Light with a frequency of 7. 30 x 10^14 hz lies in the violet region of the visible spectrum. The wavelength of this frequency of light is 4×10^-7 m.
As we know
c = λf
where
c is the speed of light, λ is wavelength and f is frequency.
Given:
c = 3×10^8
f=7.3×10^14
Putting the values
λ = c/f
λ = 3×10^8/7.3×10^14
λ = 4×10^-7 m.
The wavelength is 4×10^-7 m.
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A spherical balloon is inflated with gas at a rate of 700 cubic centimeters per minute. how fast is the radius of the balloon changing at the instant the radius is 40 centimeters?
Rate of change is used to mathematically describe the percentage change in value over a defined period of time
A spherical balloon is inflated with gas at a rate of 700 cubic centimeters per minute. the radius of the balloon will be changing at a rate of 0.035 cm/min at the instant the radius is 40 centimeters
given
Radius = 40 cm
dv / dt = 700
to find
Rate by which radius of balloon is changing = ?
volume of sphere = 4/3 π [tex]r^{3}[/tex]
dv/dt = 4/3 * π [tex]r^{3}[/tex] dr/dt
dv/dt = 4/3 * 3 * π [tex]r^{2}[/tex] dr/dt
dv/dt = 4/3 * 3 * (3.14) * ([tex]40^{2}[/tex]) dr/dt
dv/dt = 20096 dr/ dt
700 = 20096 dr/dt
dr/dt = 700 / 20096
= 0.035 cm/min
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Hurry, please i will give crown
in a short-answer response, thoroughly describe what free-body diagrams are used for and what the free-body diagram is telling us about the directions of the forces. Make sure to include which direction the object will move and give an example of what that object could be. (examples: box, soccer ball, rope in a tug-of-war, etc.)
The free body diagram is the diagram that shows us the way that forces act on an object.
What is the Freebody diagram?The free body diagram is the diagram that shows us the way that forces act on an object. We know that force is the reason for the motion of an object. An object moves when it is acted upon by unbalanced forces. If the forces that are acting on the object are balanced the object would not move,
You know that force is a vector, thus the magnitude and the direction of the force is very important when we are trying to describe the action of a system of forces that acts on a body.
For instance, a book could sit still on a table because the book is acted upon by two forces that are balanced these are the weight of the book and the reaction force of the table on the book. These balanced forces makes the book to sit still on the table.
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a fence post is 51.0 m from where you are standing, in a direction 35.0 ∘ north of east. a second fence post is due south from you. a fence post is 51.0 m from where you are standing, in a direction 35.0 ∘ north of east. a second fence post is due south from you.
Acceleration-∘ north of east. a second fence post is due south from you. a fence post is 51.0 m from where you are standing, in a direction 35.0 ∘
What is acceleration?Acceleration is the wrathful, flaming dragon of motion variables as compared to displacement and velocity. If it's big, it makes you pay attention because it might be violent, some people are afraid of it, and it can be violent. You can feel yourself accelerating while you're in a moving vehicle, such as a go kart, slamming on the brakes, or when you're seated in a moving vehicle, such as a plane, during takeoff.
Anything that causes a change in velocity is referred to as an acceleration. You can only accelerate in one of two ways—either by changing your speed or your direction, or both—since velocity is a function of both speed and direction.
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Please help if I dont get this I have two packets for hw. As the universe expands the temperature ______ and the wavelengths of photons ______.
As the universe expands the temperature decreases and the wavelengths of photons increases.
Effect of expansion of the universe on the wavelength of light?
As the universe expands, the distance between crests of the wave of light also expand, causing the wavelength to increase.
Light with a longer wavelength is redder, so light appears redshifted because of the expansion of the universe.
Also, the universe expands as a result of decrease in the temperature of surrounding air molecules according to Hubble.
Thus, we can conclude that, as the universe expands the temperature decreases and the wavelengths of photons increases.
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The speed of individual particle diffusion is influenced by temperature and particle size, not by concentration. True or false?.
Answer:
The speed of individual particle diffusion is influenced by temperature and particle size, not by concentration
Explanation:
What is the average velocity of the particle from rest to 9 seconds?
