Answer:
The atomic nucleus is the small, dense region consisting of protons and neutrons at the center of an atom, discovered in 1911 by Ernest Rutherford based on the 1909 Geiger Marsden gold foil experiment. Almost all of the mass of an atom is located in the nucleus, with a very small contribution from the electron cloud.
Which of the following would be best for absorbing light?
a A mirror
b A black shirt
c glass
d concrete
which of the following is the same as 1 second
Answer:
theres no
Explanation:
picture or anything so... ur question doesnt make sense
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:
1. According to Newton's third law of motion, how are action
and reaction forces related?
2. How is momentum conserved?
3. Suppose you and a friend, who has exactly twice your mass,
are on skates. You push away from your friend. How does the
force with which you push your friend compare to the force
with which your friend
pushes you? How do your
accelerations compare?
4. Thinking Critical be ly comparing and Contrasting
Which has more momentum, a 250-kg dolphin swimming
at 6 m/s, or a 450-kg manatee swimming at 2 m/s?
(need 1-4 answered ASAP)
Answer:
you couldn't do this on your own or search it up on google
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
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.
please help im dying
A water skier is towing by motorboat at a constant velocity of magnitude 15 km /h. The boat speed up, and after a short interval the skier is towed at a new constant velocity of magnitude 20 km/h. What is the net force on the skier when she is moving at 15 km/h? And at 20 km/h?
Answer:
The skier is experimenting a net force of 0 newtons in both cases. ([tex]v_{1} = 15\,\frac{km}{h}[/tex], [tex]v_{2} = 20\,\frac{km}{h}[/tex])
Explanation:
According to Newton's First Law, an object is in equilibrium when it is either at rest or moving at constant velocity, which means that net force is equal to 0 newtons.
Therefore, the skier is experimenting a net force of 0 newtons in both cases.
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.
What 2 gases make up most of the atmosphere?
Answer:
nitrogen oxygen carbon dioxideand argon
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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A force of 500 N acts for a time interval of 0.001 second on an object of mass 0.20 kg that was initially at rest. Calculate the final velocity of the object after the force acts
Answer:
100
Explanation:
The final velocity of the object after the force acts is 2.5 meters per second.
To calculate the final velocity of the object after the force acts, we can use Newton's second law of motion, which states that:
Force = Mass × Acceleration
We need to find the acceleration of the object first. Since the object was initially at rest, its initial velocity (u) is 0 m/s.
The force acting on the object is 500 N, and the mass of the object is 0.20 kg. We can rearrange the equation to find the acceleration (a):
Acceleration (a) = Force / Mass
a = 500 N / 0.20 kg
a = 2500 m/s²
Now, we can use the kinematic equation to find the final velocity (v) of the object:
Final velocity (v) = Initial velocity (u) + (Acceleration × Time)
The initial velocity (u) is 0 m/s (since the object was initially at rest), and the time interval (t) is 0.001 second.
Final velocity (v) = 0 m/s + (2500 m/s² × 0.001 s)
v = 0 m/s + 2.5 m/s
v = 2.5 m/s
Hence, the final velocity of the object after the force acts is 2.5 meters per second.
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Protons are found only in the atomic nucleus.
TRUE
FALSE
Answer:
true
Explanation:
Protons and neutrons are found in the nucleus
A soccer player applies a force of 48.4 N to a soccer ball while kicking it. If the ball has
a mass of 0.44 kg, what is the acceleration of the soccer ball?
A. 27.3 m/s2
B. 21.3 m/s2
C. 110 m/s2
D. 104 m/s2
Answer:
C. 110 m/s2
Explanation:
Force = Mass x Acceleration
Since we have the force and the mass, we can rearrange this equation to solve for acceleration by dividing both sides by mass:
Force/Mass = (Mass x Acceleration)/Mass
Acceleration = Force/Mass
Now we just have to plug in our values and calculate!
Acceleration = 48.4/0.44
Acceleration = 110m/s/s
It is option C. 110 m/s2
Hope this helped!
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.
A moving object with a decreasing velocity covers distance during
each new second than it covered in the previous second.
