Eclipses occur because the Moon?
A
does not create its own light and only reflects light from the Sun.
o
does not create its own light and only reflects the light from Earth.
o
с
creates its own light, which is blocked by the Sun during an eclipse.
creates its own light, which is blocked by Earth during an eclipse.
Answer:
с
creates its own light, which is blocked by the Sun during an eclipse.
Explanation:
might not be right but its what i remember
7. A car is moving at 50.0 mph when the driver applies brakes. Determine the distance it
covers before coming to a halt. Coefficient of static friction between the tires and surface of
the road is 0.514. Mass of the car is 1000 kg.
Answer:
The distance to come to a halt is approximately 49.53 meters
Explanation:
Thee given parameters are;
The speed of the car, v = 50 mph = 22.35 m/s
The mass of the car, m = 1000 kg
The coefficient of friction, = 0.514
The force of friction of the brake = Mass × Gravity × Friction = 1000 × 9.81 × 0.514 = 5042.34 N
The initial kinetic energy of the car = 1/2×m×v² = 1/2 × 1000 × 22.35² = 249761.25 J
The work done by the brake = Force of the brake × Distance, d, to come to halt
By conservation of energy, we have;
The work done by the brake = The initial kinetic energy of the car
∴ The initial kinetic energy of the car = Force of the brake × Distance, d, to come to halt
The initial kinetic energy of the car = 249761.25 J = 5042.34 N × Distance, d, to come to halt
∴The distance to come to a halt = 249761.25 J /(5042.34 N) ≈ 49.53 meters
The distance to come to a halt ≈ 49.53 meters.
The car will cover 49.65 m distance before stopping due to application of brake.
Given data:
The mass of car is, M = 1000 kg.
The initial speed of car is, u = 50.0 mph = (50)(0.447 m/s) = 22.35 m/s.
The coefficient of static friction is, [tex]\mu = 0.514[/tex].
The given problem can be solved using the third kinematic equation of motion. Which is,
[tex]v^{2}=u^{2}+2as[/tex] ....................................................(1)
Here, v is the final speed, v = 0 (Because car is finally stopping)
s is the distance covered before stopping and a is the magnitude of acceleration.
Now, since frictional force opposes the motion. Then,
[tex]F = f\\\\ma = \mu \times m \times g\\\\a = 0.514 \times 9.8\\\\a =5.03 \;\rm m/s^{2}[/tex]
Substituting the values in equation (1) as,
[tex]0^{2}=23.35^{2}+2(-5.03)s\\\\s = \dfrac{2500}{2 \times 5.03}\\\\ s=49.65 \;\rm m[/tex]
Thus, the car will cover 49.65 m distance before stopping due to application of brake.
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Question 4 of 20
Luke wraps a magnetized steel nail with several coils of insulated wire and
then connects the loose ends of the wire to a 10 W lightbulb. Which step
could he take next to get the lightbulb to light up?
O A. Replace the magnetized nail with a nonmagnetic metal to cause
an electric current to flow in the wire.
O B. Remove the magnetized nail from the wire coil to induce a
magnetic field that will flow through the wire.
C. Move the magnetized nail back and forth within the wire coil to
induce an electric current in the wire.
D. Add several more coils of the wire around the magnetized nail to
make a stronger electric current.
SUBMIT
Answer:
C. Move the magnetized nail back and forth within the wire coil to induce an electric current in the wire.
Hope this helps.
can you help me doing an essay about actual self
Answer:
yes absolutely
Explanation:
why not
Which model of the atom has electrons traveling in specific paths around the nucleus? Bohr's model Rutherford's model Thomson's model Dalton's model
Answer:
A, Bohr's model
Explanation:
Took the quiz, hope it's right for you.
(The other three did have something to do with it they just didn't create that specific model)
The model of the atom that has electrons traveling in specific paths around the nucleus is Rutherford's model.
Therefore, the correct answer is option B
J. J Thomson proposed that the atom is a sphere of positively charged matter in which negatively charged electron are embedded.
In Rutherford's model of atomic structure, the electrons moves in orbits by electrostatic attraction to the positively charged nucleus.
According to Dalton's atomic theory, atom were indestructible and indivisible solid particle.
Bohr's model made assumption that electron can only exist in circular orbits of definite quantum energy.
According to the question, the model of the atom that has electrons traveling in specific paths around the nucleus is Rutherford's model.
Therefore, the correct answer is option B
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Yodelin has fifty quarters and dimes. Their total value is $9.80. Which of these systems of equations can be used to find the number of quarters (q) and
dimes (d) Yodelin has?