201
18-
16-
Displacement (m)
86420
A.
Time (s)
1 meter/second
B.
2 meters/second
OC.
3 meters/second
OD. 4 meters/second
R₂
10
2m/s is the average velocity of the particle.
Velocity of an object can be defined as the change of distance divided by time it's SI unit is metre per second formula v= d/t.
here we have given that the time taken by the particle is 9s and the displacement is 18m/s
so by applying the given data in the formula
v=d/t
=18/9
=2m/s
so from here we can state that for a object or a paricle with a displacement of 18m and the time taken by it is 9 s the velocity of that will be v=d/t that is 18/9 which is 2m/s.
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Is Soil compound or element or mixture
Answer: mixture
Explanation:
Answer:
soil is a mixture of minerals, dead and living organisms (organic materials), air, and water. These four ingredients react with one another in amazing ways, making soil one of our planet’s most dynamic and important natural resources.
Another definition of soil
The surface mineral and/or organic layer of the earth that has experienced some degree of physical, biological and chemical weathering.
In summary, soil is a mixture....
According to the law of multiple proportions, if 12 g of carbon combine with 16 g of oxygen to form co, the number of grams of carbon that combine with 16 g of oxygen in the formation of co2 is ________.
The number of grams of carbon that combine with 16 g of oxygen in the formation of CO₂ is 6g.
When two elements combine to make more than one compound, the masses of one element combined with a fixed amount of another element are in the ratio of whole numbers, according to the law of multiple proportions.
When combined with oxygen, carbon can produce two different compounds. They are referred to as carbon dioxide (CO₂) and carbon monoxide (CO).
Carbon monoxide is formed by combining 12 g of carbon with 16 g of oxygen whereas Carbon dioxide is formed when 12 g of carbon reacts with 32 g of oxygen. The amount of carbon is fixed at 12 g in each case. The mass ratio of carbon monoxide to carbon dioxide is 16: 32, or 1: 2.
But in the given case, 16g of oxygen is reacting instead of 32g. Therefore, the number of grams of carbon reacting will be:
[tex]\frac{12}{2}=6g[/tex]
Thus, 6g of carbon will react with 16g of oxygen to form carbon dioxide.
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. Calculate the acceleration in meters per second-squared, of a car moving at 15 m/s that accelerates to 28 m/s over a time of 5.3 seconds.
The acceleration of the car is 2.45 m/s².
Acceleration is the rate of change of the velocity of an object with respect to time. Its unit is m/s² and can be given as,
a = dv/dt
Where , dv and dt is the small change in velocity and time of an object.
Velocity of an object can be said as, the rate of change of displacement with respect to time of an object . Its unit is m/s and it is a vector quantity.
V = Δx/Δt
In here , V is velocity , Δx is change in displacement and Δt is change in time of an object.
We are know that ,
Initial velocity of the car = u = 15 m/s
Final velocity of the car = v = 28 m/s
Covered time of the car = t = 5.8 s
Therefore we know the equation of motion can be given as,
v = u + at
Hence , to get the acceleration of the car we may put the values in above equation then,
28 m/s = 15 m/s + (5.3) a
a = (28 m/s - 15 m/s) /5.3s
a = 2.45 m/s²
Thus , acceleration of a car is 2.45 m/s²
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When an alternating current circuit supplies power to a ? load, the circuit voltage and current are in-phase with each other. this is also known as unity power factor.
you push a hockey puck that is initially at rest on slick ice by applying a constant force until the puck reaches a final velocity of 1 m/s. on the second attempt, you want the hockey puck to reach the same final velocity by applying a force that is twice as large.
For reaching the same final velocity as the first attempt by applying a force that is twice as large as the first attempt the time interval should be kept shorter than the first attempt.
After the hockey puck has reached the final velocity and suddenly stops, the hockey puck reduces its speed abruptly.
These solutions can be explained by Newton's Second Law of Motion. According to Newton's Second Law of Motion, the force applied to a particle is directly proportional to the product of the mass and acceleration of that particle.
F = ma; where m is the mass of the particle and a is the acceleration.