A.the same
B.more
C.less
A rocket takes off from Earth. It travels 825km in 75 seconds. What is the
speed of the rocket?
0 1 km/s
11 km/s
O 1.1 km/s
O 111 km/s
Answer:
11 km/s
Explanation:
v=s/t
v=825km/75s
v=11km/s
A moving car skids to a stop with the wheels locked across a level roadway. Of the forces listed, identify which act on the car.
Normal
Gravity
Applied
Friction
Tension
Air resistance
Answer:
Normal, Gravity, Friction, and Air Resistance.
Explanation:
When a moving car skid to stop and its wheels are locked across, then the following forces will be applied on the car:
Normal force: It will act counter to gravity that pushes an object against a surface and acts perpendicular to the contact surface.
Gravity: Gravity force acts in each and every object having mass and it can not be avoidable. So, the gravity force will also apply to the car and attract it to the earth's surface.
Friction: Friction is a force that acts opposite to the motion and stops or slows motion. Friction will be applied to the car that will oppose the motion of the car and stop it.
Air resistance: air resistance is defined as the forces exerted by air that acts opposite to the relative motion of an object. Air resistance will also be applied to the car when it will skid to stop as we are always surrounded by the air.
Hence, the correct answers are "Normal, Gravity, Friction, and Air Resistance."
In economics, _________ is the amount of a resource that firms and producers are willing and able to provide to the marketplace
sectors are the amount of resources
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 dog is 60 m away while moving at a constant velocity of 10 m/s towards you. How Long will it be before the dog is close enough to your face?
Answer:
6 seconds
Explanation:
The answer is 6 because if the dog is going 10 m/s it will take 6 seconds to get from 60 meters away to you.
A student fires a cannonball vertically upwards. The cannonball returns to the
ground after a 4.60s flight. Determine all unknowns and answer the following
questions. Neglect drag and the initial height and horizontal motion of the
cannonball. Use regular metric units (ie. meters).
How long did the cannonball rise?
unit
What was the cannonball's initial speed?
unit
What was the cannonball's maximum height?
unit
V
Answer:
(a). The distance is 207 m.
(b). The initial velocity is 45.0 m/s
(c). The maximum height is 103.3 m
Explanation:
Given that,
Time = 4.60 s
We need to calculate the initial velocity
Using equation of motion
[tex]v=u-gt[/tex]
Put the value into the formula
[tex]0=u-9.8\times4.60[/tex]
[tex]u=45.0\ m/s[/tex]
We need to calculate the distance
Using formula of distance
[tex]d=v\times t[/tex]
Put the value into the formula
[tex]d=45\times4.60[/tex]
[tex]d=207\ m[/tex]
We need to calculate the maximum height
Using equation of motion
[tex]v^2=u^2-2gh[/tex]
Put the value into the formula
[tex]0=(45.0)^2-2\times9.8\times h[/tex]
[tex]h=\dfrac{(45.0)^2}{2\times9.8}[/tex]
[tex]h=103.3\ m[/tex]
Hence, (a). The distance is 207 m.
(b). The initial velocity is 45.0 m/s
(c). The maximum height is 103.3 m
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.
At a race car driving event, a staff member notices that the skid marks left by the race car are 9.06 m long. The very experienced staff member knows that the deceleration of a car when skidding is -40.52 m/s2. What can the staff member estimate for the original speed of the race car if it came to a stop during the skid?
Answer:
27.1 m/s
Explanation:
Given that at a race car driving event, a staff member notices that the skid marks left by the race car are 9.06 m long. The very experienced staff member knows that the deceleration of a car when skidding is -40.52 m/s2.
Using third equation of motion,
V^2 = U^2 + 2aS
Since the car is decelerating, the final velocity V = 0
Substitute all the parameter into the equation above,
0 = U^2 - 2 * 40.52 * 9.06
U^2 = 734.22
U = [tex]\sqrt{734.22}[/tex]
U = 27.096
U = 27.1 m/s approximately
Therefore, the staff member can estimate for the original speed of the race car to be 27.1 m/s if it came to a stop during the skid
The moon euphoria is in a circular orbit around Jupiter. Which of the following describes the direction of the centripetal force acting on euphoria.
