Answer:
Converging (convex) lens.
Explanation:
A lens can be defined as a transparent optical instrument that refracts rays of light to produce a real image.
Basically, there are two (2) main types of lens and these includes;
I. Diverging (concave) lens.
II. Converging (convex) lens.
A converging (convex) lens refers to a type of lens that typically causes parallel rays of light with respect to its principal axis to come to a focus (converge) and form a real image. Thus, this type of lens is usually thin at the lower and upper edges and thick across the middle.
Basically, the image of the object formed by a converging (convex) lens. lens is real, enlarged and inverted.
The earth's radius is about 4000 miles. Kampala, the capital of Uganda, and Singapore are both nearly on the equator. The distance between them is 5000 miles as measured along the earth's surface.
a. Through what angle do you turn, relative to the earth, if you fly from Kampala to Singapore? Give your answer in both radians and degrees.
b. The flight from Kampala to Singapore take 9 hours. What is the plane's angular speed relative to the earth?
Answer:
a) the required angle in both radian and degree is 1.25 rad and 71.6°
b) the plane's angular speed relative to the earth is 3.86 × 10⁻⁵ rad/sec
Explanation:
Given the data in the question;
a)
we know that The expression for the angle subtended by an arc of circle at the center of the circle is,
θ = Length / radius
given that Length is 5000 miles and radius is 4000 miles
we substitute
θ = 5000 miles / 4000 miles
θ = 1.25 rad
Radian to Degree
θ = 1.25 rad × ( 180° / π rad )
θ = 71.6°
Therefore, required angle in both radian and degree is 1.25 rad and 71.6°
b)
The flight from Kampala to Singapore take 9 hours.
the plane's angular speed relative to the earth = ?
we know that, the relation between angular velocity and angular displacement is;
ω = θ / t
given that θ is 1.25 rads and time t is 9 hours or ( 9 × 3600 sec ) = 32400 sec
we substitute
ω = 1.25 rad / 32400 sec
ω = 3.86 × 10⁻⁵ rad/sec
Therefore, the plane's angular speed relative to the earth is 3.86 × 10⁻⁵ rad/sec
A copper wire 1.0 meter long and with a mass of .0014 kilograms per meter vibrates in two segments when under a tension of 27 Newtons. What is the frequency of this mode of vibration
Answer:
the frequency of this mode of vibration is 138.87 Hz
Explanation:
Given;
length of the copper wire, L = 1 m
mass per unit length of the copper wire, μ = 0.0014 kg/m
tension on the wire, T = 27 N
number of segments, n = 2
The frequency of this mode of vibration is calculated as;
[tex]F_n = \frac{n}{2L} \sqrt{\frac{T}{\mu} } \\\\F_2 = \frac{2}{2\times 1} \sqrt{\frac{27}{0.0014} }\\\\F_2 = 138.87 \ Hz[/tex]
Therefore, the frequency of this mode of vibration is 138.87 Hz
A Catapult throws a payload in a circle with an arm that is 65.0 cm long. At a certain instant, the arm is rotating at 8.0 rad/s and the angular speed is increasing at 40.0 rad/s2. For this instant, find the magnitude of the acceleration of the payload.
Answer:
The acceleration of the payload is 26 m/s2.
Explanation:
length, L = 65 cm = 0.65 m
angular acceleration = 40 rad/s^2
The acceleration is given by
a = angular acceleration x length
a = 40 x 0.65
a = 26 m/s^2
Test your prediction through calculation for the situations of the clay bob and the bouncy ball. Assume each has a mass of 100 grams, and each has an initial velocity of 20 m/s straight at the door. Ignore the effects of gravity. Calculate the change in momentum of
Answer:
a) Δp = -2.0 kgm / s, b) Δp = -4 kg m / s
Explanation:
In this exercise the change in moment of a ball is asked in two different cases
a) clay ball, in this case the ball sticks to the door and we have an inelastic collision where the final velocity of the ball is zero
Δp = p_f - p₀
Δp = 0 - m v₀
Δp = - 0.100 20
Δp = -2.0 kgm / s
b) in this case we have a bouncing ball, this is an elastic collision, as the gate is fixed it can be considered an object of infinite mass, therefore the final speed of the ball has the same modulus of the initial velocity, but address would count
v_f = - v₀
Δp = p_f -p₀
Δp = m v_f - m v₀
Δp = m (v_f -v₀)
Δp = 0.100 (-20 - 20)
Δp = -4 kg m / s
A piston-cylinder device contains 5 kg of refrigerant-134a at 0.7 MPa and 60°C. The refrigerant is now cooled at constant pressure until it exists as a liquid at 24°C. If the surroundings are at 100 kPa and-24°C, determine:
(a) the exergy of the refrigerant at the initial and the final states and
(b) the exergy destroyed during this process.
