B. The hormone erythropoeitin increases the production of red blood cells when oxygen levels are low.
which of the following is true about scientific models?
A. models are used to simplify the study of things
B.computer models are the most reliable kind of model
C. models explain past, present and future information
D.a model is accurate if it does not change over time
Answer:
A- to simplify the study of things
Explanation:
a visual reference
BIO A trap-jaw ant snaps its mandibles shut at very high speed, a good trait for catching small prey. But an ant can also slam its mandibles into the ground; the resulting force can launch the ant into the air for a quick escape. A 12 mg ant hits the ground with an average force of 47 mN for a time of 0.13 ms; these are all typical values. At what speed does it leave the ground
Answer:
Final velocity (v) = 0.509 m/s (Approx)
Explanation:
Ant use impulse power
Given:
Mass of ant = 12 mg = 12 × 10⁻⁶ kg
Average force = 47 mN = 47 × 10⁻³ N
Initial velocity(u) = 0
Time taken = 0.13 ms = 0.13 × 10⁻³ s
Find:
Final velocity (v)
Computation:
Force × Time = change in momentum
(47 × 10⁻³ N)(0.13 × 10⁻³ s) = mv - mu
(47 × 10⁻³ N)(0.13 × 10⁻³ s) = m(v - u)
6.11 × 10⁻⁶ = 12 × 10⁻⁶(v - 0)
6.11 = 12 v
Final velocity (v) = 0.509 m/s (Approx)
Which is true about a concave mirror? Incident rays that are parallel to the central axis are dispersed but will be perceived as originating from a point on the near side of the mirror.
Answer:
'Incident rays that are parallel to the central axis are sent through a point on the near side of the mirror'.
Explanation:
The question is incomplete, find the complete question in the comment section.
Concave mirrors is an example of a curved mirror. The outer surface of a concave mirror is always coated. On the concave mirror, we have what is called the central axis or principal axis which is a line cutting through the center of the mirror. The points located on this axis are the Pole, the principal focus and the centre of curvature. The focus point is close to the curved mirror than the centre of curvature.
During the formation of images, one of the incident rays (rays striking the plane surface) coming from the object and parallel to the principal axis, converges at the focus point after reflection because all incident rays striking the surface are meant to reflect out. All incident light striking the surface all converges at a point on the central axis known as the focus.
Based on the explanation above, it can be concluded that 'Incident rays that are parallel to the central axis are sent through a point on the near side of the mirror'.
A uniform stick 1.5 m long with a total mass of 250 g is pivoted at its center. A 3.3-g bullet is shot through the stick midway between the pivot and one end The bullet approaches at 250 m/s and leaves at 140 m/s
With what angular speed is the stick spinning after the collision?
Answer:
63.44 rad/s
Explanation:
mass of bullet = 3.3 g = 0.0033 kg
initial velocity of bullet [tex]v_{1}[/tex] = 250 m/s
final velocity of bullet [tex]v_{2}[/tex] = 140 m/s
loss of kinetic energy of the bullet = [tex]\frac{1}{2}m(v^{2} _{1} - v^{2} _{2})[/tex]
==> [tex]\frac{1}{2}*0.0033*(250^{2} - 140^{2} )[/tex] = 70.785 J
this energy is given to the stick
The stick has mass = 250 g =0.25 kg
its kinetic energy = 70.785 J
from
KE = [tex]\frac{1}{2} mv^{2}[/tex]
70.785 = [tex]\frac{1}{2}*0.25*v^{2}[/tex]
566.28 = [tex]v^{2}[/tex]
[tex]v= \sqrt{566.28}[/tex] = 23.79 m/s
the stick is 1.5 m long
this energy is impacted midway between the pivot and one end of the stick, which leaves it with a radius of 1.5/4 = 0.375 m
The angular speed will be
Ω = v/r = 23.79/0.375 = 63.44 rad/s
In a double-slit arrangement the slits are separated by a distance equal to 150 times the wavelength of the light passing through the slits. (a) What is the angular separation between the central maximum and an adjacent maximum
Complete Question
In a double-slit arrangement the slits are separated by a distance equal to 150 times the wavelength of the light passing through the slits. (a) What is the angular separation between the central maximum and an adjacent maximum? (b) What is the distance between these maxima on a screen 57.9 cm from the slits?
