The choices that represent the step in the oil formation process are:
Layers of sediment cover dead remains.
Organisms die and sink to the ocean floor.
Heat and pressure change dead remains.
Oil formation process:Oil and gas should be created from organic material that is deposited with respect to the sediments on the seabed and after that, it brokes down and then transformed over millions of years. The sediment layers should be covered and dead. The organisms die and then they sink to the ocean. The change to the heat and pressure should remain dead.
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Answer: 2, 3, and 4.
Organisms die and sink to the ocean floor.
Heat and pressure changes dead remains.
Layers of sediment cover dead remains.
Explanation:
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A blue ball is thrown upward with an initial speed of 24.1 m/s, from a height of 0.5 meters above the ground. 2.9 seconds after the blue ball is thrown, a red ball is thrown down with an initial speed of 7.2 m/s from a height of 32 meters above the ground. The force of gravity due to the earth results in the balls each having a constant downward acceleration of 9.81 m/s2. Take up as the positive direction.
1. What is the speed of the blue ball when it reaches its maximum height?.
2. How long does it take the blue ball to reach its maximum height?.
3. What is the maximum height the blue ball reaches?.
4. What is the height of the red ball 3.77 seconds after the blue ball is thrown?.
5. How long after the blue ball is thrown are the two balls in the air at the same height?.
Answer:
1. Speed=0
2. 2.46 s
3.30.1 m
4. 22.0 m
5.1.004 s
Explanation:
We are given that
Initial speed of blue ball, u=24.1 m/s
Height of blue ball from ground y_0=0.5 m
Initial speed of red ball , u'=7.2 m/s
Height of red from ground=y'0=32 m
Gravity, g=[tex]9.81ms^{-2}[/tex]
1.When the ball reaches its maximum height then the speed of the blue ball is zero.
2.v=0
[tex]v=u+at[/tex]
Using the formula and substitute the values
[tex]0=24.1-9.81t[/tex]
Where g is negative because motion of ball is against gravity
[tex]24.1=9.81t[/tex]
[tex]t=\frac{24.1}{9.81}=2.46s[/tex]
3.[tex]y=y_0+ut+\frac{1}{2}at^2[/tex]
Using the formula
[tex]y=0.5+24.1(2.46)-\frac{1}{2}(9.81)(2.46)^2[/tex]
[tex]y=30.1 m[/tex]
4.Time of flight for red ball=3.77-2.9=0.87s
[tex]y'=32-7.2(0.87)-\frac{1}{2}(9.81)(0.87)^2[/tex]
[tex]y'=22.0m[/tex]
Hence, the height of red ball 3.77 s after the blue ball is 22.0 m.
5.According to question
[tex]0.5+24.1(t+2.9)-\frac{1}{2}(9.81)(2.9+t)^2=32-7.2t-\frac{1}{2}(9.81)t^2[/tex]
[tex]0.5+24.1t+69.89-4.905(t^2+5.8t+8.41)=32-7.2t-4.905t^2[/tex]
[tex]0.5+24.1t+69.89-4.905t^2-28.449t-41.25105=32-7.2t-4.905t^2[/tex]
[tex]0.5+69.89-41.25105-32=-24.1t+28.449t-7.2t[/tex]
[tex]-2.86105=-2.851t[/tex]
[tex]t=\frac{2.86105}{2.851}=1.004 s[/tex]
Hence, 1.004 s after the blue ball is thrown are the two balls in the air at the same height.
A piece of aluminium with mass 1 kg and density 2700 kg/m3 is suspended from a string and then completely immersed in a container of water. Calculate the tension in the string (a) before and (b) after the metal is immersed.
(with steps please)
A piece of aluminium with mass 1 kg and density 2700 kg/m3 is suspended from a string and then completely immersed in a container of water. Calculate the tension in the string (a) before and (b) after the metal is immersed.
ANSWER: 6.296NI am unable to add the workings out, however please do message me and I will be able to provide you with them :)
An object with a mass of 5 kg is traveling ata velocity of 20 m/s. How much kinetic energy does it have?
Answer:
KE = ½mv²
KE = ½5kg×(20m/s)²
KE = ½5kg×400m²/s²
KE =5kg×200m²/s²
KE = 1000joules
a star produces energy by? ( a p e x )
A. absorbing and releasing the matter of other stars.
B. rotating quickly on its axis.
C. reflecting energy from nearby stars.
D. creating helium from hydrogen in its core.
A star produces energy by creating helium from hydrogen in its core (option D).
