Answer:
a) a = 5.03x10¹³ m/s²
b) [tex]V_{f} = 4.4 \cdot 10^{5} m/s [/tex]
Explanation:
a) The acceleration of the positron can be found as follows:
[tex] F = q*E [/tex] (1)
Also,
[tex] F = ma [/tex] (2)
By entering equation (1) into (2), we have:
[tex] a = \frac{F}{m} = \frac{qE}{m} [/tex]
Where:
F: is the electric force
m: is the particle's mass = 9.1x10⁻³¹ kg
q: is the charge of the positron = 1.6x10⁻¹⁹ C
E: is the electric field = 286 N/C
[tex] a = \frac{qE}{m} = \frac{1.6 \cdot 10^{-19} C*286 N/C}{9.1 \cdot 10^{-31} kg} = 5.03 \cdot 10^{13} m/s^{2} [/tex]
b) The positron's speed can be calculated using the following equation:
[tex] V_{f} = V_{0} + at [/tex]
Where:
[tex]V_{f}[/tex]: is the final speed =?
[tex]V_{0}[/tex]: is the initial speed =0
t: is the time = 8.70x10⁻⁹ s
[tex] V_{f} = V_{0} + at = 0 + 5.03 \cdot 10^{13} m/s^{2}*8.70 \cdot 10^{-9} s = 4.4 \cdot 10^{5} m/s [/tex]
I hope it helps you!
An experiment invilves three charges objects: A, B, and C. Object A repels object B and attracts onject C. object C ir repelled by ebonite charged with fur. What is the charge on the object?
Answer:
A and B is positive charge
C_negative
Explanation:
because when an ebonite is rubbed with fur produce negative charge due to law of electrostatic like charge repel and unlike attract
A cylindrical shell of radius 7.00 cm and length 2.44 m has its charge uniformly distributed on its curved surface. The magnitude of the electric field at a point 21.9 cm radially outward from its axis (measured from the midpoint of the shell) is 36.0 kN/C.(a) Use approximate relationships to find thenet charge on the shell.
(b) Use approximate relationships to find theelectric field at a point 4.00 cm from theaxis, measured radiallyoutward from the midpoint of the shell.
Answer:
(b) Use approximate relationships to find theelectric field at a point 4.00 cm from the axis, measured radiallyoutward from the midpoint of the shell.
Your lab instructor has asked you to measure a spring constant using a dynamic method—letting it oscillate—rather than a static method of stretching it. You and your lab partner suspend the spring from a hook, hang different masses, m, on the lower end, and start them oscillating. One of you uses a meter stick to measure the amplitude, A, and the other uses a stopwatch to time 10 oscillations, t. Your data are as follows:Mass, m(g) Amplitude, A(cm) Time, T(s) 100 6.5 7.8150 5.5 9.8200 6.0 10.9250 3.5 12.4Use the best-fit line of an appropriate graph to determine the spring constant.
Answer:
k = 6,547 N / m
Explanation:
This laboratory experiment is a simple harmonic motion experiment, where the angular velocity of the oscillation is
w = √ (k / m)
angular velocity and rel period are related
w = 2π / T
substitution
T = 2π √(m / K)
in Experimental measurements give us the following data
m (g) A (cm) t (s) T (s)
100 6.5 7.8 0.78
150 5.5 9.8 0.98
200 6.0 10.9 1.09
250 3.5 12.4 1.24
we look for the period that is the time it takes to give a series of oscillations, the results are in the last column
T = t / 10
To find the spring constant we linearize the equation
T² = (4π²/K) m
therefore we see that if we make a graph of T² against the mass, we obtain a line, whose slope is
m ’= 4π² / k
where m’ is the slope
k = 4π² / m'
the equation of the line of the attached graph is
T² = 0.00603 m + 0.0183
therefore the slope
m ’= 0.00603 s²/g
we calculate
k = 4 π² / 0.00603
k = 6547 g / s²
we reduce the mass to the SI system
k = 6547 g / s² (1kg / 1000 g)
k = 6,547 kg / s² =
k = 6,547 N / m
let's reduce the uniqueness
[N / m] = [(kg m / s²) m] = [kg / s²]
The spring-mass system forms a linear graph between the time period and mass. And the value of spring-constant from the given data is 6.46 N/m.