Answer:
a
[tex]\theta = 0.3819^o[/tex]
b
[tex]y = 0.00386 \ m[/tex]
Explanation:
From the question we are told that
The slit separation is [tex]d = 150 \lambda[/tex]
The distance from the screen is [tex]D = 57.9 \ cm = 0.579 \ m[/tex]
Generally the condition for constructive interference is mathematically represented as
[tex]dsin (\theta ) = n * \lambda[/tex]
=> [tex]\theta = sin ^{-1} [\frac{n * \lambda }{ d } ][/tex]
where n is the order of the maxima and value is 1 because we are considering the central maximum and an adjacent maximum
and [tex]\lambda[/tex] is the wavelength of the light
So
[tex]\theta = sin ^{-1} [\frac{ 1 * \lambda }{ 150 \lambda } ][/tex]
[tex]\theta = 0.3819^o[/tex]
Generally the distance between the maxima is mathematically represented as
[tex]y = D tan (\theta )[/tex]
=> [tex]y = 0.579 tan (0.3819 )[/tex]
=> [tex]y = 0.00386 \ m[/tex]
A flywheel is a mechanical device used to store rotational kinetic energy for later use. Consider a flywheel in the form of a uniform solid cylinder rotating around its axis, with moment of inertia I = 1/2 mr2.
Part (a) If such a flywheel of radius r1 = 1.1 m and mass m1 = 11 kg can spin at a maximum speed of v = 35 m/s at its rim, calculate the maximum amount of energy, in joules, that this flywheel can store?
Part (b) Consider a scenario in which the flywheel described in part (a) (r1 = 1.1 m, mass m1 = 11 kg, v = 35 m/s at the rim) is spinning freely at its maximum speed, when a second flywheel of radius r2 = 2.8 m and mass m2 = 16 kg is coaxially dropped from rest onto it and sticks to it, so that they then rotate together as a single body. Calculate the energy, in joules, that is now stored in the wheel?
Part (c) Return now to the flywheel of part (a), with mass m1, radius r1, and speed v at its rim. Imagine the flywheel delivers one third of its stored kinetic energy to car, initially at rest, leaving it with a speed vcar.
Answer:
a) 6738.27 J
b) 61.908 J
c) [tex]\frac{4492.18}{v_{car} ^{2} }[/tex]
Explanation:
The complete question is
A flywheel is a mechanical device used to store rotational kinetic energy for later use. Consider a flywheel in the form of a uniform solid cylinder rotating around its axis, with moment of inertia I = 1/2 mr2.
Part (a) If such a flywheel of radius r1 = 1.1 m and mass m1 = 11 kg can spin at a maximum speed of v = 35 m/s at its rim, calculate the maximum amount of energy, in joules, that this flywheel can store?
Part (b) Consider a scenario in which the flywheel described in part (a) (r1 = 1.1 m, mass m1 = 11 kg, v = 35 m/s at the rim) is spinning freely at its maximum speed, when a second flywheel of radius r2 = 2.8 m and mass m2 = 16 kg is coaxially dropped from rest onto it and sticks to it, so that they then rotate together as a single body. Calculate the energy, in joules, that is now stored in the wheel?
Part (c) Return now to the flywheel of part (a), with mass m1, radius r1, and speed v at its rim. Imagine the flywheel delivers one third of its stored kinetic energy to car, initially at rest, leaving it with a speed vcar. Enter an expression for the mass of the car, in terms of the quantities defined here.
moment of inertia is given as
[tex]I[/tex] = [tex]\frac{1}{2}[/tex][tex]mr^{2}[/tex]
where m is the mass of the flywheel,
and r is the radius of the flywheel
for the flywheel with radius 1.1 m
and mass 11 kg
moment of inertia will be
[tex]I[/tex] = [tex]\frac{1}{2}[/tex][tex]*11*1.1^{2}[/tex] = 6.655 kg-m^2
The maximum speed of the flywheel = 35 m/s
we know that v = ωr
where v is the linear speed = 35 m/s
ω = angular speed
r = radius
therefore,
ω = v/r = 35/1.1 = 31.82 rad/s
maximum rotational energy of the flywheel will be
E = [tex]Iw^{2}[/tex] = 6.655 x [tex]31.82^{2}[/tex] = 6738.27 J
b) second flywheel has
radius = 2.8 m
mass = 16 kg
moment of inertia is
[tex]I[/tex] = [tex]\frac{1}{2}[/tex][tex]mr^{2}[/tex] = [tex]\frac{1}{2}[/tex][tex]*16*2.8^{2}[/tex] = 62.72 kg-m^2
According to conservation of angular momentum, the total initial angular momentum of the first flywheel, must be equal to the total final angular momentum of the combination two flywheels
for the first flywheel, rotational momentum = [tex]Iw[/tex] = 6.655 x 31.82 = 211.76 kg-m^2-rad/s
for their combination, the rotational momentum is
[tex](I_{1} +I_{2} )w[/tex]
where the subscripts 1 and 2 indicates the values first and second flywheels
[tex](I_{1} +I_{2} )w[/tex] = (6.655 + 62.72)ω
where ω here is their final angular momentum together
==> 69.375ω
Equating the two rotational momenta, we have
211.76 = 69.375ω
ω = 211.76/69.375 = 3.05 rad/s
Therefore, the energy stored in the first flywheel in this situation is
E = [tex]Iw^{2}[/tex] = 6.655 x [tex]3.05^{2}[/tex] = 61.908 J
c) one third of the initial energy of the flywheel is
6738.27/3 = 2246.09 J
For the car, the kinetic energy = [tex]\frac{1}{2}mv_{car} ^{2}[/tex]
where m is the mass of the car
[tex]v_{car}[/tex] is the velocity of the car
Equating the energy
2246.09 = [tex]\frac{1}{2}mv_{car} ^{2}[/tex]
making m the subject of the formula
mass of the car m = [tex]\frac{4492.18}{v_{car} ^{2} }[/tex]
Three identical capacitors are connected in series to a battery. If a total charge of Q flows from the battery, how much charge does each capacitor carry?
