Answer:
Explanation:
a ) The velocity will never be zero . The velocity will be minimum at the highest point of projectile, which will be equal to the horizontal component of the initial velocity.
b ) The velocity will be minimum when its kinetic energy will be minimum . Kinetic energy will be minimum when its potential energy will be maximum.
Its potential energy will be maximum at the highest point so velocity will be minimum at the highest point.
c ) Velocity will never be the same as initial velocity because constant force of gravitation is acting on the projectile all the time.
d ) At the moment when the projectile returns back and hits the ground, the speed becomes equal to the initial speed ( at t = 0 ) because its kinetic energy becomes the same as initial energy , the height becoming zero.
In projectile motion on level ground, the velocity is never zero. The velocity is minimum at the apex of the trajectory and maximum at the launch and impact points. The velocity can never be the same as the initial velocity at a time other than t=0, but the speed can be the same as the initial speed at a time other than t=0 when the projectile lands.
Explanation:a) In projectile motion on level ground with negligible air resistance, the velocity is never zero. The horizontal velocity remains constant throughout the motion, while the vertical velocity changes. However, the total velocity (magnitude of the velocity vector) is always non-zero.
b) The velocity is minimum when the projectile reaches its highest point, called the apex of the trajectory. The velocity is maximum at both the launch and impact points.
c) No, the velocity can never be the same as the initial velocity at a time other than t = 0. This is because the horizontal velocity remains constant during the motion, while the vertical velocity changes.
d) Yes, the speed (magnitude of the velocity vector) can be the same as the initial speed at a time other than t = 0. This occurs when the projectile lands, as the horizontal velocity is constant and the vertical velocity becomes opposite in direction but equal in magnitude to the initial vertical velocity.
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A student at the top of building of height h throws one ball upward with the initial speed V and then throws a second ball downward with the same initial speed. How do the final speeds of the balls compare when they reach the ground?
Answer:
They are the same (assuming there is no air friction)
Explanation:
Take a look at the picture.
When the first ball (the one thrown upward) gets to the point marked as A, the speed will has the exact same value V but the velocity will now point downward (just like the second ball).
So if you think about it, the first ball, from point A to the ground, will behave exactly like the second ball (same initial speed, same height).
That is why the speeds will be the same when they reach the ground.
Answer:
Velocity is same
Explanation:
Case I:
When the ball throws upwards
Let the velocity of the ball as it hits the ground is V'.
Initial velocity, u = V
Final velocity, v = V'
height = h
acceleration due to gravity = g
Use third equation of motion
[tex]v^{2}=u^{2}+2as[/tex]
By substituting the values
[tex]V'^{2}=V^{2}+2(-g)(-h)[/tex]
[tex]V'=\sqrt{V^{2}+2gh}[/tex] .... (1)
Case II:
When the ball throws downwards
Let the velocity of the ball as it hits the ground is V''.
Initial velocity, u = V
Final velocity, v = V''
height = h
acceleration due to gravity = g
Use third equation of motion
[tex]v^{2}=u^{2}+2as[/tex]
By substituting the values
[tex]V''^{2}=V^{2}+2(-g)(-h)[/tex]
[tex]V''=\sqrt{V^{2}+2gh}[/tex] .... (2)
By comparing the equation (1) and equation (2), we get
V' = V''
Thus, the velocity of balls in both the cases is same as they strikes the ground.
A ball is thrown horizontally from the top of a 60 m building and lands 100 m from the base of the building. How long is the ball in the air? What must have been the initial horizontal component of the velocity? What is the vertical component of the velocity just before the ball hits the ground? What is the velocity of the ball just before it hits the ground?