Given data:
Mass suspended by spring is, [tex]m=100 \;\rm g =0.1 \;\rm kg[/tex].
Number of oscillations is, [tex]n =10\;\rm oscillations[/tex].
Time period of oscillation is, [tex]T=7.8 \;\rm s[/tex].
The expression for the angular frequency of spring-mass system is,
[tex]\omega =\drac \sqrt{\dfrac{k}{m} }[/tex] ......................................................(1)
Here, k is the spring constant.
Angular frequency is also expressed as,
[tex]\omega = 2 \pi f[/tex] .........................................................(2)
here, f is the linear frequency of spring-mass system.
And linear frequency is,
[tex]f=\dfrac{n}{T}\\f=\dfrac{10}{7.81}\\f=1.28 \;\rm cycles/sec[/tex]
Then substitute equation (2) in equation (1) as,
[tex]2 \pi f=\drac \sqrt{\dfrac{k}{m} }\\2 \pi \times 1.28=\drac \sqrt{\dfrac{k}{0.1} }\\(2 \pi \times 1.28)^{2}= \dfrac{k}{0.1}\\k = 6.46 \;\rm N/m[/tex]
Thus, the value of spring constant is 6.46 N/m. And the suitable graph for the spring-mass system is given below.
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Use Hooke's Law to determine the work done by the variable force in the spring problem. Nine joules of work is required to stretch a spring 0.5 meter from its natural length. Find the work required to stretch the spring an additional 0.40 meter.
Answer:
29.16 J
Explanation:
From Hook's law,
W = 1/2(ke²)..................... Equation 1
Where W = work done, k = Spring constant, e = extension.
Given: W = 9 J, e = 0.5 m.
Substitute into equation 1
9 = 1/2(k×0.5²)
Solve for k
k = 18/0.5²
k = 72 N/m.
The work done required to stretch the spring by additional 0.4 m is
W = 1/2(72)(0.4+0.5)²
W = 36(0.9²)
W = 29.16 J.
Suppose a stone is thrown vertically upward from the edge of a cliff on Mars (where the acceleration due to gravity is only about 12 ft/s2 with an initial velocity of 64 ft/s from a height of 192 ft above the ground. The height s of the stone above the ground after t seconds is given by
s=−6t2+64t+192
a. Determine the velocity v of the stone after t seconds. b. When does the stone reach its highest point? c. What is the height of the stone at the highest point? d. When does the stone strike the ground? e. With what velocity does the stone strike the ground?
Answer:
a) v = -12t + 64
b) t = 5.33s
c) s = 362.66ft
d) t = 13.10s
e) v = 93.2ft/s
Explanation:
You have the following equation for the height of a stone thrown in Mars:
[tex]s(t)=-6t^2+64t+192[/tex] (1)
a) The velocity of the stone after t seconds is obtained with the derivative of s in time:
[tex]v=\frac{ds}{st}=-12t+64[/tex] (2)
The equation for the speed of the stone is v = -12t + 64
b) The highest point is obtained when the speed of the stone is zero. Then, from the equation (2) equal to zero, you can obtain the time when the stone is at its maximum height:
[tex]-12t+64=0\\\\t=5.33s[/tex]
The time in which the stone is at the maximum height is 5.33s
c) For this time the stone is at the maximum height. Then, you replace t in the equation (1):
[tex]s(1)=-6(5.33)^2+64(5.33)+192=362.66ft[/tex]
the maximum height is 362.66 ft
d) To find the time when the stone arrive to the ground you equal the equation (1) to zero and you solve for t:
[tex]0=-6t^2+64t+192[/tex]
you use the quadratic formula:
[tex]t_{1,2}=\frac{-64\pm\sqrt{64^2-4(-6)(192)}}{2(-6)}\\\\t_{1,2}=\frac{-64\pm 93.29}{-12}\\\\t_1=13.10s\\\\t_2=-2.44s[/tex]
You use the result with positive values because is the onlyone with physical meaning.
The time for the stone hits the ground is 13.10 s
e) You replace 13.10s in the equation (2) to obtain the velocity of the stone when it strike the ground:
[tex]v=-12t+64=-12(13.10)+64=-93.2\frac{ft}{s}[/tex]
The minus sign is because the stone's direction is downward.