Answer:
Each of the capacitor carries the same charge, Q
Explanation:
When capacitors are connected in series, the battery voltage is divided equally across the capacitors. The total voltage across the three identical capacitors is calculated as follows;
[tex]V_T = V_1 + V_2 + V_3[/tex]
We can also calculate this voltage in terms of capacitance and charge;
[tex]V = \frac{Q}{C} \\\\V_T = V_1 + V_2 +V_3 \\\\(given \ total \ charge \ as \ Q, then \ the \ total \ voltage \ V_T \ can \ be \ written \ as)\\\\V_T = \frac{Q}{C_1} + \frac{Q}{C_2} + \frac{Q}{C_3} \\\\V_T = Q(\frac{1}{C_1 } +\frac{1}{C_2} + \frac{1}{C_3 })\\\\[/tex]
Therefore, each of the capacitor carries the same charge, Q
Suppose a 185 kg motorcycle is heading toward a hill at a speed of 29 m/s. The two wheels weigh 12 kg each and are each annular rings with an inner radius of 0.280 m and an outer radius of 0.330 m.
Randomized Variables
m = 185 kg
v = 29 m/s
h = 32 m
A. How high can it coast up the hill. if you neglect friction in m?
B. How much energy is lost to friction if the motorcycle only gains an altitude of 33 m before coming to rest?
Answer:
a) Height reached before coming to rest is 42.86 m
b) Energy lost to friction is 17902.45 J
Explanation:
mass of the motorcycle = 185 kg
speed of the towards the hill = 29 m/s
The wheels weigh 12 kg each
Wheels are annular rings with an inner radius of 0.280 m and outer radius of 0.330 m
a) To go up the hill, the kinetic energy of motion of the motorcycle will be converted to the potential energy it will gain in going up a given height
the kinetic energy of the motorcycle is given as
[tex]KE[/tex] = [tex]\frac{1}{2}mv^{2}[/tex]
where m is the mass of the motorcycle
v is the velocity of the motorcycle
[tex]KE[/tex] = [tex]\frac{1}{2}*185*29^{2}[/tex] = 77792.5 J
This will be converted to potential energy
The potential energy up the hill will be
[tex]PE[/tex] = mgh
where m is the mass
g is acceleration due to gravity 9.81 m/s^2
h is the height reached before coming to rest
[tex]PE[/tex] = 185 x 9.81 x m = 1814.85h
equating the kinetic energy to the potential energy for energy conservation, we'll have
77792.5 = 1814.85h
height reached before coming to rest = 77792.5/1814.85 = 42.86 m
b) if an altitude of 33 m was reached before coming to rest, then the potential energy at this height is
[tex]PE[/tex] = mgh
[tex]PE[/tex] = 185 x 9.81 x 33 = 59890.05 J
The energy lost to friction will be the kinetic energy minus this potential energy.
energy lost = 77792.5 - 59890.05 = 17902.45 J
A) The motorcycle can coast up the hill by ; 42.86m
B) The amount of energy lost to friction : 17902.45 J
A) Determine how high the motorcycle can coast up the hill when friction is neglected
apply the formula for kinetic and potential energies
K.E = 1/2 mv² ---- ( 1 )
P.E = mgH ---- ( 2 )
As the motorcycle goes uphiLl the kinetic energy is converted to potential energy.
