An equipotential line is like the contour line of a map that had lines of equal altitude. In this case the "altitude"is the electrical potential or voltage. Equipotential lines are always perpendicular to the electric field. In three dimensions these lines form equipotential surfaces. The movement along an equipotential surface does not perform work, because that movement is always perpendicular to the electric field.
A straight nonconducting plastic wire 9.5 cm long carries a charge density of 100 nc/m distributed uniformly along its length. It is lying on a horizontal tabletop. A. Find the magnitude and direction of the electric field this wire produces at a point 4.50 cm directly above its midpoint. B. If the wire is now bent into a circle lying flat on the table, find the magnitude and direction of the electric field it produces at a point 4.50 cm directly above its center.
Answer:
Explanation:
Given that,
Length of wire = 9.5 cm =
Charge density = 100 NC/m
Distance = 4.50 cm
(A). We need to calculate the the magnitude and direction of the electric field
Using formula of magnetic field
[tex]E=\dfrac{k\lambda\times l}{x\sqrt{x^2+a^2}}[/tex]
Here, x = distance from mid point of the wire
Put the value into the formula
[tex]E=\dfrac{9\times10^{9}\times100\times10^{-9}\times9.5\times10^{-2}}{4.50\times10^{-2}\sqrt{(4.50)^2+(\dfrac{9.5\times10^{-2}}{2})^2}}[/tex]
[tex]E=2.90\times10^{4}\ N/C[/tex]
The electric field is [tex]2.90\times10^{4}\ N/C[/tex].
The direction of electric field is upward.
(B). If the wire is now bent into a circle lying flat on the table,
We need to calculate the the magnitude and direction of the electric field
Using formula of electric field
[tex]E=\dfrac{k\lambda lx}{(x^2+(\dfrac{L}{\2\pi})^2)^{\dfrac{3}{2}}}[/tex]
Put the value into the formula
[tex]E=\dfrac{9\times10^{9}\times100\times10^{-9}\times9.5\times10^{-2}\times4.50\times10^{-2}}{((4.50\times10^{-2})^2+(\dfrac{9.5\times10^{-2}}{2\pi})^2)^{\dfrac{3}{2}}}[/tex]
[tex]E=3.59\times10^{4}\ N/C[/tex]
The electric field is [tex]3.59\times10^{4}\ N/C[/tex]
The direction of electric field is upward.
Hence, This is the required solution.
The electric field at a point above a charged wire depends on the configuration of the wire. For a straight wire, only the vertical components of the electric field produced by the charges along the wire add up, resulting in a field pointing upwards. If the wire is reshaped into a circle, the field still points upwards due to the circular symmetry.
Explanation:Part A: Electric Field of a Straight Wire
The electric field E produced by a straight nonconducting plastic wire with a length of 9.5 cm carrying a uniform charge density λ of 100 nc/m can be calculated using Coulomb's law. Considering a small charge dq located at the midpoint of the wire, the electric field dE at a point P located 4.5 cm directly above this midpoint will have both a vertical (dE cos θ) and a horizontal component (dE sin θ). Given the symmetry of the problem, the horizontal components for all the dq's will cancel out, leaving only the resultant vertical component. Integrating this over the length of the wire, we can calculate the magnitude of the electric field E (directed vertically upwards).
Part B: Electric Field of a Circular Wire
If the wire is reshaped into a flat circle lying on the table, the situation changes. Now, because of the circular symmetry, the horizontal components of the electric field produced by the small charge elements dq around the circle do not cancel out, and instead add up to create a resultant electric field pointing vertically upwards at a point located 4.5 cm above the center of the circle.
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Calculate the kinetic energy in units of electron volts for an electron emitted from a surface with a work function of 1.5 eV that has been illuminated with632.8-nm (red) helium neon laser, such as those used in a supermarket scanner. Perform the same calculation for a 543.5-nm (green) helium-neon laser.
Answer:
The kinetic energies are 0.46 eV and 0.78 eV.
Explanation:
Given that,
Work function = 1.5 eV
Wavelength = 632.8 nm
We need to calculate the energy for red helium neon laser
Using formula of energy
[tex]E=\dfrac{hc}{\lambda}[/tex]
Put the value into the formula
[tex]E=\dfrac{4.136\times10^{-15}\times3\times10^{8}}{632.8\times10^{-9}}[/tex]
[tex]E=1.96\ eV[/tex]
We need to calculate the kinetic energy
Using formula of K.E
[tex]K.E=E-\phi[/tex]
[tex]\phi[/tex]=work function
E = energy
Put the value into the formula
[tex]K.E=1.96-1.5[/tex]
[tex]K.E=0.46\ eV[/tex]
We need to calculate the energy for green helium neon laser
Using formula of energy
[tex]E=\dfrac{4.136\times10^{-15}\times3\times10^{8}}{543.5\times10^{-9}}[/tex]
[tex]E=2.28\ eV[/tex]
We need to calculate the kinetic energy
Using formula of K.E
[tex]K.E=2.28-1.5[/tex]
[tex]K.E=0.78\ eV[/tex]
Hence, The kinetic energies are 0.46 eV and 0.78 eV.
A cannon tilted up at a 33.0° angle fires a cannon ball at 74.0 m/s from atop a 18.0 m -high fortress wall. What is the ball's impact speed on the ground below?
Answer:
The cannon ball hits the ground with a speed of 76.35 m/s.
Explanation:
We shall use the conservation of energy principle to solve the problem
The initial energy of the cannon ball is the sum of the potential and the kinetic energies.
We know that potential energy = [tex]mgh[/tex]
Applying the given values we obtain the initial potential energy as
[tex]P.E=m\times g\times 18.0[/tex]
Similarly the initial kinetic energy of the ball equals
[tex]K.E_{initial}=\frac{1}{2}mv_{o}^{2}[/tex]
Applying values we get
[tex]K.E_{initial}=\frac{1}{2}m(74.0)^{2}[/tex]
Thus the initial energy is thus [tex]E_{initial}=m\times g\times 18.0+\frac{1}{2}\times m\times (74.0)^{2}[/tex]
Now upon hitting the ground the only energy that the cannon ball will posses is kinetic energy since the potential energy of any object upon touching the surface of earth equals zero
Thus we have
[tex]Energy_{final}=\frac{1}{2}mv_{f}^{2}[/tex]
Equating initial and final energies we get
[tex]m\times g\times 18+\frac{1}{2}\times m\times (74.0)^{2}=\frac{1}{2}\times m\times v_{f}^{2}\\\\v_{f}^{2}=36g+(74.0)^{2}\\\\\therefore v_f=\sqrt{36\times 9.81+(74.0)^{2}}\\\\v_f=76.35m/s[/tex]
Three equal charges are placed at the corners of an equilateral triangle 0.50 m on a side. What are the magnitude of the force on each charge if the charges are each -3.1 x 10^-9 C?
