a convex lens has a focal length of 0.5 it is combined with a second lens so combination has power 2.5 diopter which of the following could be second lens
A) concave lens focal length 2m
B) convex lens focal length 0.5m
C) concave lens 0.5m
D) convex lens 2m
Answer:
concave lens focal length 2m
An EMT raceway contains two 3-phase, 480-volt branch circuits. What minimum size THHN branch circuit conductor is required for one of the circuits when it supplies a noncontinuous load of 27 kW?
Answer:8 AWG
Explanation: 6 current carrying conductors with 2 - 3 phase branch circuits. Divide 27000 by (480v × 1.732 or square root of 3) and you get 32.46 amps. Now using the code book, divide 32.46 by the adjustment factor of 6 CCC's which is .8 and you get 40.61 amps. Now go to Table 310.16 in the 2020 NEC book and you'll see that in the 90° column that 40.61 amps is above #10 AWG so you size up to #8 AWG.
An aircraft flying at a steady velocity of 70 m/s eastwards at a height of 800 m drops a package of supplies.
a) Express the initial velocity of the package as a vector. What
observer on the ground.
b) How long will it take for the package to reach the ground?
c) How fast will it be going as it lands? Express your answer as
a vector.
d) Describe the path of the package as seen by a stationary
assumptions have you made about the frame of reference?
e) Describe the path of the package as seen by someone in the
aeroplane.
The motion of the package can be described as the motion of a projectile,
given that it has an horizontal velocity and it is acted on by gravity.
a) [tex]\vec{v}[/tex] = 70·ib) The package will reach ground in approximately 12.77 seconds.c) The speed of the package as it lands is approximately 145.51 m/s.d) The path of the package based on a stationary frame of reference is parabolice) The path of the package as seen from the plane is directly vertical downwards
Reasons:
Velocity of the aircraft = 70 m/s
Direction of flight of the aircraft = Eastward
Height from which the aircraft drops the package, h = 800 m
a) The initial velocity of the package, [tex]\vec{v}[/tex] = 70·i
b) The time it will take the package to reach the ground, t, is given by the formula;
[tex]\displaystyle h = \mathbf{\frac{1}{2} \cdot g \cdot t^2}[/tex]
Where;
g = The acceleration due to gravity ≈ 9.81 m/s²
Therefore;
[tex]\displaystyle t = \mathbf{\sqrt{ \frac{2 \cdot h}{g} }}[/tex]
Which gives;
[tex]\displaystyle t = \sqrt{ \frac{2 \times 800}{9.81} } \approx \mathbf{12.77}[/tex]
The time it will take the package to reach the ground, t ≈ 12.77 seconds
c) The vertical velocity just before the package reaches the ground, [tex]v_y[/tex], is given as follows;
[tex]v_y^2[/tex] = 2·g·h
Therefore;
[tex]v_y[/tex] = √(2·g·h)
Which gives;
[tex]v_y[/tex] = √(2 × 9.81 × 800) ≈ 125.28
[tex]v_y[/tex] ≈ 125.28 m/s
Which gives; [tex]\vec{v}[/tex] = 70·i - 125.28·j
Therefore, |v| = √(70² + (-125.28)²) ≈ 143.51
The speed of the package as it lands, |v| ≈ 143.51 m/s
d) The motion of the package that includes both horizontal and vertical motion is a projectile motion.
Therefore;
The path of the package is the path of a projectile, which is a parabolic shape.
e) As seen by someone on the aeroplane, the horizontal velocity will be
zero, therefore, the package will appear as accelerating directly vertical
downwards.
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Suppose you take a piece of hard wax from an unlit candle. After you roll the wax between your fingers for a while, it becomes soft. What state of matter is the softer wax? Is it possible for the wax to change without changing its state?
Answer:
An amorphous solid does not have a definite melting point; instead, it melts gradually over a range of temperatures, because the bonds do not break all at once. This means an amorphous solid will melt into a soft, malleable state (think candle wax or molten glass) before turning completely into a liquid.
Explanation:
Hope this helps
May I get braineist pls?
If you we're in charge of designing a wire to cart electricity across your city, state or provide, which of the following properties would be most important for your wire to have?
Should it be thick or thin, aluminum orchrome, buried underground or installed out in the sun?
Write a paragraph defending your choice.
PLEASE HELP!!