∴ K.E = P.E
1/2 * mv² = mgH
∴ H = ( 1/2 * mv² ) / mg ---- ( 3 )
where ; m = 185 kg , v = 29 m/s , g = 9.81
Insert values into equation ( 3 )
H ( height travelled by motorcycle neglecting friction ) = 42.86m
B) Determine how much energy is lost to friction if the motorcycle attains 33m before coming to rest
P.E = mgh = 185 * 9.81 * 33 = 59890.05 J
where : h = 33 m , g = 9.81
K.E = 1/2 * mv² = 77792.5 J ( question A )
∴ Energy lost ( ΔE ) = ( 77792.5 - 59890.05 ) = 17902.45 J
Hence we can conclude that The motorcycle can coast up the hill by ; 42.86m , The amount of energy lost to friction : 17902.45 J.
Learn more : https://brainly.com/question/3586510
You stand near the edge of a swimming pooland observe through the water an object lying on the bottom of thepool. Which of the following statements correctly describes whatyou see?
a. The apparent depth of the object is less than thereal depth.
b. The apparent depth of the object is greater thanthe real depth.
c. There is no difference between the apparent depth and the actual depth of the object.
Answer:
a
Explanation:
The correct answer would be that the apparent depth of the object is less than the real depth.
The refractive property of light as it passes from air to water would make the depth of the pool appear less shallow than the actual depth to an observed. Hence, an object placed at the bottom of the pool will have an apparent depth that is shallower than its actual depth.
Due to the difference in the density of air and that of water, as the ray of light from an observer standing at the edge of a swimming pool travels from air into the water, it becomes refracted by bending away from the original traveling angle.
The same refraction occurs when light rays from an object inside the pool travel from water into the air. Hence, due to the refraction of the ray of light coming from the object at the bottom of the pool, the depth appears shallower than the actual depth.
Correct option: a
"On a movie set, an alien spacecraft is to be lifted to a height of 32.0 m for use in a scene. The 260.0-kg spacecraft is attached by ropes to a massless pulley on a crane, and four members of the film's construction crew lift the prop at constant speed by delivering 135 W of power each. If 18.0% of the mechanical energy delivered to the pulley is lost to friction, what is the time interval required to lift the spacecraft to the specified height?"
Answer:
The time interval required to lift the spacecraft to this specified height is 123.94 seconds
Explanation:
Height through which the spacecraft is to be lifted = 32.0 m
Mass of the spacecraft = 260.0 kg
Four crew member each pull with a power of 135 W
18.0% of the mechanical energy is lost to friction.
work done in this situation is proportional to the mechanical energy used to move the spacecraft up
work done = (weight of spacecraft) x (the height through which it is lifted)
but the weight of spacecraft = mg
where m is the mass,
and g is acceleration due to gravity 9.81 m/s
weight of spacecraft = 260 x 9.81 = 2550.6 N
work done on the space craft = weight x height
==> work = 2550.6 x 32 = 81619.2 J
this is equal to the mechanical energy delivered to the system
18.0% of this mechanical energy delivered to the pulley is lost to friction.
this means that
0.18 x 81619.2 = 14691.456 J is lost to friction.
Total useful mechanical energy = 81619.2 J - 14691.456 J = 66927.74 J
Total power delivered by the crew to do this work = 135 x 4 = 540 W
But we know tat power is the rate at which work is done i.e
[tex]P = \frac{w}{t}[/tex]
where p is the power
where w is the useful work done
t is the time taken to do this work
imputing values, we'll have
540 = 66927.74/t
t = 66927.74/540
time taken t = 123.94 seconds
How many electrons circulate each second through the cross section of a conductor, which has a current intensity of 4A.
Answer:
2.5×10¹⁹
Explanation:
4 C/s × (1 electron / 1.60×10⁻¹⁹ C) = 2.5×10¹⁹ electrons/second
A man has vocal cords of length 22 mm, with a mass per length of 0.0042 kg/m. What tension is required in the vocal cords in order to produce a tone of middle C (261.62 Hz)?
Answer:
Tension, T = 0.556 N
Explanation:
It is given that,
Length of vocal cords, l = 22 mm = 0.022 m
Mass per unit length, [tex]\mu=0.0042\ kg/m[/tex]
We need to find the tension is required in the vocal cords in order to produce a tone of middle C of frequency 261.62 Hz. The frequency in terms if tension is given by :
[tex]f=\dfrac{1}{2l}\sqrt{\dfrac{T}{\mu}}[/tex]
T = tension in the vocal cords
[tex]f^2=\dfrac{1}{4l^2}\times \dfrac{T}{\mu}\\\\T=4l^2\mu f^2\\\\T=4\times (0.022)^2\times 0.0042 \times (261.62 )^2\\\\T=0.556\ N[/tex]
So, the tension in the vocal cords is 0.556 N.