How does a star produce energy?A star is a luminous celestial body, made up of plasma (particularly hydrogen and helium) and having a spherical shape.
Nuclear fusion is the energy-producing process in stars. The "proton-proton chain," a series of events that turns four hydrogen atoms into one helium atom, dominates this process for the majority of stars.
Nuclear fusion is the combining of the nuclei of small atoms to form the nuclei of larger ones, with a resulting release of large quantities of energy.
Hydrogen atoms fuse in the core of the star to form a helium atom, which generates the energy for the star.
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24. Which wave below has the highest amplitude? Why?
As you sit in a fishing boat, you noticed that 22 waves pass the boat every 60 seconds . If the distance from one crest to the next is 0.5 m, what is the speed of these waves?
Answer:
0.1835m/s
Explanation:
The formula for calculating the speed of wave is expressed as;
v = fλ
f is the frequency - The number of oscillations completed in one seconds
If 22 waves pass the boat every 60 seconds,
number of wave that passes in 1 seconds = 22/60 = 0.367 waves
Therefore the frequency f of the wave is 0.367Hertz
λ (wavelength) is the distance between successive crest and trough of a wave
λ = 0.5m
Substitute the given values into the formula
v = fλ
v = 0.367 * 0.5
v = 0.1835
Hence the speed of the waves is 0.1835m/s
A ball is dropped from a top of a 70 m tall building. How fast will the velocity be when it hits the
ground?
Answer:
take gravaiy and times it by 70m to find max v
Explanation:
ps i am olny 10 cool ya
A rock thrown vertically upward from the surface of the moon at a velocity of 32m/sec reaches a height of sequals32 t minus 0.8 t squaredmeters in t sec.a. Find the rock's velocity and acceleration at time t.b. How long does it take the rock to reach its highest point?c. How high does the rock go?d. How long does it take the rock to reach half its maximum height?e. How long is the rock aloft?
Answer:
A) v(t) = (32 - 1.6t) m/s
a(t) = -1.6 m/s²
B) t = 20 seconds
C) 320 m
D) t = 5.85 seconds going up and 34.14 seconds going down
E) 40 seconds
Explanation:
Height is given by the equation;
S(t) = 32t - 0.8t²
A) Velocity after time t is gotten from first derivative of the distance.
Thus;
v(t) = dS/dt = 32 - 1.6t
Acceleration at time t is gotten from derivative of the velocity.
Thus;
a(t) = d²S/dt² = -1.6 m/s²
B) At highest point, velocity is zero.
Thus;
32 - 1.6t = 0
1.6t = 32
t = 32/1.6
t = 20 seconds
C) To find how high the rock goes, it means we are looking for maximum height.
This will be at t = 20 seconds.
Thus;
S(20) = 32(20) - 0.8(20)²
S(20) = 640 - 320
S(20) = 320 m
D) we want to find the time it will take to reach half its maximum height.
since maximum height is 320 m, then half the maximum height is; S_½ = 320/2 = 160
Thus;
160 = 32t - 0.8t²
0.8t² - 32t + 160 = 0
Using quadratic formula, we will get;
t = 5.85 seconds going up and 34.14 seconds going down
E) time the rock is aloft = twice the time it took to reach maximum height.
Thus; t_aloft = 2 × 20 = 40 seconds
Kinetic energy depends on which two things
Answer:
Explanation: relationship between the object and the observer's frame of reference.
A small block with mass 0.200 kg is released from rest at the top of a frictionless incline. The block travels a distance 0.796 m down the incline in 2.00 s. The 0.200 kg block is replaced by a 0.400 kg block. If the 0.400 kg block is released from rest at the top of the incline, how far down the incline does it travel in 2.00 s
Answer:
distance travelled by the block is 0.796 m
{ acceleration is independent of mass, so both the masses travel equal distance in 2 s }
Explanation:
Given that;
mass of block m = 0.200 kg
distance travelled d = 0.796 m
time t = 2.00 s
m₂ = 0.400 kg
If the 0.400 kg block is released from rest at the top of the incline, how far down the incline does it travel in 2.00 s?