Also, a = v/t; where v is the velocity of the particle and t is the time interval.
Therefore, the force applied is inversely proportional to the time interval.
F = mv/t
So, to maintain the same velocity keeping the force at double the time interval should be halved.
Therefore, for reaching the same final velocity as the first attempt by applying a force that is twice as large as the first attempt the time interval should be kept shorter than the first attempt.
Also, the Force applied to the particle is directly proportional to the velocity of the particle. So, if the force applied to the hockey puck is suddenly stopped its velocity will reduce abruptly.
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Complete Question:
You push a hockey puck that is initially at rest on slick ice by applying a constant force until the puck reaches a final velocity of 1 m/s. On the second attempt, you want the hockey puck to reach the same final velocity by applying a force that is twice as large.
1. Therefore, you must exert the force for a time interval that is
A. shorter than the time interval of your first attempt.
B. longer than the time interval of your first attempt.
C. the same as the time interval of your first attempt.
2. After the hockey puck has reached the final velocity, you suddenly stop pushing it. The hockey puck:
A. stops abruptly
B. reduces speed gradually
C. continues at constant velocity
D. increases speed gradually
E. reduces speed abruptly
To find your Target Heart Rate (THR), you need to know your age and your Resting Heart Rate (RHR).
True
False
Sally runs around a 400 meter track 2 times. What is Sally's distance traveled?
Answer:
0.5 miles or 800 meters!
Explanation:
Sally traveled 0.5 miles or 800 meters.
I'm not too sure if this is what you were asking but if you need any more help let me know! :D
The mass of the car is 1200 kg.
(a) Calculate, for the first 20s of the motion,
(1) the distance travelled by the car
The distance travelled by the car in the first 20 seconds of the motion is 130 m.
What is the area under speed time graph?
The area under the speed-time graph is the distance the particle travels.
The distance traveled by the car in the first 20 seconds is calculated as follows;
From the speed - time graph, at 20 seconds, the speed of the car is 13 m/s.
Distance = area of the graph = area of triangle formed from time, 0 s to 20 seconds.
Distance = ¹/₂bh
where;
b is the base of the triangle formed = 20 s - 0 s = 20 sh is the height of the triangle formed = 13 m/sSubstitute the given parameters and solve for the distance traveled by the car.
Distance = ¹/₂bh
Distance = ¹/₂(20)(13)
Distance = 130 m
Thus, the distance travelled by the car in the first 20 seconds of the motion is 130 m.
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The missing graph is in the image attached.
an observer in a moving car with 80 km/h was observing a moving car with 90km/h in the same direction do the observed speed of the second car is
Answer:
.......
Explanation:
A point charge of -3. 00 μC is located in the center of a spherical cavity of radius 6. 80 cm inside an insulating spherical charged solid. The charge density in the solid is 7. 35 x 10^-4 C/m^3
Calculate the magnitude of the electric field inside the solid at a distance of 9. 30 cm from the center of the cavity
A point charge of -3. 00 μC is located in the center of a spherical cavity of radius 6. 80 cm inside an insulating spherical charged solid. The charge density in the solid is 7. 35 x 10^-4 C/m^3. The magnitude of the electric field inside the solid at a distance of 9.20 cm from the center of the cavity is 394.6857x10⁻⁶ N/C.
Charge inside the cavity = -3μC = -3x10⁻⁶ C
Charge density of the sphere, ρ = 7.35x 10⁻⁴ C/m³
Radius of he cavity = 6.90 cm = 0.0690 m
We have to find the electric field at a radius of 9.20 cm, therefore, the effective radius,
Effective radius = 0.092-0.069 = 0.023 m
Effective Volume = [tex]\frac{4}{3}[/tex] π[tex]r^{3}[/tex]= 0.5093 x10⁻⁴ m³
The charge, Q = volume x charge density = 374.33 x 10⁻⁶ C
The net charge enclosed,
[tex]Q_{in} =(374.33-3)x10^{-6} =371.33 x10^{-6}[/tex]C
We know according to the Gauss law,
⇒E=[tex]\frac{Q_{in} }{e_{o} }[/tex]
E=394.6857 X[tex]10^{-6}[/tex]N/C
Hence, the magnitude of the electric field inside the solid at a distance of 9.20 cm from the center of the cavity is 394.6857x10⁻⁶ N/C.