The speed of the stone just when it strikes the ground is 93.2ft/s
Three sheets of plastic have unknown indices of refraction. Sheet 1 is placed on top of sheet 2, and a laser beam is directed onto the sheets from above so that it strikes the interface at an angle of 26.50 with the normal. The refracted beam in sheet 2 makes an angle of 31.70 with the normal. The experiment is repeated with sheet 3 on top of sheet 2, and with the same angle of incidence, the refracted beam makes an angle of 36.70 with the normal. If the experiment is repeated again with sheet 1 on top of sheet 3, determine the expected angle of refraction in sheet 3? Assume the same angle of incidence.
Answer:
The angle of refraction of sheet 3 when sheet 1 is on top of it is [tex]\theta_{r_s } = 23.1 ^o[/tex]
Explanation:
From the question we are told that
The angle of incidence is [tex]\theta _i = 26.50 ^o[/tex]
The angle of refraction angle for sheet 1 is [tex]\theta _{r_1}} = 31.70 ^o[/tex]
The angle of refraction for sheet 3 is [tex]\theta _{r_3}} = 36.70 ^o[/tex]
According to Snell's law
[tex]\frac{n_2}{n_1} = \frac{sin (\theta_1)}{sin (\theta_{r_1})}[/tex]
Where [tex]n_1 \ and \ n_2[/tex] are refractive index of sheet 1 and sheet 2
=> [tex]n_2 = n_1 \frac{sin(\theta_i)}{sin (\theta _{r_1})}[/tex]
Also when sheet 3 in on top of sheet 2
[tex]\frac{n_2}{n_3} = \frac{sin \theta_i}{sin \theta_{r_3}}[/tex]
substituting for [tex]n_2[/tex]
[tex]n_1 \frac{sin(\theta_i)}{sin (\theta _{r_1})} = n_3 \frac{sin \theta_i}{sin \theta_{r_3}}[/tex]
[tex]n_1 \frac{sin(\theta_i)}{sin (\theta _{r_1})} = n_3 \frac{sin \theta_i}{sin \theta_{r_3}}[/tex]
=> [tex]n_3 = n_1 * \frac{sin(\theta_{r_3})}{sin(\theta_{r_1})}[/tex]
when sheet 1 in on top of sheet 3
[tex]\frac{n_3}{n_1} = \frac{sin(\theta_i)}{\theta_{r_s}}[/tex]
where [tex]r_s[/tex] is the angle of refraction when sheet 1 is on top of sheet 3
substituting for [tex]n_3[/tex]
[tex]\frac{ n_1 * \frac{sin(\theta_{r_3})}{sin(\theta_{r_1})}}{n_1} = \frac{sin(\theta_i)}{\theta_{r_s}}[/tex]
=> [tex]sin (\theta _{r_s}) = n_1 * sin (\theta_i) * \frac{sin (\theta_{r_1})}{ n_1 * sin(\theta_{r_3})}[/tex]
substituting values
[tex]sin (\theta _{r_s}) = n_1 * sin (26.50) * \frac{sin (31.70)}{ n_1 * sin(36.70)}[/tex]
=> [tex]\theta_{r_s } = sin^{-1} (0.3923)[/tex]
=> [tex]\theta_{r_s } = 23.1 ^o[/tex]
Shrinking Loop. A circular loop of flexible iron wire has an initial circumference of 165 cmcm , but its circumference is decreasing at a constant rate of 14.0 cm/scm/s due to a tangential pull on the wire. The loop is in a constant uniform magnetic field of magnitude 0.800 TT , which is oriented perpendicular to the plane of the loop. Assume that you are facing the loop and that the magnetic field points into the loop.
(a) Find the emf induced in the loop at the instant when 9.0 s have passed.
(b) Find the direction of the induced current in the loop as viewed looking along the direction of the magnetic field.
