The speed of the beam of light moving along the shoreline when it is 1 km from p 62.83 km/min revolution.
A revolution is the motion of 1 object around a center or some other object, a forceful overthrow of a government by using the people, or any sudden or grand trade. An example of revolution is the movement of the earth around the solar.
dθ/dt = 3 rev/min
= 3×2π = 6π rad/min
tanθ = x/3
[tex]\frac{d}{dt}[/tex] tanθ = [tex]\frac{d}{dt}[/tex] (x/3)
sec²θ dθ/dt = 1/3 dx/dt
dx/dt = 3sec²θdθ/dt
At x = 1km; tanθ =x/3 =1/3
sec²θ = 1 + tan²θ = 1+(1/3)²
sec²θ =10/9
dx/dt = 3sec²θ dθ/dt
dx/dt = 3×10/9×6π
dx/dt = 62.83 km/min
Hence, the beam of light along the shoreline when it is 1 km away from point P is 62.83 km/min
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. overdriving your headlights is a. driving with only your parking lights on. b. driving at a speed that makes your stopping distance longer than the distance illuminated by your headlights. c. having your headlights aimed improperly. d. driving with your high-beam headlights on.
Overdriving is driving at a speed making the stopping distance longer than the distance illuminated by headlights.
Typically, one can utilize their headlights to schedule stops in dimly lit or low visibility locations. Typically, one can measure their stopping distance using the visible light cone in the headlights. Exceeding the stopping distance that the headlights highlight is overdriving. Thus, driving too fast and leaving little room to stop in an emergency is known as overdriving the headlights.
When the lights are far enough ahead of a person to illuminate the road, one should often stop so that they can see any approaching impediments. When one is traveling at a rate of speed that prevents them from stopping in that lit region, that rate of speed is taken into account.
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its hard it have time help
Answer:
i think it's a
Explanation:
A force of 30.ON is acted on a block of mass 15.0kg placed on a horizontal surface but the boxdoes not move. Calculate the magnitude of the frictional force between the surfaces when theapplied force is along the horizontal and when it makes angle 60° with the horizontal.
We will have the following:
1. In the first case, we will calculate the normal force:
[tex]N=(15\operatorname{kg}\cdot9.8m/s^2)\Rightarrow N=147N[/tex]Then, we will have that the fricion coefficient will be:
[tex]\mu_f=\frac{30.0N}{147N}\Rightarrow\mu_f=\frac{10}{49}\Rightarrow\mu_f\approx0.20[/tex]So the magnitude of the friccional force when applied horizontally will be of 30.0N.
2. And the magnitude when applied horizontal but the block is on an inclined surface of 60°, we will have that:
[tex]F=(10/49)(15.0\operatorname{kg}\cdot9.8m/s^2)\cos (60)\Rightarrow F=15N[/tex]So, the friction force when the force is applied, is 15N.
A concave mirror has a focal length of 0.50 m. If an object produces a virtual image 0.19 m from the mirror, where is it located?
In order to calculate the object's position, we can use the formula below:
[tex]\frac{1}{f}+\frac{1}{d_o}+\frac{1}{d_i}[/tex]Where f is the focal length, do is the object's position and di is the image's position.
Using f = 0.5 m and di = -0.19 m (we use a negative value because the image is virtual), we have:
[tex]\begin{gathered} \frac{1}{0.5}=\frac{1}{d_o}+\frac{1}{-0.19}\\ \\ 2=\frac{1}{d_o}-5.263\\ \\ \frac{1}{d_o}=2+5.263\\ \\ \frac{1}{d_o}=7.263\\ \\ d_o=\frac{1}{7.263}\\ \\ d_o=0.14\text{ m} \end{gathered}[/tex]Therefore the object is at 0.14 meters from the mirror.
suppose you observe a small star orbiting a much larger star. you observe that the small star is orbiting at a distance of 20 au (in terms of the semi-major axis of its orbit), and with an orbital period of 2 years. how massive is the large star? (hint: for this problem you can assume that the small star has a much smaller mass than the large star.)
