinitial distance traveled by the lion, = 1.760 km = 1,760 mWhen the pig meets the lion, both have covered a total distance of 1,760 m
The remaining distance to be covered = 2,200 m - 1,760 m = 440 m
Let the time both animals finished the remaining distance = tApply relative velocity concept; as the pig is moving, the lion is closing the gap between them until the finish line.
[tex](V_l - V_p)t = 440\\\\(18- 2.7)t = 440\\15.3 t = 440\\\\t = \frac{440}{15.3} \\\\t = 28.76 \ s[/tex]
If the lion had maintained a constant motion without stopping, the time it would have finished the remaining race is calculated as;
[tex]t_l = \frac{440}{18} = 24.44 \ s[/tex]
(b) This show that the lion was stationary for (28.76 - 24.44 )s = 4.32 s
(a) The distance traveled by the pig during the 4.32 s that the lion was stationary is calculated as;
d = 4.32 x 2.7 = 11.7 m
Thus, the distance of the pig from the finish line is (440 - 11.7)m = 428.3 m.
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if the power developed in an electric circuit is doubled the energy used in one second is
Answer:
Energy is doubled.
Explanation:
Power developed in an electric is the rate of change in time of electric energy travelling throughout circuit. The most common units is the amount of energy used in a second. Therefore, if power is double, the energy used in one second is also doubled.
You must determine the length of a long, thin wire that is suspended from the ceiling in the atrium of a tall building. A 2.00-cm-long piece of the wire is left over from its installation. Using an analytical balance, you determine that the mass of the spare piece is 14.5 μg . You then hang a 0.400-kg mass from the lower end of the long, suspended wire. When a small-amplitude transverse wave pulse is sent up that wire, sensors at both ends measure that it takes the wave pulse 24.7 ms to travel the length of the wire.
Answer:
Explanation:
Let L be the length of the wire.
velocity of pulse wave v = L / 24.7 x 10⁻³ = 40.48 L m /s
mass per unit length of the wire m = 14.5 x 10⁻⁶ x 10⁻³ / 2 x 10⁻² kg / m
m = 7.25 x 10⁻⁷ kg / m
Tension in the wire = Mg , M is mass hanged from lower end.
= .4 x 9.8
= 3.92 N
expression for velocity of wave in the wire
[tex]v = \sqrt{\frac{T}{m} }[/tex] , T is tension in the wire , m is mass per unit length of wire .
40.48 L = [tex]\sqrt{\frac{3.92}{7.25\times10^{-7}} }[/tex]
1638.63 L² = 3.92 / (7.25 x 10⁻⁷)
L² = 3.92 x 10⁷ / (7.25 x 1638.63 )
L² = 3299.64
L = 57.44 m /s
Which factors affect the resistance of a material?
Explanation:
The resistance of a material is given by the formula as follows :
[tex]R=\rho \dfrac{L}{A}[/tex]
Here,
[tex]\rho[/tex] = resistivity of a material
L is the length of the material
A is the area of cross section of the material
So, the factors on which the resistance of a material depends are :
Length and area of cross section
How will you determine the direction
of a torque? Explain.
Which factor indicates the amount of charge on the source charge?
A. the number of field lines on the test charge
B. the number of field lines on the source charge
C. the direction of lines on the source charge
D. the direction of lines on the test charge
Answer:
B. the number of field lines on the source charge
Explanation:
As we know that electric flux is defined as the number of electric field lines passing through a given area.
So here electric flux due to a point charge "q" is given by
so here we know that flux depends on the magnitude of charge and hence we can say that number of filed lines originating from a point charge will depends on the magnitude of the charge.
The factor indicates the amount of charge on the source charge is the number of field lines on the source charge.
What is electric flux?The electric flux is defined as the number of electric field lines passing through a given area.
The electric flux due to a point charge q is given by the number of filed lines through particular closed area.
We know that flux depends on the magnitude of charge and number of field lines starting from a point charge will depends on the magnitude of the charge.
Thus, the correct option is B.
Learn more about electric flux.
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What does the speed of sound depend on ?
Answer:
As with any wave the speed of sound depends on the medium in which it is propagating.
Explanation:
Which shows evidence of active transport?
A scientist places four identical cells into four different
liquids, each with different concentrations of magnesiuni.
Celll
w
Description of Liquid
Slightly more magnesium than the
cell
The least amount of magnesium
O
O
O
O
cells W and Z
cell W only
cell Y only
cells X and Y
Result
Took in
magnesium
Took in
magnesium
Took in
magnesium
Took in
magnesium
Slightly less magnesium than the
cell
The most amount of magnesium
Answer: D
X and Y
Explanation:
X The least amount of magnesium Took in magnesium
Y Slightly less magnesium than the cell Took in magnesium
Because active transport occurs when ions or molecules move from less concentration region to high concentration region through semi membrane with the help of some energy.
Answer:
Cells X and Y
Explanation:
Active transport occurs when a substance moves across a membrane against its concentration gradient.
Cells W and Z were placed in a liquid containing more magnesium than the cells. Magnesium therefore, diffuses passively down it's concentration gradient into the cells.
However, cells X and Y were placed in a solution containing less magnesium than the cells, yet these cells took in magnesium against this concentration gradient. This, shows that active transport had taken place.
Consider two cars, a 700kg Porsche and a 600kg Honda Civic. The Porsche is speeding along at 40 m/s (mph) and the Civic is going half the speed at 20 m/s. If the two cars brake to a stop with the same constant acceleration, lets look at whether the amount of time required to come to a stop or the distance traveled prior to stopping is influenced by their initial velocity.
1. A car traveling 5m/s slams on its brakes, creating an acceleration of -2 m/s^2. How far did the car travel after it applied its brakes?
2. The same car traveling for 10m/s applies the same acceleration of -2 m/s^2. How far did the car travel after it applied its brakes?
Answer:
Explanation:
To find the distance covered by the car after it applied brakes, we use 3rd equation of motion.
2as = Vf² - Vi²
s = (Vf² - Vi²)/2a
1.
We have:
Vi = Initial Velocity = 5 m/s
Vf = Final Velocity = 0 m/s (Since, car finally stops)
a = deceleration = - 2 m/s²
s = distance covered by the car = ?
Therefore,
s = [(0 m/s)² - (5 m/s)²]/2(- 2 m/s²)
s = 6.25 m
2.
We have:
Vi = Initial Velocity = 10 m/s
Vf = Final Velocity = 0 m/s (Since, car finally stops)
a = deceleration = - 2 m/s²
s = distance covered by the car = ?
Therefore,
s = [(0 m/s)² - (10 m/s)²]/2(- 2 m/s²)
s = 25 m
Hence, the distance traveled by the car is affected by the initial speed in accordance with a direct relationship.