Answer:
The vertical velocity of the skater upon landing is 10.788 meters per second.
Explanation:
Skateboarder experiments a parabolic movement. As skateboarder jumps horizontally off the top of the staircase, it means that vertical component of initial velocity is zero and accelerates by gravity, the final vertical speed is calculated by the following expression:
[tex]v = v_{o} + g\cdot t[/tex]
Where:
[tex]v_{o}[/tex] - Initial vertical speed, measured in meters per second.
[tex]v[/tex] - Final vertical speed, measured in meters per second.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]t[/tex] - Time, measured in seconds.
Given that [tex]v_{o} = 0\,\frac{m}{s}[/tex], [tex]g = -9.807\,\frac{m}{s^{2}}[/tex] and [tex]t = 1.10\,s[/tex], the final velocity of the skater upon landing is:
[tex]v = 0\,\frac{m}{s} + \left(-9.807\,\frac{m}{s^{2}} \right)\cdot (1.10\,s)[/tex]
[tex]v = -10.788\,\frac{m}{s}[/tex]
The vertical velocity of the skater upon landing is 10.788 meters per second.
Answer:
-10.8
Explanation:
Answer on Khan Academy
"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
A thin film with an index of refraction of 1.60 is placed in one of the beams of a Michelson interferometer. If this causes a shift of 8 bright fringes in the pattern produced by light of wavelength 580 nm, what is the thickness of the film
Answer:
3.867 μm
Explanation:
The index of refraction, μ = 1.6
Wavelength of the light, λ = 580 nm
N2 - N1 = (2L / λ) (n2 - n1), Making L subject of formula, we have
(N2 - N1) λ = 2L (n2 - n1)
L = [(N2 - N1) * λ] / 2(n2 - n1)
L = (8 * 580) / 2(1.6 - 1.0)
L = 4640 nm / 1.2
L = 3867 nm or 3.867 μm
Therefore we can come to the conclusion that the thickness of the film is 3.867 nm
The intensity of sunlight at the Earth's distance from the Sun is 1370 W/m2. (a) Assume the Earth absorbs all the sunlight incident upon it. Find the total force the Sun exerts on the Earth due to radiation pressure. N (b) Explain how this force compares with the Sun's gravitational attraction.
Answer:
F= 3.56e22N
Explanation:
Using the force of radiation acting on the earth which is
force = radiation pressure x area = (intensity/c)xpi R^2
force = 1370W/m^2 x pi x( 6.37x10^6m)^2/3x10^8m/s
force = 5.82x10^8 N
But the sun's gravitational attraction means the magnitude of the solar gravitational force on earth: If that's the case, the answer is approx 10^22 N:
F=GMm/r^2
G=6.67x10^(-11)=6.67e-11
M=mass sun = 2x10^30kg=2e30
m=mass earth = 6x10^24kg
r=earth sun distance = 1.5x10^11m
F=(6.6e-11)(2e30)(6e24)/(1.5e11)^2 =
F= 3.56e22N
Recall that the voltages VL(t) and VC(t) across the inductor and capacitor are not in phase with the respective currents IL(t) and IC(t). In particular, which of the following statements is true for a sinusoidal current driver?
VL(t) and VC(t):
a) both lag their respective current
b) both lead their respective currents
c) VL(t) lags IL(t) and VC(t) leads IC(t)
d) VL(t) leads IL(t) and VC(t) lags IC(t)
Answer:
D) VL(t) leads IL(t) and VC(t) lags IC(t)
Explanation:
This is because The phase angle between voltage and current for inductors and capacitors is 90 degrees, or radians, also, this means that no power is dissipated in either the inductor or the capacitor, since the time average of current times voltage,
( I(t), V(t)), is zero.
What is the average flow rate in cm3 /s of gasoline to the engine of a car traveling at 100 km/h if it averages 10.0 km/L
Answer:
2.78 cm³/s
Explanation:
From the question,
Q = v/A'.................... Equation 1
Where Q = Average flow rate, A' = inverse of Area, v = velocity of the car.
