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
a 25 kg cart has 125 kg*m/s of momentum, how fast is the car going
Answer:
5m/s
Explanation:
equation for momentum is mass times velocity
25 x 5 = 125
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
What is the wavelength of an AM radio wave in a vacuum if its frequency is 810 kHz?
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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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)
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 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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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.
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
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.
A hockey player hits a rubber puck from one side of the rink to the other. It has a mass of .170 kg, and is hit at an initial speed of 6 m/s. If the rink is 61 meters long, how fast is the puck moving when it hits the far wall?
By using third law of equation of motion, the final velocity V of the rubber puck is 8.5 m/s
Given that a hockey player hits a rubber puck from one side of the rink to the other. The parameters given are:
mass m = 0.170 kg
initial speed u = 6 m/s.
Distance covered s = 61 m
To calculate how fast the puck is moving when it hits the far wall means we are to calculate final speed V
To do this, let us first calculate the kinetic energy at which the ball move.
K.E = 1/2m[tex]U^{2}[/tex]
K.E = 1/2 x 0.17 x [tex]6^{2}[/tex]
K.E = 3.06 J
The work done on the ball is equal to the kinetic energy. That is,
W = K.E
But work done = Force x distance
F x S = K.E
F x 61 = 3.06
F = 3.06/61
F = 0.05 N
From here, we can calculate the acceleration of the ball from Newton second law
F = ma
0.05 = 0.17a
a = 0.05/0.17
a = 0.3 m/[tex]s^{2}[/tex]
To calculate the final velocity, let us use third equation of motion.
[tex]V^{2}[/tex] = [tex]U^{2}[/tex] + 2as
[tex]V^{2}[/tex] = [tex]6^{2}[/tex] + 2 x 0.3 x 61
[tex]V^{2}[/tex] = 36 + 36
[tex]V^{2}[/tex] = 72
V = [tex]\sqrt{72}[/tex]
V = 8.485 m/s
Therefore, the puck is moving at the rate of 8.5 m/s (approximately) when it hits the far wall.
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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:
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.
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
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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why is acceleration not constant near the speed of light
Answer:
because when an object approaches the speed of light, it's mass starts to increase exponentially, and would be infinite at the speed of light. It would therefore require MORE than an infinite amount of energy to accelerate even a single electron to the speed of light
Which impact would be more forceful, decreasing speed from 60mph to 0 mph over 2 seconds or 0.5 seconds? Explain
The average distance between the Earth and the Moon is 384,000,000 meters. What is the distance in kilometers?
Answer:
384000
Explanation:
it would be 384000 in km
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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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.
Help me answer this question please!
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:
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
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
A 1000 kg car and a 1500 kg car are driving in opposite directions at 20 m/s and 10 m/s respectively. If the cars collide head-on and stick together, determine their combined speed after the collision.
Hi there!
Recall the conservation of momentum for an inelastic collision:
[tex]\large\boxed{m_1v_1 + m_2v_2 = (m_1 + m_2)v_f}}[/tex]
Remember, velocity is a VECTOR and direction must be accounted for. Let the 1000 kg car have a positive velocity and the 1500 kg car have a negative velocity (opposite direction).
Plug in the given values:
[tex](1000)(20) + (1500)(-10) = (1000 + 1500)v_f}}[/tex]
Solve:
[tex]20000 -15000 = 2500v_f}}\\\\5000 = 2500v_f\\\\v_f =\boxed{ 2 m/s}[/tex]
1. A block with a mass of 5.0 kg is pushed on a frictionless surface by applying a horizontal force of 80.0 N. The block starts from rest, and its final velocity is 12.6 m/s.
Answer:
397 j
Explanation:
Because 5.0kg yuh
The kinetic energy of the block with a mass of 5.0 kg is pushed on a frictionless surface by applying a horizontal force of 80.0 N. The block starts from rest, and its final velocity is 12.6 m/s is 396.9 J.
What is kinetic energy?A moving item or particle has a certain kind of power called kinetic energy. When an item exerts a net force to perform work, which involves the transfer of energy, the object accelerates and as a result, gains kinetic energy.
The amount of kinetic energy that a moving object or particle possesses is determined by both its mass and rate of motion. Any combination of vibration, axis rotation, translation (or movement along a path from one location to another), and translation are all possible motion types.
Given:
The mass of the block, m = 5 kg,
The horizontal force, F = 80 N,
The velocity of the block, v = 12.6 m / s,
Calculate the kinetic energy by the formula given below,
[tex]KE = 1/2mv^2[/tex]
KE = 1 / 2 × 5 × 12.6²
KE = 1 / 2 × 5 × 158.6
KE = 396.9 J
Therefore, the kinetic energy of the block with a mass of 5.0 kg is pushed on a frictionless surface by applying a horizontal force of 80.0 N. The block starts from rest, and its final velocity is 12.6 m/s is 396.9 J.
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The complete question is A block with a mass of 5.0 kg is pushed on a frictionless surface by applying a horizontal force of 80.0 N. The block starts from rest, and its final velocity is 12.6 m/s. Find its kinetic energy?
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?
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
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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