The mass of the large star based on orbital period is 6912 x 10³⁰ kg.
We need to know about the orbital period to solve this problem. The orbital period depends on the semi-major axis, gravitational constant, and the mass of the massive object. It can be written as
T² = 4π².a³ / GM
where T is the orbital period, a is the semi-major axis, G is the gravitational constant and M is the mass.
From the question above, the given parameters are
a = 24 au = 24 . ³√(G.Msun yr²/ 4π²)
T = 2 years
Msun = 2 x 10³⁰ kg
By substituting the given parameters, we get
T² = 4π².a³ / GM
2² = 4π². (24 . ³√(G.Msun yr²/ 4π²))³ / GM
4 = 13824 Msun / M
M = 6912 x 10³⁰ kg
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he radius of a circular oil slick expands at a rate of 4 m/min. (a) how fast is the area of the oil slick increasing when the radius is 58 m? (use symbolic notation and fractions where needed.)
It is 24 pi fast area of oil sick increasing when the radius is 60m.
This question we are given the rate of change is the radius T. R. D. T. Is equal to two m per minute. And we know that the area since we're talking about the radius theory of the seco which is pi R squared. And our D A t. R. Would be the first derivative of A. That will be two pi R. And our D A D. T. Would be equal to T. A. T. R. Multiplied by the era T. T. And already we can plug in our numbers. Here we have two pi R. A. Multiplied by two and we get four pi hi and this season meters squared per minute.
Now we want to find the value of our rate of change of area with time At R. A. is equal to 56. And therefore our D. A. T. T. At our escort 56 is equal to four pi my play by 56. And here we get 224 pi square meters per minute. This is our answer to part A no for part B. We are giving that At time is equal to zero is equal to zero. And our rate of change of radius the time is still equal to two meters per minute and Therefore from 0 to 3 we have a tiara Is equal to two d. T. And we integrate Both sides from 0 to 3. So on the left hand side we have our era on the right hand side we have to t From 0 to 3 and here we get six So our radius is six m.
Therefore our D a t t will be a quarter full pie. Sarah it R is equal to six and Plugging. In our 6th year. We have four pay And our era is six. We get 24 pi in this season square meters per minute. That's it. We had them.
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two people push a 2,159. kg car to get it started. an acceleration of at least 3.4 m/s2 is required to start the car. assuming both people apply the same magnitude force in the same horizontal direction, how much force will each need to apply if friction between the car and the road is 310 n? express your answer in scientific notation. n
The force needed to be applied by each person is 5.150×10^3 N.
The required acceleration is 3.4 m/s2 and the mass of the car is 2,159 kg.
According to Newton's second law:
F = ma
F = 2,159 × 3.4.
F = 7340.6 N
Adding the force of friction into it
F = 7340 + 310
F = 7650.6 N
Since two people need to apply the force
F = 7650 N / 2
F = 3,825.3 N
or
F = 3.825×10^3 N
The force applied by each person to push the car is 3.825×10^3 N.
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A cosmic-ray electron moves at 7. 5 × 10^6 m/s perpendicular to earth’s magnetic field at an altitude where the field strength is 1. 0 × 10^-5t. What is the radius of the circular path the electron follows?.
4.266 m is the radius of the circular path the electron follows
Electron's speed = 7.50 × 10⁶ m/s
Magnetic field, B = 1 × 10⁻⁵ T
Electron's mass. m = 9.11 × 10⁻³¹
Charge of electron, q = 1.6 × 10⁻¹⁹
We know the relation of radius of curvature, = 4.266 m
What is a circle's radius of curvature?Every point on a circle has a curvature that is equal to the reciprocal of the radius; for other curves (and straight lines, which can be thought of as circles of infinite radius), the curvature is equal to the reciprocal of the radius of the circle that at that point most closely approximates the curve.