Answer:
(a) emf = 1.18 mV
(b) counter-clockwise sense
Explanation:
(a) The induced emf is given by the following formula:
[tex]emf=-\frac{d\Phi_B}{dt}[/tex] (1)
where:
ФB: magnetic flux = AB = (area of the loop)*(magnitude of the magnetic field)
A = πr^2
B = 0.800 T
You replace the expression for the magnetic flux in the equation (1):
[tex]emf=-B\frac{\Delta A}{\Delta t}=-B\frac{A_2-A_1}{t_2-t_1}[/tex]
A1: initial area
A2: final area
t2-t1: time interval = 9.0s
Then you have to calculate the change in the area of the loop, by using the information about the circumference of the loop. First you calculate the radius of the loop for a circumference of 165 cm = 1.65m
[tex]s=1.65m=2\pi r\\\\r=\frac{1.65m}{2\pi}=0.262m[/tex]
You calculate the initial area A1:
[tex]A_1=\pi (0.262m)^2=0.215m^2[/tex]
After 9.0 second the circumference will be:
[tex]s'=1.65m-0.14\frac{m}{s}(9.0s)=0.39m[/tex]
the new radius and the final area is:
[tex]r=\frac{0.39m}{2\pi}=0.062m[/tex]
[tex]A_2=\pi(0.062m)^2=0.012m^2[/tex]
Finally, you replace in the equation (1):
[tex]emf=-(0.800T)\frac{0.012m^2-0.215m^2}{9.0s}=1.8*10^{-3}V=1.8mV[/tex]
The induced emf in the circular loop is 1.18mV
(b) The induced emf generates an electric current, which produces a magnetic field that is opposite to the direction of the constant magnetic field of 0.800T. Due to this magnetic field point into the loop. The current has to have a direction in a counter-clockwise sense.
g Tether ball is a game children play in which a ball hangs from a rope attached to the top of a tall pole. The children hit the ball, causing it to swing around the pole. What is the total initial acceleration of a tether ball on a 2.0 m rope whose angular velocity changes from 13 rad/s to 7.0 rad/s in 15 s
Answer:
a = -0.8 m/s²
Here, negative sign indicates that the acceleration has opposite direction to the direction of motion.
Explanation:
First we find the angular acceleration of the ball from the following formula:
α = (ωf - ωi)/t
where,
α = angular acceleration = ?
ωf = final angular velocity = 7 rad/s
ωi = initial angular velocity = 13 rad/s
t = Time taken = 15 s
Therefore,
α = (7 rad/s - 13 rad/s)/15 s
α = - 0.4 rad/s
negative sign shows that acceleration is in opposite direction to the direction of motion.
Now, for the linear acceleration, we use the formula:
a = rα
where,
a = linear acceleration = ?
r = radius of circular path = length of rope = 2 m
therefore,
a = (2 m)(- 0.4 rad/s²)
a = -0.8 m/s²
Here, negative sign indicates that the acceleration has opposite direction to the direction of motion.
wha is amplitde in sound
Answer:
The number of molecules displaced in a vibration makes the amplitude of a sound.
You measure the power delivered by a battery to be 1.15 W when it is connected in series with two equal resistors. How much power will the same battery deliver if the resistors are now connected in parallel across it
Answer:
The power is [tex]P_p = 4.6 \ W[/tex]
Explanation:
From the question we are told that
The power delivered is [tex]P_{s} = 1.15 \ W[/tex]
Let it resistance be denoted as R
The resistors are connected in series so the equivalent resistance is
[tex]R_{eqv} = R+ R = 2 R[/tex]
Considering when it is connected in series
Generally power is mathematically represented as
[tex]P_s = V * I[/tex]
Here I is the current which is mathematically represented as
[tex]I = \frac{V}{2R}[/tex]
The power becomes
[tex]P_s = V * \frac{V}{2R}[/tex]
[tex]P_s = \frac{V^2}{2R}[/tex]
substituting value
[tex]1.15 = \frac{V^2}{2R}[/tex]
Considering when resistance is connected in parallel
The equivalent resistance becomes
[tex]R_{eqv} = \frac{R}{2}[/tex]
So The current becomes
[tex]I = \frac{V}{\frac{R}{2} } = \frac{2V}{R}[/tex]
And the power becomes
[tex]P_p = V * \frac{2V}{R} = \frac{2V^2}{R} = \frac{4 V^2}{2 R} = 4 * P_s[/tex]
substituting values
[tex]P_p = 4 * 1.15[/tex]
[tex]P_p = 4.6 \ W[/tex]
please help! i will be giving 50 points, this is for my psychology class.
Iris has been ahead of her classmates for as long as she has been in school. Lately, her classmates have started making fun of her for being a “teacher’s pet,” and they mock her whenever she raises her hand to answer a question.