Answer:
The ball is in the air for 3.5 seconds
The initial horizontal component of velocity is 28.6 m/s
The vertical component of the final velocity is 34.3 m/s downward
The final velocity is 44.7 m/s in the direction 50.2° below the horizontal
Explanation:
A ball is thrown horizontally
That means the vertical component of the initial velocity [tex]u_{y}=0[/tex]
The initial velocity is the horizontal component [tex]u_{x}[/tex]
The ball is thrown from the top of a 60 m
That means the vertical displacement component y = 60 m
→ y = [tex]u_{y}[/tex] t + [tex]\frac{1}{2}[/tex] gt²
where g is the acceleration of gravity and t is the time
y = -60 m , g = -9.8 m/s² , [tex]u_{y}=0[/tex]
Substitute these values in the rule
→ -60 = 0 + [tex]\frac{1}{2}[/tex] (-9.8)t²
→ -60 = -4.9t²
Divide both sides by -4.9
→ 12.2449 = t²
Take √ for both sides
∴ t = 3.5 seconds
* The ball is in the air for 3.5 seconds
The initial velocity is the horizontal component [tex]u_{x}[/tex]
The ball lands 100 meter from the base of the building
That means the horizontal displacement x = 100 m
→ x = [tex]u_{x}[/tex] t
→ t = 3.5 s , x = 100 m
Substitute these values in the rule
→ 100 = [tex]u_{x}[/tex] (3.5)
Divide both sides by 3.5
→ [tex]u_{x}[/tex] = 28.57 m/s
The initial horizontal component of velocity is 28.6 m/s
The vertical component of the final velocity is [tex]v_{y}[/tex]
→ [tex]v_{y}[/tex] = [tex]u_{y}[/tex] + gt
→ [tex]u_{y}[/tex] = 0 , g = -9.8 m/s² , t = 3.5 s
Substitute these values in the rule
→ [tex]v_{y}[/tex] = 0 + (-9.8)(3.5)
→ [tex]v_{y}[/tex] = -34.3 m/s
The vertical component of the final velocity is 34.3 m/s downward
The final velocity v is the resultant vector of [tex]v_{x}[/tex] and [tex]v_{y}[/tex]
→ Its magnetude is [tex]v=\sqrt{(v_{x})^{2}+(v_{y})^{2}}[/tex]
→ Its direction [tex]tan^{-1}\frac{v_{y}}{v_{x}}[/tex]
→ [tex]v_{y}[/tex] = 28.6 , [tex]v_{y}[/tex] = -34.3
Substitute this values in the rules above
→ [tex]v=\sqrt{(28.6)^{2}+(-34.3)^{2}}=44.66[/tex]
→ Its direction [tex]tan^{-1}\frac{-34.3}{28.6}=-50.18[/tex]
The negative sign means the direction is below the horizontal
The final velocity is 44.7 m/s in the direction 50.2° below the horizontal
The ball remained in the air for 3.49 seconds. The initial horizontal velocity was approximately 28.65 m/s. Just before hitting the ground, the ball had a vertical velocity of approximately 34.21 m/s downwards and an overall velocity of approximately 45.1 m/s at an angle of roughly 49.8 degrees below the horizontal.
Explanation:The time a projectile, such as the ball in this case, is in the air is completely determined by vertical motion. To calculate this, we can use the equation, y = y0 + v0yt - 0.5gt^2, where y is the final position, y0 is the initial position (the height of the building, 60 m in this case), v0y is the initial vertical velocity (which is 0 since the ball is thrown horizontally), g is acceleration due to gravity (approximated as 9.8 m/s^2), and t is the time we are looking for. By setting y to 0 (the base of the building where the ball lands), we can solve for t which gives us 3.49 seconds.
Next, the initial horizontal component of the velocity can be calculated using distance/time, which in this case is 100 m (distance traveled by the ball) divided by 3.49 seconds (the time the ball was in the air), giving us approximately 28.65 m/s.
The vertical component of the velocity just before the ball hit the ground can be calculated using v = v0y + gt, where v0y is still 0 and t is the time the ball is in the air. This calculation gives us approximately 34.21 m/s, in the downward direction, taking into account that acceleration due to gravity acts downward.
By combining the horizontal and vertical components, we can determine the velocity of the ball just before it hits the ground. The magnitude of this velocity can be found by using the Pythagorean theorem (v = sqrt((v0x)^2 + (v0y)^2)) which will give us approximately 45.1 m/s and the angle it makes with the horizontal can be calculated through the formula theta = tan^-1(v0y/v0x) giving us approximately 49.8 degrees below the horizontal.
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A car accelerates at a constant rate from zero to 33.7 m/s in 10 seconds and then slows to 17.6 m/s in 5 seconds. What is its average acceleration to the nearest tenth of a m/s2 during the 15 seconds?
Answer:
[tex]a_{avg} = 1.17 m/s^{2}[/tex]
Given:
initial velocity, u = 0
final velocity, v = 33.7 m/s
t = 10 s
final velocity, v' = 17.6 m/s
t' = 5 s
Total time, T = 10 + 5 = 15 s
Solution:
The rate of change of velocity of an object is referred to as the acceleration of that object.
Average accelaeration, [tex]a_{avg} = \frac{\Deta v}{\Delta t}[/tex]
Now,
Initial acceleration of the body, [tex]a = \frac{v - u}{t}[/tex]
[tex]a = \frac{33.7 - 0}{10} = 3.37 m/s^{2}[/tex]
Now, average acceleration during the 15 seconds:
[tex]a = \frac{v' - u}{T}[/tex]
[tex]a = \frac{17.6 - 0}{15} = 1.17 m/s^{2}[/tex]
Vectors Have
A.magnitude only.
B.magnitude and direction.
C.direction.
D.neither magnitude nor direction.
Answer:
B.magnitude and direction
Explanation:
Vector is moved from one place to another.