The curvature's reciprocal, R, is the radius of curvature in differential geometry. It is equal to the radius of the circular arc that, at that moment, most closely resembles the curve for a curve. The radius of curvature for surfaces is the diameter of a circle that, individually or in combination, best fits a normal section.
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n sussex county, delaware, a post-halloween tradition is punkin chunkin, a competition in which contestants build cannons, catapults, trebuchets, and other devices to launch pumpkins to the greatest distance they can. tough hard to believe, pumpkins have been projected a distance of 1245 m (over 3 4 mi) in this contest. what is the minimum initial speed needed for such a shot?
110.53m/s is the minimum initial speed or initial velocity needed for such a shot.
The lowest speed at which a particular range is attained is determined. It is stated the distance that the pumpkins were hurled. By inserting the range of the pumpkin and the acceleration owing to gravity, the minimal velocity at which they were launched to go this distance was determined.
[tex]R=u_m^2/g[/tex]
[tex]1245.41m=(u_m^2/9.80m/s^2[/tex]
[tex]u_m^2=12217.47[/tex]
[tex]u_m=110.53m/s[/tex]
Any object susceptible to forces will move more quickly. This acceleration causes a change in the object's velocity. The speed of the item prior to the change brought on by acceleration is therefore known as the initial velocity. The speed will be the final speed once the object has been accelerating for some time.
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Compare and contrast oxygen-16 and
oxygen-17.
Answer:
oxygen 16 has 8 protons and 8 neutrons, oxygen 17 has 8 protons and 9 neutrons
Explanation:
isotope's number is equal to the sum of its protons and neutrons
Answer:Oxygen 16 is number 16 while oxygen 17 is number 17 and oxygen 16 is also oxygen like oxygen 17.
Explanation: Common Knowledge in my brain....
Johnny is driving North at 42 mph. Then drives west at 37 mph. What is the resultant velocity?
A) 55.9 mph North
B) 5 mph West
C) 55.9 mph Northwest
D) 79 mph Northwest
If Johnny is driving North at 42 miles per hour and drives west at 37 miles per hour, then the resultant velocity would be 55.9 mph north, therefore the correct answer is option A.
What is Velocity?The total displacement covered by any object per unit of time is known as velocity. It depends on the magnitude as well as the direction of the moving object.
As given in the problem Johnny is driving North at 42 mph. Then drives west at 37 mph.
The resultant velocity of the Jhonny = √ ( 42² + 37² )
= 55.97 miles per hour
Thus, the resultant velocity of Jhonny would be 55.9 mph north, therefore the correct answer is option A.
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An engineer is troubleshooting trying to figure out the value of a capacitor. She knows that the charge value is 480 coulombs and the power supply feeds 110 V. What's the capacitor value?Question 19 options:A) 128B) 4.36C) 0.88D) 5,400
We can use the formula of the capacitance
[tex]C=\frac{Q}{V}[/tex]where C is the capacitance, Q is the charge and V is the voltage.
In our case
Q=480 C
V= 110 V
[tex]C=\frac{480}{110}=4.36\text{ F}[/tex]The value of the capacitor is 4.36 F
ANSWER B) 4.36F
Explain some endless capabilities of our vast universe?
ANSWER:
STEP-BY-STEP EXPLANATION:
Answer:
It is impossible to answer that as we could discover more than we know now in years to come with advanced technologies
Explanation:
Pick the answer that correctly completes the sentence: as a sound wave approaches an observer the frequency will _____, the wavelength will____, and the speed of sound will______.A) decrease; increase; stay the same B) increase; increase; increase C) increase; decrease; stay the same D) decrease; decrease; increase
To find
Pick the answer correctly that fits the sentence.