Now, using the second equation of motion;
d = ut + ([tex]\frac{1}{2}[/tex] × at²)
as the object started from rest, u=0
so, we substitute
0.796 = 0×2 + ([tex]\frac{1}{2}[/tex] × a(2)²)
0.796 = 0 + ([tex]\frac{1}{2}[/tex] × 4a)
0.796 = 2a
a = 0.796 / 2
a = 0.398 m/s²
using first equation of motion
[tex]V_{f}[/tex] = u + at
we substitute
[tex]V_{f}[/tex] = 0 + 0.398 × 2
[tex]V_{f}[/tex] = 0.796 m/s
now, average velocity is given as;
[tex]V_{avg}[/tex] = ( 0.796 m/s + 0 ) / 2
[tex]V_{avg}[/tex] = ( 0.796 m/s + 0 ) / 2
now, distance as the block moves in 2s will be;
D = [( 0.796 m/s + 0 ) / 2 ] × 2
D = 0.796 m
Therefore, distance travelled by the block is 0.796 m
{ acceleration is independent of mass, so both the masses travel equal distance in 2 s }
The distance traveled by the object with the uniform motion can be given by the second equation of the motion.
The distance traveled by the big block with weight in 2 seconds is.
What is second equation of motion?
The distance traveled by the object with the uniform motion can be given by the second equation of the motion. It can be given as,
[tex]s=ut+\dfrac{1}{2} at^2[/tex]
Given information-
The mass of the small block is 0.200 kg.
The total distance traveled by the block is 0.796.
Initial velocity of small block is zero.
Total time taken by the block to travel this distance is 2 seconds.
Put the values in the above equation as,
[tex]0.796=0\times 2+\dfrac{1}{2} a\times 2^2\\0.796=2a\\a=0.398[/tex]
Thus the acceleration of the small block is 0.398 meter per second.
Now the mass is doubled which is, 0.400 kg. As the acceleration does not depends on the mass, thus the acceleration for both cases is 0.398.
The velocity of the big block can be given as,
[tex]V=u+0.398\times2\\V=0+0.796\\V=0.796[/tex]
The velocity of the big block is 0.796.
The average velocity of the big block is given by,
[tex]V_{avg}=(\dfrac{0.796+0}{2} })\\V_{avg}=0.398[/tex]
The distance traveled by the object is the ratio of the velocity of the body to the time taken by it. Thus the distance traveled by the big block in 2 seconds is,
[tex]d={{0.398} \times2}\\d=0.796[/tex]
Thus the distance traveled by the big block with weight in 2 seconds is.
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A train 350 m long is moving on a straight track with a speed of 84.1 km/h. The engineer applies the brakes at a crossing, and later the last car passes the crossing with a speed of 15.8 km/h. Assuming constant acceleration, determine how long the train blocked the crossing. Disregard the width of the crossing.
Answer:
t = 25.0 s
Explanation:
Assuming that the engineer applies the brakes just over the crossing, the train moves exactly 350 m at a constant acceleration, with a final speed (when the last car of the train leaves the crossing) of 15.8km/h.Since we know the initial and final speeds, and the horizontal distance traveled (the length of the train) we can use the following kinematic equation to get the acceleration:[tex]v_{f}^{2} - v_{o}^{2} = 2*a* \Delta x (1)[/tex]
Since we need to find the time in seconds, it is advisable to convert vf and vo to m/s first, as follows:[tex]v_{o} = 84.1 km/h*\frac{1h}{3600s} *\frac{1000m}{1km} = 23.4 m/s (2)[/tex]
[tex]v_{f} = 15.8 km/h*\frac{1h}{3600s} *\frac{1000m}{1km} = 4.4 m/s (3)[/tex]
Replacing (2) and (3) in (1), since Δx =350m, we can solving for a:[tex]a = \frac{(4.4m/s)^{2} - (23.4m/s)^{2}}{2*350m} = -0.76 m/s2 (4)[/tex]
In order to get the time, we can simply use the definition of acceleration, and rearrange terms:[tex]t =\frac{v_{f}-v_{o}}{a} = \frac{(4.4m/s)-(23.4m/s)}{-0.76m/s2} = 25.0 s (5)[/tex]
What is light? Let's tallk about anything
Patrick the pole vaulter is running with a 20m long pole toward a 10m long garage with doors at each end. Patrick is running so fast that to an observer O in the rest frame of the garage, the pole appears only 10m long (i.e. the pole would just fit inside the garage).
a. In the frame of observer O, how fast is Patrick running?
b. The observer O closes very quickly the doors on both ends of the garage just as Patrick and the pole are inside, and then immediately opens them again so that Patrick can run out on the other end. However, according to Patrick, the garage is moving towards him and is only 5m long. How can Patrick and his 20m pole possibly make it through without hitting a door?