(a) Find the speed of waves on a violin string of mass 717 mg and length 24.3 cm if the fundamental frequency is 980 Hz. (b) What is the tension in the string? For the fundamental, what is the wavelength of (c) the waves on the string and (d) the sound waves emitted by the string? (Take the speed of sound in air to be 343 m/s.)
Answer:
a)v = 476.28 m / s , b) T = 6.69 10⁵ N , c) λ = 0.486 m , d) λ = 0.35 m
Explanation:
a) The speed of a wave on a string is
v = √T /μ
also all the waves fulfill the relationship
v = λ f
they indicate that the fundamental frequency is f = 980 Hz.
The wavelength that is fixed at its ends and has a maximum in the center
L = λ / 2
λ = 2L
we substitute
v = 2 L f
let's calculate
v = 2 0.243 980
v = 476.28 m / s
b) The tension of the rope
T = v² μ
the density of the string is
μ = m / L
T = v² m / L
T = 476.28² 0.717 / 0.243
T = 6.69 10⁵ N
c) λ = 2L
λ = 2 0.243
λ = 0.486 m
d) The violin has a resonance process with the air therefore the frequency of the wave in the air is the same as the wave in the string. Let's find the wavelength in the air
v = λ f
λ= v / f
λ = 343/980
λ = 0.35 m
How much heat is needed to melt 2.5 KG of water at its melting point? Use Q= mass x latent heat of fusion.
Answer:
Q = 832 kJ
Explanation:
It is given that,
Mass of the water, m = 2.5 kg
The latent heat of fusion, L = 333 kJ/kg
We need to find the heat needed to melt water at its melting point. The formula of heat needed to melt is given by :
Q = mL
[tex]Q=2.5\ kg\times 333\ kJ/kg\\\\Q=832.5\ kJ[/tex]
or
Q = 832 kJ
So, the heat needed to melt the water is 832 kJ.
Suppose that a 117.5 kg football player running at 6.5 m/s catches a 0.43 kg ball moving at a speed of 26.5 m/s with his feet off the ground, while both of them are moving horizontally.
(a) Calculate the final speed of the player, in meters per second, if the ball and player are initially moving in the same direction.
(b) Calculate the change in kinetic energy of the system, in joules, after the player catches the ball.
(c) Calculate the final speed of the player, in meters per second, if the ball and player are initially moving in opposite directions.
(d) Calculate the change in kinetic energy of the system, in joules, in this case.
Answer:
a) 6.57 m/s
b) 53.75 J
c) 6.37 m/s
d) -98.297 J
Explanation:
mass of player = [tex]m_{p}[/tex] = 117.5 kg
speed of player = [tex]v_{p}[/tex] = 6.5 m/s
mass of ball = [tex]m_{b}[/tex] = 0.43 kg
velocity of ball = [tex]v_{b}[/tex] = 26.5 m/s
Recall that momentum of a body = mass x velocity = mv
initial momentum of the player = mv = 117.5 x 6.5 = 763.75 kg-m/s
initial momentum of the ball = mv = 0.43 x 26.5 = 11.395 kg-m/s
initial kinetic energy of the player = [tex]\frac{1}{2} mv^{2}[/tex] = [tex]\frac{1}{2}[/tex] x 117.5 x [tex]6.5^{2}[/tex] = 2482.187 J
a) according to conservation of momentum, the initial momentum of the system before collision must equate the final momentum of the system.
for this first case that they travel in the same direction, their momenta carry the same sign
[tex]m_{p}[/tex][tex]v_{p}[/tex] + [tex]m_{b}[/tex][tex]v_{b}[/tex] = ([tex]m_{p}[/tex] +[tex]m_{b}[/tex])v
where v is the final velocity of the player.