Explanation
In a doppler effect, when a sound source apporaches a observer, its frequency increases. As wavelength is inversely proportional so the wavelength decreases. Speed of sound ina given medium is constant, so the speed of sound doesnot change
Thus,
as a sound wave approaches an observer the frequency will increase, the wavelength will decrease, and the speed of sound will stay the same.
Conclusion
The correct option is C) increase; decrease; stay the same.
Two objects with the same mass have the same force applied to them. What can be said of their acceleration?
A. Their acceleration will be the same.
B. Their acceleration will be different.
C. Their acceleration will depend on their net force, which is not provided.
D. Their acceleration will depend on their weight, which is not provided.
A force of 30 N to the right is applied to an object. An opposite force of 20 N to the left is applied to the same object. What is the net force applied to the object?
A. 10 N to the left
B. 50 N to the right
C. 50 N to the left
D. 10 N to the right
Answer:
Explanation:
A. Their acceleration will be the same. F=ma, a = F/m. If F and m are the same, acceleration will be the same.
B. Their acceleration will be different.
C. Their acceleration will depend on their net force, which is not provided.
D. Their acceleration will depend on their weight, which is not provided.
======
A force of 30 N to the right is applied to an object. An opposite force of 20 N to the left is applied to the same object. What is the net force applied to the object?
A. 10 N to the left
B. 50 N to the right
C. 50 N to the left
D. 10 N to the right
Add the two vectors: 30R = +30, 20L = -20 Sum = +10 or 10R
The force acting on two bodies will be the same if they have same mass and same acceleration. But the acceleration is dependant on the net force of the body and thus option C is correct. The net force of the body is 10 N to the right.
What is force?Force is an external agent or factor which change an object from its state of rest or motion. The force to be applied depends on the mass and acceleration of the body by the equation F = ma.
Acceleration of a moving body is dependant on the direction and thus the direction of force and net force. Thus the two objects having the same mass and same force f applied to them, the acceleration may not be the same. Thus option C is correct.
The two force acting on a body in opposite direction will cancel each other in magnitude. Thus 20N of force to the right will be cancel by 20 N to the left. Remaining 10N will be there to the right direction. Hence, option D is correct.
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Calculate the magnitude of the resultant velocity of a pair of 100 km/h velocity vectors that are at right angles to each other
The magnitude of the resultant velocity of a pair of 100 km/h velocity vectors that are at right angles to each other is 141.42 km/h.
What is velocity?When an item is moving, its velocity is the rate at which its direction is changing as seen from a certain point of view and as measured by a specific unit of time.
Given that in the question velocity of a pair of 100 km/h velocity vectors that are at right angles to each other,
Resultant velocity,
v = [tex]\sqrt{v^{2} +v^{2} }[/tex]
= [tex]100\sqrt{2}[/tex]
= 141.42 km/h
The resultant velocity of a pair of 100 km/h velocity vectors that are at right angles to each other is 141.42 km/h.
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a satellite moves in a circular earth orbit that has a radius of 8.75 x 106 m. a model airplane is flying on a 20.3-m guideline in a horizontal circle. the guideline is nearly parallel to the ground. find the speed of the plane such that the plane and the satellite have the same centripetal acceleration.
The speed of the plane is 32.519 m/s
Given radius of satellite orbit is [tex]r_{1} = 8.75[/tex] × [tex]10^{6} m[/tex]
Also, Orbital velocity is given by
[tex]v = \sqrt{\frac{GM}{r} }[/tex]
where M is the mass of the earth which is equal to [tex]5.98[/tex] × [tex]10^{24}[/tex] kg
G = [tex]6.67[/tex] × [tex]10^{-11}[/tex]
[tex]v = \sqrt{\frac{(6.67 * 10^{-11} )(5.98*10^{24}) }{8.75 * 10^{6} } }[/tex]
[tex]v=21.35[/tex] × [tex]10^{3}[/tex] m/s
Centripetal acceleration is given as
[tex]a_{c} = \frac{v^{2} }{r}[/tex]
[tex]a_{c} = \frac{(21.35 * 10^{3} )^{2} }{8.75 * 10^{6} }[/tex]
[tex]a_{c} = 52.094 m/s^{2}[/tex]
For model airplane
[tex]a_{c} = \frac{v^{2} }{r}[/tex]
[tex]52.094 = \frac{v^{2} }{20.3}[/tex]
v = 32.519 m/s
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A forklift raises a crate weighing 8.35 × 102 newtons to a height of 6.0 meters. What amount of work does the forklift do?