Answer:
Yes sure, keep it going, and never give up because your dreams are so important
A) The exergy of the refrigerant at the initial and final states are :
Initial state = - 135.5285 kJ Final state = -51.96 kJB) The exergy destroyed during this process is : - 1048.4397 kJ
Given data :
Mass ( M ) = 5 kg
P1 = 0.7 Mpa = P2
T1 = 60°C = 333 k
To = 24°C = 297 k
P2 = 100 kPa
A) Determine the exergy at initial and final states
At initial state :
U = 274.01 kJ/Kg , V = 0.034875 m³/kg , S = 1.0256 KJ/kg.k
exergy ( Ф ) at initial state = M ( U + P₂V - T₀S )
= 5 ( 274.01 + 100* 10³ * 0.034875 - 297 * 1.0256)
≈ - 135.5285 kJ
At final state :
U = 84.44 kJ / kg , V = 0.0008261 m³/kg, S = 0.31958 kJ/kg.k
exergy ( ( Ф ) at final state = M ( U + P₂V - T₀S )
= -51.96 kJ
B) Determine the exergy destroyed
exergy destroyed = To * M ( S2 - S1 )
= 297 * 5 ( 0.31958 - 1.0256 )
= - 1048.4397 KJ
Hence we can conclude that A) The exergy of the refrigerant at the initial and final states are : Initial state = - 135.5285 kJ, Final state = -51.96 kJ and The exergy destroyed during this process is : - 1048.4397 kJ
Learn more about exergy : https://brainly.com/question/25534266
Which number has four significant figures?
A. 4000
B. 3.008
C. 86.012
D. 0.0001
a. 4000
This has 4-digits.
Answer:
in my opinion letter d.
Explanation:
Sana pi tama
Suppose that a ball decelerates from 8.0 m/s to a stop as it rolls up a hill, losing 10% of its kinetic energy to friction. Determine how far vertically up the hill the ball reaches when it stops. Show your work.(2 points)
Answer:
The maximum height is 0.33 m.
Explanation:
initial velocity, u = 8 m/s
final velocity, v = 0 m/s
10% of kinetic energy is lost in friction.
The kinetic energy used to move up the top,
KE = 10 % of 0.5 mv^2
KE = 0.1 x 0.5 x m x 8 x 8 = 3.2 m
Let the maximum height is h.
Use conservation of energy
KE at the bottom = PE at the top
3.2 m = m x 9.8 x h
h = 0.33 m
The height traveled vertically up the hill by the ball when it stops is 0.327 meter.
Given the following data:
Velocity = 8.0 m/sKinetic energy = 10% lost to friction.Scientific data:
Acceleration due to gravity = 9.8 [tex]m/s^2[/tex]To determine how far (height) vertically up the hill the ball reaches when it stops:
By applying the law of conservation of energy, we have:
Kinetic energy lost at the bottom = Potential energy gained at the top.
Mathematically, the above expression is given by the formula:
[tex]0.1 \times \frac{1}{2} mv^2 = mgh\\\\0.1 \times \frac{1}{2} v^2 = gh\\\\h=\frac{0.1v^2}{2g}[/tex]
Substituting the given parameters into the formula, we have;
[tex]h=\frac{0.1 \times 8^2}{2\times 9.8} \\\\h=\frac{0.1 \times 64}{19.6} \\\\h=\frac{6.4}{19.6}[/tex]
Height, h = 0.327 meter.