Answer:
It should be fibre optic cable , buried underground
1. An 80 kg skydiver uses a parachute to produce an applied force of 700 N while falling with an initial
velocity of 40 m/s
Time taken by the skydiver to land on the ground is 4.57 seconds
The applied force, F = 700 N
The mass of the skydiver, m = 80 kg
The velocity, v = 40 m/s
To calculate the time taken by the skydiver to fall to the ground will be calculated using the formula
[tex]F=\frac{mv}{t}[/tex]
Substitute F = 700, m = 80, and v = 40 to solve for t
[tex]700=\frac{80(40)}{t}\\\\700t=3200\\\\t=\frac{3200}{700}\\\\t=4.57 s[/tex]
Time taken by the skydiver to land on the ground is 4.57 seconds
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A student holds an egg outside of an open window and let's go. The window is 40 meters above the ground and the egg is falling under the acceleration of gravity, which is 9.8 m/s2. There are two equations that can be used to describe its motion over time:
How far will the egg travel after 2 seconds? Be sure to use the 5 steps to solve the problem. Show which equation you used, what each variable equals, and answer the question in a complete sentence.
The distance traveled by the egg after 2 seconds is 19.6 m.
The given parameters:
Height of the window, h = 40 mThe distance traveled by the egg after 2 seconds is calculated as follows;
[tex]h = v_0_y t + \frac{1}{2} gt^2\\\\[/tex]
where;
[tex]v_0_y[/tex] is the initial vertical velocity of the egg = 0g is acceleration due to gravity = 9.8 m/s²t is the time of motion, = 2 s[tex]h = 0 \ + \ \frac{1}{2} \times 9.8 \times (2)^2\\\\h = 19.6 \ m[/tex]
Thus, the distance traveled by the egg after 2 seconds is 19.6 m.
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What is the net force on a car with a mass of 1000 kg if its
acceleration is 35 m/s^2?
Answer:
3000N
Explanation:
divided to get answer
the force needed to accelerate the 1000kg car by 3m/s2 is 3000N
1. A ball rolls off a desk at a speed of 3.0 m/s, lands 0.40 seconds later, and lands 1.2 m away from the desk
b) how high is the desk
Answer:
0.784m
Explanation:
This is a projectile motion problem so we analyze the horizontal and vertical motions separately. Since the question is asking for the height of the desk, it shows we need to analyze the vertical motion.
The mass of a sample of sodium bicarbonate is 2. 1 kilograms (kg). There are 1,000 grams (g) in 1 kg, and 1 Times. 109 nanograms (ng) in 1 g. What is the mass of this sample in ng? 2. 1 Times. 103 ng 2. 1 Times. 106 ng 2. 1 Times. 109 ng 2. 1 Times. 1012 ng.
2.1 kg of sodium bicarbonate is equal to the 2.1 x 10¹² ng of sample. Option D is correct.
Mass is the quantity of the substance in the body or object. The SI unit of mass is Kilogram.
There are other units of measure,
Milligram: 1 g is equal to the [tex]\bold {10^3 \ mg}[/tex]Micro-gram: 1 g is equal to [tex]\bold {10^{6} \ \mu g}[/tex]Nano-gram: 1 g is is equal to[tex]\bold {10^{9} \ ng}[/tex]First convert kg to gram,
Since, 1 Kg = 1000 g
2.1 kg = grams of sample
So,
Do the cross multiplication,
[tex]\rm mass\ of\ sample = \dfrac {2.1\ kg \times 1000\ g }{ 1 kg}\\\\\rm mass\ of\ sample =2100 g[/tex]
Now, convert 2100 g to nano-grams
Since, 1 g = 1 x 10⁹ ng
2100 g = ng of sample
So,
Do the cross multiplication,
[tex]\rm mass\ of\ sample = \dfrac {2100 g \times 1 \times 10^9 ng }{1\ g}\\\\\rm mass\ of\ sample = 2.1 \times 10^1^2 ng[/tex]
Therefore, 2.1 kg of sodium bicarbonate is equal to the 2.1 x 10¹² ng of sample.
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A 0.5 kg rock is thrown straight downward off a cliff. If the rock only experience a gravitation force for 4.3 seconds, and ends with a downward velocity of 46.1 m/s, what was rock initial velocity? What is the gravitational force?
Answer:Answer:
See diagram for answers. The force and acceleration vectors are directed inwards towards the center of the circle and the velocity vector is directed tangent to the circle.
Explanation:
Can someone please give me the (Answers) to this? ... please ...
Answer:
1. 60 kg m/s
2. 2.4 kg
3. none they both have same momentum
help please I really need this
Answer:
Explanation:
Height(h) = 4m; depth of the half filled tank 4m/2 =2m
Acceleration due to Gravity (g)=9.8m/s^2
Density (d)=1000kg/m^3
Pressure = hdg
Pressure = 2m × 9.8m/s^2 × 1,000kg/m^3
Pressure = 19,600Pa
Under the Big Top elephant. Ella [2500 kg]. is attracted to Phant, the 3,000 kg elephant. They are separated by 8 m. What is the gravitational attraction between them? G=6.67×10^-11 (-11 is an exponent)
Hi there!