Example: acceleration
only scalar have magnitude only
Example: speed , temperature
The correct answer is B. Magnitude and direction
Explanation:
In physics, vectors are used to represent motion, forces acting on a body, among other common concepts. A vector is usually represented as a line that connects to different points and has a specific direction. Due to this, all vectors have a magnitude (specific size or length representing a concept) and a direction (position of forces, movements, etc). For example, in physics, it is common to use vectors to represent the forces that act on a body such as gravity or friction and each of these ave specific magnitudes represented with the length of the vector and specific directions depending on the position of the vector. Thus, vectors have both magnitude and direction.
A soccer ball is kicked from the ground with an initial speed of 20.1 m/s at an upward angle of 47.8˚. A player 53.0 m away in the direction of the kick starts running to meet the ball at that instant. What must be his average speed if he is to meet the ball just before it hits the ground? Neglect air resistance.
Answer:3.95 m/s
Explanation:
Given
Initial velocity(u)[tex]=20.1 m/s[/tex]
launch angle[tex]=47.8^{\circ}[/tex]
Player is 53 m away
Range of projectile
[tex]R=\frac{u^2sin2\theta }{g}[/tex]
[tex]R=\frac{20.1^2\times sin95.6}{9.8}[/tex]
R=41.02 m
so he need to run 53-41.02=11.98 m
Time of flight of projectile [tex]=\frac{2usin\theta }{g}[/tex]
T=3.03 s
thus average speed of boy [tex]s_{avg}=\frac{11.98}{3.03}=3.95 m/s[/tex]
What happens to dry ice when its temperature rises
When heated, dry ice undergoes sublimation, transitioning directly from a solid to a gas. It bypasses the liquid phase, turning directly into carbon dioxide gas.
Explanation:When the temperature of dry ice, which is the solid form of carbon dioxide, rises, it undergoes a process called sublimation. This is a phase transition where a substance moves directly from the solid phase to the gas phase, without passing through the liquid phase. So instead of melting and becoming liquid when heated, dry ice turns directly into carbon dioxide gas.
This is quite a unique characteristic, as most forms of matter have to pass through a liquid state before becoming a gas. The phenomenon of sublimation can also be observed in other substances like iodine and naphthalene (the main ingredient in mothballs).
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Starting from home, you bicycle 25 km north in 2.4 h and then turn around and pedal straight home in 1.4 h . What are your (a) displacement at the end of the first 2.4 h , (b) average velocity over the first 2.4 h , (c) average velocity for the homeward leg of the trip, (d) displacement for the entire trip, and (e) average velocity for the entire trip
Answer:
Part a)
[tex]d = 25 km[/tex]
Part b)
[tex]v_{avg} = 10.42 km/h[/tex]
Part c)
[tex]v_{avg} = 17.86 km/h[/tex]
Part d)
Displacement for entire trip = 0
Part e)
Average velocity for entire trip will be zero
Explanation:
Part a)
Displacement after t = 2.4 hours is the straight line distance between initial and final positions
so we have
[tex]d = 25 km[/tex]
Part b)
Average velocity is defined as
[tex]v_{avg} = \frac{displacement}{time}[/tex]
[tex]v_{avg} = \frac{25 km}{2.4 h}[/tex]
[tex]v_{avg} = 10.42 km/h[/tex]
Part c)
During his return journey the displacement will be same
[tex]displacement = 25 km[/tex]
[tex]time = 1.4 h[/tex]
so average velocity is defined as
[tex]v_{avg} = \frac{25 km}{1.4 h}[/tex]
[tex]v_{avg} = 17.86 km/h[/tex]
Part d)
Displacement for entire trip = 0
as initial and final position will be same
Part e)
Average velocity for entire trip will be zero
Final answer:
The displacement after 2.4 hours of cycling is 25 km north. The average velocity is 10.42 km/h north for the first 2.4 hours, 17.86 km/h south for the return trip, and the average velocity for the entire trip is zero because the displacement is zero.
Explanation:
To address the questions posed by the student, let's examine each part of the cycling trip:
(a) Displacement at the end of the first 2.4 h
The displacement after the first 2.4 hours is the straight-line distance from the starting point to the turnaround point, which would be 25 km directly north. Displacement is a vector quantity that refers to the change in position of an object.
(b) Average velocity over the first 2.4 h
The average velocity is the displacement divided by the time taken. Since the cyclist traveled 25 km north in 2.4 hours, we calculate the average velocity as 25 km / 2.4 h = 10.42 km/h north.
(c) Average velocity for the homeward leg of the trip
On the way back, the cyclist travels 25 km to the south (back home) in 1.4 h. The average velocity for this would be 25 km / 1.4 h = 17.86 km/h south.
(d) Displacement for the entire trip
Since the cyclist returns straight home, the displacement for the entire trip is zero, because there is no net change in position from the starting point.
(e) Average velocity for the entire trip
The total distance travelled is 25 km out and 25 km back, for a total of 50 km. The total time taken is 2.4 h for the first leg and 1.4 h for the return, which adds up to 3.8 h. However, the displacement is zero, therefore the average velocity for the entire trip is zero.