inserting calculated momenta of ball and player from above, we have
763.75 + 11.395 = (117.5 + 0.43)v
775.145 = 117.93v
v = 775.145/117.93 = 6.57 m/s
b) the player's new kinetic energy = [tex]\frac{1}{2} mv^{2}[/tex] = [tex]\frac{1}{2}[/tex] x 117.5 x [tex]6.57^{2}[/tex] = 2535.94 J
change in kinetic energy = 2535.94 - 2482.187 = 53.75 J gained
c) if they travel in opposite direction, equation becomes
[tex]m_{p}[/tex][tex]v_{p}[/tex] - [tex]m_{b}[/tex][tex]v_{b}[/tex] = ([tex]m_{p}[/tex] +[tex]m_{b}[/tex])v
763.75 - 11.395 = (117.5 + 0.43)v
752.355 = 117.93v
v = 752.355/117.93 = 6.37 m/s
d) the player's new kinetic energy = [tex]\frac{1}{2} mv^{2}[/tex] = [tex]\frac{1}{2}[/tex] x 117.5 x [tex]6.37^{2}[/tex] = 2383.89 J
change in kinetic energy = 2383.89 - 2482.187 = -98.297 J
that is 98.297 J lost
Sally who weighs 450 N, stands on a skate board while roger pushes it forward 13.0 m at constant velocity on a level straight street. He applies a constant 100 N force.
Work done on the skateboard
a. Rodger Work= 0J
b. Rodger work= 1300J
c. sally work= 1300J
d. sally work= 5850J
e. rodger work= 5850J
Answer:
b. Rodger work = 1300 J
Explanation:
Work done: This can be defined as the product of force and distance along the direction of the force.
From the question,
Work is done by Rodger using a force of 100 N in pushing the skateboard through a distance of 13.0 m.
W = F×d............. Equation 1
Where W = work done, F = force, d = distance.
Given: F = 100 N, d = 13 m
Substitute these values into equation 1
W = 100(13)
W = 1300 J.
Hence the right option is b. Rodger work = 1300 J
EXAMPLE 5 Find the radius of gyration about the x-axis of a homogeneous disk D with density rho(x, y) = rho, center the origin, and radius a. SOLUTION The mass of the disk is m = rhoπa2, so from these equations we have 2 = Ix m = 1 4πrhoa4 rhoπa2 = a2 4 .
Answer:
Radius of gyration = a/2.
Explanation:
So, from the question above I can see that the you are already answering the question and you are stuck up or maybe that's how the problem is set from the start. Do not worry, you are covered in any of the ways. So, from the question we have that;
"The mass of the disk is m = ρπa^2, so from these equations we have y^2 = Ix/m."
(NB: I changed the "rho" word to its symbol).
Thus, the radius of gyration with respect to x-axis = (1/4 πρa^4)/ πρa^2 = a^2/4.
Therefore, the Radius of gyration = a/2.
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.
Required:
a. 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?
b. If a farsighted person has a near point that is 0.600 m from the eye, what is the focal length f2 of the contact lenses that the person would need to be able to read a book held at 0.350 m from the person's eyes?
Answer:
a) f₁ = 3.50 m , b) f₂ = 0.84 m
Explanation:
For this exercise we must use the constructor equation
1 / f = 1 / p + 1 / q
where f is the focal length, p is the distance to the object and q is the distance to the image
a) the distance where the image should be placed is q = 3.50 m and the object is located at infinity p = ∞
1 / f₁ = 1 /∞ + 1 / 3.50
f₁ = 3.50 m
b) in this case the image is at q = -0.600 m and the object p = 0.350 m
1 / f₂ = 1 / 0.350 -1 / 0.600
the negative sign, is because the image is in front of the object
1 / f₂ = 1,1905
f₂ = 1 / 1,1905
f₂ = 0.84 m
Uses of pressure and the uses of density
Answer:
Pressure is a scalar quantity defined as per unit area.
Density is the objects ,times its the acceleration due to gravity.
Explanation:
Pressure is the alternative object increases the area of contact decrease .
Pressure is the force component to the surface used to calculate pressure.
pressure is that collisions of the gas to container as the per unit time .
pressure is an physical important quantity to play the solid and fluid .
Pressure is the expressed in a number of units depend the context use, pressure exerted by the liquid alone.
Density is the objects, times, volume of the object that times acceleration objects.
Density is the used to the system complex objects and materials.
Density force is the weight of a region or objects static fluid.
A 3 kg rock is swung in a circular path and in a vertical plane on a 0.25 m length string. At the top of the path, the angular velocity is 11 rad/s. What is the tension in the string at that point
Answer:
The tension in the string at that point is 90.75 N
Explanation:
Given;
mass of the object, m = 3 kg
length of string, r = 0.25 m
the angular velocity, ω = 11 rad/s
The tension on string can be equated to the centrifugal force on the object;
T = mω²r
Where;
T is the tension in the string
m is mass of the object
ω is the angular velocity
r is the radius of the circular path
T = 3 x (11)² x 0.25
T = 90.75 N
Therefore, the tension in the string at that point is 90.75 N
Solve 3* +5-220t = 0
Answer:
t = 27.5
Explanation:
[tex]3 + 5 -220t = 0[/tex]
Well to solve for t we need to combine like terms and seperate t.