We can use the same equation for Gravitational Force:
[tex]\large\boxed{F_g = G\frac{m_1m_2}{r^2}}[/tex]
Fg = force due to gravity (N)
G = gravitational constant
m1,m2 = masses of objects (kg)
r = distance between objects (m)
Plug in the values provided:
[tex]F_g = (6.67*10^{-11})\frac{(2500)(3000)}{8^2} = \large\boxed{7.814 * 10^{-6}N }[/tex]
[tex]\huge\bf\underline{\underline{\pink{A}\orange{N}\blue{S}\green{W}\red{E}\purple{R:-}}}[/tex]
Here we've been given,
Universal gravitational constant (G) = [tex] \sf{6.67 \times {10}^{ - 11} }[/tex]Mass of object 1 (m1) = 2500 kg Mass of object 2 (m2) = 3000 kg Distance between two objects (r) = 8 mWe have to find the gravitational attraction force (Fg) = ?
The standard formula to solve is given by,
[tex]:\implies\tt{F_g = g \frac{m_1m_2}{ {r}^{2} } } [/tex]
[tex]:\implies\tt{F_g = 6.67 \times {10}^{ - 11} \times \frac{(2500 )(3000)}{ {8}^{2} } }[/tex]
[tex]:\implies\tt{F_g = 6.67 \times {10}^{ - 11} \times \frac{7500000}{64} }[/tex]
[tex]:\implies\tt{F_g = 7.814 \times {10}^{ - 6} }[/tex]
Gravitational force of attraction is 7.814 × 10^-6 N.What is the energy of a 5 kg object that is held at a height of 3 m above the ground?
I really need the Formula, substitute, answer
Answer:
a 5 kilogram mass, at a height of 3 meters, while acted on by Earth's gravity would have 147.15 Joules of potential energy, PE = 3kg * 9.81 m/s2 * 5m = 147.15 J.
Explanation:
what is the cost of monthly (30 days) electric bill of ana if her city's cost of electricity is 0.05$ per kwh and she uses three refrigerators running in 600-watt power rating and open 24 hours
Answer:
1kW = 1000W
600W = 0.6kW
Cost of electric bill = 0.6kWh × 24 × 30 × $0.05
= $21.60
Student 1 lifts a box with a force of 500 N and sets it on a tabletop 1.2 m high. Student 2 pushes an identical box up a 5 m ramp from the floor to the top of the same table. Which student did the MOST work?
The student who did the most work is student 2 with 2500 Joules.
Given the following data:
Force 1 = 500 NewtonDistance 1 = 1.2 meterForce 2 = 500 NewtonDistance 2 = 5 meterTo determine which of the students did the most work:
Mathematically, the work done by an object is given by the formula;
[tex]Work\;done = Force \times distance[/tex]
For student 1:
[tex]Work\;done = 500 \times 1.2[/tex]
Work done = 600 Joules
For student 2:
[tex]Work\;done = 500 \times 5[/tex]
Work done = 2500 Joules.
Therefore, the student who did the most work is student 2 with 2500 Joules.
Read more: Read more: brainly.com/question/13818347
Answer:student 2 did the most work
Explanation:student 2 did more work because he pushed it up 5 m instead of 1.2.
What part of the reactor is used to control the speed of the nuclear reaction? How does it work?
Control rods
____________________
Control rods are inserted into the core of a nuclear reactor and adjusted in order to control the rate of the nuclear chain reaction
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How to do this question plz
Answer:
Let L be the length of the wire.
N λ = M L where N is the number of wavelengths λ. Also, M and are integers and L is the length of the wire
There must be nodes at the ends of the wire and anti-nodes at the points of zero displacement - λ/2, λ, 3λ/ 2, etc.
The simplest would be 2 λ = L or L = λ / 2
In such a situation one would have
N-A-N or node-antinode-node
There is a 180 degree phase change upon reflection of the wave creating the standing wave
The distance between nodes or anti-nodes must be λ/2 and there must be nodes at the ends of the wire
What is the wavelength of an AM radio wave in a vacuum if its frequency is 810 kHz?
If a car can go from 0 to 60 km/h in 8.0 seconds, what would be its final speed after 5.0 seconds if its starting speed were 50 km/h?