Your companion on a train ride through Illinois notices that telephone poles near the tracks appear to be passing by very quickly, while telephone poles in the distance are passing by much more slowly. This is an example of______________.
Answer:
Motion Parallax
Explanation:
According to my research on different scientific anomalies or terminology, I can say that based on the information provided within the question this is an example of Motion Parallax. This term refers to a depth cue where an individual views objects that are closer to us as moving faster than those objects further from us. Which is what the companion in this situation is describing.
I hope this answered your question. If you have any more questions feel free to ask away at Brainly.
what is the mass of a dog that weighs 382 N?(unit=kg)
Explanation:
If I assume you are talking about on Earth, then by using the equation
Weight Force=mg
382=(10)m
m=382/10
m=38.2 kg
Note that regardless of whether you are on Earth or any other planet or body the mass of something does not change. Only the weight of it changes as the gravitational acceleration (g) varies from planet to planet.
Answer : The mass of a dog that weighs 382 N is, 38.93 kg
Explanation :
Formula used :
[tex]F=m\times g[/tex]
where,
F = force = 382 N
m = mass of a dog = ?
g = acceleration due to gravity = [tex]9.8m/s^2[/tex]
Now put all the given values in the above formula, we get:
[tex]382N=m\times 9.8m/s^2[/tex]
[tex]382\frac{kg.m}{s^2}=m\times 9.8m/s^2[/tex]
[tex]m=38.98kg[/tex]
Thus, the mass of a dog that weighs 382 N is, 38.93 kg
The Bohr model was determined ultimately to be flawed. Knowing that "opposites attract (and 'likes' repel)," can you think of any flaws in the Bohr model that disobey these statements?
Answer:
Yes, there are 2 flaws
Explanation:
Electromagnetics say that, given two particles with charge, they will be attracted if they have opposite charge or repelled if they have identical charge.
Then, we can find two flaws in the Bohr model. The first one is that the electrons move in energy layers (or just layers to get the idea) far from the proton compared with the distance between the electrons themselves. So, if the model was right, how can it be that they don't repel each other? With the same logic, the protons don't repel each other either even though they are all together in the nucleus.
The second flaw, related to what we've just said, is that the electrons can move from one layer to another, but they will always stay at a minimum distance to the proton. How can it be so if it is known that they attract because of their charge signs?
A flask weights 45.4 g when it is empty and 121.8 g when filled with water. When the same flask is filled with another liquid, the mass is 91.39 g. What is the density of the second liquid?
Answer:
The answer to your question is: 1.19 gr/ml
Explanation:
empty flask weighs = 45.4 g
filled with water = 121.8 g
filled with a different substance = 91.39 g
density of the second liquid =
First we must calculate the mass of water in the flask by subtracting the flask filled with water and the flask empty.
mass = flask filled with water - the flask empty.
mass = 121.8 - 45.4 gr
mass = 76.4 gr
The density of water is always 1 gr/ml
and density is mass/volume
from here we calculate the volume = mass/density
volume = 76.4/1 = 76.4 ml
Now we calculate the density of the other liquid
density = 91.39/76.4 = 1.19 g/ml
Earth is slightly closer to the Sun in January than in July. How does the area swept out by Earth's orbit around the Sun during the 31 days of January compare to the area swept out during the 31 days of July?
a. Both areas are the same.
b. The area swept out in January is larger.
c. The area swept out in July is larger.
Answer: Option a.
Explanation:
Kepler's 2nd law of planetary motion states:
A line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time.
It tells us that it doesn't matter how far Earth is from the Sun, at equal times, the area swept out by Earth's orbit it's always the same independently from the position in the orbit.
A body of mass m moves in a horizontal direction such that at time t its position is given byx(t)=at4+bt3+ct,where a, b, and c are constants. (a) What is the acceleration of the body? (b) What is the time-dependent force acting on the body?
Answer:
The acceleration is given by de second derivative of x(t) which is equal to [tex]\frac{d^f{2}f(x) }{d^{2}x }=12at^{2} + 6bt[/tex] m/s^2
Explanation:
a) We have the equation x(t)=at^4+bt^3+ct which is the position of the body of mass m at a time t
Where a, b and c are constants
From the rules of differenciation we have that the first derivative of the position is the velocity and the second derivative is the acceleration.