Read more on kinetic energy here: https://brainly.com/question/17081653
b) Assume the rod is 0.60 m long and has a mass of 0.50 kg, and the clay blob has a mass of 0.20 kg and moves at an initial velocity of 8.0 m/s. Calculate the final angular velocity of the rod. Be sure to put units in your calculation and show the resulting units in your answer.
Answer:
The correct answer is "6.96 rad/s".
Explanation:
The given values are:
Length,
L = 0.6 m
Mass,
m₁ = 0.5 kg
m₂ = 0.2 kg
Initial velocity,
V = 8 m/s
Now,
The final angular velocity will be:
⇒ [tex]\omega =\frac{6m_1V}{(4m_1+3m_2)L}[/tex]
By substituting the values, we get
⇒ [tex]=\frac{6\times 0.2\times 8}{(4\times 0.2+3\times 0.5)0.6}[/tex]
⇒ [tex]=\frac{9.6}{1.38}[/tex]
⇒ [tex]=6.96 \ rad/s[/tex]
Question 2 of 10
Which of the following statements is true of an isolated system?
A.The system has energy but no matter.
B.Energy separates the matter in the system from outside matter.
C.The matter within the system does not interact with matter outside the system
D.The matter within the system does not interact with other matter in the system
Answer:
D
Explanation:
because it is system removed from system and it dosent interact with them
When an object is in free fall, ____________________.
Answer:
Objects that are said to be undergoing free fall, are not encountering a significant force of air resistance; they are falling under the sole influence of gravity.
Explanation:
Under such conditions, all objects will fall with the same rate of acceleration, regardless of their mass.
A 200-lb man carries a 10-lb can of paint up a helical staircase that encircles a silo with radius 30 ft. If the silo is 60 ft high and the man makes exactly two complete revolutions, how much work is done by the man against gravity in climbing to the top
Answer:
17.07 kJ
Explanation:
The work done against gravity by the man W equals the potential energy change of the man and can of paint, ΔU
W = ΔU = mgΔy where m = mass of man and can of paint = 200 lb + 10 lb = 210 lb = 210 × 1 kg/2.205 lb, g = acceleration due to gravity = 9.8 m/s² and Δy = height of silo = 60 ft = 60 × 1m/3.28 ft
Since W = mgΔy, we substitute the values of the variables into the equation.
So,
W = mgΔy
W = 210 lb × 1 kg/2.205 lb × 9.8 m/s² × 60 ft × 1m/3.28 ft
W = 123480/7.2324 J
W = 17073.2 J
W = 17.0732 kJ
W ≅ 17.07 kJ
Explain what a circuit breaker is and how it helps protect your house?
Explanation:
A circuit breaker is an electrical switch designed to protect an electrical circuit from damage caused by overcurrent/overload or short circuit. Its basic function is to interrupt current flow after protective relays detect a fault.
Circuit breakers have been designed to detect when there is a fault in the electricity, so it will “trip” and shut down electrical flow. ... This detection is key to preventing surges of electricity that travel to appliances or other outlets, which can cause them to break down
two point charges with charge q are initially separated by a distance d. if you double the charge on both charges, how far should the charges be separated for the potential energy between them to remain the same
Answer:
r ’= 4 r
Explanation:
Electric potential energy is
U = [tex]k \frac{q_1q_2}{r_{12}}[/tex]k q1q2 / r12
in this exercise
q₁ = q₂ = q
U = k q² / r
for when the charge change
U ’= k q’² / r’
indicate that
q ’= 2q
U ’= U
we substitute
U = k (2q) ² / r ’
U = 4 k q² / r ’
we substitute
[tex]k \ \frac{q^2}{r} = 4 k \ \frac{q^2}{r'}[/tex]k q² / r = 4 k q² / r ’
r ’= 4 r
A planet of mass m moves around the Sun of mass M in an elliptical orbit. The maximum and minimum distance of the planet from the Sun are r1 and r2, respectively. Find the relation between the time period of the planet in terms of r1 and r2.