Given: v = 100 km/h, A' = 10 km/L
Substitute this value into equation 1
Q = 100/10
Q = 10 L/h.
Now, we convert L/h to cm³/s.
Since,
1 L = 1000 cm³, and
1 h = 3600 s
Therefore,
Q = 10(1000/3600) cm³/s
Q = 2.78 cm³/s
Interactive Solution 8.29 offers a model for this problem. The drive propeller of a ship starts from rest and accelerates at 2.24 x 10-3 rad/s2 for 2.80 x 103 s. For the next 1.57 x 103 s the propeller rotates at a constant angular speed. Then it decelerates at 2.01 x 10-3 rad/s2 until it slows (without reversing direction) to an angular speed of 2.99 rad/s. Find the total angular displacement of the propeller.
Answer:
θ = 26.19 x 10³ radians
Explanation:
FOR ACCELERATED MOTION:
we use 2nd equation of motion for accelerated motion:
θ₁ = ωi t + (1/2)αt²
Where,
θ₁ = Angular Displacement covered during accelerated motion = ?
ωi = Initial Angular Speed = 0 rad/s
t = Time Taken = 2.8 x 10³ s
α = Angular Acceleration = 2.24 x 10⁻³ rad/s²
Therefore,
θ₁ = (0 rad/s)(2.8 x 10³ s) + (1/2)(2.24 x 10⁻³ rad/s²)(2.8 x 10³ s)²
θ₁ = 8.78 x 10³ radians
Now we find final angular velocity (ωf) by using 1st equation of motion:
ωf = ωi + αt
ωf = 0 rad/s + (2.24 x 10⁻³ rad/s²)(2.8 x 10³ s)
ωf = 6.272 rad/s
FOR UNIFORM ANGULAR SPEED:
For uniform angular speed we use following equation:
θ₂ = ωt
where,
θ₂ = Angular Displacement during uniform motion = ?
ω = Uniform Angular Speed = ωf = 6.272 rad/s
t = Time Taken = 1.57 x 10³ s
Therefore,
θ₂ = (6.272 rad/s)(1.57 x 10³ s)
θ₂ = 9.85 x 10³ radians
FOR DECELERATED MOTION:
Now, we use 3rd equation of motion for decelerated motion:
2αθ₃ = ωf² - ωi²
where,
α = Angular deceleration = - 2.01 x 10⁻³ rad/s²
θ₃ = Angular Displacement during decelerated motion = ?
ωf = Final Angular Speed = 2.99 rad/s
ωi = Initial Angular Speed = 6.272 rad/s
Therefore,
2(-2.01 x 10⁻³ rad/s²)θ₃ = (2.99 rad/s)² - (6.272 rad/s)²
θ₃ = (- 30.4 rad²/s²)/(-4.02 x 10⁻³ rad/s²)
θ₃ = 7.56 x 10³ radians
FOR TOTAL ANGULAR DISPLACEMENT:
Total Angular Displacement = θ = θ₁ + θ₂ + θ₃
θ = 8.78 x 10³ radians + 9.85 x 10³ radians + 7.56 x 10³ radians
θ = 26.19 x 10³ radians
When dots are placed on a page from a laser printer, they must be close enough so that you do not see the individual dots of ink. To do this, the separation of the dots must be less than Raleigh's criterion. Take the pupil of the eye to be 3.2 mm and the distance from the paper to the eye of 42 cm; find the maximum separation (in cm) of two dots such that they cannot be resolved. (Assume the average wavelength of visible light is 550 nm.)
Answer:
y <8 10⁻⁶ m
Explanation:
For this exercise, they indicate that we use the Raleigh criterion that establishes that two luminous objects are separated when the maximum diffraction of one of them coincides with the first minimum of the other.