Answer:
This question assumes that the car accelerates at the same rate as when it went from 0 to 60km/h
24.29m/s or 87.4km/h
Explanation:
Let's find the acceleration of the car:
let vi=0, vf=60km/h (16.67m/s), Δt = 8.0s
a = (vf-vi)/Δt
a = (16.67m/s-0)/8.0
a = 2.08m/s^2
Now we can use this acceleration to find vf in the second part:
50km/h is 13.89m/s
a = (vf-vi)Δt
vf = aΔt + vi
vf = 2.08m/s^2*5.0+13.89m/s
vf = 24.29m/s (87.4km/h)
Pleaseeeee HELPPPP THIS IS TIMED ALSO,
A book slides along a table and comes to a stop. Explain, in detail, all the forces acting on the book.
Answer:
Friction, normal force, and weight
Explanation:
If the book slows down, it means that there must be friction acting in the opposite direction of the direction the book is moving in.
Weight is caused by the gravitational pull of the Earth on the book, and normal force is the table pushing the book up because the book is pushing down on the table (3rd law.)
Note that weight and normal force is not the 3rd law action-reaction pair. The pair is the force of the book on the table and the force of the table on the book.
Connor rode an inner tube down a river. For 4.6 minutes, he moved downriver at 15 meters per minute. During this time, how far did he move?
answers:
3.26 meters
3.26 minutes
69 meters
69 minutes
Answer:
69 meters
Explanation:
how to put in velocity into a graphing calculator
Answer:
You need a graphing calculator that implemented vectors. I don't know any that have it. Sorry.
Answer:
Provided an object traveled 500 meters in 3 minutes , to calculate the average velocity you should take the following steps:
Change minutes into seconds (so that the final result would be in meters per second). 3 minutes = 3 * 60 = 180 seconds ,
Divide the distance by time: velocity = 500 / 180 = 2.77 m/s .
Explanation:
Before a rifle is fired, the linear momentum of the bullet-rifle system is zero.
After the rifle is fired:
I. kinetic energy of the system is zero
II. linear momentum of the system is zero
Hi there!
II. Linear momentum of the system is zero.
This is an example of a RECOIL collision. With the Law of Conservation of Momentum, momentum remains constant before and after the collision.
Thus, the total momentum would also be equivalent to zero after the collision.
What is the radius of a circular space station that spins with a linear speed of 90.0 m/s I’m order for its walls to supply a centripetal acceleration equal to the acceleration provided by gravity on the surface of the earth?
Answer:
a = v^2 / R centripetal acceleration
R = v^2 / a a at the surface of earth is 9.8 m/s^2
R = (90 m/s)^2 / 9.8 m/s^2 = 827 m
Which impact would be more forceful, decreasing speed from 60mph to 0 mph over 2 seconds or 0.5 seconds? Explain
7 point
The heat energy from convection currents in magma also drives the rock
cycle. Convection and movement of the plates causes rock to move
between Earth's surface and interior. It provides the heat that changes
rocks inside Earth to form metamorphic rocks. Igneous rocks form when
magma cools and solidifies. The older rocks that are broken down and
converted to sedimentary rocks might come from igneous and
metamorphic rocks or even older sedimentary rocks. TRUE or FALSE: The
rock cycle is directly related to convection currents.
True
False
how is the thermal efficiency of a heat engine defined?
A massless string connects a 1.00 kg mass to a 3.00 kg cart which is resting on a frictionless horizontal surface. The mass hangs over a
frictionless pulley. When the mass is released, the cartaccelerates to the right
2.45 m/s2
4.90 m/s
9.80 m/s?
19.6 m/s
Both masses will have the same acceleration. The cart accelerates to the right with a magnitude of 4.9 m/[tex]s^{2}[/tex]. The correct answer is 4.90 m/[tex]s^{2}[/tex]
Given that a massless string connects a 1.00 kg mass to a 3.00 kg cart which is resting on a frictionless horizontal surface.
Let M = 1kg and m = 3 kg
Since the horizontal surface is frictionless, the tension in the string will be the same. when the mass is hanged over a frictionless pulley, the tension will also be the same.
When the mass is released, the cart accelerates to the right can be calculated from Newton' second law of motion. That is,
M( g + a) = m(g - a)
1(9.8 + a) = 3( 9.8 - a)
9.8 +a = 29.4 - 3a
collect the like terms
4a = 19.6
a = 19.6/4
a = 4.9 m/[tex]s^{2}[/tex]
Therefore, the cart accelerates to the right with a magnitude of 4.9 m/[tex]s^{2}[/tex]. The correct answer is 4.90 m/[tex]s^{2}[/tex]
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