Hence the first derivative of the function is equal to [tex]4at^{3} +3bt^{2}+c[/tex][/tex] m/s
Don´t forget to write down the unities
Then we have to derivate again this equation, so we have
[tex][tex]\frac{d^{2}f(x) }{d^{2}x }=12at^{2} + 6bt[/tex] m/s^2[/tex]
b) Remembering the Newton´s laws we know that
[tex]F=ma[/tex]
where:
F is the force
m is the mass
and a is the acceleration
From the first part we know the value of the acceleration which is
[tex]\frac{d^{2}f(x) }{d^{2}x }=12at^{2} + 6bt[/tex] m/s^2
So using the second law formula and replacing the values we have that
F=m([tex]\frac{d^{2}f(x) }{d^{2}x }=12at^{2} + 6bt[/tex] ) N
Remember the that N= Newton which is kg*m/s^2
After robbing a bank, a criminal tries to escape from the police by driving at a constant speed of 55 m/s (about 125 mph). A police officer, initially at rest, starts chasing the criminal when he drives by, accelerating at a constant rate of 6.1m/s2. How long will it take the officer to catch the criminal?
Answer:
18.03 s
Explanation:
We have two different types of motions, the criminal moves with uniform motion while the police do it with uniformly accelerated motion. Therefore we will use the equations of these cases. We know that by the time the police reach the criminal they will have traveled the same distance.
[tex]x=vt\\x=x_{0}+v_{0}t+\frac{a}{2}t^2[/tex]
The distance between the police and the criminal when the first one starts the persecution is 0, its initial speed is also zero. So:
[tex]x=(55m/s)t\\x=\frac{6.1m/s^2}{2}t^2=(3.05m/s^2)t^2[/tex]
Equalizing these two equations and solving for t:
[tex](55m/s)t=(3.05m/s^2)t^2\\(3.05m/s^2)t^2-(55m/s)t=0\\t((3.05m/s^2)t-55m/s)=0\\t=0 \\(3.05m/s^2)t-55m/s=0\\t=\frac{55m/s}{3.05m/s^2}=18.03 s[/tex]
Suppose a NASCAR race car rounds one end of the Martinsville Speedway. This end of the track is a turn with a radius of approximately 57.0 m . If the track is completely flat and the race car is traveling at a constant 27.5 m/s (about 62 mph ) around the turn, what is the race car's centripetal (radial) acceleration?
What is the force responsible for the centripetal acceleration in this case?
friction, normal, gravity, or weight?
Answer:
[tex]a_c = 13.26 m/s^2[/tex]
Friction Force
Explanation:
As we know that centripetal force is the product of mass and centripetal acceleration
so we know that
[tex]a_c = \frac{v^2}{R}[/tex]
so here we have
[tex]v = 27.5 m/s[/tex]
[tex]R = 57 m[/tex]
so we have
[tex]a_c = \frac{27.5^2}{57}[/tex]
[tex]a_c = 13.26 m/s^2[/tex]
This acceleration is given by the force which may be towards the center of the circular path
Here in the above case it is possible due to friction force.
Final answer:
The centripetal acceleration of the race car is 3.45 m/s². Centripetal force is calculated to be 1725 N, and the normal force is responsible for the centripetal acceleration.
Explanation:
The centripetal acceleration of the race car is 3.45 m/s². Centripetal force on the race car can be calculated using the formula F = m * a, where m is the mass of the car and a is the centripetal acceleration. This makes the centripetal force 1725 N. In this case, the force responsible for the centripetal acceleration is the normal force.
High‑speed ultracentrifuges are useful devices to sediment materials quickly or to separate materials. An ultracentrifuge spins a small tube in a circle of radius 13.1 cm at 50000.0 rev/min. What is the centripetal force experienced by a sample that has a mass of 0.00310 kg?
The force needed to move a body in a curved way is understood as centripetal force. The centripetal force experienced by a sample will be 7.14 N.
What is centripetal force?The force needed to move a body in a curved way is understood as centripetal force. This is a force that can be sensed from both the fixed frame and the spinning body's frame of concern.
The direction of centripetal force is always in the path of the center of the course.
The given data in the problem;
r is the radius =13.1 cm=0.13 m
n is the rpm = 50000.0 rev/min
To find velocity first we have to go to angular velocity
[tex]\omega = \frac{2\pi n}{60} \\\\ \omega = \frac{2\times3.14\times50000}{60} \\\\\omega = 132.63\; rad/sec.[/tex]
The centripetal force is given as
[tex]\rm F_C= m\omega^{2} r\\\\F_C= 0.00310(132.63)^2 \times 0.131\\\\\rm{F_C=7.14 N[/tex]
Hence The centripetal force experienced by a sample will be 7.14 N.
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Which of the following have quantized values? Explain your selections. a. the pitch of a note played on a slide trombone b. the pitch of a note played on a flute c. the wavelengths of light produced by the heating elements in a toaster d. the wind speed at the top of Mt. Everest
Answer:
Option b and c
Explanation:
Quantized values are discrete in nature and are not continuous.
These values can be expressed as integral multiples or integers.