Therefore the diffraction equation for slits with m = 1 remains
a sin θ = λ
in general these experiments occur for oblique angles so
sin θ = θ
θ = λ / a
in the case of circular openings we must use polar coordinates to solve the problem, the solution includes a numerical constant
θ = 1.22 λ / a
The angles in these measurements are taken in radians, therefore
θ = s / R
as the angle is small the arc approaches the distance s = y
y / R = 1.22 λ / s
y = 1.22 λ R / a
let's calculate
y = 1.22 500 10⁻⁹ 0.42 / 0.032
y = 8 10⁻⁶ m
with this separation the points are resolved according to the Raleigh criterion, so that it is not resolved (separated)
y <8 10⁻⁶ m
Two spherical objects at the same altitude move with identical velocities and experience the same drag force at a time t. If Object 1 has twice (2x) the diameter of Object 2, which object has the larger drag coefficient? Explain your answer using the drag equation.
Answer:
The object with the twice the area of the other object, will have the larger drag coefficient.
Explanation:
The equation for drag force is given as
[tex]F_{D} = \frac{1}{2}pu^{2} C_{D} A[/tex]
where [tex]F_{D}[/tex] IS the drag force on the object
p = density of the fluid through which the object moves
u = relative velocity of the object through the fluid
p = density of the fluid
[tex]C_{D}[/tex] = coefficient of drag
A = area of the object
Note that [tex]C_{D}[/tex] is a dimensionless coefficient related to the object's geometry and taking into account both skin friction and form drag. The most interesting things is that it is dependent on the linear dimension, which means that it will vary directly with the change in diameter of the fluid
The above equation can also be broken down as
[tex]F_{D}[/tex] ∝ [tex]P_{D}[/tex] A
where [tex]P_{D}[/tex] is the pressure exerted by the fluid on the area A
Also note that [tex]P_{D}[/tex] = [tex]\frac{1}{2}pu^{2}[/tex]
which also clarifies that the drag force is approximately proportional to the abject's area.
In this case, the object with the twice the area of the other object, will have the larger drag coefficient.
In a wire with a 1.05 mm2 cross-sectional area, 7.93×1020 electrons flow past any point during 3.97 s. What is the current ???? in the wire?
Answer:
The current in the wire is 31.96 A.
Explanation:
The current in the wire can be calculated as follows:
[tex] I = \frac{q}{t} [/tex]
Where:
q: is the electric charge transferred through the surface
t: is the time
The charge, q, is:
[tex] q = n*e [/tex]
Where:
n: is the number of electrons = 7.93x10²⁰
e: is the electron's charge = 1.6x10⁻¹⁹ C
[tex] q = n*e = 7.93 \cdot 10^{20}*1.6 \cdot 10^{-19} C = 126.88 C [/tex]
Hence, the current in the wire is:
[tex] I = \frac{126.88 C}{3.97 s} = 31.96 A [/tex]
Therefore, the current in the wire is 31.96 A.
I hope it helps you!
Use Kepler's third law to determine how many days it takes a spacecraft to travel in an elliptical orbit from a point 6 965 km from the Earth's center to the Moon, 385 000 km from the Earth's center.
Answer:
0.0665 days
Explanation:
We are given;
The mean distance from the Earth's center to the moon;a1 = 385000 km
The mean distance from the Earth's center to the space craft;a2 = 6965 km
Formula for kepplers third law is;
T² = 4π²a³/GM
However, the proportion of both distances would be;
(T1)²/(T2)² = (a1)³/(a2)³
Where;
T1 is the period of orbit of the moon around the earth. T1 has a standard value of 27.322 days
T2 is the period of the space craft orbit.
Making T2 the subject, we have;
T2 = √((T1)²×(a2)³)/(a1)³)
Thus, plugging in the relevant values;
T2 = √(27.322² × 6965³)/(385000)³
T2 = 0.0665 days
A pickup truck has a width of 79.0 in. If it is traveling north at 42 m/s through a magnetic field with vertical component of
40 ut, what magnitude emf is induced between the driver and passenger sides of the truck?
Answer:
The magnitude of the induced Emf is [tex]0.003371V[/tex]
Explanation:
The width of the truck is given as 79inch but we need to convert to meter for consistency, then
The width= 79inch × 0.0254=2.0066 metres.