Iris is most likely being negatively stereotyped as being __________.
A.
below average
B.
normal
C.
intellectually disabled
D.
gifted
Answer:
D
Explanation:
the other students are making fun of her most likely because they are jealous that she is successing in school. hope this helps :)
Answer:
D
Explanation:
describe Piaget's four stages of cognitive development. Include the major hallmarks of each stage.
Answer:
Explanation:
Sensorimotor Infants "think" by acting on the world with their eyes, ears, hands, and mouth.
Preoperational. Development of language and make-believe play takes place.
Concrete Operational children think in a logical, organized fashion only when dealing with concrete information they can perceive directly.
Formal Operational. Adolescences can also evaluate the logic of verbal statements without referring to real-world circumstances.
Sensorimotor, preoperational, concrete operational, and formal operational are Piaget's four phases of cognitive development.
What is cognitive development?The way youngsters think, investigate, and figure things out is referred to as cognitive development.
Piaget defined four stages of cognitive development:
1. Sensorimotor. From birth through the age of 18-24 months.
2. Preoperational.From infancy (18-24 months) until toddlerhood (age 7)
3. Operational concrete. 7 to 11 years old
4. Formal operational. From adolescence to adulthood
Hence, sensorimotor, preoperational, concrete operational, and formal operational are Piaget's four phases of cognitive development.
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You're driving a vehicle of mass 850 kg and you need to make a turn on a flat road. The radius of curvature of the turn is 80 m. The maximum horizontal component of the force that the road can exert on the tires is only 0.22 times the vertical component of the force of the road on the tires (in this case the vertical component of the force of the road on the tires is , the weight of the car, where as usual 9.8 N/kg, the magnitude of the gravitational field near the surface of the Earth). The factor 0.22 is called the "coefficient of friction" (usually written "", Greek "mu") and is large for surfaces with high friction, small for surfaces with low friction.(a) What is the fastest speed you can drive and still make it around the turn? Invent symbols for the various quantities and solve algebraically before plugging in numbers.
maximum speed = ____ m/s
(b) Which of the following statements are true about this situation?
The net force is nonzero and points away from the center of the kissing circle.The momentum points toward the center of the kissing circle.The net force is nonzero and points toward the center of the kissing circle.The rate of change of the momentum is nonzero and points toward the center of the kissing circle.The centrifugal force balances the force of the road, so the net force is zero.The rate of change of the momentum is nonzero and points away from the center of the kissing circle.
(c) Look at your algebraic analysis and answer the following question. Suppose your vehicle had a mass 3 times as big (5250 kg). Now what is the fastest speed you can drive and still make it around the turn?
maximum speed = ____ m/s
(d) Look at your algebraic analysis and answer the following question. Suppose you have the original 1750 kg vehicle but the turn has a radius twice as large (166 m). What is the fastest speed you can drive and still make it around the turn?
maximum speed = ____m/s
Answer:
(a) v = 13.13 m/s
(b) The centrifugal force balances the force of the road, so net force is zero.
(c) v = 13.13 m/s
(d) v = 18.92 m/s
Explanation:
(a)
To make it around the turn without skidding the frictional force on cat must balance the centrifugal force. Therefore:
Frictional Force = Centrifugal Force
μR = mv²/r
where,
R = Normal Reaction = Weight of Car = mg
Therefore,
μmg = mv²/r
μg = v²/r
v = √μgr
where,
v = maximum possible velocity of car = ?
μ = coefficient of friction = 0.22
g = 9.8 m/s²
r = radius of curvature = 80 m
Therefore,
v = √[(0.22)(9.8 m/s²)(80 m)
v = 13.13 m/s
(b)
In order for the car to move without skidding around the turn, all the forces in horizontal direction must be equal. Hence, the centrifugal force and the frictional force (force of the road) must balance each other. So the true statement is:
The centrifugal force balances the force of the road, so net force is zero.
(c)
v = √μgr
Since the formula for speed is independent of mass. Therefore, the speed will remain same.
v = 13.13 m/s
(d)
v = √μgr
v = √[(0.22)(9.8 m/s²)(166 m)
v = 18.92 m/s
A 1KW electric heater is switched on for ten minutes
How much heat does it produce?