Answer:
the magnitude of the force is 192.29 N
Solution:
As per the question:
Charges present on the corner of the triangle are same, Q = [tex]- 3.1\times 10^{- 9} C[/tex]
Since, its an equilateral triangle, distance between the charges, l = 0.50 m
Now,
The Coulomb force on a charge due to the other is:
[tex]F_{C} = K\frac{Q^{2}}{l^{2}}[/tex]
where
K = Coulomb constant = [tex]9\times 10^{9} C^{2}/m^{2}[/tex]
[tex]F_{C} = (9\times 10^{9})\frac{(3.1\times 10^{- 9})^{2}}{0.5^{2}}[/tex]
[tex]F_{C} = 111.6 N[/tex]
The the net force on the charges in an equilateral triangle on all the charges due to each other:
[tex]F_{eq} = \sqrt{3}F_{C} = \sqrt{3}\times 111.6 = 193.29 N[/tex]
The magnitude of the force on each charge of -3.1 x 10^-9 C at the corners of an equilateral triangle 0.50 m on a side is 1.088 x 10^-7 N.
For one charge, the force due to the other two charges will be the vector sum of the forces due to each charge separately. Since the triangle is equilateral, the forces will have the same magnitude but different directions. Using Coulomb's law, we find the magnitude of the force between any two charges:
F = k × |q1 × q2| / r2
Where:
k is the Coulomb's constant (8.988 x 109 Nm2/C2)q1 and q2 are the magnitudes of the two chargesr is the distance between the chargesSubstituting the given values:
F = (8.988 x 109 Nm2/C2) × (3.1 x 10-9 C)2 / (0.50 m)2
This results in:
F = 1.088 x 10-7 N
As all sides are equal, and charges are equal, the net force will be the same in magnitude for each charge. The direction of the resulting force on each charge will be along the perpendicular bisector of the side opposite to each charge due to the symmetry of the configuration.
Describe how the motion of a charged particle changes when it is subjected to an accelerating voltage? ( This particle is in an electric field)
Answer:
An electric field has a direction, +ve to -ve. This is the trajectory in which the electric field will bring about to accelerate a positive charge.
Explanation:
An electric field has a direction, +ve to -ve. This is the trajectory in which the electric field will bring about to accelerate a positive charge.
if the +ve charge moves in the very same direction as the electrical field vector then the velocity of the particle will increase. It will slow down if it moves in the reverse direction.
if the adverse charge moves in the very same direction as the electrical field vector, the particle velocity will decelerate. If it moves in the reverse direction, it will speed up.
Answer and Explanation:
When a particle is placed in an Electric field having a charge Q and mass M, then it experiences an electric force on it. and it is given by:
[tex]F = QE[/tex]
This force is the overall force which is responsible for the acceleration of the particle.
We know, from Newton's second law:
F = Ma
and the force on a particle in electric field:
[tex]F_{E} = QE[/tex]
Therefore, from these two forces:
Ma = QE
a = [tex]\frac{QE}{M}[/tex]
In case of uniform Electric field, the particle moves with constant acceleration.
This acceleration is also capable of bending the object and to move it in a circle.
Two wooden crates rest on top of one another. The smaller top crate has a mass of m1 = 24 kg and the larger bottom crate has a mass of m2 = 86 kg. There is NO friction between the crate and the floor, but the coefficient of static friction between the two crates is μs = 0.79 and the coefficient of kinetic friction between the two crates is μk = 0.62. A massless rope is attached to the lower crate to pull it horizontally to the right (which should be considered the positive direction for this problem). 1)The rope is pulled with a tension T = 234 N (which is small enough that the top crate will not slide). What is the acceleration of the small crate?
2)In the previous situation, what is the magnitude of the frictional force the lower crate exerts on the upper crate?
3)What is the maximum tension that the lower crate can be pulled at before the upper crate begins to slide?
4)The tension is increased in the rope to 1187 N causing the boxes to accelerate faster and the top box to begin sliding. What is the acceleration of the upper crate?m/s2
5)As the upper crate slides, what is the acceleration of the lower crate?
Answer:
The sum of all forces for the two objects with force of friction F and tension T are:
(i) m₁a₁ = F
(ii) m₂a₂ = T - F
1) no sliding infers: a₁ = a₂= a
The two equations become:
m₂a = T - m₁a
Solving for a:
a = T / (m₁+m₂) = 2.1 m/s²
2) Using equation(i):
F = m₁a = 51.1 N
3) The maximum friction is given by:
F = μsm₁g
Using equation(i) to find a₁ = a₂ = a:
a₁ = μs*g
Using equation(ii)
T = m₁μsg + m₂μsg = (m₁ + m₂)μsg = 851.6 N
4) The kinetic friction is given by: F = μkm₁g
Using equation (i) and the kinetic friction:
a₁ = μkg = 6.1 m/s²
5) Using equation(ii) and the kinetic friction:
m₂a₂ = T - μkm₁g
a₂ = (T - μkm₁g)/m₂ = 12.1 m/s²
1) a = 2.13 m/s²
2) F = 51.12 N
3) T_max = 851.62 N
4)a₁ = 6.08 m/s²
5) a₂ = 12.11 m/s²
This is based on the concept of frictional motion with coefficients of friction.
We are given;
Mass of smaller top crate; m₁ = 24 kg
Mass of larger bottom crate; m₂ = 86 kg
Coefficient of static friction; μ_s = 0.79
Coefficient of kinetic friction; μ_k = 0.62
For the top crate, the sum of forces will be expressed as;
F = m₁a₁ - - - (eq 1)
For the bottom crate, the sum of forces will be expressed as;
T - F = m₂a₂ - - - (eq 2)
Where F is frictional force exerted by small crate on big crate and T is tension.
1) We are told that the rope is pulled with a tension T = 234 N. And that the top crate will not slide. Thus, the acceleration of the small crate will be gotten from a combination of equation 1 and 2 to get the formula;
T = m₂a + m₁a
Making a which is the acceleration the subject gives;
a = T/(m₁+m₂)
Plugging in the relevant values gives;
a = 234/(24 + 86)
a = 2.13 m/s²
2) From eq(1) above, we see that the formula for the frictional force the small crate exerts on the big one is;
F = m₁a₁
Thus, Plugging in the relevant values gives;
F = 24 × 2.12
F = 51.12 N
3) maximum tension that the lower crate can be pulled at before the upper crate begins to slide will occur at the maximum friction. The formula for mx friction is;
(F₂)_max = μ_s•m₂•g
Using the concept of eq(1) We can say that;
F₂ = m₂a
m₂a = μ_s•m₂•g
m₂ will cancel out to give;
a = μ_s • g
From first answer earlier, we saw that;
T = m₂a + m₁a
T = (m₂ + m₁)a
Plugging in the relevant values gives;
T_max = (m₁ + m₂)μ_s*g
T_max = (24 + 86) × 0.79 × 9.8
T_max = 851.62 N
4) To get the acceleration of the upper crate, we will make use of the formula for kinetic friction which is:
F = μ_k*m₁g
From earlier, we saw that; F = m₁a₁
Thus;
m₁a₁ = μ_k*m₁g
m₁ will cancel out to get;
a₁ = μ_k*g
a₁ = 0.62 × 9.8
a₁ = 6.08 m/s²
5) the acceleration of the lower crate as the upper crate slides will be gotten by putting μ_k*m₁*g for m₁a₁ in eq(2) to get;
m₂a₂ = T - (μ_k*m₁*g)
Making a₂ the subject gives us;
a₂ = (T - μ_k•m₁•g)/m₂
Plugging in the relevant values;
a₂ = (1187 - (0.62 × 24 × 9.8))/86
a₂ = 12.11 m/s²
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A .5 kg air puck moves to the right at 3 m/s, colliding with a 1.5kg air puck that is moving to the left at 1.5 m/s.