Answer:
q + d = 50
25q + 10d = 980
Explanation:
The equation that can be employed to determine the number of quarters(q), as well as, dimes(d) that Yodelin has would be:
q + d = 50...(1)
25q + 10d = 980...(2)
Multiplying (1) by 25
25q + 25d = 1250 ...(3)
25q + 10d = 980...(4)
subtracting (4) from (3)
15d = 270
d = 18
q = 50 - 18
= 32
The system of the equations could be
q + d = 50
25q + 10d = 980
Calculation of the equation and the number of quarters and dimes:Since Yodelin has fifty quarters and dimes. Their total value is $9.80.
So here the equations be
q + d = 50...(1)
25q + 10d = 980...(2)
Now we have to Multiplying (1) by 25
So,
25q + 25d = 1250 ...(3)
25q + 10d = 980...(4)
Now
subtracting (4) from (3)
15d = 270
d = 18
So,
q = 50 - 18
= 32
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how much force is needed to accelerate 300 kg at a rate of 4 m/s/s
Answer:
1200
Explanation:
Force = Mass x Acceleration
Answer:
[tex]\boxed {\tt 1,200 \ Newtons }[/tex]
Explanation:
Force can be found by multiplying the mass by the acceleration.
[tex]F=m*a[/tex]
The mass is 300 kilograms.
The acceleration is 4 meters per second squared.
[tex]m= 300 \ kg\\a= 4 \ m/s^2[/tex]
Substitute the values into the formula.
[tex]F= 300 \ kg * 4 \ m/s^2[/tex]
Multiply.
[tex]F= 1200 \ kg * m/s^2[/tex]
1 Newton is equal to 1 kilograms meters per second squared, so our current answer of 1200 kg m/s² is equal to 1200 Newtons.
[tex]F= 1200 \ N[/tex]
The force needed is 1,200 Newtons.
How are velocity, wavelength, and frequency related
Answer:
The wave velocity and the wavelength are related to the wave's frequency and period by vw=λT or vw=fλ. The time for one complete wave cycle is the period T. The number of waves per unit time is the frequency ƒ. The wave frequency and the period are inversely related to one another.
Explanation:
Will was riding his bike when a dog ran out in front of him. He slammed on his brakes. During this quick stop, some of the mechanical energy (his motion) was changed into
A) heat energy.
B) light energy.
C) kinetic energy.
D) gravitational energy.
Answer:
A
Explanation:
big brain
During this quick stop, some of the mechanical energy (his motion) was changed into heat energy.
What is energy?Energy is the ability to do work. There are different types of energy such as Heat energy, light energy, kinetic energy, and gravitational energy.
Heat is the energy that moves from one body to another when temperatures are different. Heat passes from the hotter to the colder body when two bodies with differing temperatures are brought together.
The joule is a unit of energy that serves as the SI unit for heat (J). The calorie (cal), which is defined as "the amount of heat necessary to raise the temperature of one gram of water from 14.5 degrees Celsius to 15.5 degrees Celsius," is another common unit of heat measurement.
Therefore, During this quick stop, some of the mechanical energy (his motion) was changed into heat energy.
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List out Units For: Acceleration
Unit for Acceleration is m/s² or ms raised to -2
What is the name of the kind of stretch that involves stretching as far as you can and then holding for 10-30 seconds
Question 2 options:
PNF
ballistic
dynamic
static
Answer:
Static stretching.
Explanation:
It is static stretching because it is a form of stretching which u can do actively for a period of time and you hold position for about 30 to 60 seconds which allow the muscles and connective tissues to lengthen. It is done after work out with out movement in order to loosen up muscles so as to gain flexibility.
Three toy boats with the same mass were in a lake. Two boats were moving and one was stopped. Each boat got bumped by another boat, but not in the same direction. All the boats changed speed as a result of being bumped. Use the information in the diagram to answer. Which toy boat exprienced the strongest force when it was bumped ? How do you know ?
The answer is " The orange and gray toy boats experienced the strongest force because both gained the same force as they sped up the blue boat lost force so it slow down
Explanation:
Trust me I did this one for an assignment
The boat that will experience the strongest force is the boat that has the highest speed.
According to Newton's second law of motion, the force applied to an object is directly proportional to the product of mass and acceleration of the object.
F = ma
[tex]F = \frac{mv}{t}[/tex]
where;
F is the force experienced by the objectv is the velocity of the objectt is the time of motionThe force experienced by the boats is directly proportional to the velocity of their motion. Since the boats have the same mass, the force experienced by each boat will depend on the speed with which the boat moves.
Thus, the boat that will experience the strongest force is the boat that has the highest speed.
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please help
Which equipment is best to hold and to measure liquids?
A). flask and graduated cylinder
B). scale and ruler
C). caliper and thermometer
D). stopwatch and beaker
Answer:
I'm pretty sure that the answer is A
1. Which of the following provides a correct answer for a problem that can be solved using the kinematic equations?
A A body starts from rest and accelerates at 8 m/s/s for 3 s. The body's final
position is 32 m.