Now we can calculate the induced Emf using expresion below;
Then the [tex]induced EMF= B L v[/tex]
Where B= magnetic field component
L= width
V= velocity
=(40*10^-6) × (42) × (2.0066)
=0.003371V
Therefore, the magnitude emf that is induced between the driver and passenger sides of the truck is 0.003371V
A 5 kg block is sliding on a horizontal surface while being pulled by a child using a rope attached to the center of the block. The rope exerts a constant force of 28.2 N at an angle of \theta=θ = 30 degrees above the horizontal on the block. Friction exists between the block and supporting surface (with \mu_s=\:μ s = 0.25 and \mu_k=\:μ k = 0.12 ). What is the horizontal acceleration of the block?
Answer:
The horizontal acceleration of the block is 4.05 m/s².
Explanation:
The horizontal acceleration can be found as follows:
[tex] F = m \cdot a [/tex]
[tex] Fcos(\theta) - \mu_{k}N = m\cdot a [/tex]
[tex] Fcos(\theta) - \mu_{k}[mg - Fsen(\theta)] = m\cdot a [/tex]
[tex] a = \frac{Fcos(\theta) - \mu_{k}[mg - Fsen(\theta)]}{m} [/tex]
Where:
a: is the acceleration
F: is the force exerted by the rope = 28.2 N
θ: is the angle = 30°
[tex]\mu_{k}[/tex]: is the kinetic coefficient = 0.12
m: is the mass = 5 kg
g: is the gravity = 9.81 m/s²
[tex] a = \frac{28.2 N*cos(30) - 0.12[5 kg*9.81 m/s^{2} - 28.2 N*sen(30)]}{5 kg} = 4.05 m/s^{2} [/tex]
Therefore, the horizontal acceleration of the block is 4.05 m/s².
I hope it helps you!
You are fixing a transformer for a toy truck that uses an 8.0-V emf to run it. The primary coil of the transformer is broken; the secondary coil has 40 turns. The primary coil is connected to a 120-V wall outlet.
(a) How many turns should you have in the primary coil?
(b) If you then connect this primary coil to a 240-V source, what emf would be across the secondary coil?
Comments: The relevant equation is N1/N2 = V1/V2 where N is the number of turns and V is the voltage. I'm just not sure how to get the voltage of the secondary coil using emf.
Answer:
a. The primary turns is 60 turns
b. The secondary voltage will be 360 volts.
Explanation:
Given data
secondary turns N2= 40 turns
primary turns N1= ?
primary voltage V1= 120 volts
secondary voltage V2= 8 volts
Applying the transformer formula which is
[tex]\frac{N1}{N2} =\frac{V1}{V2}[/tex]
we can solve for N1 by substituting into the equation above
[tex]\frac{N1}{40} =\frac{120}{8} \\\ N1= \frac{40*120}{8} \\\ N1= \frac{4800}{8} \\\ N1= 60[/tex]
the primary turns is 60 turns
If the primary voltage is V1 240 volts hence the secondary voltage V2 will be (to get the voltage of the secondary coil using emf substitute the values of the previously gotten N1 and N2 using V1 as 240 volts)
[tex]\frac{40}{60} =\frac{240}{V2}\\\\V2= \frac{60*240}{40} \\\\V2=\frac{ 14400}{40} \\\\V2= 360[/tex]
the secondary voltage will be 360 volts.
(a) In the primary coil, you have "60 turns".
(b) The emf across the secondary coil would be "360 volts".
Transformer and VoltageAccording to the question,
Primary voltage, V₁ = 120 volts
Secondary voltage, V₂ = 8 volts
Secondary turns, N₂ = 40 turns
(a) By applying transformer formula,
→ [tex]\frac{N_1}{N_2} = \frac{V_1}{V_2}[/tex]
or,
N₁ = [tex]\frac{N_2\times V_1}{V_2}[/tex]
By substituting the values,
= [tex]\frac{40\times 120}{8}[/tex]
= [tex]\frac{4800}{8}[/tex]
= 60
(2) Again by using the above formula,
→ V₂ = [tex]\frac{60\times 240}{40}[/tex]
= [tex]\frac{14400}{40}[/tex]
= 360 volts.