1. Determine the direction and speed of the pucks' motion if they stick to each other after the collision.
2. Suppose instead that the pucks do not stick together. If the 1.5kg puck moves to the right at .231 m/s after the collision, was the collision elastic? Provide evidence!
3. Determine the size and direction of the average force acting on the .5kg puck during the second collision, if the collision lasted for 25 ms.
Answer:
Explanation:
Total momentum of the system before the collision
.5 x 3 - 1.5 x 1.5 = -0.75 kg m/s towards the left
If v be the velocity of the stuck pucks
momentum after the collision = 2 v
Applying conservation of momentum
2 v = - .75
v = - .375 m /s
Let after the collision v be the velocity of .5 kg puck
total momentum after the collision
.5 v + 1.5 x .231 = .5v +.3465
Applying conservation of momentum law
.5 v +.3465 = - .75
v = - 2.193 m/s
2 ) To verify whether the collision is elastic or not , we verify whether the kinetic energy is conserved or not.
Kinetic energy before the collision
= 2.25 + 1.6875
=3.9375 J
kinetic energy after the collision
= .04 + 1.2 =1.24 J
So kinetic energy is not conserved . Hence collision is not elastic.
3 ) Change in the momentum of .5 kg
1.5 - (-1.0965 )
= 2.5965
Average force applied = change in momentum / time
= 2.5965 / 25 x 10⁻³
= 103.86 N
A room contains 75 kg of air at 100 kPa and 15°C. The room has a 250-W refrigerator (the refrigerator consumes 250 W of electricity when running), a 120-W TV, a 1.8-kW electric resistance heater, and a 50-W fan. During a cold winter day, it is observed that the refrigerator, the TV, the fan, and the electric resistance heater are running continuously but the air temperature in the room remains constant. The rate of heat loss from the room that day is _____ kJ/h.
Answer:
Rate of heat loss=7992 kJ/h
Explanation:
First, we can consider the room as a closed system, so we can use the first law of the thermodynamic:
[tex]Qnet,in-Wnet,out = Esystem[/tex]
The incoming net heat is the incoming heat less the leaving heat, the leaving net work is the leaving work less the incoming work and the system energy is only the change of the intern energy because we have a stationary system, so:
[tex](Qin-Qout)-(Wout-Win) =U2-U1[/tex]
From the problem we consider that there is no change of the temperature thus there is no change in the intern energy(U2=U1), moreover there is no incoming heat neither exit work, all the work is made to the system as a result the leaving heat is equal to the incoming work:
[tex](0-Qout)-(0-Win) =0[/tex]
[tex]-Qout+Win =0[/tex]
[tex]Qout=Win[/tex]
Substituting the known values we can get the rate of heat loss (exit heat):
[tex]Q_{out}=Q_{loss}=W_{ref}+W_{TV}+W_{res}+W_{fan}[/tex]
[tex]Q_{loss}=250 [W]+120 [W]+1800 [W]+50 [W][/tex]
[tex]Q_{loss}=2220 [W]=2220 [\frac{J}{s} ][/tex]
Converting to Joules/hours:
[tex]Q_{loss}=2220 [\frac{J}{s} ][\frac{3600 s}{1 h} ][/tex]
[tex]Q_{loss}=7992000 [\frac{J}{h} ][/tex]
Finally the rate of heat loss is:
[tex]Q_{loss}=7992 [\frac{kJ}{h} ][/tex]
Final answer:
The rate of heat loss from a room with several electrical appliances running and constant air temperature is 7992 kJ/h, calculated by summing the power of all appliances and converting to kJ/h.
Explanation:
To find the rate of heat loss from the room, we need to consider the energy input from all the electrical appliances and the fact that the air temperature in the room remains constant. The total power consumption of the appliances is the sum of the power of the refrigerator, TV, electric resistance heater, and fan, which is 250 W + 120 W + 1.8 kW + 50 W = 2.22 kW.
Since the temperature is constant, the rate of energy input equals the rate of heat loss. This input is the combined power consumption of the appliances and is entirely converted into heat (assuming 100% efficiency for simplicity).
To express this rate in kilojoules per hour (kJ/h), multiply the power in kilowatts by the number of hours and by 1000 to convert from kW to kJ since 1 kW = 1 kJ/s:
2.22 kW = 2.22 kJ/s
So, the rate of heat loss is:
2.22 kJ/s × 3600 s/h = 7992 kJ/h
A rock is thrown vertically upward from some height above the ground. It rises to some maximum height and falls back to the ground. What one of the following statements is true if air resistance is neglected? The acceleration of the rock is zero when it is at the highest point. The speed of the rock is negative while it falls toward the ground. As the rock rises, its acceleration vector points upward. At the highest point the velocity is zero, the acceleration is directed downward. The velocity and acceleration of the rock always point in the same direction.
Answer:
At the highest point the velocity is zero, the acceleration is directed downward.
Explanation:
This is a free-fall problem, in the case of something being thrown or dropped, the acceleration is equal to -gravity, so -9.80m/s^2. So, the acceleration is never 0 here.
I attached an image from my lecture today, I find it to be helpful. You can see that because of gravity the acceleration is pulled downwards.
At the highest point the velocity is 0, but it's changing direction and that's why there's still an acceleration there.
Final answer:
The correct statement is that c) at the highest point of its trajectory, a rock thrown vertically has a velocity of zero and an acceleration directed downward at -9.80 m/s². The acceleration due to gravity is constant throughout the motion, whether the rock is rising or falling.
Explanation:
Among the given statements regarding a rock thrown vertically upward and neglecting air resistance, the true statement is: c) At the highest point the velocity is zero, the acceleration is directed downward. This is because at the maximum height, the rock stops for an instant (velocity = 0 m/s), but the acceleration due to gravity (acceleration due to gravity) is still acting on it, which is approximately -9.80 m/s² towards the Earth. This acceleration remains constant whether the rock is moving up or down.
It's important to note that as the rock rises, its acceleration vector always points downward, opposite to the direction of its velocity. Once the rock reaches its maximum height and begins its descent, the velocity becomes negative, which indicates the rock is falling towards the ground, yet still the acceleration is -9.80 m/s² downwards. The velocity and acceleration are in the same direction during the fall, both pointing toward the Earth.
New York and Los Angeles are about 3000 mi apart, the time difference between these two cities is 3 h. Calculate the circumference of the Earth.
Answer:
circumference of earth = 24000 miles
Explanation:
given data
New York and Los Angeles distance = 3000 mi
time difference = 3 h
to find out
circumference of the Earth
solution
we find speed first that is
speed = [tex]\frac{distance }{time}[/tex]
speed = [tex]\frac{3000}{3}[/tex]
speed = 1000 mph
now we know speed of earth is
speed = [tex]\frac{circumference of earth}{time of rotation}[/tex]
so
circumference = speed × time of rotation
and we know earth take 24 hours to rotate at its axis
so
circumference of earth = 1000 × 24
circumference of earth = 24000 miles
Final answer:
To calculate the Earth's circumference, we establish that a 3-hour difference in time between New York and Los Angeles represents a 45° longitudinal difference. Since 3000 miles correspond to this 45°, we extend this to a full 360° rotation to find that the Earth's estimated circumference is about 24,000 miles.
