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
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.
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
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
the system shown above is released from rest. if friction is negligible, the acceleration of the 4.0 kg block sliding on the table shown above is most nearly
The acceleration of the first block (4 kg) is -9.8 m/s².
The given parameters:
Mass of the first block, m₁ = 4.0 kgMass of the second block, m₂ = 2.0 kgThe net force on the system of the two blocks is calculated as follows;
[tex]m_2 g - T = m_1 a[/tex]
where;
T is the tension in the connecting string due weight of the first block[tex]m_2 g - m_1 g = m_1 a\\\\a = \frac{m_2 g - m_1g}{m_1} \\\\a = \frac{g(m_2 - m_1)}{m_1} \\\\a = \frac{9.8(2-4)}{2} \\\\a = -9.8 \ m/s^2[/tex]
Thus, the acceleration of the first block (4 kg) is -9.8 m/s².
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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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Shanika is an engineer at an amusement park who is experimenting with changes to the setup for a magnetic roller coaster ride. In one ride, there are two identical roller coaster cars (orange and green) that start on opposite sides of a large magnet located at the center of a station. Shanika wants to get the largest increase in potential energy she can by moving one car one space to the left or the right.
Shanika can move the orange car to point A or point B, or she can move the green car to point C or point D. Which movement should she make? Why will that movement result in the largest increase in potential energy? Describe the magnetic force that will act on the roller coaster car she moves.
Answer:
I believe it might be point A since the question ask what will result in the ln a largest increase in potential energy
Explanation:
point A since the question ask what will result in the ln a largest increase in potential energy
What are the types of energy ?The energy can be defined as ability to work or produce action and / or movement and manifests itself in many different ways, such as body movement, heat, electricity, etc.
The different types of energy include Kinetic energy which is associated with the movement of bodies. Potential energy which is stored by virtue of a body's position relative to its surface is also called gravitational potential energy.
Thermal Energy or Heat can be defined as the energy associated with the kinetic energy of the molecules that make up an element, it can be manifested if there is a temperature difference between two bodies.
Chemical energy released or formed from chemical reactions, Solar energy from sunlight. This form of energy is used to generate electricity through photovoltaic plates, for example.
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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
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 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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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 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
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.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
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.
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 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]
what is the magnitude of the force exerted on the bottom of the bottle?
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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which is the following is a modern method used to pressure food
1. sprey drying
2.salting
3. smoking
4. Honey
Answer:
1. Spray Drying
Explanation:
it is the most modern
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
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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What is the wavelength of an AM radio wave in a vacuum if its frequency is 810 kHz?
Help me answer this question please!
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
Which model below shows the positions of the Sun, Moon, and Earth that have the greatest effect on ocean tides?
Answer:
Below!
Explanation:
Referring to the picture, we can conclude that the picture J will have the most effect on ocean tides.
Whichever model is similar to model J in the picture will be your answer.
Hoped this helped.
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?
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 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
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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