Thus the above approach is correct.
Find out more information about voltage here:
https://brainly.com/question/4389563
collision occurs betweena 2 kg particle traveling with velocity and a 4 kg particle traveling with velocity. what is the magnitude of their velocity
Answer:
metre per seconds
Explanation:
because velocity = distance ÷ time
Two 15-Ω and three 25-Ω light bulbs and a 24 V battery are connected in a series circuit. What is the current that passes through each bulb?
Answer:
0.229AExplanation:
Before we determine the amount of current in each bulb, we must first know that the same current flows in a series connected resistors. Since the Two 15-Ω and three 25-Ω light bulbs are connected in series, same current will flow in all of them.
According to Ohm's law, E = IRt where;
E is the supply voltage = 24V
I is the total current flowing in the circuit
Rt is the total equivalent resistance.
First, we need to calculate Rt.
Rt = 15Ω+15Ω+25Ω+25Ω+25Ω (Two 15-Ω and three 25-Ω light bulb in series)
Rt = 105Ω
From ohms law formula, I = E/Rt
I = 24/105
I = 0.229Amp
Since the total current in the circuit is 0.229A, therefore the amount of current that passes through each bulb is the same as the total current i.e 0.229A
The current that passes via each bulb is 0.229A
Ohm law:According to the above law,
E = IRt
Here
E should be the supply voltage = 24V
I should be the total current flowing in the circuit
Rt should be the total equivalent resistance.
Now Rt should be
Rt = 15Ω+15Ω+25Ω+25Ω+25Ω
Rt = 105Ω
Now the current is
I = E/Rt
I = 24/105
I = 0.229Amp
Therefore, The current that passes via each bulb is 0.229A.
Learn more about current here: https://brainly.com/question/21075693
(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
g When attempting to determine the coefficient of kinetic friction, why is it necessary to move the block with constant velocity
Answer:
This is because motion is intended to occur but at zero acceleration. It means at a constant velocity, henceFor that to happen the pulling force F must exactly equal the frictional force Fk .
Suppose you sit on a rotating piano stool and hold a 2kg mass in each outstreched hand. If without your arms relative to your body you now drop these masses,
a) your angular velocity remains unchanged
b) your angular velocity increases
c) your angular velocity decreases but your kinetic energy increases.
d) your kinetic energy and angular velocity increases.
Answer:
C is the best answer for this question
9. You are given a number of 10 Ω resistors, each capable of dissipating only 1.0 W without being destroyed. What is the minimum number of such resistors that you need to combine in series or in parallel to make a 10 Ω resistance that is capable of dissipating at least 5.0 W?
Answer:
here are 9 resistors, forming a group of 3 resistors in parallel and each group in series with the other.
Explanation:
Let's work carefully this exercise, they indicate that the total resistance 10 ohm and dissipates 5W, so we can use the power equation to find the circuit current
P = Vi = i² R
i = √ P / R
i = √ (5/10)
i = 0.707 A
This is the current that must circulate in the circuit.
Let's build a circuit with three resistors in series and each resistor in series has three resistors in parallel
The equivalent resistance is
1 /[tex]R_{equi}[/tex] = 1/10 + 1/10 + 1/10 = 3/10
Requi = 10/3
Requi = 3.3 Ω
The current in the three series resistors is I = 0.707 A, and this is divided into three equal parts for the parallel resistors
current in each residence in parallel
i_P = 0.707 / 3
I_p = 0.2357 A
now let's look at the power dissipated in each resistor
P = R i²
P = 10 0.2357²
P = 0.56 W
the power dissipated by each resistance is within the range of 1 A, let's see the total power that the 9 resistors dissipate
P_total = 9 P
P = total = 9 0.56
P_total = 5 W
we see that this combination meets the specifications of the problem.