Answer:5110.2 joule
Explanation:Weight of the crate:- 8.35 x 102 N
Height to which the forklift lift the crate:- 6.0 M
Work done by the forklift=F.d
F=Gravitational Force =m.g
d=height=h
Work w=m.g.h
m.g=8.35 x 102 N
h=6.0 M
w=8.35x102N x 6.0M
w=5110.2NM
w=5110.2joule
When inflating a balloon, which of the following represents an equal but opposite force acting on the balloon?
A.) releasing the balloon
B) the air entering the balloon
C) tying the balloon
D) the expansion of the balloon
When inflating a balloon, the expansion of the balloon represents an equal but opposite force acting on the balloon; option D.
What are equal and opposite forces?Forces are agents which cause a change in the motion of an object moving in a straight line or an object at rest to change its position.
According to the third law of motion, reaction and action are equal and opposite forces.
This law is seen in the inflation of balloons.
When a balloon is being inflated, the force applied in inflating the balloon pushes against the wall of the balloon. The balloon acts in an equal opposite direction by forcing the air molecules together, thus increasing the pressure of the air inside the balloon.
Therefore, an equal and opposite force is exerted by the air molecules o the air molecules.
When the balloon is released, the air is forcefully forced out in the opposite direction.
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Answer:
its D
Explanation:
the person above is correct too
a car has two headlights, and their power is derived from the car's battery. the filament in one burns out, but the other headlight stays on. are the headlights connected in series or in parallel?
It is necessary to collimate the light source before using the prisim to disperse the light so that the light rays entering the prisim are parallel series.
What is colimate?
collimator is a tool for converting a point source's divergent radiation into a parallel beam. To perform specific measurements in spectroscopy, geometrical optics, and physical optics, the light must be collimated.
An optical collimator is made up of a tube with a convex lens at one end and a variable aperture at the other end that is located in the lens' focal plane. To examine the image without parallax, radiation entering the aperture exits the collimator as a parallel beam.
Therefore collimation is needed for the light rays entering the prism to be parallel .
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if you have a sharper curve and youa re 60 mph if you decrease the radius of an object traveling in a circle, but keep the velocity the same, what happens to the centripetal acceleration experienced by the object?
Given a constant speed, the centripetal acceleration of an item travelling in a circle is inversely proportional to the radius of the curve (a=v2/r).
As a result, the acceleration is smaller for a gradual curve with a large radius and greater for a sharp curve with a short radius.
Centripetal acceleration: What is it?The speed at which a body rotates around a circle is known as centripetal acceleration. When a body moves in a circle, its d is equal to its velocity since velocity is a vector quantity (i.e., it has both a magnitude, the speed, and a direction) direction is continually changing, which causes a change in velocity, which results in an acceleration.
What is the centripetal acceleration's direction?The acceleration is brought about by a force that is focused at the, It is referred to as a centripetal force.
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pls the answer.........................................................................................................................
a) The motion of the motorcyclist is a uniform motion
b) The distance is 600 m
c) The speed is 20 m/s.
What is the motion?We now that motion has to do with the change in the position of an object with time. In this case, we have the graph that described the motion of the motorcyclist . It is clear that the distance is increasing in equal amounts, in equal time intervals thus this is a uniform motion.
Looking at the graph, we can see that the motorcyclist covers 600 meters of distance within 30 seconds.