Answer:
the relation between the time period of the planet is
T = 2π √[( r1 + r2 )³ / 8GM ]
Explanation:
Given the data i the question;
mass of sun = M
minimum and maximum distance = r1 and r2 respectively
Now, using Kepler's third law,
" the square of period T of any planet is proportional to the cube of average distance "
T² ∝ R³
average distance a = ( r1 + r2 ) / 2
we know that
T² = 4π²a³ / GM
T² = 4π² [( ( r1 + r2 ) / 2 )³ / GM ]
T² = 4π² [( ( r1 + r2 )³ / 8 ) / GM ]
T² = 4π² [( r1 + r2 )³ / 8GM ]
T = √[ 4π² [( r1 + r2 )³ / 8GM ] ]
T = 2π √[( r1 + r2 )³ / 8GM ]
Therefore, the relation between the time period of the planet is
T = 2π √[( r1 + r2 )³ / 8GM ]
1. On each of your equipotential maps, draw some electric field lines with arrow heads indicating the direction of the field. (Hint: At what angle do field lines intersect equipotential lines?) Draw sufficient field lines that you can "see" the electric field.
Answer:
The angle between the electric field lines and the equipotential surface is 90 degree.
Explanation:
The equipotential surfaces are the surface on which the electric potential is same. The work done in moving a charge from one point to another on an equipotential surface is always zero.
The electric field lines are always perpendicular to the equipotential surface.
As
[tex]dV = \overrightarrow{E} . d\overrightarrow{r}\\\\[/tex]
For equipotential surface, dV = 0 so
[tex]0 = \overrightarrow{E} . d\overrightarrow{r}\\\\[/tex]
The dot product of two non zero vectors is zero, if they are perpendicular to each other.
A bag of cement of Weight 1000N hangs from ropes. Two of the ropes make angles of 1=60 and 2=30 with the horizontal.if the system is in equilibrium,find the tension T1,T2andT3 in the ropes
Answer:
T1 = 499.9N, T2 = 865.8N, T3 = 1000N
Explanation:
To find the tensions we need to find the vertical and horizontal components of T1 and T2
T1x = T1 cos60⁰, T1y = T1 sin60⁰
Also, T2x = T2 cos30⁰, T2y = T2 sin30⁰
For the forces to be in equilibrium,
the sum of vertical forces must be zero and the sum of horizontal forces must also be zero
Sum of Fx = 0
That is, T1x - T2x=0
NB: T2x is being subtracted because T1x and T2x are in opposite directions
T1 cos60⁰ - T2 cos30⁰ = 0
0.866T1 - 0.5T2 = 0 ............ (1)
Sum of Fy = 0
T1y + T2y - 1000 = 0
T1 sin60⁰ + T2 sin30⁰ - 1000 = 0
NB: The weight of the bag of cement is also being subtracted because it's in an opposite direction.
0.5T1 - 0.866T2 - 1000 = 0 ........(2)
From (1)
make T1 the subject
T1 = 0.5T2/0.866
Substitute T1 into (2)
0.5 (0.5T2/0.866) - 0.866T2 = 1000
(0.25/0.866)T2 - 0.866T2 = 1000
0.289T2 - 0.866T2 = 1000
1.155T2 = 1000
T2 = 865.8N
Then T1 = 0.5 x 865.8 / 0.866
T1 = 499.9N
T3 = 1000N
NB: The weight of the bag is the Tension above the rope, which is T3
A single-turn square loop of wire, 2.00 cm on each edge, carries a clockwise current of 0.240 A. The loop is inside a solenoid, with the plane of the loop perpendicular to the magnetic field of the solenoid. The solenoid has 30.0 turns/cm and carries a clockwise current of 15.0 A. Find:
a. the force on each side of the loop
b. the torque acting on the loop.