Therefore,
The pitch of notes played on trombone are of continuous form and are not quantized.The pitch of the notes that are played on a flute are dependent on the number of open and closed holes on the instrument, clearly discrete in nature thus quantized.The wavelength of light is produced as a result of transition of electron in between energy levels which occurs in discrete steps thus quantized.The wind speed at Mt. Everest's top is continuous and thus not quantized.In the previous question, the person had an initial velocity of 8m/s and a constant acceleration of −4m/s2. How would the maximum distance he travels to the right of the origin change if instead his initial velocity were doubled (vx,0=16m/s)?
Final answer:
The maximum distance traveled to the right of the origin will change if the initial velocity is doubled.
Explanation:
To find the maximum distance traveled by the person to the right of the origin, we can use the equation:
d = v0t + 0.5at2
Given the initial velocity (vx,0) of 16 m/s, we can substitute it into the equation along with the constant acceleration (-4 m/s2) and the time (t) to calculate the maximum distance.
d = (16 m/s)(t) + 0.5(-4 m/s2)(t2)
The maximum distance will change as we vary the time, but it will be greater than the distance traveled with an initial velocity of 8 m/s.
If an automobile had a 100%-efficient engine, transferring all of the fuel's energy to work, would the engine be warm to your touch? Would its exhaust heat the surrounding air? Would it make any noise? Would it vibrate? Would any of its fuel go unused? Discuss.
Answer:
The engine would be warm to touch, and the exhaust gases would be at ambient temperature. The engine would not vibrate nor make any noise. None of the fuel entering the engine would go unused.
Explanation:
In this ideal engine, none of these events would happen due to the nature of the efficiency.
We can define efficiency as the ratio between the used energy and the potential generable energy in the fuel.
n=W, total/(E, available).
However, in real engines the energy generated in the combustion of the fuel transforms into heat (which heates the exhost gases, and the engine therefore transfering some of this heat to the environment). Also, there are some mechanical energy loss due to vibrations and sound, which are also energy that comes from the fuel combustion.
Final answer:
This answer addresses the implications of a hypothetical 100%-efficient engine in an automobile, discussing its heat production, exhaust impact, noise and vibration levels, and fuel usage.
Explanation:
If an automobile had a 100%-efficient engine:
Would the engine be warm to your touch? - Yes, the engine would still produce heat as a byproduct of its operation, even with 100% efficiency, due to thermodynamics.Would its exhaust heat the surrounding air? - The exhaust gases would still carry thermal energy and heat the air around.Would it make any noise or vibration? - While noise and vibration are not related to engine efficiency, they stem from mechanical components and combustion processes.Would any of its fuel go unused? - With 100% efficiency, all the fuel's energy would convert to work, leaving no unused fuel.The electrical resistance of dry skin is 100 kΩ, but can be lowered to 20 Ω if electrode contact area is large and conducting gel is used on the skin. If Vmax of the defibrillatror 500 V and lasts 0.01 s, what is the maximum possible current delivered to the heart during defibrillation?
Answer:
25 Ampers
Explanation:
As the current in a defibrillation is DC (Direct current), we can use simply the Ohm's law which shows:
V=I*R
Then if we want to know the current, we can follow the next step for I:
I=V/R
Using the information, and asumming electrode contact area is large and conducting gel is used on the skin (as is normal) , we can solve:
Imax=Vmax/R
Imax= 500 V / 20 ohms
Imax= 25 Ampers
*Fact: The time in this case doesn't make any difference in the result.
If a child pulls a sled through the snow on a level path with a force of 60 N exerted at an angle of 39° above the horizontal, find the horizontal and vertical components of the force.
Explanation:
Given that,
Force with which a child pulls a sled, F = 60 N
It is at an angle of 39° above the horizontal. We need to find the horizontal and vertical components of the force.
The horizontal component is given by :
[tex]F_x=30\ cos(39)=23.31\ N[/tex]
The vertical component is given by :
[tex]F_y=30\ sin(39)=18.87\ N[/tex]
So, the horizontal and vertical components of the force are 23.31 N and 18.87 N. Hence, this is the required solution.
To find the horizontal and vertical components of the force, use the formulas Fx = F * cos(angle) and Fy = F * sin(angle), respectively.
Explanation:To find the horizontal and vertical components of the force, we can use trigonometry. The horizontal component of the force can be found using the formula: Fx = F * cos(angle), where F is the magnitude of the force and angle is the angle above the horizontal. Plugging in the values, we get Fx = 60 N * cos(39°) = 45.85 N.
The vertical component of the force can be found using the formula: Fy = F * sin(angle), where F is the magnitude of the force and angle is the angle above the horizontal. Plugging in the values, we get Fy = 60 N * sin(39°) = 36.34 N.