We can obtain the speed as the ratio of the distance to the time or the slope of the graph. Hence the speed is obtained from;
Speed = distance / time
Using the slope concept;
[tex]y_{2} - y_{1} /x_{2} - x_{1}[/tex]
= 600 - 0/30 - 0
= 20 m/s
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how to sketch the following:Sketch a ray diagram for a spherical convex lens with an object at Do = 2f. (Submit a file with a maximum size of 1 MB.)
Do = 2f means that object distance is 2 x focal length.
2 x focal length = radius of curvature
When an object is placed at the radius of curvature, a real, inverted image is formed at the other radius of curvature. The sizes of the image and the object are the same. The sketch is shown below
a fish looking straight up toward the smooth surface of a pond receives a cone of rays and sees a circle of light filled with the images of sky and birds and whatever else is up there. this bright circular field is surrounded by darkness. explain what is happening and compute the cone angle.
The phenomenon that occurs here is known as the snell's window, and the cone angle is 98 degrees.
To calculate the cone angle,
The cone angle is equal to two times the critical angle which is say, Ф
To calculate the critical angle Ф take the equation sinФ = n1÷n2 where
n1 is the refractive index of air, and n2 is the refractive index of water.
The refractive index of air, n1 = 1,
The refractive index of water, n2 = 1.33,
Substituting these values in the equation sinФ = n1÷n2, the equation becomes
sinФ = 1÷1.33
solving for Ф
Ф = [tex]sin^{-1}[/tex](1÷1.33)
Ф = 49°
Cone angle = 2×Ф
Cone angle = 2×49
Cone angle = 98°.
The phenomenon taking place here is known as snell's window. In this phenomenon, the person under the water sees everything above the surface through a cone of light whose width is about 96°-98°. Whatever is outside this cone appears dark to the person who is under the water. This is caused due to the refraction of light in different mediums.
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Two students, Jenny and Cho, are investigating motion.
Jenny walks in a straight line.
Cho measures the distance Jenny has walked at 10 s intervals.
State two measuring instruments the students should use.
Answer:
1: a measuring instruments the students should use for time is a stopwatch
2: a measuring instruments the students should use for distance is a measuring tape
Explanation:
pls mark brainliest
Question 6:A particle of mass 0.2 kg is moving in a straight line at speed 2.5ms, a constant force of magnitude FN is applied to it in the direction in which it is moving. The speed of the particle 5 seconds later is 4.5ms Find the value of F, magnitude of the force applied.
First, let's calculate the particle acceleration, using the formula below:
[tex]\begin{gathered} V=V_0+a\cdot t \\ 4.5=2.5+a\cdot5 \\ 5a=4.5-2.5 \\ 5a=2 \\ a=\frac{2}{5}=0.4\text{ m/s}^2 \end{gathered}[/tex]Now, to find the force applied, let's use the second law of Newton:
[tex]\begin{gathered} F=m\cdot a \\ F=0.2\cdot0.4 \\ F=0.08\text{ N} \end{gathered}[/tex]Therefore the value of F is 0.08 N.
a spotlight on the ground is shining on a wall away. if a woman tall walks from the spotlight toward the building at a speed of how fast is the length of her shadow on the building decreasing when she is from the building?
His shadow is shortening at a rate of dy/dt = 0.6m/s at the time.
What is triangle total degrees?The triangle's three angles add up to a total of 180 degrees
The individual is standing at point D and has his head at point E at the time indicated in the problem.
At that moment, his shadow is visible on the wall at y = BC.
The two right triangles ABC and ADE both contain similar triangles. The ratios on their respective sides in the similar triangles are equal:
AD/AB = DE/BC
8/12 = 2/y
y = 3
If we suppose that the man is x distance from the building, then the distance from the spotlight to the man is 12 x.
12-x/12 = 2/y
1-1/12x = 2 1/y
Let's take derivatives of both sides:
-1/12dx = -2 .1/y²dy
Divide both sides by dt as follows.