Answer:
Explanation:
a )
Magnetic field inside solenoid B = μ₀ NI ,
μ₀ = 4π x 10⁻⁷ ; N is no of turns per meter length in solenoid and I is current B= 4π x 10⁻⁷ x 30 x 10² x 15
= .0565 T .
Force on each side of square loop = B i L
B is external magnetic field , i is current in loop and L is length of side
Force on each side of square loop = .0565 x .24 x 2 x 10⁻²
= 2.7 x 10⁻⁴ N .
b )
Torque on the loop = F x d
F is force on one side , d is distance between two sides , that is side of the square loop
= 2.7 x 10⁻⁴ x 2 x 10⁻² N.m
= 5.4 x 10⁻⁶ N.m .
The gravitational force Asteroid A experiences is the gravitational force Asteroid C experiences
Answer:
The gravitational force Asteroid A experiences is greater than the gravitational force Asteroid C experiences
A pump lifts 400 kg of water per hour a height of 4.5 m .
Part A
What is the minimum necessary power output rating of the water pump in watts?
Express your answer using two significant figures.
Part B
What is the minimum necessary power output rating of the water pump in horsepower?
Express your answer using two significant figures.
Answer:
Power = Work / Time
P = 400 kg * 9.8 m/s * 4.5 m / 3600 sec = 4.9 J/s = 4.9 Watts
Also, 4.9 Watts / (746 Watts / Horsepower) = .0066 Hp
A mass attached to the end of a spring is oscillating with a period of 2.25 s on a horizontal frictionless surface. The mass was released from rest at
t = 0
from the position
x = 0.0480 m.
Determine the location of the mass at
t = 5.85 s?
Answer:
[tex]X=0.0389m[/tex]
Explanation:
From the question we are told that:
Period of spring [tex]T_s=2.25s[/tex]
Initial Position of Mass [tex]x=0.0480m[/tex]
Final Mass period [tex]T_f=5.85s[/tex]
Generally the equation for the Mass location is mathematically given by
[tex]X=xcos*\frac{2\pi T_s}{T_f}[/tex]
[tex]X=0.048*cos*\frac{2\pi 5.85}{2.25}[/tex]
[tex]X=0.0389m[/tex]
It takes the elevator in a skyscraper 4.0 s to reach its cruising speed of 10 m/s. A 60 kg passenger gets aboard on the ground floor.
1. What is the passenger's apparent weight before the elevator starts moving?
2. What is the passenger's apparent weight whilethe elevator is speeding up?
3. What is the passenger's apparent weight afterthe elevator reaches its cruising speed?
Answer:
1. 588 N
2. 738 N
3. 588 N
Explanation:
time, t = 4 s
initial velocity, u = 0
final velocity, v = 10 m/s
mass, m= 60 kg
1.
Weight of passenger before starts
W =m g = 60 x 9.8 = 588 N
2.
When the elevator is speeding up
v = u + a t
10 = 0 + a x 4
a = 2.5 m/s2
Now the weight is
W' = m (a + g) = 60 (9.8 + 2.5) = 738 N
3.
When he reaches the cruising speed, the weight is
W = 588 N
A truck moves 70 m east, then moves 120 m west, and finally moves east again a distance of 90 m. If east is chosen as the positive direction, what is the truck's resultant displacement
Answer:
140m east
Explanation:
If East is positive then lets rephrase the problem into integers
A truck moves +70 m, then moves -120m, and finally moves +90m.
So totally Displacement = +70-120+90= +140m
Since east is positive, the trucks resultant displacement is 140 m east of origin
When an apple falls towards the earth,the earth moves up to meet the apple. Is this true?If yes, why is the earth's motion not noticeable?
Answer:
because the mass of the apple is very less compared to the mass of earth. Due to less mass the apple cannot produce noticable acceleration in the earth but the earth which has more mass produces noticable acceleration in the apple. thus we can see apple falling on towards the earth but we cannot see earth moving towards the apple.
12) If, after viewing a specimen at low power, you switch to high-dry power and, after using fine focus, cannot find the specimen, what things could you do to help yourself (before calling me over to assist you?)