Therefore there are 9 resistors, forming a group of 3 resistors in parallel and each group in series with the other.
A 50 g ball of clay traveling at speed v0 hits and sticks to a 1.0 kg brick sitting at rest on a frictionless surface.
Required:
a What is the speed of the block after the collision?
b. What percentage of the ball's initial energy is "lost"?
Answer:
(a) The speed of the block after the collision is 0.0476v0.
(b) The percentage of the ball's initial energy lost, is 0 % (energy is conserved)
Explanation:
Given;
mass of ball of clay, m₁ = 50 g = 0.05 kg
mass of brick, m₂ = 1 kg
initial velocity of the ball of clay, u₁ = v0
initial velocity of the brick, u₂ = 0
Since the clay ball sticks with the brick after collision, it is inelastic collision.
Therefore, let their final velocity = v
(a) What is the speed of the block after the collision?
Apply the principle of conservation linear momentum
m₁u₁ + m₂u₂ = v (m₁ + m₂)
0.05v₀ + 1(0) = v( 0.05 + 1)
0.05v₀ = 1.05v
v = 0.05v₀ / 1.05
v = 0.0476v₀
Thus, the speed of the block after the collision is 0.0476 of its initial velocity.
(b). What percentage of the ball's initial energy is "lost"?
Initial kinetic energy of the ball = ¹/₂mv₀²
= ¹/₂ x 0.05 x v₀²
= 0.025v₀²
Final kinetic energy of the ball, = ¹/₂(m₁ + m₂)v²
= ¹/₂ x 1.05 x 0.0476v₀²
= 0.025v₀²
Change in kinetic energy = 0.025v₀² - 0.025v₀²
= 0
percentage change in the initial kinetic energy of the ball;
= (0 / 0.025v₀²) x 100%
= 0 x 100%
= 0 %
Therefore, the percentage of the ball's initial energy lost, is 0 % (energy is conserved)
Calculate the wavelengths of the first five members of the Lyman series of spectral lines, providing the result in units Angstrom with precision one digit after the decimal point.
Answer:
Explanation:
The formula for hydrogen atomic spectrum is as follows
energy of photon due to transition from higher orbit n₂ to n₁
[tex]E=13.6 (\frac{1}{n_1^2 } - \frac{1}{n_2^2})eV[/tex]
For layman series n₁ = 1 and n₂ = 2 , 3 , 4 , ... etc
energy of first line
[tex]E_1=13.6 (\frac{1}{1^2 } - \frac{1}{2 ^2})[/tex]
10.2 eV
wavelength of photon = 12375 / 10.2 = 1213.2 A
energy of 2 nd line
[tex]E_2=13.6 (\frac{1}{1^2 } - \frac{1}{3 ^2})[/tex]
= 12.08 eV
wavelength of photon = 12375 / 12.08 = 1024.4 A
energy of third line
[tex]E_3=13.6 (\frac{1}{1^2 } - \frac{1}{4 ^2})[/tex]
12.75 e V
wavelength of photon = 12375 / 12.75 = 970.6 A
energy of fourth line
[tex]E_4=13.6 (\frac{1}{1^2 } - \frac{1}{5 ^2})[/tex]
= 13.056 eV
wavelength of photon = 12375 / 13.05 = 948.3 A
energy of fifth line
[tex]E_5=13.6 (\frac{1}{1^2 } - \frac{1}{6 ^2})[/tex]
13.22 eV
wavelength of photon = 12375 / 13.22 = 936.1 A
Which has more mass electron or ion?
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
A plane progressive
the expression
in time, ys
where you
progressivo ware is no presented by
(At + A
y- 5 sin
in metre, t es in time the doplicensel
Calculate
the amplitude of the wave.