Explanation:
To calculate the circumference of the Earth using the time difference between New York and Los Angeles, we use the relationship between time and longitude. Since the Earth rotates 360° in 24 hours, it rotates 15° per hour. Thus, a 3 hour time difference corresponds to a 45° difference in longitude (because 3 hours x 15° per hour = 45°).
If New York and Los Angeles are approximately 3000 miles apart, and this represents 45° of the Earth's rotation, we can find the total circumference of the Earth by setting up the proportion:
3000 miles / 45° = Circumference / 360°
Circumference = (3000 miles / 45°) x 360°
Circumference = 24,000 miles
Therefore, the estimated circumference of the Earth is approximately 24,000 miles.
A boat sounds a fog horn on a day when both the sea water and the air temperature are 25.0° C. The speed of sound in sea water is 1,533 m/s. How much earlier (in s) does a dolphin 1000 m from the source hear the sound than a person in a boat that is also 1000 m distant? (Ignore the time it takes the sound to reach the water surface.) Please show work/ explanations
(A) 0.652
(B) 2.12
(C) 2.24
(D) 2.77
(E) 2.90
The sound would reach the dolphin in the water about 2.12 seconds earlier than the person in the boat, using given speeds of sound in air and water, and the distance provided.
Explanation:The time it takes for sound to travel is calculated using the formula Time = Distance/Speed. Given, distance is 1000m in both air and sea water. We need to find the speed of sound in the air to calculate how long it will take the sound to reach the person in the boat.
The speed of sound in the air at 25°C is approximately 346.13 m/s. Using the formula: Time = Distance/Speed, the time it takes for the sound to reach the person in the boat is about 2.889 seconds (1000m/346.13 m/s).
The speed of sound in seawater is given as 1,533 m/s. Using the same formula, the time it takes for the sound to reach the dolphin in the water is about 0.652 seconds (1000m/1,533 m/s).
Therefore, the difference in time it takes for the sound to travel to the person and the dolphin is approximately 2.237 seconds (2.889 seconds - 0.652 seconds). Hence, the correct answer would be (B) 2.12. Although, it's close to the calculated difference, none of the available options are an exact match.
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g List the following types of electromagnetic radiation in order of increasing frequency. radio waves, microwaves, visible light, gamma rays radio waves, visible light, microwaves, gamma rays visible light, radio waves, microwaves, gamma rays visible light, radio waves, gamma rays, microwaves Request Answer Part B List the following types of electromagnetic radiation in order of decreasing energy per photon. List the following types of electromagnetic radiation in order of decreasing energy per photon. microwaves, visible light, gamma rays, radio waves microwaves, gamma rays, visible light, radio waves gamma rays, visible light, microwaves, radio waves gamma rays, microwaves, visible light, radio waves
Answer:
a) radio < microwaves < visible < gamma
Explanation:
Electromagnetic waves are transverse waves produced by oscillations of the electric and magnetic fields, so light speeds and must comply with the relationship of the wave speed
v = λ f
Name Wavelength Frequency
Radio waves Km 10⁵
Microwaves cm 10¹⁰
Visible light μm 10¹⁴
Gamma rays pm 10¹⁹
All waves also meet the Planck energy equation
E = h f
h is the constant of Planck 6.6 10 -34 J s (4.136 10-15 eV-s), the unit of electron volts is very like for the waves of greater energy we substitute the frequencies and we calculate
Name Frequency Energy( J) Energy (eV)
Gamma ray 10¹⁹ 10⁻¹⁵ 10⁺⁴
Visible 10¹⁴ 10⁻²⁰ 10⁻¹
Microwave 10¹⁰ 10⁻¹⁴ 10⁻⁵
Radio waves 10⁵ 10⁻²⁹ 10⁻¹⁰
You can see that frequency and energy are proportional
Answer:
In increasing order of frequency:
radio waves < microwaves< visible light < gamma rays
In increasing order of energy per photon:
radio waves < microwaves< visible light < gamma rays
Explanation:
Electromagnetic waves can transfer energy in the space as well. Electromagnetic spectrum is a range of radiation having different frequency. The radiation which has more frequency has more energy but smaller wavelength. Gamma rays have maximum frequency where as radio waves have smaller frequency and radiation.
The ink drops have a mass m = 1.00×10^−11 kg each and leave the nozzle and travel horizontally toward the paper at velocity v = 23.0 m/s . The drops pass through a charging unit that gives each drop a positive charge q by causing it to lose some electrons. The drops then pass between parallel deflecting plates of length D0 = 1.80 cm , where there is a uniform vertical electric field with magnitude E = 8.20×10^4 N/C If a drop is to be deflected a distance d = 0.290 mm by the time it reaches the end of the deflection plate, what magnitude of charge q must be given to the drop? Assume that the density of the ink drop is 1000 kg/m^3
The magnitude of the charge q that must be given to the ink drop is [tex]7.5 \times 10^{-14}[/tex] coulombs.
The vertical displacement of the ink drop as it passes between the deflecting plates is due to the electric force acting on the charged drop. The force of gravity on the drop is negligible compared to the electric force.
Given:
Mass of the ink drop [tex](m) = 1.00 \times 10^{-11}\ kg[/tex]
Vertical displacement[tex](d) = 0.290\ mm[/tex]
Horizontal velocity [tex](v) = 23.0\ m/s[/tex]
Electric field magnitude [tex](E) = 8.20 \times 10^4\ N/C[/tex]
Plate length [tex](D_o) = 1.80\ cm[/tex]
The electric force on a charged particle in an electric field is given by Coulomb's law:
[tex]F_{electric} = q \times E[/tex]
Where:
[tex]F_{electric}[/tex] is the electric force,
q is the charge of the particle,
E is the magnitude of the electric field.
The vertical displacement (d) of the drop is caused by this electric force and is given by the equation:
[tex]d = (1/2) \times a \times t^2[/tex]
Since the drop is initially moving horizontally and is not influenced by gravity, the vertical acceleration (a) is due solely to the electric force
[tex]F_{electric} = m \times a[/tex]
Equate the two expressions for the electric force:
[tex]q \times E = m \times a\\a = (q \times E) / m\\d = (1/2) \times ((q \times E) / m) \times t^2[/tex]
The time it takes to travel through the plates using the horizontal distance [tex](D_o)[/tex] and the horizontal velocity:
[tex]t = D_o/ v[/tex]
[tex]d = (1/2) \times ((q \times E) / m) \times (D_o / v)^2\\q = (2 \times m \times d \times v^2) / (E \times D_o^2)\\q = (2 \times (1.00 \times 10^{-11}) \times (0.290 \times 10^{-3} ) \times (23.0 )^2) / ((8.20 \times 10^4) \times (1.80 \times 10^{-2})^2)\\q = 7.5 \times 10^{-14}[/tex]
Therefore, the magnitude of the charge q that must be given to the ink drop is [tex]7.5 \times 10^{-14}[/tex] coulombs.
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To find the charge necessary for an ink drop to be deflected a certain distance in an electric field, apply the equation q = Fe/E, where Fe is derived from the kinematic equations for the ink drop's motion in the field.