B A body starts from rest and accelerates at 8 m/s/s for 3 s. The body's final
velocity is 16 m/s
C A body starts from rest and accelerates at 8 m/s/s for 3 s. The body's travels a
distance of 16 m
A body starts from rest and accelerates at 8 m/s/s for 3 s. The body's final
velocity is 24 m/s
E A body starts from rest and accelerates at 8 m/s/s for 3 s. The body's final
position is 30 m.
1 of 14
2
3
5 6 7 8
9
Answer:
12
Explanation:
12
16. True/False: Protons are exchanged between objects by induction
#7 If the Moon’s mass was suddenly tripled, what would occur to the gravitational force between the Earth and the Moon?
(A) Gravitational force would increase by a factor of 3.
(B) Gravitational force would decrease by a factor of 3.
(C) Gravitational force would increase by a factor of 9.
(D) Gravitational force would decrease by a factor of 9.
Answer:
well it would be (A) Gravitational force would increase by a factor of 3.
Explanation:
because the moon is the only thing that gained mass not the Earth
An ant is crawling along a yardstick that is pointed with the 0-inch mark to the east and the 36-inch mark to the west. It starts at the 14-inch mark, crawls to the 20-inch mark, then moves to the 16-inch mark. What was the total distance the ant traveled?
Given :
An ant is crawling along a yardstick that is pointed with the 0-inch mark to the east and the 36-inch mark to the west.
It starts at the 14-inch mark, crawls to the 20-inch mark, then moves to the 16-inch mark.
To Find :
The total distance the ant traveled.
Solution :
Total distance travelled by ant = (distance between 14 and 20 inch mark) +
(distance between 20 and 16 inch mark)
Total distance = (20-14 ) + ( 20-16) = 6 + 4 = 10 inch.
Therefore, total distance the ant traveled is 10 inch.
Hence, this is the required solution.
Give reasons.
a. A stretched catapult has potential energy.
b A stone on ground has zero energy.
C. Rolling a drum is work against friction.
d. Lifting a stone is work against gravity.
e. Work and energy both have same unit.
matter makes up all ? and ?
A student is on a skateboard facing a wall. The student and skateboard have a mass of 75 kilograms. The student pushes off of the wall with a force of 100 Newtons. What is the force of the wall pushing back on the student? What is the acceleration of the student as she moves away from the wall?
Answer:
F=-100N; a=1.3m/s^2
Explanation:
Force is being made by student, so wall counteracts that force by not moving so it is equally opposite.
The force of the wall pushing back on the student would be 100 Newtons, and the acceleration of the student as she moves away from the wall would be 1.33 m/s²
What is Newton's third law of motion?Newton's third law of motion state that for every action force there exists a complementary reaction force that balances it.
As given in the problem a student is on a skateboard facing a wall. The student and skateboard have a mass of 75 kilograms. The student pushes off of the wall with a force of 100 Newtons,
The force of the wall pushing back the student = 100 Newtons
The acceleration of the student = 100/75
=1.33 m/s²
Thus, The student would be pushed back against the wall by a force of 100 Newtons, and she would accelerate away from the wall at a rate of 1.33 m/s².
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Which words help you to determine the meaning of the word decelerate in paragraph 3
Answer:
the words
Explanation:
Answer:
Pressing the brake pedal
Explanation:
Decelerate sounds a lot like the word accelerate, but we know that the speed is decreasing, so it would have to be going down, getting slower, so whichever sounds like a good example of speed decreasing should be a good answer!
Protons are found only in the atomic nucleus.
TRUE
FALSE
Answer:
true
Explanation:
Protons and neutrons are found in the nucleus
Find the resultant of an easterly force of 100 N and a southeast force of 80 N acting at 65 degrees to the 100 N force
Answer:
Resultant is 152 N at 28.5 degrees south to the 100 N force
Explanation:
Calculate the TOTAL mechanical energy of pendulum is it swings from his highest point to its lowest point. Pendulum mass is 4 kg. Use your equations for gravitational potential energy and kinetic energy to determine these values based on the data given below. Total energy is the sum of gravitational potential energy and kinetic energy. In this problem, round gravity to: g = 10 m/s^2.
Answer:
its should be 2.0 and 4.5 on it
Which event is an example of condensation?
A. Wet clothes are drying on a clothesline.
B. A mirror fogs up when someone takes a hot shower.
C. Water drips from an icicle on the edge of a roof.
O D. Rain turns to sleet as it nears the ground.
Answer:
the answer to your question is b a mirror fogs up when someone takes a hot shower.