Answer:
See the answer below
Explanation:
After seeing an object on a slide at the low-power objective of the microscope and it disappears on changing to high power, the following can be done to resolve the problem
1. Drop a few drops of immersion oil on the slide and view again under high the power objective.
2. If the object is still not visible after the action above, return the microscope to the low-power objective and make sure the object is refocused and centered. Then carefully change back to the high power objective and use the fine adjustment to bring it into focus.
A 31 kg block is initially at rest on a horizontal surface. A horizontal force of 83 N is required to set the block in motion. After it is in motion, a horizontal force of 55 N i required to keep it moving with constant speed. From this information, find the coefficients of static and kinetic friction
Answer:
The static and kinetic coefficients of friction are 0.273 and 0.181, respectively.
Explanation:
By Newton's Laws of Motion and definition of maximum friction force, we derive the following two formulas for the static and kinetic coefficients of friction:
[tex]\mu_{s} = \frac{f_{s}}{m\cdot g}[/tex] (1)
[tex]\mu_{k} = \frac{f_{k}}{m\cdot g}[/tex] (2)
Where:
[tex]\mu_{s}[/tex] - Static coefficient of friction, no unit.
[tex]\mu_{k}[/tex] - Kinetic coefficient of friction, no unit.
[tex]f_{s}[/tex] - Static friction force, in newtons.
[tex]f_{k}[/tex] - Kinetic friction force, in newtons.
[tex]m[/tex] - Mass, in kilograms.
[tex]g[/tex] - Gravitational constant, in meters per square second.
If we know that [tex]f_{s} = 83\,N[/tex], [tex]f_{k} = 55\,N[/tex], [tex]m = 31\,kg[/tex] and [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], then the coefficients of friction are, respectively:
[tex]\mu_{s} = \frac{83\,N}{(31\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}[/tex]
[tex]\mu_{s} = 0.273[/tex]
[tex]\mu_{k} = \frac{55\,N}{(31\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}[/tex]
[tex]\mu_{k} = 0.181[/tex]
The static and kinetic coefficients of friction are 0.273 and 0.181, respectively.
Which parts of The Action Potential Are Represented On The ECG?
Answer:
The phases of the cardiac action potential correspond to the surface ECG (ECG) (Figure). The P wave reflects atrial depolarization (phase 0), the PR interval reflects the conduction velocity through the AV node, the QRS complex the ventricular depolarization and QT interval the duration potential ventricular action.
A 6.0-cm-diameter horizontal pipe gradually narrows to 4.0 cm. When water flows through this pipe at a certain rate, the gauge pressure in these two sections is 32.0 kPa and 24.0 kPa, respectively. What is the volume rate of flow?
Answer:
a n c
Explanation:
Differences between angle of twist and angle of shear
Answer:
idek
Explanation:
An automobile engine has an efficiency of 22.0% and produces 2510 J of work. How much heat is rejected by the engine
Answer:
If efficiency is .22 then W = .22 * Q where Q is the heat input
Heat Input Q = 2510 / .22 = 11,400 J
Heat rejected = 11.400 - 2510 = 8900 J of heat wasted
Also, 8900 J / (4.19 J / cal) = 2120 cal
An efficiency is the measure of productivity of an engine. The heat rejected by the engine is 8900 Joules.
What is efficiency?An efficiency of a heat engine is the ratio of the work done and heat supplied.
Given is the automobile engine has the efficiency 22% and Work done is 2510 Joules.
The efficiency is written as,
η= W / Qs.
The work done is W= Qs - Qr, where Qr is the rejected heat.
The heat rejected can be represented as
Qr = W ( 1/η -1)
Substituting the value into the equation, we get the rejected heat.
Qr = 2510 (1/0.22 -1)
Qr = 8900 Joules.
Thus, the heat rejected by the engine is 8900 Joules.
Learn more about efficiency.
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g As they reach higher temperatures, most semiconductors... Selected Answer: have an increased resistance. Answers: have a constant resistance. have an increased resistance. have a decreased resistance.
Answer:
have an increased resistance