Explanation:To calculate the magnitude of charge q that must be given to an ink drop such that it is deflected by a distance d = 0.290 mm by the time it reaches the end of the deflection plate of length D0 = 1.80 cm in a uniform electric field with magnitude E = 8.20×104 N/C, we can use the concepts of electric force and motion.
Firstly, the electric force Fe acting on the ink drop is equal to qE. This force will cause the ink drop to accelerate. Since the drops are moving horizontally at velocity v = 23.0 m/s and they need to be deflected by a certain distance d while travelling through the electric field, we can use the kinematic equations of motion to find the vertical displacement due to acceleration.
The time t it takes for the ink drop to pass through the plates is t = D0/v. From this, the vertical displacement d can be given as d = 0.5 * a * t^2 where a is the acceleration due to the electric force. Knowing d, t, and E, we can find a and subsequently use Fe = mdrop × a to solve for the charge q as q = Fe/E.
Using the given mass m = 1.00×10−11 kg for the ink drop, we can rearrange and solve the appropriate equations to find the value of the charge q necessary to cause the required deflection d.
You mount a ball launcher to a car. It points toward the back of the car 39 degrees above the horizontal and launches balloons at 9.7 m/s. When you drive the car straight ahead at 2.2 m/s. Assume the ball is launched from ground level. For the ball, what is the separation between the ball and the car in m when the ball hits the ground?
Answer:
9.4 m
Explanation:
We can use a moving frame of reference with the same speed as the car. From this frame of reference the car doesn't move. The origin is at the back of the car, the positive X axis points back and the positive Y axis points up.
If the ballon is launched at 9.7 m/s at 39 degrees of elevation.
Vx0 = 9.7 * cos(39) = 7.5 m/s
Vy0 = 9.7 * sin(39) = 6.1 m/s
If we ignore air drag, the baloon will be subject only to the acceleration of gravity. We can use the equation of position under constant acceleration.
Y(t) = Y0 + Vy0 * t + 1/2 * a * t^2
Y0 = 0
a = -9.81 m/s^2
It will fall when Y(t) = 0
0 = 6.1 * t - 4.9 * t^2
0 = t * (6.1 - 4.9 * t)
t1 = 0 (this is when the balloon was launched)
0 = 6.1 - 4.9 * t2
4.9 * t2 = 6.1
t2 = 6.1 / 4.9 = 1.25 s
The distance from the car will be the horizonta distance it travelled in that time
X(t) = X0 + Vx0 * t
X(1.25) = 7.5 * 1.25 = 9.4 m
A charge of +3.00 μC is located at the origin, and a second charge of -2.00 μC is located on the x-y plane at the point (50.0 cm, 60.0 cm). 1) Determine the x-component of the electric force exerted by the -2.00 μC charge on the +3.00 μC charge. (Express your answer to three significant figures.)
Answer:
0.0567 N
Explanation:
q1 = 3 micro coulomb
q2 = - 2 micro coulomb
OB = 50 cm
AB = 60 cm
By using Pythagoras theorem in triangle OAB
[tex]OA^{2}=OB^{2}+AB^{2}[/tex]
[tex]OA^{2}=50^{2}+60^{2}=6100[/tex]
OA = 78.1 cm
By using the Coulomb's law, the force on 3 micro coulomb due to -2 micro coulomb is
[tex]F=\frac {Kq_{1}q_{2}}{OA^{2}}= \frac{9 \times 10^{9}}\times 3\times 10^{-6} \times 2 \times 10^{-6}{0.781^{2}}[/tex]
F = 0.0885 N
The horizontal component of force is
= F CosФ = [tex]F\times \frac{OB}{OA}[/tex]
= 0.0885 x 50 / 78.1 = 0.0567 N
Coal is lifted out of a mine a vertical distance of 50 m by an engine that supplies 500 W to a conveyer belt. How much coal per minute can be brought to the surface? Ignore the effects of friction.
Answer:
mass = 61.16 kg
Explanation:
given data
distance = 50 m
energy = 500 W = 500 × 60sec = 30000 J/min
to find out
mass
solution
we will apply here energy equation
that is
energy = m×g×h .............1
put here all value m mass here and g = 9.81 and h is distance
energy = m×g×h
30000 = m×9.81×50
mass = 61.16 kg
Answer:
61 kg
Explanation:
Given data
distance (d): 50 m
power (P): 500 W = 500 J/s
mass (m): ?
time (t): 1 min = 60 s
The engine must do some work (w) to lift the coal. The work done in 1 minute is:
P = w / t
w = P . t = 500 J/s × 60 s = 3.0 × 10⁴ J
The work is equal to the force exerted (F) times the distance (d).
w = F × d
F = w / d = 3.0 × 10⁴ J / 50 m = 6.0 × 10² N
The force exerted is equal to the mass lifted (m) times the acceleration. Here, the acceleration to surpass is that of gravity (g = 9.8 m/s²).
F = m × g
m = F / g = 6.0 × 10² N / 9.8 m/s² = 61 kg
If the temperature of a volume of ideal gas increases for 100°C to 200°C, what happens to the average kinetic energy of the molecules?
Answer:[tex]\Delta E=2.0715\times 10^{-21} J[/tex]
Explanation:
Given
Temperature of the gas is increased from 100 to 200
Also we know that average kinetic energy of the molecules is
[tex]E=\frac{3}{2}\cdot \frac{R}{N_A}T[/tex]
Where
R=Gas constant
[tex]N_A[/tex]=Avogadro's number
T=Temperature in kelvin
[tex]\frac{R}{N_A}=1.381\times 10^{-23}[/tex]
So kinetic energy increases by
[tex]\Delta E=\frac{3}{2}\times 1.381\times 10^{-23}\left ( 200-100\right )[/tex]
[tex]\Delta E=2.0715\times 10^{-21} J[/tex]
A truck moves 70 m east, then moves 120 m west, andfinally
moves east again a distance of 90 m. If east ischosen as the
positive direction, what is the truck's
resultantdisplacment?
Answer:
40m (40m east)
Explanation:
The fist move is 70m east, and east is the positive direction so the truck initially is at +70m.
The second move is 120m to the west, since east is the positive direction, west must be the negative direction, this means the truck now is at:
[tex]+70m -120m = -50m[/tex]
The third move is 90m to the east, again, this is the positive direction, so the new position is:
[tex]-50m + 90m = +40m.[/tex]
The truck is at + 40m, so it ended up 40m away from its initial position and this is the resultant dispacement.
The 5-mm-thick bottom of a 200-mm-diameter pan may be made from aluminum (k = 240 W/m K) or copper (k = 390 W/m K). When used to boil water, the surface of the bottom exposed to the water is nominally at 110 °C. If heat is transferred from the stove to the pan at a rate of 600 W, what is the temperature of the surface in contact with the stove for each of the two materials?