The process of conversion of gaseous water to its liquid form is called condensation. The fogs formed on the mirror from the hot shower is an example of condensation.
What is condensation?Condensation is the process of cooling up of water vapor to liquid water. Water vapor condenses when it cools. Condensation can be best understood from the reason behind raining.
When water vaporizers from resources and cools from the sky, and the vapor condenses to form liquid droplets. Similarly we can observe water droplets in the window pane due to the similar effect.
The water vapor arises from the hot shower condenses in the atmosphere and forms the drops on the mirror. Therefore, the mirror fogs up by the hot shower is an example of condensation.
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Which property describes the number of waves that move past a point each second.
A) Speed
B) Frequency
C) Period
D) Wavelength
Answer:B) Frequency
Explanation:
Answer:
B) Frequency
Explanation:
The photo shows dew drops that formed on the grass when temperatures
cooled during the night. Which change of state happened to form the dew?
A. Liquid to gas
B. Gas to liquid
C. Solid to gas
D. Gas to solid
protons are present in sodium atom.
a 11
b 10
C 12
d 9
11
Explanation:
There are 11 protons in a sodium atom
A force of 75 N at an angle of 15° to the direction of motion moves a chair 3 m. Which change would result in more work being done on the chair?
Answer:
Decreasing the angle to 10
Explanation:
Edge 2020
Dos pilotos suicidas, que están inicialmente a una distancia de 500 m entre sí, deciden chocar directamente de frente arrancando ambos desde el reposo. Ambos autos pueden desarrollar una aceleración máxima constante de 15 m/s2 . Si el piloto A arranca un segundo antes que el piloto B, encuentra: a) la posición donde los autos chocan, medida a partir de la posición donde arranca el piloto A, y b) la rapidez relativa de la colisión (la rapidez de B con respecto a A justo antes de la colisión, ó viceversa).
Answer:
a) El punto de colisión de los dos automóviles desde donde parte el conductor del automóvil A es de aproximadamente 293,14 metros.
b) La velocidad del automóvil A, en relación con la velocidad del automóvil B, es de 15 m / s
Explanation:
Los parámetros del movimiento son;
La distancia entre ambos coches = 500 m
La aceleración de ambos coches = 15 m / s²
La dirección de movimiento de ambos coches = uno hacia el otro
La hora de inicio del conductor A = Un segundo antes de la hora de inicio del conductor B
Por lo tanto, de la ecuación de movimiento, tenemos;
s = u · t + 1/2 · a · t²
v² = u² + 2 · a · s
Dónde;
u = La velocidad inicial de los autos = 0 (los autos parten del reposo)
t = El tiempo de movimiento de una aceleración dada
a = La aceleración = 15 m / s²
s = La distancia recorrida en el tiempo t
Por lo tanto, para el controlador A, tenemos;
s₁ = 0 × (t + 1) + 1/2 × 15 × (t + 1) ² = 7.5 × (t + 1) ²
s₁ = 7.5 × (t + 1) ²
Para el conductor B, tenemos;
s₂ = 0 × t + 1/2 × 15 × t² = 7.5 × t²
s₂ = 7,5 × t²
Dado que ambos chocan a lo largo del camino de 500 m, tenemos;
s₁ + s₂ = 500 metros
∴ 7.5 × (t + 1) ² + 7.5 × t² = 500
∵ s₁ + s₂ = 7.5 × (t + 1) ² + 7.5 × t²
Lo que da;
15 · t² + 15 · t + 7.5 = 500
15 · t² + 15 · t - 492,5 = 0
Resolver usando la aplicación en línea da; t = -6,25181 o t = 5,25181
Dado que t es un número natural, tenemos el valor correcto para t = 5.25181 segundos
a) Por tanto, el punto de colisión de los dos coches desde donde parte el conductor del coche A es;
s₁ = 7.5 × (t + 1) ² = 7.5 × (5.25181 + 1) ² ≈ 293.14 metros
El punto de colisión de los dos automóviles desde donde parte el conductor del automóvil A ≈ 293,14 metros
b) La semilla de A en el punto de colisión se da de la siguiente manera
Velocidad, v₁ = u + a × (t + 1) = 0 + 15 × (5.25181 + 1) ≈ 93.78
v₁ ≈ 93,78 m / s
La semilla de B en el punto de colisión también se da de la siguiente manera
Velocidad, v = u + a × (t) = 0 + 15 × 5.25181 ≈ 78.78
v₁ ≈ 78,78 m / s
Por lo tanto, la velocidad del automóvil A, en relación con la velocidad del automóvil B, [tex]v_{relative}[/tex] se da como sigue;
= v₁ - v₂ = 93,78 m / s - 78,78 m / s = 15 m / s
la velocidad del automóvil A, relativa a la velocidad del automóvil B = 15 m/s.