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a satellite is orbiting at a distance of 4.2x10⁶ m from the surface of the Earth. The radius of the Earth is 6.4x10⁶ m. What is the ratio of gravitational force on the satellite in orbit to gravitational force on the surface of the Earth?
Answer:
372 252
Explanation:
At a certain instant a moving object comes to
momentary rest.
Is that object accelerating at that moment?
Answer: no
Explanation:
momentary rest = object is NOT in motion at that instant of time
In the moment that the object is at rest, the velocity is 0 meaning that there is no acceleration.
a car is travelling to the right with a speed of 42 m/s when the driver slams on the brakes. the car skids for 4 s with constant acceleration before it comes to a stop. what distance does the car travel as it skids to a stop?
The car that travels to the proper with a speed of 42 m/s, skids 84 meters for 4 seconds before it involves a stop.
Solving :
The distance traveled by car before coming to a stop can be calculated with the following equation:
[tex]v^{2} f_{} =v^{2} i_{} + 2ad[/tex]
Where:
[tex]v_{f}[/tex]: is that the final speed = 0 (it stops)
[tex]v_{i}[/tex]: is that the initial speed = 42 m/s
a: is that the acceleration
d: is that the distance =?
We need to find the acceleration. we will use the next equation:
[tex]v_{f} = v_{i} +at[/tex] ...........2
Where:
t: is that the time = 4.0 s
Hence, the acceleration is:
a = [tex]\frac{v_{f} -v_{i} }{t}[/tex]
a = [tex]\frac{0-42m/s}{4.0s}[/tex]
a = - 10.5 m/s²
Now, the car skid the subsequent meters before coming to a stop (eq 1).
d = [tex]\frac{v^{2}_{f} - v^{2} _{f} }{2a}[/tex]
d = 84 m
Therefore, the car skids 84 meters before coming to a stop.
Constant acceleration :
Constant acceleration refers to motion where the speed increases by the identical amount each second. the foremost notable and important example is free fall. When an object is thrown or dropped, it experiences a continuing acceleration due to gravity, which features a constant value of approximately 10 meters per second squared
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an ink-jet printer steers charged ink drops vertically. each drop of ink has a mass of 10-11 kg, and a charge due to 751896 extra electrons. it goes through two electrodes that gives a vertical acceleration of 104 m/s2. the deflecting electric field is mv/m.
The deflecting electric field is 9.50 x 10^14 mv/m.
The force on the drop of ink due to vertical acceleration = ma
And the force on the drop due to electrodes = qE
To calculate the deflecting electric field, we need to equate both these forces,
Hence, qE = ma
As there are 751896 extra electrons, the charge due to them will be = 751896 x 1.6 x 10^-19
M = mass of each drop of ink = 11 kg
A = acceleration of the drop of ink = 104 m/s^2
Putting the above values in the equation, qE = ma
1.6 x 10^-19 x 751896 x E = 11 x 104
On solving the above equation
E = 9.50 x 10^14 mv/m
Hence, the deflecting electric field is 9.50 x 10^14 mv/m.
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A 4. 0-g string is 0. 36 m long and is under tension. The string vibrates at 500 hz in its third harmonic. What is the wavelength of the standing wave in the string?.
the wavelength of the standing wave in the string of third harmonic is 0.24 meters.
We need to know about the third harmonic wavelength to solve this problem. When the string is vibrating at its third harmonic, it should follow the rule
λ = 2/3 L
where λ is wavelength and L is the length of string
From the question above, we know that
f = 500 Hz
L = 0.36 m
m = 4 g
By substituting to the following equation, we can calculate the wavelength
λ = 2/3 . L
λ = 2/3 . 0.36
λ = 0.24 m
Hence, the wavelength at its third harmonic is 0.24 meters.
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How do signals from sensory neurons reach motor neurons?
Responses:
A-Signals move from dendrite to dendrite towards motor neurons.