Answer:
Explanation:
Conduction of heat per second Q through a metal sheet having thermal conductivity K , thickness d , area A and temperature difference between cold and hot surface ( T₂ -T₁ ) is given by the following equation
Q = [tex]\frac{KA(T_2-T_1)}{d}[/tex]
Area A = π R² = 3.14 X 10000 X 10⁻⁶
= 3.14 X 10⁻2 m²
For aluminium plate :-
T₂ - T₁ = [tex]\frac{Q\times d}{KA}[/tex]
= [tex]\frac{600\times5\times10^{-3}}{240\times3.14\times10^{-2}}[/tex]
T₂ - T₁ = .4
T₂ = 110.4 ° C
For copper plate :
[tex][tex]\frac{600\times5\times10^{-3}}{390\times3.14\times10^{-2}}[/tex][/tex]
T₂ - T₁ = .25
T₂ = 110.25 ° C
The temperature of the surface in contact with the stove is approximately 113.0 °C for aluminum and 111.3 °C for copper, considering the thermal conductivity and dimensions of the pan.
Explanation:This is a heat transfer problem involving conduction through different materials. The heat conduction through a flat wall, like the bottom of a pan, can be calculated using Fourier's law of heat conduction: q = -kA(T₂ - T₁)/d, where q is the heat transfer rate (600 W in this case), 'k' is the thermal conductivity, 'A' is the area across which heat is transferring, 'T₁' and 'T₂' represent the temperatures on either side of the wall, and 'd' represents the wall thickness.
First, convert the diameter of the pan to the radius in meters, which is (200mm/2)/1000 = 0.1 m. The area, A, is then π*(0.1 m)² = 0.0314 m². The thickness d given is already in meters: 5 mm = 0.005 m.
For aluminum, rearrange Fourier's law to find T₂, the temperature on the stove side of the pan: T₂ = T₁ + qd/(kA), where T₁ = 110°C and k for aluminum = 240 W/m K. Plugging in all values we get: T₂ = 110 + (600*0.005)/(240*0.0314) ≈ 113.0 °C.
For copper, using k for copper = 390 W/m K in the above formula will give: T₂ = 110 + (600*0.005)/(390*0.0314) ≈ 111.3 °C.
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A proton and an electron are released from rest, with only the electrostatic force acting. Which of the following statements must be true about them as they move toward each other? (There could be more than one correct choice.) Check all that apply. Check all that apply. Their electric potential energy keeps increasing. Their kinetic energy keeps increasing. Their acceleration keeps decreasing. Their electric potential energy keeps decreasing. Their kinetic energy keeps decreasing.
Answer:
Their kinetic energy will increase but potential energy will decrease.
Explanation:
Given that
Initial velocities of electron and proton is zero.
We know that ,electron have negative charge and proton have positive charge it means that they will attract to each other.We know that opposite charge attract to each other and same charge repels to each other.
It means that the velocities of proton and electron will increase and that leads to increase in the kinetic energy of proton and electron.
We know that potential energy U
[tex]U\alpha -\dfrac{1}{r}[/tex]
So when r will decrease then U will increase but in negative direction it means that U will decrease.
So we can say that their kinetic energy will increase but potential energy will decrease.
There is a person who throws a coin vertically downward with an initial speed of 11.8 m/s from the roof of a building, 34.0 m above the ground. How long does it take the coin to reach the ground? Answer in s.
Answer:
Time taken by the coin to reach the ground is 1.69 s
Given:
Initial speed, v = 11.8 m/s
Height of the building, h = 34.0 m
Solution:
Now, from the third eqn of motion:
[tex]v'^{2} = v^{2} + 2gh[/tex]
[tex]v'^{2} = 11.8^{2} + 2\times 9.8\times 34.0 = 805.64[/tex]
[tex]v' = \sqrt{805.64} = 28.38 m/s[/tex]
Now, time taken by the coin to reach the ground is given by eqn (1):
v' = v + gt
[tex]t = \frac{v' - v}{g} = \frac{28.38 - 11.8}{9.8} = 1.69 s[/tex]
An object moving on a horizontal, frictionless surface makes a glancing collision with another object initially at rest on the surface. In this case which of the following is true about momentum and kinetic energy? 4A) Momentum is always conserved, and kinetic energy may be conserved. (B) Kinetic energy is always conserved, and momentum may be conserved. (C) Momentum is always conserved, and kinetic energy is never conserved. (D) Both momentum and kinetic energy are always conserved. (E) Neither momentum nor kinetic energy is conserved.
Answer:
The correct answer is option 'a' 'The momentum is always conserved while as the kinetic energy may be conserved'
Explanation:
The conservation of momentum is a basic principle in nature which is always valid in an collision between 'n' number of objects if there are no external forces on the system. It is valid for both the cases weather the collision is head on or glancing or weather the object is elastic or inelastic.
The energy is only conserved in a collision that occurs on a friction less surface and the objects are purely elastic. Since in the given question it is mentioned that only the surface is friction less and no information is provided regarding the nature of the objects weather they are elastic or not hence we cannot conclusively come to any conclusion regarding the conservation of kinetic energy as the objects may be inelastic.
A tortoise and a hare engage in a race. A tortoise can run with a speed of 0.15 m/s. A hare can run 25 times as fast as the tortoise. In the race, they start at the same time; however, the hare being very proud of his much higher speed stops for a nap for 5.0 minutes. The tortoise wins the race by 35 cm. (a) What is the length of the race? (b) How long does the race take?
Answer:
Explanation:
The speed of hare = .15 x 25 = 3.75 m /s . Let tortoise took t second to complete the race .
Distance traveled by it = .15 t
Distance traveled by hare = .15 t - .35 m
Time taken by hare to complete this distance
= t - 5 x 60 s
Speed of hare
= Distance / time
( .15t-.35 ) / t - 300 , so
[tex]\frac{.15t-.35}{t-300} = 3.75[/tex]
t = 312.40
= 5 minutes 12.4 seconds
Distance of race
312.4 x speed of tortoise
= 312.4 x .15
= 46.85 m
(a) The length of the race is 91.1 meters.
(b) The race takes 605.0 seconds.
The tortoise moves at a speed of 0.15 m/s while the hare moves at a speed of 25 * 0.15 = 3.75 m/s.
The hare stops for 5 minutes, which is equal to 5 * 60 = 300 seconds.
In the 300 seconds that the hare was taking a nap, the tortoise was able to move a distance of 0.15 * 300 = 45 meters.
When the hare woke up, he started running at his top speed. He was able to catch up to the tortoise and surpass him by 35 cm. This means that the hare was able to run a distance of 45 + 0.35 = 45.35 meters.
Therefore, the length of the race is 45.35 meters.
The total time it took for the race to finish is 300 + 605 = 905 seconds.
So the answer is (a) 91.1 meters (b) 605.0 seconds
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According to Archimedes’ principle, the mass of a floating object equals the mass of the fluid displaced by the object. Use this principle to solve the following problems. (a) A wooden cylinder 30.0 cm high floats vertically in a tub of water (density 1:00 g/cm3). The top of the cylinder is 13.5 cm above the surface of the liquid. What is the density of the wood? (b) The same cylinder floats vertically in a liquid of unknown density. The top of the cylinder is 18.9 cm above the surface of the liquid. What is the liquid density? (c) Explain why knowing the length and width of the wooden objects is unnecessary in solving Parts (a) and (b). 2.15.
Answers:
a) [tex]\rho_{cylinder}= 0.55 g/cm^{3}[/tex]
b) [tex]\rho_{liq}= 1.48 g/cm^{3}[/tex]
c) When we divided both volumes (sumerged and displaced) the factor [tex]\pi r^{2}[/tex] is removed during calculations.
