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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compare and contrast speed and strength
Answer:
Down Below
Explanation:
As nouns the difference between speed and strength
is that speed is the state of moving quickly or the capacity for rapid motion; rapidity while strength is the quality or degree of being strong.
Answer:
Explanation:is that speed is the state of moving quickly or the capacity for rapid motion; rapidity while strength is the quality or degree of being strong.
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:
21000 km/hr how far will it travel in 5 hours
Answer:
105000 is the answer your welcome
Explanation:
105000
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
3. A person who is initially stationary is eventually walking
at a speed of 1.5 m/s after an acceleration of 0.5 m/s?,
calculate the distance it takes them to reach this speed.
4. A lorry pulls forward after initially being stationary, it
takes the lorry 40 m to reach a speed of 8 m/s, calculate the
lorry's acceleration.
7. A person begins moving after initially being stationary, the person
accelerates at 0.5m/s2 over a distance of 9m what is their final speed
Explanation:
3. A person who is initially stationary is eventually walking
at a speed of 1.5 m/s after an acceleration of 0.5 m/s?,
calculate the distance it takes them to reach this speed.
Given parameters:
Initial velocity = 0m/s
Final velocity = 1.5m/s
Acceleration = 0.5m/s
Unknown:
Distance = ?
Solution:
To solve this problem, we apply the relevant motion equation;
V² = U² + 2aS
where V is the final velocity
U is the initial velocity
a is the acceleration
S is the distance
Input the parameters;
1.5² = 0 + 2 x 0.5 x S
2.25= S
The distance to reach this speed is 2.25m
4. A lorry pulls forward after initially being stationary, it
takes the lorry 40 m to reach a speed of 8 m/s, calculate the
lorry's acceleration.
Given parameters:
Initial velocity = 0m/s
Final velocity = 8m/s
Distance = 40m
Unknown:
Acceleration = ?
Solution:
V² = U² + 2aS
Input the parameters and solve;
8² = 0² + 2 x a x 40
64 = 80a
a = [tex]\frac{64}{80}[/tex] = 0.8m/s²
7. A person begins moving after initially being stationary, the person
accelerates at 0.5m/s2 over a distance of 9m what is their final speed
Given parameters:
Initial velocity = 0m/s
acceleration = 0.5m/s²
distance = 9m
Unknown:
Final velocity = ?
Solution:
V² = U² + 2aS
Input the parameters and solve;
V² = 0² + 3 x 0.5 x 9
V² = 13.5
V = 3.7m/s
conversion
18230 mJ to kJ?
Answer:
0.01823
Explanation:
big brain
a tennis ball is thrown straight up with an initial velocity of 22.5 m/s how much total time is the ball in the air
Answer:
4.6s
Explanation:
v=u+at
0=22.5+(-9.8)t
-22.5=-9.8t
t=-22.5/-9.8
t=2.295 s
The total time will double
2.295×2=4.59s
=4.6s
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."
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
please help me this is due in 30 minutes. If you can do three that would be so great.
Answer:
1. 3000 N
Explanation:
you multiply 1000kg and 3
Answer:
2. acceleration will double
The acceleration will increase for example: 4/2=2 m/s^2 and if you double 4 it will be 8/2=4 m/s^2. Which means the acceleration also doubles.
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
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
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
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!
a car travels at a speed of 50 m/s for 13 seconds. what is the distance that the car travels?
How many meters did the car go in the first 4 seconds?
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.
can we add 2 atoms together? 3? How do particles combine to form the variety of matter one observes?
Where are streams located?
Answer:
Larger seasonal streams are more common in dry areas. Rain-dependent streams (ephemeral) flow only after precipitation. Runoff from rainfall is the primary source of water for these streams. Like seasonal streams, they can be found anywhere but are most prevalent in arid areas.
Explanation:
Carlos gets tired of pushing and instead begins to pull with force Fpull at an angle to the horizontal.
The block slides along the rough horizontal surface at a constant speed. A free-body diagram for the
situation is shown below. Blake makes the following claim about the free-body diagram:
Blake: “The velocity of the block is constant, so the net force exerted on the block must be zero.
Thus, the normal force FN equals the weight Fmg, and the force of friction Ff equals the applied
force Fpull.”
What, if anything, is wrong with this statement? If something is
wrong, identify it and explain how to correct it. If this statement is
correct, explain why.
Answer:
The wrong items are;
1) The normal for FN equals the weight Fmg
2) The force of friction, Ff, equals the applied force Fpull
The corrected statements are;
1) The normal force is weight less the vertical component of the applied force Fpull
FN = Fmg - Fpull × sin(θ)
2) The force of friction equals the horizontal component of the applied force Fpull
Ff = Fpull × cos(θ)
Explanation:
The given statement was;
The velocity of the block is constant, so the net force exerted on the block must be zero. Thus, the normal force FN equals the weight Fmg, and the force of friction Ff equals the applied force Fpull
By the equilibrium of forces actin on the system, given that the applied force acts at an angle, θ, with the horizontal, we have;
The normal force is equal to the weight less the vertical component of the applied force;
That is we have, FN = Fmg - Fpull × sin(θ)
The friction force similarly, is equal to the horizontal component of the applied force;
Ff = Fpull × cos(θ)
The wrong items are therefore as follows;
1) The normal for FN equals the weight Fmg
1 i) The normal force is weight less the vertical component of the applied force Fpull
FN = Fmg - Fpull × sin(θ)
2) The force of friction, Ff, equals the applied force Fpull
2 i) The force of friction equals the horizontal component of the applied force Fpull
Ff = Fpull × cos(θ).
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.
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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What is the molar mass of the molecule C8H18, crude oil?
Answer:
114.23 g/mol
Explanation:
Consider a buggy being pulled by a horse.
Which is correct?
1. The horse can pull the buggy forward only
if the horse weighs more than the buggy.
2. The force on the buggy is as strong as the
force on the horse. The horse is joined to the
Earth by flat hoofs, while the buggy is free to
roll on its round wheels.
3. The horse pulls before the buggy has time
to react so they move forward.
4. The horse pulls forward slightly harder
than the buggy pulls backward on the horse,
so they move forward.
Answer:
Option 2
Explanation:
Answer:B
Explanation:because she started of with no money and now she made her money up
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
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 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 present in sodium atom.
a 11
b 10
C 12
d 9
11
Explanation:
There are 11 protons in a sodium atom
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