So 3+5= 8
8 - 220t = 0
We do +220 to both sides
8 = 220t
And now we divide 220 by 8 which is 27.5
Hence, t = 27.5
Two long, parallel, current-carrying wires lie in an xy-plane. The first wire lies on the line y = 0.300 m and carries a current of 26.0 A in the +x direction. The second wire lies along the x-axis. The wires exert attractive forces on each other, and the force per unit length on each wire is 295 µN/m. What is the y-value (in m) of the line in the xy-plane where the total magnetic field is zero?
Answer:
The y-value is z = 0.759 m
Explanation:
From the question we are told that
The position of the first y-axis is [tex]y_1 = 0.300 \ m[/tex]
The current on the first wire is [tex]I_ 1 = 26.0 \ A[/tex]
The force per unit length on each wire is [tex]\frac{F}{l} = 295 \mu N/m = 295 * 10^{-6} \ N/m[/tex]
Generally the force per unit length on first wire is mathematically represented as
[tex]\frac{F}{l} = \frac{\mu_o * I_1 * I_2 }{2*\pi* y_1}[/tex]
Where [tex]\mu _o[/tex] is the permeability of free space with value [tex]\mu_o = 4\pi * 10^{-7} N/A^2[/tex]
substituting values
[tex]295 *10^{-6} = \frac{ 4\pi * 10^{-7} * 26.0 * I_2 }{2 *3.142* 0.300}[/tex]
[tex]I_2 = \frac{295 *10^{-6 } * 0.300 * 2* 3.142 }{ 4\pi * 10^{-7} * 26 }[/tex]
[tex]I_2 = 17.0 \ A[/tex]
Now the at the point where the magnetic field is zero the magnetic field of each wire are equal , let that point by z meters from the second wire on the y-axis so
[tex]\frac{\mu_o I_2}{2 * \pi * y_1} = \frac{\mu_o I_1}{2 * \pi * (y_1-z)}[/tex]
[tex]I_2 (y_1 - z) = I_1 * y_1[/tex]
substituting values
[tex]17.0 ( 0.300 - z) = 26 * 0.300[/tex]
z = 0.759 m
A toboggan is sliding down an icy slope. As it goes down, _________ does work on the toboggan and ends up converting __________ energy to _________ energy.
Answer:
As it goes down, weight does work on the toboggan and it ends up converting gravitational potential energy to kinetic energy.
1. weight
2. gravitational potential energy to kinetic energy.
Explanation:
As it goes down, weight does work on the toboggan and it ends up converting gravitational potential energy to kinetic energy.
work done by toboggan = weight × distance
W = mg and the distance is down the icy slope
By using law of conservation of energy, energy can neither be created nor destroyed, but can be conserve from one form to another in a closed system.
Toboggan converts gravitational potential energy (mgh) to kinetic energy(¹/₂mv²)
A person is nearsighted with a far point of 75.0 cm. a. What focal length contact lens is needed to give him normal vision
Complete Question
The complete question is shown on the first uploaded image
Answer:
a
[tex]f= -75 \ cm = - 0.75 \ m[/tex]
b
[tex]P = -1.33 \ diopters[/tex]
Explanation:
From the question we are told that
The image distance is [tex]d_i = -75 cm[/tex]
The value of the image is negative because it is on the same side with the corrective glasses
The object distance is [tex]d_o = \infty[/tex]
The reason object distance is because the object father than it being picture by the eye
General focal length is mathematically represented as
[tex]\frac{1}{f} = \frac{1}{d_i} - \frac{1}{d_o}[/tex]
substituting values
[tex]\frac{1}{f} = \frac{1}{-75} - \frac{1}{\infty}[/tex]
=> [tex]f= -75 \ cm = - 0.75 \ m[/tex]
Generally the power of the corrective lens is mathematically represented as
[tex]P = \frac{1}{f}[/tex]
substituting values
[tex]P = \frac{1}{-0.75}[/tex]
[tex]P = -1.33 \ diopters[/tex]
Someone help find centripetal acceleration plus centripetal force!
Answer:Centripetal force that acts an object keep it along a moving circular path.
Explanation:Centripetal force along a path circular of radius(r) with velocity(V) acceleration the center of the path.
a=v/r
object will along moving continue a straight path unless by the external force.External force is the centripetal force.
Centripetal force is to moving in horizontal circle,Centripetal force is not a fundamental force.Gravitational force satellite and orbit of centripetal force.