Explanation:
a) According to Archimedes’ Principle:
A body totally or partially immersed in a fluid at rest, experiences a vertical upward thrust equal to the mass weight of the body volume that is displaced.
In this case we have a wooden cylinder floating (partially immersed) in water. This object does not completely fall to the bottom because the net force acting on it is zero, this means it is in equilibrium. This is due to Newton’s first law of motion, that estates if a body is in equilibrium the sum of all the forces acting on it is equal to zero.
Hence:
[tex]W_(cylinder)=B[/tex] (1)
Where:
[tex]W_(cylinder)=m.g[/tex] is the weight of the wooden cylinder, where [tex]m[/tex] is its mass and [tex]g[/tex] gravity.
[tex]B[/tex] is the Buoyant force, which is the force the fluid (water in this situation) exert in the submerged cylinder, and is directed upwards.
We can rewrite (1) as follows:
[tex]m_{cylinder}g=m_{water}g[/tex] (2)
On the other hand, we know density [tex]\rho[/tex] establishes a relationship between the mass of a body andthe volume it occupies. Mathematically is expressed as:
[tex]\rho=\frac{m}{V}[/tex] (3)
isolating the mass:
[tex]m=\rho V[/tex] (4)
Now we can express (2) in terms of the density and the volume of cylinder and water:
[tex]\rho_{cylinder} V_{cylinder} g=\rho_{water} V_{water} g[/tex] (5)
In this case [tex]V_{water}[/tex] is the volume of water displaced by the wooden cylinder (remembering Archimedes's Principle).
At this point we have to establish the total volume of the cylinder and the volume of water displaced by the sumerged part:
[tex]V_{cylinder}=\pi r^{2} h[/tex] (6)
Where [tex]r[/tex] is the radius and [tex]h=30 cm[/tex] the total height of the cylinder.
[tex]V_{water}=\pi r^{2} (h-h_{top})[/tex] (7)
Where [tex]h_{top}=13.5 cm[/tex] is the height of the top of the cylinder above the surface of water and [tex](h-h_{top})[/tex] is the height of the sumerged part of the cylinder.
Substituting (6) and (7) in (5):
[tex]\rho_{cylinder} \pi r^{2} h g=\rho_{water} \pi r^{2} (h-h_{top}) g[/tex] (8)
Clearing [tex]\rho_{cylinder}[/tex]:
[tex]\rho_{cylinder}=\frac{\rho_{water}(h-h_{top})}{h}[/tex] (9)
Simplifying;
[tex]\rho_{cylinder}=\rho_{water}(1-\frac{h_{top}}{h}[/tex] (10)
Knowing [tex]\rho_{water}=1g/cm^{3}[/tex]:
[tex]\rho_{cylinder}=1g/cm^{3}(1-\frac{13.5 cm}{30cm})[/tex] (11)
[tex]\rho_{cylinder}= 0.55 g/cm^{3}[/tex] (12) This is the density of the wooden cylinder
b) Now we have a different situation, we have the same wooden cylinder, which density was already calculated ([tex]\rho_{cylinder}= 0.55 g/cm^{3}[/tex]), but the density of the liquid [tex]\rho_{liq}[/tex] is unknown.
Applying again the Archimedes principle:
[tex]\rho_{cylinder} V_{cylinder} g = \rho_{liq} V_{liq} g[/tex] (13)
Isolating [tex]\rho_{liq}[/tex]:
[tex]\rho_{liq}= \frac{\rho_{cylinder} V_{cylinder}}{V_{liq}}[/tex] (14)
Where:
[tex]V_{cylinder}=\pi r^{2} h[/tex]
[tex]V_{liq}=\pi r^{2} (h-h_{top})[/tex]
Then:
[tex]\rho_{liq}= \frac{\rho_{cylinder} \pi r^{2} h}{\pi r^{2} (h-h_{top})}[/tex] (15)
[tex]\rho_{liq}= \frac{\rho_{cylinder} h}{h-h_{top}}[/tex] (16)
[tex]\rho_{liq}= \frac{0.55 g/cm^{3} (30 cm)} {30 cm - 18.9 cm}[/tex] (17)
[tex]\rho_{liq}= 1.48 g/cm^{3}[/tex] (18) This is the density of the liquid
c) As we can see, it was not necessary to know the radius of the cylinder (we did not need to knoe its length and width), we only needed to know the part that was sumerged and the part that was above the surface of the liquid.
This is because in this case, when we divided both volumes (sumerged and displaced) the factor [tex]\pi r^{2}[/tex] is removed during calculations.
A small spherical insulator of mass 6.68 x 10^-2 kg and charge +0.600 C is hung by a thin wire of negligible mass. A charge of -0.900 C is held 0.150 m away from the sphere and directly to the right of it, so the wire makes an angle with the vertical (see the drawing). Find (a) the angle and (b) the tension in the wire.
Final answer:
To find the angle and tension in the wire, we can use the concept of electrostatic force and Coulomb's Law.
Explanation:
To find the angle and tension in the wire, we can use the concept of electrostatic force. The electrostatic force between the two charges can be calculated using Coulomb's Law:
FE = k * (q1 * q2) / r^2
where FE is the electrostatic force, k is the electrostatic constant (8.99 x 10^9 N·m^2/C^2), q1 and q2 are the charges, and r is the distance between the charges.
By equating this force to the tension in the wire and solving for the angle, we can find the angle:
tan(angle) = FE / T
where T is the tension in the wire.
A glittering glass globe is given a net electric charge of 5.31 × 10^-6 C. Does the globe now have more or fewer electrons than it does in its neutral state? How many more or fewer?
Final answer:
The glass globe with a net electric charge of 5.31 × 10^-6 C has 3.32 × 10^13 fewer electrons than it did in its neutral state, because positive net charge indicates electron removal.
Explanation:
When a glittering glass globe is given a net electric charge of 5.31 × 10^-6 C, it means that electrons have been added or removed to achieve this net charge. The charge on an electron is known to be approximately -1.6 × 10^-19 C. To determine if the globe has more or fewer electrons than in its neutral state, we compare the sign of the given charge. Here, the positive charge indicates that electrons have been removed. To calculate the quantity of electrons removed, we use the formula:
Number of electrons (n) = Net charge (Q) / Elementary charge (e).
Plugging in the values:
n = 5.31 × 10^-6 C / 1.6 × 10^-19 C/e-
n = 3.32 × 10^13 electrons.
Therefore, the globe now has 3.32 × 10^13 fewer electrons than it did in its neutral state.
A gray kangaroo can bound across level ground with each jump carrying it 9.1 m from the takeoff point. Typically the kangaroo leaves the ground at a 26° angle. If this is so: 1) What is its takeoff speed
2) What is its maximum height above the ground?
Answer:
u = 10.63 m/s
h = 1.10 m
Explanation:
For Take-off speed ..
by using the standard range equation we have
[tex]R = u² sin2θ/g[/tex]
R = 9.1 m
θ = 26º,
Initial velocity = u
solving for u
[tex]u² = \frac{Rg}{sin2\theta}[/tex]
[tex]u^2 = \frac{9.1 x 9.80}{sin26}[/tex]
[tex] u^2 = 113.17[/tex]
u = 10.63 m/s
for Max height
using the standard h(max) equation ..