A) pointing hand into the path of its orbit
B) pointing towards the sun
C)Pointing away from Jupiter
D) pointing towards Jupiter
can we add 2 atoms together? 3? How do particles combine to form the variety of matter one observes?
7. Your teacher has mixed salt, pepper, and water. Describe a procedure that you could use to separate this mixture. Be sure to list all of the materials you would need, your set up, and your expected results.
Answer:
hope this could help you
2
How is acceleration related to force when mass is constant, according to Newton's second law of motion?
A. The acceleration is directly proportional to the net force,
OB
The acceleration is inversely proportional to the net force.
C. The acceleration is inversely proportional to the square root of the net force.
OD
The acceleration is directly proportional to the square root of the net force.
Reset
Next Question
Answer:
A
Explanation:
The acceleration of an object is directly proportional to its net force.
[tex]a = \frac{f}{m} [/tex]
Answer:
Correct choice: A. The acceleration is directly proportional to the net force
Explanation:
Newton's Second Law of Motion
According to Newton's second law of motion, the acceleration of an object is directly proportional to the net force F and inversely proportional to the object's mass m:
[tex]\displaystyle a=\frac{F}{m}[/tex]
Correct choice: A. The acceleration is directly proportional to the net force
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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A freshly prepared sample of radioactive isotope has an activity of 10 mCi. After 4 hours, its activity is 8 mCi. Find: (a) the decay constant and half-life T1/2; (b) How many atoms of the isotope were contained in the freshly prepared sample? (c) What is the sample's activity 30 hours after it is prepared?
Answer:
(a). The decay constant is [tex]1.55\times10^{-5}\ s^{-1}[/tex]
The half life is 11.3 hr.
(b). The value of N₀ is [tex]2.38\times10^{11}\ nuclei[/tex]
(c). The sample's activity is 1.87 mCi.
Explanation:
Given that,
Activity [tex]R_{0}=10\ mCi[/tex]
Time [tex]t_{1}=4\ hours[/tex]
Activity R= 8 mCi
(a). We need to calculate the decay constant
Using formula of activity
[tex]R=R_{0}e^{-\lambda t}[/tex]
[tex]\lambda=\dfrac{1}{t}ln(\dfrac{R_{0}}{R})[/tex]
Put the value into the formula
[tex]\lambda=\dfrac{1}{4\times3600}ln(\dfrac{10}{8})[/tex]
[tex]\lambda=0.0000154\ s^{-1}[/tex]
[tex]\lambda=1.55\times10^{-5}\ s^{-1}[/tex]
We need to calculate the half life
Using formula of half life
[tex]T_{\dfrac{1}{2}}=\dfrac{ln(2)}{\lambda}[/tex]
Put the value into the formula
[tex]T_{\dfrac{1}{2}}=\dfrac{ln(2)}{1.55\times10^{-5}}[/tex]
[tex]T_{\dfrac{1}{2}}=44.719\times10^{3}\ s[/tex]
[tex]T_{\dfrac{1}{2}}=11.3\ hr[/tex]
(b). We need to calculate the value of N₀
Using formula of [tex]N_{0}[/tex]
[tex]N_{0}=\dfrac{3.70\times10^{6}}{\lambda}[/tex]
Put the value into the formula
[tex]N_{0}=\dfrac{3.70\times10^{6}}{1.55\times10^{-5}}[/tex]
[tex]N_{0}=2.38\times10^{11}\ nuclei[/tex]
(c). We need to calculate the sample's activity
Using formula of activity
[tex]R=R_{0}e^{-\lambda\times t}[/tex]
Put the value intyo the formula
[tex]R=10e^{-(1.55\times10^{-5}\times30\times3600)}[/tex]
[tex]R=1.87\ mCi[/tex]
Hence, (a). The decay constant is [tex]1.55\times10^{-5}\ s^{-1}[/tex]
The half life is 11.3 hr.
(b). The value of N₀ is [tex]2.38\times10^{11}\ nuclei[/tex]
(c). The sample's activity is 1.87 mCi.