B-Through interneurons in the brain and spinal cord that connect sensory neurons and motor neurons.
C-Motor neurons signal for the body to move the signal toward them.
D-Signals from sensory neurons do not get sent to motor neurons.
B. Signals from sensory neurons reach motor neurons through interneurons in the brain and spinal cord that connect sensory neurons and motor neurons.
What is sensory neurons?Sensory neurons are the nerve cells that work based on the sensory input from outside stimulus.
What is motor neurons?Motor neurons are special type of neurons or never cells that are found within the spinal cord and the brain.
How do signals from sensory neurons reach motor neurons?The sensory neurons carry afferent impulse to the central interneuron, which connects with the motor neuron.
The motor neuron carries efferent impulses to the effector, and cause the response observed.
Thus, signals from sensory neurons reach motor neurons through interneurons in the brain and spinal cord that connect sensory neurons and motor neurons.
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If the speedometer of a car reads a constant speed of of 50 km/h can you say that the car has a constant velocity why or why not
No. You can say that the car had a constant speed, but not constant velocity if the speedometer consistently displays 50 km/h.
The car's velocity would change if its direction changed, even though it was moving at the same pace.
What is velocity?The speed at which something moves in a specific direction is known as its velocity.
Velocity is the pace and direction of an object's movement, whereas speed is the time rate at which an object is traveling along a path.
Consider the speed of a car driving north on a highway as an example. Because the velocity vector is scalar, its absolute value magnitude will always equal the motion's speed.
The standard unit of velocity magnitude is the m/s.
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a string or rope will break apart if it is placed under too much tensile stress. thicker ropes can withstand more tension without breaking because the thicker the rope, the greater the cross-sectional area and the smaller the stress. one type of steel has density 7710 kg/m3 and will break if the tensile stress exceeds 7.0×108n/m2. you want to make a guitar string from a mass of 4.2 g of this type of steel. in use, the guitar string must be able to withstand a tension of 900 n without breaking. your job is the following.
The maximum length the strig can have is 1.89 x 10^10 m.
Tensile stress may be defined as the external forces acting per unit area on a material which cause the material to stretch. This stress acts along the axis of the material.
Density is given by the formula:
Density= mass/volume
So, volume is given by;
Volume= mass/density
Mass of string in kg= 4.2/1000= 0.0042 kg.
Density of steel is in kg/m^3.
Then volume V will be;
V= 0.0041/7710 = 5.31 x 10-7m^3
Now we know that stress = Force/Area
Thus, Area is given by; Area = force/stress => 900/7*10^8= 2795.95m^2
Length can be determined by= volume/Area =>5.31^10-7/2795.95
Therefore, the maximum Length of string is = 1.89 x 10^10 m.
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Complete Question:
If A string or rope will break apart if it is placed under too much tensile stress. Thicker ropes can withstand more tension without breaking because the thicker the rope, the greater the cross-sectional area and the smaller the stress. One type of steel has a density of 7710kg/m3 and will break if the tensile stress exceeds 7.0×108N/m2. You want to make a guitar string from a mass of 4.2 g of this type of steel. In use, the guitar string must be able to withstand a tension of 900 N without breaking. Your job is the following.
(a) Determine the maximum length the string can have
Of brick
(a)
the pushing force does not make the brick move. explain why.
The pushing force does not make the brick move as there is not enough power being exerted.
What is Force?A force is an effect that can alter an object's motion according to physics. An object with mass can change its velocity, or accelerate, as a result of a force. An obvious way to describe force is as a push or a pull. A force is a vector quantity since it has both magnitude and direction.For instance, the Moon is subject to the gravitational pull of Earth. A force that acts perpendicular to the surface is called a normal force. In more detail, it is a contact force that pulls back on a surface-placed object. For instance, a book placed on a tabletop is affected by an upward normal force.A force is the result of the interaction between two things and can be either a push or a pull. Due to the fact that it is a vector quantity, it possesses both magnitude and direction. The object's state of motion, direction, size, and shape could all change as a result.The pushing force does not make the brick move as there is not enough power being exerted.