Explanation:
P=W/T ==> 1000w = Q/600 ==> Q=600000j
If a 1 - kilowatt electric heater is switched on for ten minutes then the heat produced by the electric heater would be 600 - kilo Joules .
What is thermal energy ?It can be defined as the form of the energy in which heat is transferred from one body to another body due to their molecular movements, thermal energy is also known as heat energy .
As given in the problem , we have to find out the heat produced by the 1 - kilo watts electric heater if it is switched on for ten minutes ,
The heat produced by the electric heater = Power × time
= 1000 × 600 Joules
= 600 kilo - Joules
Thus , the heat produced by the electric heater would be 600 - kilo Joules .
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A sound level of 96 dB is how many times as intense as one of 90 dB?
Answer:
A sound level of 96 dB is 4 times as intense as one of 90 dB
Explanation:
The formula of the intensity level of sound in decibels is given as follows:
Intensity Level = 10 log₁₀(I/I₀)
where,
I = Intensity of Sound
I₀ = Reference Intensity Level = 10⁻¹² W/m²
Therefore, for 96 dB sound level:
96 = 10 log₁₀(I₁/10⁻¹²)
log₁₀(I₁/10⁻¹²) = 96/10
I₁/10⁻¹² = 10^9.6
I₁ = (10⁻¹²)(4 x 10⁹)
I₁ = 0.004 W/m²
For 90 dB sound level:
90 = 10 log₁₀(I₂/10⁻¹²)
log₁₀(I₂/10⁻¹²) = 90/10
I₂/10⁻¹² = 10^9
I₂ = (10⁻¹²)(10⁹)
I₂ = 0.001 W/m²
Therefore,
I₁/I₂ = 0.004/0.001
I₁ = 4 I₂
Hence, the sound level of 96 dB is 4 times as intense as one of 90 dB.
In which situation is chemical energy being converted to another form of energy?
Answer:
A burning candle. (chemical energy into energy of heat and light, i.e. thermal and wave)
Explanation:
A particle moving along the x-axis has a position given by m, where t is measured in s. What is the magnitude of the acceleration of the particle at the instant when its velocity is zero
Question:
A particle moving along the x-axis has a position given by x=(24t - 2.0t³)m, where t is measured in s. What is the magnitude of the acceleration of the particle at the instant when its velocity is zero
Answer:
24 m/s
Explanation:
Given:
x=(24t - 2.0t³)m
First find velocity function v(t):
v(t) = ẋ(t) = 24 - 2*3t²
v(t) = ẋ(t) = 24 - 6t²
Find the acceleration function a(t):
a(t) = Ẍ(t) = V(t) = -6*2t
a(t) = Ẍ(t) = V(t) = -12t
At acceleration = 0, take time as T in velocity function.
0 =v(T) = 24 - 6T²
Solve for T
[tex] T = \sqrt{\frac{-24}{6}} = \sqrt{-4} = -2 [/tex]
Substitute -2 for t in acceleration function:
a(t) = a(T) = a(-2) = -12(-2) = 24 m/s
Acceleration = 24m/s
A politician holds a press conference that is televised live. The sound picked up by the microphone of a TV news network is broadcast via electromagnetic waves and heard by a television viewer. This viewer is seated 2.7 m from his television set. A reporter at the press conference is located 5.5 m from the politician, and the sound of the words travels directly from the celebrity's mouth, through the air, and into the reporter's ears. The reporter hears the words exactly at the same instant that the television viewer hears them. Using a value of 343 m/s for the speed of sound, determine the maximum distance between the television set and the politician. Ignore the small distance between the politician and the microphone. In addition, assume that the only delay between what the microphone picks up and the sound being emitted by the television set is that due to the travel time of the electromagnetic waves used by the network.
Answer:
Explanation:
Time taken by sound waves to cover distance between politician and reporter = time taken by em waves to travel distance between politician and the television viewer.
5.5 / 343 = d / 3 x 10⁸ + 2.7 / 343
d is distance between politician and television set + time taken by sound to travel distance between television and its viewer.
.0160349 = d / 3 x 10⁸ + .0078717
d / 3 x 10⁸ = .0081632
d = 2448960 m
= 2448.96 km
= 2449 km .