Answer:
Amplitude, A = 5 m
Explanation:
Let a progressive wave is given by equation :
[tex]y=5\sin (100\pi t-0.4\pi x)[/tex] .....(1)
The general equation of a progressive wave is given by :
[tex]y=A\sin (\omega t-kx)[/tex] ....(2)
Here,
A is the amplitude of the wave
[tex]\omega[/tex] is the angular frequency
k is propagation constant
We need to find the amplitude of the wave.
If we compare equations (1) and (2), we find that the amplitude of the given plane progressive wave is 5 m.
When a certain gas under a pressure of 4.65 106 Pa at 21.0°C is allowed to expand to 3.00 times its original volume, its final pressure is 1.06 106 Pa. What is its final temperature?
Answer:
-72.0°C
Explanation:
PV = nRT
Since n, number of moles, is constant:
PV / T = PV / T
(4.65×10⁶ Pa) V / (21 + 273.15) K = (1.06×10⁶ Pa) (3V) / T
T = 201.16 K
T = -72.0°C
The maximum amount of pulling force a truck can apply when driving on
concrete is 10,560 N. If the coefficient of static friction between a trailer and
concrete is 0.8, what is the heaviest that the trailer can be and still be pulled
by the truck?
Answer:
Explanation:
Let the weight of the truck be W . reaction force R = W
Maximum frictional force = μ R
= .8 x W
So for movement of truck
Pulling force = frictional force
10560 = .8W
W = 13200 N
weight of heaviest truck required = 13200 N .
An object on a rope is lowered steadily decreasing speed. Which is true?
A) The tope tensions is greater than the objects weight
B) the rope tension equals the objects weight
C)the rope tension is less than the objects weight
D) the rope tension can’t be compared to the objects weight
Answer:
C) the rope tension is less than the objects weight
Explanation:
According to Newton's Second Law, when an unbalanced or net force is applied to a body, it produces an acceleration in the body in the direction of the net force itself.
In this scenario, we have two forces acting on the object. First is the weight of object acting downward. Second is the tension in the rope acting upwards.
Since, the object is being lowered in the direction of weight. Therefore, weight of the object must be greater than the tension in the rope. So, the net force has the downward direction and the object is lowered. Hence, the correct option is:
C) the rope tension is less than the objects weight
What is the minimum magnitude of an electric field that balances the weight of a plasticsphere of mass 5.4 g that has been charged to -3.0 nC
Answer:
E = 17.64 x 10⁶ N/C = 17.64 MN/C
Explanation:
The electric field is given by the following formula:
E = F/q
E= W/q
E = mg/q
where,
E = magnitude of electric field = ?
m = mass of plastic sphere = 5.4 g = 5.4 x 10⁻³ kg
g = acceleration due to gravity = 9.8 m/s²
= charge = 3 nC = 3 x 10⁻⁹ C
Therefore,
E = (5.4 x 10⁻³ kg)(9.8 m/s²)/(3 x 10⁻⁹ C)
E = 17.64 x 10⁶ N/C = 17.64 MN/C
A typical home uses approximately 1600 kWh of energy per month. If the energy came from a nuclear reaction, what mass would have to be converted to energy per year to meet the energy needs of the home
Answer:
7.68×10^25kg
Explanation:
The formula for energy used per year is calculated as
Energy used per year =12 x Energy used per month
By substituting Energy used per month in the above formula, we get
Energy used per year =12 x 1600kWh
= 19200kWh
Conversion:
From kWh to J:
1 kWh=3.6 x 10^6 J
Therefore, it is converted to J as
19200 kWh =19200 x 3.6 x 10^6 J
= 6.912×10^10 J
Hence, energy used per year is 6.912×10^10 J
To find the mass that is converted to energy per year.
E = MC^2 ............1
E is the energy used per year
C is the speed of light = 3.0× 10^8m/s
Where E= 6.912×10^10 J
Substituting the values into equation 1
6.912×10^10 J = M × 3.0× 10^8m/s
M = 6.912×10^10 J / (3.0× 10^8m/s)^2
M = 6.912×10^10 J/9×10^16
M = 7.68×10^25kg
Hence the mass to be converted is
7.68×10^25kg
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.