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The driver of a car traveling at a speed of 20.0 m/s slams on the brakes and comes to a stop in 5.8 s. If we assume that the car's speed changed at a constant rate (constant net force), what was theaverage speed during this 5s interval?
average speed = m/s
Answer:
10 m/s
Explanation:
u = 20 m/s
v = 0 m/s
t = 5.8 s
Let a be the acceleration and s be the distance traveled by the car in time t.
use first equation of motion
v = u + a t
0 = 20 + a x 5.8
a = -3.45 m/s^2
Use third equation of motion
[tex]v^{2}=u^{2}+2as[/tex]
[tex]0^{2}=20^{2}-2 \times3.45 \times s[/tex]
s = 58 m
The average speed is defined as the ratio of distance traveled to the time taken.
[tex]Average speed = \frac{58}{5.8} = 10 m/s[/tex]
thus, the average speed of the car is 10 m/s.
Suppose oil spills from a ruptured tanker and spreads in a circular pattern. If the radius of the oil spill increases at a constant rate of 1 m/s, how fast is the area of the spill increasing when the radius is 30 m?
Answer:
[tex]\frac{dA}{dt} = 188.5 m^2/s[/tex]
Explanation:
As we know that area of the circle at any instant of time is given as
[tex]A = \pi r^2[/tex]
now in order to find the rate of change in area we will have
[tex]\frac{dA}{dt} = 2\pi r\frac{dr}{dt}[/tex]
here we know that
rate of change of radius is given as
[tex]\frac{dr}{dt}= 1 m/s[/tex]
radius of the circle is given as
[tex]r = 30 m[/tex]
now we have
[tex]\frac{dA}{dt} = 2\pi (30)(1)[/tex]
[tex]\frac{dA}{dt} = 60\pi[/tex]
[tex]\frac{dA}{dt} = 188.5 m^2/s[/tex]
Mike's car, which weighs 1,000 kg, is out of gas. Mike is trying to push the car to a gas station, and he makes the car go 0.05 m/s2. Using Newton's Second Law, you can compute how much force Mike is applying to the car.
Answer:
The answer to your question is: F = 50 N
Explanation:
Data
mass = 1000 kg
acceleration = 0.05m/s2
F = ?
Formula
F = m x a
Substitution
F = 1000 kg x 0.05 m/s2 = 50 kgm/s2 = 50 N
Mike is applying a force of 50 N to the car.
The magnitude of applied force on the car by Mike is 50 N.
Given data:
The mass of car is, m = 1000 kg.
The magnitude of acceleration of car is, [tex]a = 0.05 \;\rm m/s^{2}[/tex].
According to Newton's second law of motion, the force applied on the object is expressed as the product of mass of object and magnitude of acceleration caused by the applied force on the object.
Therefore,
[tex]F = m \times a[/tex]
Here, F is magnitude of applied force on car.
Solving as,
[tex]F = 1000 \times 0.05\\F = 50 \;\rm N[/tex]
Thus, we can conclude that Mike is applying 50 N of force on his car.
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diver uses a sonar device to determine the distance to her diving partner. The device sends a sound wave and records the time it takes for the wave to reach the diving partner and return to the device. Suppose the wave travels at a rate of about 4000 feet per second. The wave returns 0.8 second(s) after it was sent. How far away is the diving partner?
Answer:
Assuming the divers are not moving. The diving partner is at 1600 feet.
Explanation:
The sound wave have to go to the diving partner and return, so its traveling the double of the total distance.
Been X= distance, V= speed, and t= time
[tex]x=v*t\\[/tex]
[tex]WaveTravelDistance=4000*0.8\\\\Distance to the partner=WaveTravelDistance/2\\\\WaveTravelDistance=3200feet\\\\Distance to the partner=1600feet[/tex]
A model of a helicopter rotor has four blades, each of length 3.80 m from the central shaft to the blade tip. The model is rotated in a wind tunnel at a rotational speed of 450 rev/min.a. What is the linear speed of the blade tip?b. What is the radial acceleration of the blade tip expressed as a multiple of the acceleration of gravity?
Answer:
a)V= 179.056 m/s
b)[tex]a_{r}=843.71 g \ m/s^2[/tex]
Explanation:
Given that
Length of blade = 3.8 m
Rotational speed N= 450 rev/min
We know that
[tex]\omega =\dfrac{2\pi N}{60}\ \frac{rad}{s}[/tex]
[tex]\omega =\dfrac{2\pi \times 450}{60}\ \frac{rad}{s}[/tex]
ω=47.12 rad/s
Linear velocity
We know that linear velocity V = ω x r
Here r = 3.8 m
So by putting the values
V = ω x r
V = 47.12 x 3.8 m/s
V= 179.056 m/s
Radial acceleration
[tex]a_{r}=\omega ^2r\ m/s^2[/tex]
[tex]a_{r}=47.12^2\times 3.8 \ m/s^2[/tex]
[tex]a_{r}=8437.12 \ m/s^2[/tex]
[tex]a_{r}=843.71 g \ m/s^2[/tex]
The linear speed of the blade tip is approximately 178.6 m/s. The radial acceleration of the blade tip is approximately 864 times the acceleration of gravity.