If the radius of curvature of the cornea is 0.75 cm when the eye is focusing on an object 36.0 cm from the cornea vertex and the indexes of refraction are as described before, what is the distance from the cornea vertex to the retina
Answer:
The distance from the cornea vertex to the retina is 2.36 cm
Explanation:
The question is incomplete.
The complete question is as follows;
A Nearsighted Eye. A certain very nearsighted person cannot focus on anything farther than 36.0 cm from the eye. Consider the simplified model of the eye. In a simplified model of the human eye, the aqueous and vitreous humors and the lens all have a refractive index of 1.40, and all the refraction occurs at the cornea, whose vertex is 2.60 cm from the retina.
If the radius of curvature of the cornea is 0.65 cm when the eye is focusing on an object 36.0 cm from the cornea vertex and the indexes of refraction are as described before, what is the distance from the cornea vertex to the retina?
Solution.
We use image-object reaction to calculate the distance from the cornea vertex to the retina.
Mathematically;
n1/s + n2/s’ = n2-n1/R
From the question, we identify the following;
n1 ; Refractive index of air = 1
n2 ; Refractive index of lens = 1.4
S ; Object Distance = 36 cm
S’ = ?
R ; Radius of curvature of the cornea = 0.65
Substituting these values into the equation above;
1/36 + 1.4/S’ = (1.4-1)/0.65
{S’+ 36(1.4)}/36S’ = 0.4/0.65
{S’ + 50.4}/36S’ = 0.62
S’ + 50.4 = 22.32S’
50.4 = 22.32S’ -S’
21.32S’ = 50.4
S’ = 50.4/21.32
S’ = 2.36 cm
You rub a balloon on your head and it becomes negatively charged. The balloon will be most attracted to what?
Answer:
To things that are positive charged
Volume of an block is 5 cm3. If the density of the block is 250 g/cm3, what is the mass of the block ?
Answer:
1.25kg
Explanation:
Simply multiply volume and density together
Identify the five categories of stressors.
Answer:
The five kinds of stressors are:
Acute time-limited
Brief naturalistic
Stressful events sequences
Chronic
Distant
Explanation:
yeah
Photoelectric effect:
A. What is the maximum kinetic energy of electrons ejected from barium (W0=2.48eV) when illuminated by white light, lambda=410-750nm?
B. The work functions for sodium, cesium, copper, and iron are 2.3, 2.1, 4.7, and 4.5eV, respectively. Which of these metals will not emit electrons when visible light shines on it?
Answer:
A. K = 0.546 eV
B. cooper and iron will not emit electrons
Explanation:
A. This is a problem about photoelectric effect. Then you have the following equation:
[tex]K=h\nu-\Phi=h\frac{c}{\lambda} -\Phi[/tex] (1)
K: kinetic energy of the ejected electron
Ф: Work function of the metal = 2.48eV
h: Planck constant = 4.136*10^{-15} eV.s
λ: wavelength of light = 410nm - 750nm
c: speed of light = 3*10^8 m/s
As you can see in the equation (1), higher the wavelength, lower the kinetic energy. Then, the maximum kinetic energy is obtained with the lower wavelength (410nm). Thus, you replace the values of all variables :
[tex]K=(4.136*10^{-15}eV)\frac{3*10^8m/s}{410*10^{-9}m}-2.48eV\\\\K=0.546eV[/tex]
B. First you calculate the energy of the photon with wavelengths of 410nm and 750nm
[tex]E_1=(4.136*10^{-15}eV)\frac{3*10^{8}m/s}{410*10^{-9}m}=3.02eV\\\\E_2=(4.13610^{-15}eV)\frac{3*10^{8}m/s}{750*10^{-9}m}=1.6544eV[/tex]
You compare the energies E1 and E2 with the work functions of the metals and you can conclude:
sodium = 2.3eV < E1
cesium = 2.1 eV < E1
cooper = 4.7eV > E1 (this metal will not emit electrons)
iron = 4.5eV > E1 (this metal will not emit electrons)
b. A locomotive of a train exerts a constant force of 280KN on a train while pulling
it at 50 km/h along a level track. What is:
[4 marks)
i. Workdone in quarter an hour and
[4 marks]
Answer:
Work-done in quarter an hour = 3.5 × 10⁶ J
Explanation:
Given:
Force (F) = 280 KN = 280,000 N
Velocity (V) = 50 km / h
Time (t) = 1 / 4 = 0.25 hour
Find:
Work-done in quarter an hour
Computation:
⇒ Displacement = Velocity (V) × Time
⇒ Displacement = 50 × 0.25
⇒ Displacement = 12.5 km
⇒ Work-done = Force (F) × Displacement
⇒ Work-done in quarter an hour = 280,000 × 12.5
⇒ Work-done in quarter an hour = 3,500,000
Work-done in quarter an hour = 3.5 × 10⁶ J
A skater wearing in – line skates (no friction) is standing in the middle of the aisle inside a bus and is not holding on to anything. Which way would the skater move in reaction to the bus as it pulls away from the bus stopA skater wearing in – line skates (no friction) is standing in the middle of the aisle inside a bus and is not holding on to anything. Which way would the skater move in reaction to the bus as it pulls away from the bus stop
Before the bus starts moving, the bus and the skater are both standing still.