[tex]v^2 = (v_osin\theta)^2 -2gh[/tex]
[tex]h =\frac{(v_o^2sin\theta)^2}{2g}[/tex]
[tex]h = \frac{(113.17)(sin26)^2}{(2 x 9.80)}}[/tex]
h = 1.10 m
We have that for the takeoff speed and maximum height above ground is
u=10.69m/sHmax=1.072mFrom the question we are told
A gray kangaroo can bound across level ground with each jump carrying it 9.1 m from the takeoff point. Typically the kangaroo leaves the ground at a 26° angle. If this is so:
1) What is its takeoff speed
2) What is its maximum height above the ground?
takeoff speed& maximum heightGenerally the equation for the Projectile motion is mathematically given as
[tex]R=\frac{u^2sin 2\theta}{g}\\\\Therefore\\\\9.1=\frac{u^2sin 52}{9.8}[/tex]
u=10.69m/s
And
Generally the equation for the Projectile Height is mathematically given as
[tex]Hmax=\frac{u^2sin^2 2\theta}{2g}\\\\Therefore\\Hmax=\frac{114.27*0.184}{19.6}[/tex]
Hmax=1.072m
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When an inductor is connected to a 60.0 Hz source it has an inductive reactance of 59.2 Ω. Determine the maximum current in the inductor (in A) if it is connected to a 45.0 Hz source that produces a 120 V rms voltage.
Answer:
Maximum current in the inductor will be 3.824 A
Explanation:
In first case inductive reactance = 59.2 ohm
Frequency = 60 Hz
We know that inductive reactance is given by [tex]X_L=\omega L[/tex]
[tex]59.2=2\pi f\times L[/tex]
[tex]59.2=2\times 3.14\times 60\times L[/tex]
[tex]L=0.157H[/tex]
In second case frequency f = 45 Hz
Now inductive reactance [tex]X_L=\omega L =2\times 3.14\times 45\times .157=44.368ohm[/tex]
Now current [tex]i=\frac{V}{X_L}=\frac{120}{44.368}=2.70A[/tex]
Maximum current [tex]i_{max}=\sqrt{2}i=1.414\times 2.70=3.824A[/tex]
To find the maximum current when the inductor is connected to a 45Hz source, one needs to first calculate the inductance of the inductor using the provided inductive reactance at 60Hz. Then, with the inductance determined, the inductive reactance at 45Hz can be computed. Finally, Arnold's law (I = V/X₁) is used to determine the current.
Explanation:In this problem, the concept of inductive reactance and Ohm's law are involved. First, we know that the inductive reactance (X₁) is directly proportional to the frequency, and it is given by the formula X₁ = 2πfL. From the information provided, we can find the inductance (L) of the inductor by using this formula at 60Hz, then looking at how X₁ changes when we use a 45Hz source. Once we determine L, we can find the new inductive reactance at 45Hz.
Next, we use Ohm's Law (I = V/X₁) to find the current at 45Hz, where V is the rms voltage across the inductor and X₁ is the inductive reactance. This would give us the maximum current in the inductor.
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A runner ran V1 km/hr for the first half of the race. The runner then ran V2 km/h for the second half of the race. What was the average speed of this runner? It is okay for the answer to be in variables.
Answer:
[tex]v=\frac{v_{1}v_{2}}{\left ( v_{1}+v_{2} \right )}[/tex]
Explanation:
Let the distance traveled in first half is d and then in the next half is also d.
Let the time taken in first half is t1 and the time taken in the second half is t2.
[tex]t_{1}=\frac{d}{v_{1}}[/tex]
[tex]t_{2}=\frac{d}{v_{2}}[/tex]
Total time taken
[tex]t=t_{1}+t_{2}=\frac{d}{v_{1}}+\frac{d}{v_{2}}=\frac{d\left ( v_{1}+v_{2} \right )}{v_{1}v_{2}}[/tex]
The average speed of a body is defined as the total distance traveled by teh body to the total time taken.
[tex]Average speed = \frac{total distance}{total time}[/tex]
[tex]v=\frac{2d}{t}[/tex]
[tex]v=\frac{2d}{\frac{2d\left ( v_{1}+v_{2} \right )}{v_{1}v_{2}}}[/tex]
[tex]v=\frac{v_{1}v_{2}}{\left ( v_{1}+v_{2} \right )}[/tex]
The average speed of a runner who ran at different speeds for two equal halves of a race is calculated as 2V1V2 / (V1+V2), where V1 and V2 are the speeds for the first and second halves respectively.
Explanation:The question pertains to calculating average speed in a two-part race which essentially is a concept in Mathematics. We'll start by understanding the definition of average speed. It is calculated as the total distance travelled divided by the total time taken.
In this case, as the race is divided into two halves, the distance covered in the first half is equal to the distance covered in the second half. Hence, the total distance covered by the runner is 2D (where D is the distance of one half of the race).
However, the time to cover each half of the race is different. The time taken to cover the first and second halves would be D/V1 and D/V2 respectively. Therefore, the total time taken for the whole race is D/V1+D/V2.
Now we can calculate the average speed by dividing the total distance by the total time, which gives: Average Speed = 2D / (D/V1+D/V2). This can be simplified to: Average Speed = 2V1V2 / (V1+V2).
Learn more about Average Speed here:https://brainly.com/question/12322912
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What is the difference between average speed and instantaneous speed? What’s the difference between speed and velocity?
Explanation:
The average speed is the displacement of a particle divided into a time in which the displacement occurs. Instead, instantaneous speed is the limit of the average speed as the time interval approaches zero.
In physics expressions
[tex]Average speed=\frac{x_{2}-x_{1} }{t_{2}-t_{1} }[/tex]
Instantaneous speed=[tex]lim_{t \to \infty} \frac{x}{t} =\frac{dx}{dt}[/tex]
b) Speed denotes distance traveled divided by time, and instantaneous speed measures how quickly and in which direction does it move
The difference between average speed and instantaneous speed is that average speed is the total distance traveled divided by the total time taken, whereas instantaneous speed is the speed of an object at a particular moment in time. Speed is a scalar quantity which means it only has magnitude and no direction, while velocity is a vector quantity which includes both magnitude and direction.
Average speed is calculated over a stretch of time and it does not consider the direction of the motion. For instance, if you drove for 100 kilometers and it took you 2 hours, your average speed would be 50 kilometers per hour. On the other hand, instantaneous speed is the speed of an object at a specific instant. It can be seen as the speedometer reading in a vehicle at a particular moment. To illustrate, while driving, your speedometer might read 60 km/h at one instant, which would be your instantaneous speed at that time.
Moreover, velocity includes direction as part of its definition and it describes the rate of change of an object's position. Average velocity takes into account the total displacement (change in position) and the total time. If the same 100 kilometers were travelled in a direct line to the north and it took 2 hours, then the average velocity would be 50 km/h north. If, however, the trip involved a return to the starting point (making the total displacement zero), the average velocity would also be zero, despite having a non-zero average speed. Therefore, while they might have the same magnitude under certain circumstances, speed and velocity are not interchangeable terms due to velocity's directional component.
Understanding the distinction between speed and velocity is crucial in physics because it determines how motion is described and analyzed.