-1/12.dx/dt = 2/y²dy/dt
At the time specified:
dx/dy = 1.6m/s
y = 3
Let's connect them
- 1/2 (1.6) = -2.9 dy/dt
dy/dt = 1.6/12×9/2 = 0.6m/s.
At the time in question, this is the pace at which his shadow's length is shortening.
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The complete question is:
"A spotlight on the ground shines on a wall 12 m away. If a man 2 m tall walks from the spotlight toward the building at a speed of 1.6 m/s, how fast is the length of his shadow on the building decreasing when he is 4 m from the building?"
Figure 5 shows a ray of light in an optical fibre.
Figure 5
Inside of optical fibre
What is the name given to the dotted line on Figure 5?
[1 mark]
The technology that sends information as light pulses via a glass or plastic fiber is known as fiber optics, or optical fiber. These glass fibers can range in quantity from a few to several hundred in a fiber optic cable. The glass fiber core is encircled by yet another glass layer known as cladding.
What fiber optics does and how it works?In fiber optics, data is sent as pulses of light, or photons, that travel over a fiber optic cable. Incoming light is bent at a specific angle due to the differing refractive indices of the glass fiber core and cladding.
Total internal reflection is the method by which light signals traveling through fiber optic cable bounce off the core and cladding in a series of zigzag motions. Due to the thick glass layers, the light signals move around 30% slower than the speed of light instead of at the speed of light.
Fiber optic transmission sometimes needs repeaters at a distance to renew, or amplify, the signal throughout its path. By converting the optical signal to an electrical signal, processing that electrical signal, and then retransmitting the optical signal, these repeaters regenerate the optical signal.
Up to 10 Gbps transmissions can currently be supported by fiber optic connections. A fiber optic cable often gets more expensive as its bandwidth capacity increases.
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you are designing a diving bell to withstand the pressure of seawater at a depth of 210 m. part a what is the gauge pressure at this depth? (you can ignore changes in the density of the water with depth.)
The gauge pressure at the depth of 210 m will be 2.11974 x 10^{6} Pa .
What is gauge pressure ?
The amount of pressure measured in relation to the applied atmospheric pressure is known as gauge pressure. A diaphragm sensor can be used to measure gauge pressure, with one side of the diaphragm exposed to the pressure medium that has to be measured and the other exposed to the surrounding atmospheric pressure.
The pressure from the weight of the atmosphere is included in gauge pressure readings because gauge pressure is zero-referenced against ambient air (or atmospheric) pressure. This indicates that gauge pressure varies in response to both weather and height above sea level. Gauge pressure measurement is adequate for the majority of industrial applications because every operation in a refinery or manufacturing facility operates at the same air pressure.
The gauge pressure is ;
P = density x g x depth
where , density is of sea water .
Calculate the gauge pressure as -:
P = density x g x depth
= (1030 kg/m^{3})(9.8 m/s^{2})(210 m)
= 2.11974 x 10^{6} Pa .
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a young weightlifter raises a weight from a height of 0.6 m to a height of 1.3 m in a time of 3.2 seconds. his applied force has put out an average power of 43 w. if the weight does not accelerate at all in the process, what must be its mass?
The mass is 20.038 Kg.
Mass is the amount of count in a physical body. it's also a degree of the frame's inertia, the resistance to acceleration while a net pressure is carried out. An item's mass additionally determines the power of its gravitational attraction to different bodies. The SI base unit of mass is the kilogram.
F = mg
Displacement of weight = h
= 0.7 m
work done by weight lifter W = F × h
W = (mgh)
= 0.7mg
work done in 3.2 seconds
power P = W/Δt = 0.7 mg/3.2
Given,
P = 43 W = 43 J/s
⇒ 0.7 mg/3.2 = 43 j/s
⇒ M = (43 × 3.2)/(0.7 × 9.8)
M = 20.038 Kg ans.
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