When the bus starts moving and pulls away from the bus-stop, the skater stays right where she is.
The people outside on the sidewalk see her standing still, and they see the bus moving out from under her.
The other passengers on the bus see her rolling backwards down the aisle, toward the back of the bus.
What type of device forms images by changing the speed at which light travels?
Answer:
A lens
Explanation:
A lens forms images when light passes Through it bending the rays of in the process.A phenomena called refraction and the speed of light changes in the process because it enters a medium since it's wavelength is reduced.
The type of device that forms images by changing the speed at which light travels is the lens.
What is refraction through the lens?
A lens bends a light beam at an aimed perspective and converges or diffuses bundles of rays by taking benefit of refraction taking vicinity while the mild travels from air into glass or plastic. For that purpose, the aspect geometry of a lens has a spherical parent, which can be kind of divided into sorts.
A lens bends a mild beam at an aimed perspective and converges or diffuses bundles of rays through taking gain of refraction taking area whilst the mild travels from air into glass or plastic. For that motive, the facet geometry of a lens has a round parent, which may be kind of divided into sorts.
Learn more about the speed of light here:-https://brainly.com/question/104425
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To practice Problem-Solving Strategy 6.1: Circular motion A highway curve with radius R = 274 m is to be banked so that a car traveling v = 25.0 m/s will not skid sideways even in the absence of friction. At what angle should the curve be banked?
Answer:
The curve should be banked at an angle of 13 degrees.
Explanation:
We have,
Radius of a highway curve is 274 m
Speed of car on this curve is 25 m/s
Let [tex]\theta[/tex] is the banking angle. On a banked curve, the angle of safe diving is given by following expression.
[tex]\tan\theta=\dfrac{v^2}{Rg}[/tex]
g = 10 m/s²
Plugging all the values in above formula,
[tex]\tan\theta=\dfrac{(25)^2}{274\times 9.8}\\\\\theta=\tan^{-1}\left(\dfrac{(25)^{2}}{274\times9.8}\right)\\\\\theta=13^{\circ}[/tex]
So, the curve should be banked at an angle of 13 degrees.
A whistle of frequency 516 Hz moves in a circle of radius 64.3 cm at an angular speed of 17.9 rad/s. What are (a) the lowest and (b) the highest frequencies heard by a listener a long distance away, at rest with respect to the center of the circle
Answer:
(a) 498.6 Hz
(b) 534.6 Hz
Explanation: Please see the attachments below
A heavy copper ball of mass 2 kg is dropped from a fiftieth-floor apartment window. Another one with mass 1 kg is dropped immediately after 1 second. Air resistance is negligible. The difference between the speeds of the two balls:__________.
a. increases over time at first, but then stays constant.
b. decreases over time.
c. remains constant over time.
d. increases over time.
Answer:
C
Explanation:
Because everything on Earth falls at the same speed, the masses of the balls do not matter. Since the acceleration due to gravity is constant, their speeds will both be increasing at the same rate, and therefore the difference in speeds would remain constant until they hit the ground. Hope this helps!
Which is the best description of the scientific theory
Explanation:
a scientific theory is a well substantiated explanation of some aspect of the nature world, based on a body of facts that have been repeatedly confirmed through observation and experiment. search fact-supported theories are not "guesses" but reliable account of the real world .