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a lever is 8ft long. a force of 80lb is applied to one end of the lever and a force of 560lb is applied to the other end. where is the fulcrum located when the system balances
When a lever is 8ft long. a force of 80lb is applied to one end of the lever and a force of 560lb is applied to the other end. The fulcrum located when the system balances 6 feet from the greater force.
Define force.
X = length from greater weight to fulcrum
132X = 88(15-X)
132X = 1320-88X
220X = 1320
X = 6
The fulcrum located when the system balances 6 feet from the greater force.
An object experiences a push or pull as a result of interacting with another object. Each object is subject to a force whenever two objects interact. The two items no longer feel the force after the interaction ends.
In physics, a force is an effect with the capacity to change the velocity of an object. A force can cause an object with mass to move at a different speed or accelerate. To describe force, a push or a pull makes intuitive sense. Forces have both magnitude and direction because they are vector quantities.
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problem 11.021 - particle motion - dependent multi-part problem - assign all parts skip to question note: this is a multi-part question. once an answer is submitted, you will be unable to return to this part. the acceleration of a particle is defined by the relation a
The velocity of particle when x = -2m is 4.7 m/s.
Acceleration of a particle, a = k ( 1 - [tex]e^{-x}[/tex])
Velocity of the particle, v = +9 when x = -3m
Velocity of the particle, v = 0 when x = 0
We konw that acceleration is:
[tex]a = \frac{dv}{dt}[/tex]
[tex]a= \frac{dx}{dt}.\frac{dv}{dx}[/tex]
[tex]a=v[\frac{dv}{dx}][/tex]
Now,
v dv = a dv .........(i)
Putting a = k ( 1 - [tex]e^{-x}[/tex]) in (i), we get,
[tex]\frac{vdv}{dx}=k (1-e^{-x} )[/tex]
[tex]\int\limits^9_0 {v} \, dv = k\int\limits^3 _0 {(1-e^{-x}) } \, dx[/tex]
[tex]\frac{81}{2} = k(-3 + e^{3}-e^{0} )[/tex]
[tex]\frac{81}{2}=k(-9 + e^{3})[/tex]
[tex]\frac{81}{2}= k[/tex] × [tex]16.0855[/tex]
[tex]k = 2.52[/tex]
Again,
[tex]\int\limits^v_0 {v} \, dx=k\int\limits^2_0 {(1-e^{-x)} } \, dx[/tex]
[tex]\frac{v^{2} }{2}= 2.52 [ (-2 +e^{2}-(0+e^{0})[/tex]
[tex]v= 4.70 m/s[/tex]
Therefore, the velocity of particle when x = -2m is 4.7 m/s.
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The acceleration of a particle is defined by the relation a= k (1 - [tex]e^{-x}[/tex]), where k is a constant. The velocity of the particle v =+9 m/s when x = -3m and it comes to rest at the origin. Obtaining velocity from a position-dependent acceleration by integration. Also, determine the velocity of the particle when x = -2 m.
The speed of a projectile when it reaches its maximum height is one-half its speed when it is at half its maximum height. What is the initial projection angle of the projectile?.
According to uniform motion, the initial projection angle is 73.89⁰.
We need to know about uniform motion to solve this problem. The uniform motion is an object's motion under acceleration. It should follow the rule
vt = vo + a . t
vt² = vo² + 2a . s
s = vo . t + 1/2 . a . t²
where vt is final velocity, vo is initial velocity, a is acceleration, t is time and s is displacement.
From question above, we know that
vpeak = 1/2 vt
2vpeak = vt
when projectile at maximum height should equal
vpeak = vx = vo.cosθ
When the projectile reaches a half its maximum height, the time taken is
t = 1/2 tpeak
t = 1/2 (vo sinθ/g)
t = vo sinθ/2g
Find the speed at its half maximum height.
vty = voy + a . t
vty = vo.sinθ - g.vo sinθ/2g
vty = vo sinθ/2
vtx = vo cosθ
vt = √vty² + vtx²
2vpeak = √(vo sinθ/2)² + (vo cosθ)²
2 . cosθ = √(sinθ/2)² + (cosθ)²
4cos²θ = 1/4 . sin²θ + cos²θ
3cos²θ = 1/4 . sin²θ
sin²θ / cos²θ = 12
tan²θ = 12
tanθ = √12
θ = 73.89⁰
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The change in internal energy for a given system is –325 j. The system loses 155 j of energy as heat at the same time that it expands from an initial volume of 15. 0 l at 1. 00 atm of pressure. What is the final volume of the system in liters? enter your response without units to the nearest 0. 1 l.