Centripetal acceleration and centripetal force are used to calculate the motion of objects in circular motion. The main answer to the question is given below:The centripetal force is given by:F = mv²/rwhere m is the mass of the object, v is the speed of the object and r is the radius of the circle. The unit of centripetal force is Newtons (N).The centripetal acceleration is given by:a = v²/rThe unit of centripetal acceleration is meters per second squared
(m/s²).Explanation:When an object moves in a circular motion, there is a force that acts upon it. This force is called the centripetal force. This force always points towards the center of the circle. It is responsible for keeping the object moving in a circular motion.The centripetal force is related to the centripetal acceleration.
The centripetal acceleration is the acceleration of an object moving in a circle. It is always directed towards the center of the circle.The magnitude of the centripetal force is given by:F = mv²/rwhere F is the force, m is the mass of the object, v is the speed of the object and r is the radius of the circle.The magnitude of the centripetal acceleration is given by:a = v²/rwhere a is the acceleration, v is the speed of the object and r is the radius of the circle.
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A particle with a charge of 4.0 μC has a mass of 5.0 × 10 -3 kg. What electric field directed upward will exactly balance the weight of the particle?
Answer:
E = 12.25 x 10³ N/C = 12.25 KN/C
Explanation:
In order to balance the weight of the object the electrostatic force due to the electric field must be equal to the weight of the body or charge. Therefore,
Electrostatic Force = Weight
E q = mg
where,
E = Electric Field = ?
m = Mass of the Charge = 5 x 10⁻³ kg
g = acceleration due to gravity = 9.8 m/s²
q = magnitude of charge = 4 μC = 4 x 10⁻⁶ C
Therefore,
E(4 x 10⁻⁶ C) = (5 x 10⁻³ kg)(9.8 m/s²)
E = 0.049 N/4 x 10⁻⁶ C
E = 12.25 x 10³ N/C = 12.25 KN/C
Huygens claimed that near the surface of the Earth the velocity downwards of an object released from rest, vy, was directly proportional to the square root of the distance it had fallen, . This is true if c is equal to
Answer:
the expression is False
Explanation:
From the kinematics equations we can find the speed of a body in a clean fall
v = v₀ - g t
v² = V₀² - 2 g y
If the body starts from rest, the initial speed is zero (vo = 0)
v= √ (2g y)
In the first equation it gives us the relationship between speed and time.
With the second equation we can find the speed in which the distance works, this is the expression, see that speed is promotional at the height of a delicate body.
Therefore the expression is False
3. According to Hund's rule, what's the expected magnetic behavior of vanadium (V)?
O A. Ferromagnetic
O B. Non-magnetic
C. Diamagnetic
O D. Paramagnetic
Answer:
Diamagnetic
Explanation:
Hunds rule states that electrons occupy each orbital singly first before pairing takes place in degenerate orbitals. This implies that the most stable arrangement of electrons in an orbital is one in which there is the greatest number of parallel spins(unpaired electrons).
For vanadium V ion, there are 18 electrons which will be arranged as follows;
1s2 2s2 2p6 3s2 3p6.
All the electrons present are spin paired hence the ion is expected to be diamagnetic.
Answer:
its paramagnetic
Explanation:
i took this quiz
A horizontal uniform meter stick is supported at the 50.0 cm mark. It has a mass of 0.52 kg, hanging from it at the 20.0 cm mark and a mass of 0.31 kg mass hanging from the 60.0 cm mark. Determine the position on the meter stick, at which one would hang a third mass of 0.61 kg, to keep the meter stick in balance. Group of answer choices
Answer: 70.5 cm
Explanation:
The position on the meter stick, at which one would hang a third mass of 0.61 kg, to keep the meter stick in balance will be at the side of 0.31kg.
You will use the moment techniques.
That is,
Sum of the clockwise moment = sum of anticlockwise moments
Please find the attached file for the remaining explanation and solution.
Before you start taking measurements though, we’ll first make sure you understand the underlying concepts involved. By what method is each of the spheres charged?
Answer:
If they are metallic spheres they are connected to earth and a charged body approaches
non- metallic (insulating) spheres in this case are charged by rubbing
Explanation:
For fillers, there are two fundamental methods, depending on the type of material.
If they are metallic spheres, they are connected to earth and a charged body approaches, this induces a charge of opposite sign and of equal magnitude, then it removes the contact to earth and the sphere is charged.
If the non- metallic (insulating) spheres in this case are charged by rubbing with some material or touching with another charged material, in this case the sphere takes half the charge and when separated each sphere has half the charge and with equal sign.