Explanation:The two parts of this question relate to basic concepts in physics, specifically relating to circular motion and the relationship between linear speed, radial acceleration, and the acceleration of gravity. The model of a helicopter rotor in question has four blades, each of length 3.80 m, and rotates at a speed of 450 rev/min.
a. What is the linear speed of the blade tip? The linear velocity (v) of the tip of the blade can be calculated using the formula v = rω, where r is the radial distance (half the length of the blade) and ω is the angular velocity. The angular velocity can be converted to rad/s by multiplying the rotational speed (450 rev/min) by 2π rad/rev and then by 1/60 min/s, giving approximately 47 rad/s. Then, v = (3.8 m) * (47 rad/s) gives a linear velocity of approximately 178.6 m/s.
b. What is the radial acceleration of the blade tip expressed as a multiple of the acceleration of gravity? The radial (or centripetal) acceleration (a_r) can be calculated using the formula a_r = ω²r. Substituting the known values gives a_r = (47 rad/s)² * 3.8 m = approximately 8460 m/s². As the acceleration of gravity is approximately 9.8 m/s², the radial acceleration is therefore about 864 times the acceleration of gravity.
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Calculate the amount of heat transferred when 710 grams of water warms from an initial temperature of 4.0 ºC to a final temperature of 25.0 ºC. The specific heat capacity of liquid water is 4.184 J/g ºC.
Answer:
Q = 62383.44 Joules
Explanation:
Given that,
Mass of water, m = 710 gm
Initial temperature of water, [tex]T_i=4^{\circ} C[/tex]
Final temperature of water, [tex]T_f=25^{\circ} C[/tex]
The specific heat capacity of liquid water is, [tex]c=4.184\ J/g\ ^oC[/tex]
Heat transferred is given by :
[tex]Q=mc(T_f-T_i)[/tex]
[tex]Q=710\times 4.184\times (25-4)[/tex]
Q = 62383.44 Joules
So, the amount of heat transferred is 62383.44 Joules. Hence, this is the required solution.
The flywheel of a steam engine runs with a constant angular speed of 158 rev/min. When steam is shut off, the friction of the bearings and the air brings the wheel to rest in 2.6 h. What is the magnitude of the constant angular acceleration of the wheel in rev/min^2? Do not enter the units.
Answer:
α=1.013rev/min^2
Explanation:
Wo=initial angular speed=158rev/min
t=time=2.6h=156min
Wf=final angular speed=0rev /min , this happens because in the end the wheel stops
this is a circular motion with uniform acceleration therefore we can use the following equation
α=angular aceleration=(Wf-Wo)/t
α=(0-158)/156=1.013rev/min^2
A baseball is thrown at an angle of 20° relative to the ground at a speed of 25 m/s if the ball was caught 50 m from the thrower how long was it in the air ?
2.1 s
0.5 s
10 s
5 s
Answer:
2.1 s
Explanation:
The motion of the ball is a projectile motion. We know that the horizontal range of the ball is
[tex]d = 50 m[/tex]
And that the initial speed of the ball is
[tex]u=25 m/s[/tex]
at an angle of
[tex]\theta=20^{\circ}[/tex]
So, the horizontal speed of the ball (which is constant during the entire motion) is
[tex]u_x = u cos \theta = 25 \cdot cos 20^{\circ} = 23.5 m/s[/tex]
And since the horizontal range is 50 m, the time taken for the ball to cover this distance was
[tex]t=\frac{d}{u_x}=\frac{50}{23.5}=2.1 s[/tex]
which is the time the ball spent in air.
A straight, nonconducting plastic wire 8.00 cm long carries a charge density of 100 nC/m distributed uniformly along its length. It is lying on a horizontal tabletop.
(a) Find the magnitude and direction of the Electric Field that is produced 6.0cm directly above its midpoint.(b) If the wire is now bent into a circle lying flat on the table, find the magnitude and direction of the Electric field on a point 6.0cm directly above its center.
To find the electric field at a point directly above the midpoint of the plastic wire, use the formula for the electric field due to a charged line. When the wire is bent into a circle, the electric field at a point directly above its center is zero.
Explanation:To find the electric field at a point directly above the midpoint of the plastic wire, you can use the formula for the electric field due to a charged line. The formula is given by E = (k * λ) / r, where E is the electric field, k is the Coulomb's constant (9 * 10^9 Nm²/C²), λ is the charge density, and r is the distance from the wire.
In this case, the charge density is 100nC/m and the distance is 6.0cm. Plug in the values to calculate the electric field. Since the wire is nonconducting, it does not affect the direction of the electric field, so the magnitude is the only value you need to find.
(b) When the wire is bent into a circle lying flat on the table, the electric field at a point directly above its center is zero. This is because the electric field due to each small segment of the wire cancels out by the symmetry of the circular shape.