According to internal energy, the final volume is 19.75 liters.
We need to know about internal energy to solve this problem. For a closed system, with matter transfer excluded, the changes in internal energy are due to heat transfer and due to thermodynamic work done by the system on its surroundings. It can be determined as
ΔU = Q - W
where ΔU is the change in internal energy, Q is heat and W is work done.
From the question above, we know that
ΔU = -325 J
Q = 155 J
Vo = 15 L = 0.015 m³
P = 1 atm = 1.01 x 10⁵ Pa
By substituting the given parameter, we can calculate the work done
ΔU = Q - W
-325 = 155 - W
W = 480 joule
Find the final volume
W = P . (Vi - Vo)
480 = 1.01 x 10⁵ (Vi - 0.015)
Vi = 0.01975 m³
Vi = 19.75 liters
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you throws two stones from the top edge of a building with a speed of 48.0 m/s. you throw one straight down and the other straight up. the first one hits the street in a time t1. how much later is it before the second stone hits?
Answer:
H1 = V t1 - 1/2 g t1^2 where upward is chosen as positive
H2 = -V t2 - 1/2 g t2^1 second stone thrown downwards
V t1 - 1/2 g t1^2 = -V t2 - 1/2 g t2^2
V (t1 + t2) = 1/2 g (t1^2 - t2^2)
V = 1/2 g (t1 - t2)
t2 = 2 V / g = 96 / 9.8 = 9.8 sec
Check:
One can also note the stone 1 will return to the starting point with a time delay of V / g * 2 which is 96 / 9.8 = 9.8 sec because the time for stone 1 to go to zero is 48 / 9.80 = 4.90 sec when its velocity reaches zero
Approximately 9.8 seconds when rounded to one decimal place. Therefore option F is correct.
Calculate the time it takes for the first stone to hit the street (t1) and the time it takes for the second stone to hit the street (t2).
For the first stone (thrown straight down):
Using the equation of motion:
[tex]\rm \[ d = \frac{1}{2} g t_1^2 \][/tex]
where:
d = distance traveled (height of the building)
g = acceleration due to gravity = [tex]\rm 9.8\ m/s^2[/tex] (downward)
Since the stone is thrown from the top of the building, the distance traveled (d) is equal to the height of the building.
Now, let's find t1:
[tex]\[ h = \frac{1}{2} g t_1^2 \]\\\ \\t_1^2 = \frac{2h}{g} \]\\\\\ t_1 = \sqrt{\frac{2h}{g}} \][/tex]
For the second stone (thrown straight up):
The initial velocity of the second stone is also 48.0 m/s, but it is directed upward. We can use the equation of motion for upward motion:
[tex]\rm \[ v = u - gt_2 \][/tex]
where:
v = final velocity = 0 m/s (at the peak of the motion, the velocity becomes 0)
u = initial velocity = 48.0 m/s (upward)
g = acceleration due to gravity = 9.8 [tex]\rm m/s^2[/tex] (downward)
Now, let's find t2:
[tex]\[ 0 = 48.0 - 9.8t_2 \]\\\\\ t_2 = \frac{48.0}{9.8} \]\\\\\ t_2 = 4.898 \][/tex]
Since the second stone takes time [tex]t_2[/tex] to reach the peak and another time [tex]t_2[/tex] to fall back down, the total time before the second stone hits the street is [tex]\( 2 \times 4.898 \)[/tex].
[tex]\[ t_{\text{total}} = 2 \times 4.898 \]\\\ t_{\text{total}} = 9.796 \][/tex]
The correct answer is approximately 9.8 seconds (F) when rounded to one decimal place.
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