Answer with Explanation:
We are given that
Length of pole(l0)=20 m
Length of garage(l)=10 m
a. We have to find the speed by which the Patrick is running.
We know that
[tex]l=l_0\sqrt{1-\frac{v^2}{c^2}}[/tex]
Substitute the values
[tex]10=20\sqrt{1-\frac{v^2}{c^2}}[/tex]
[tex]\frac{10}{20}=\sqrt{1-\frac{v^2}{c^2}}[/tex]
[tex]\frac{1}{2}=\sqrt{1-\frac{v^2}{c^2}}[/tex]
[tex]\frac{1}{4}=1-\frac{v^2}{c^2}[/tex]
[tex]\frac{v^2}{c^2}=1-\frac{1}{4}=\frac{3}{4}[/tex]
[tex]v^2=0.75c^2[/tex]
[tex]v=0.87c[/tex]
b.
l0=10 m
[tex]l=10\sqrt{1-\frac{3}{4}}[/tex]
[tex]l=5m[/tex]
Hence, the length of garage appear to him=5 m long.
Microtubules are filamentous structures in cells that maintain cell shape and facilitate the movement of molecules within the cell. They are long, hollow cylinders with a diameter of about 25 nm. It is possible to incorporate fluorescent molecules into microtubules; when illuminated by an ultraviolet light, the fluorescent molecules emit visible light that can be imaged by the optical system of a microscope.
Required:
If the emitted light has a wavelength of 550nm and the NA of the microscope objective is 1.3, what is the smallest resolvable separation between two objects in the microscope?
Answer:
the smallest resolvable separation between two objects in the microscope is 2.58 × 10⁻⁷ m
Explanation:
Given the data in the question;
the smallest resolvable separation distance between two objects in the microscope can be determined using the following quation;
[tex]d_{min}[/tex] = Dλ/NA = (0.61)λ /NA
where λ is the wavelength of emitted light = 550nm = 5.5 × 10⁻⁷ m
NA = microscope objective = 1.3
so we substitute
[tex]d_{min}[/tex] = (0.61)5.5 × 10⁻⁷ m) / 1.3
[tex]d_{min}[/tex] = 0.0000003355 / 1.3
[tex]d_{min}[/tex] = 2.58 × 10⁻⁷ m
Therefore, the smallest resolvable separation between two objects in the microscope is 2.58 × 10⁻⁷ m
Through what potential difference should electrons be accelerated so that their speed is 1.0 % of the speed of light when they hit the target
Answer:
Explanation:
Considering non - relativistic approach : ----
Speed of electron = 1 % of speed of light
= .01 x 3 x 10⁸ m /s
= 3 x 10⁶ m /s
Kinetic energy of electron = 1/2 m v²
= .5 x 9.1 x 10⁻³¹ x ( 3 x 10⁶ )²
= 40.95 x 10⁻¹⁹ J
Kinetic energy in electron comes from lose of electrical energy equal to
Ve where V is potential difference under which electron is accelerated and e is electronic charge .
V x e = kinetic energy of electron
V x 1.6 x 10⁻¹⁹ = 40.95 x 10⁻¹⁹
V = 25.6 Volt .
A friend claims that as long as he has his seat belt on, he can hold on to a 14 kg child in a 64 mi/h head-on collision with a brick wall in which the car passenger compartment comes to a stop in 0.05 s. Show that the violent force during the collision will tear the child from his arms. A child should always be in a toddler seat secured with a seat belt in the back seat of a car.
Answer:
Force is too large for a person to exert.
Explanation:
m = Mass of child = 14 kg
u = Initial velocity of compartment = 64 mi/h = [tex]\dfrac{64\times 1609.34}{3600}=28.61\ \text{m/s}[/tex]
v = Final velocity of compartment = 0
t = Time taken by the compartment to stop = 0.05 s
Force is given by
[tex]F=ma\\\Rightarrow F=m\dfrac{v-u}{t}\\\Rightarrow F=14\times \dfrac{0-28.61}{0.05}\\\Rightarrow F=-8010.8\ \text{N}[/tex]
The person has to exert a force of [tex]-8010.8\ \text{N}[/tex] in the opposite direction to keep holding the child. This is a huge amount of force and cannot be done by a person's physical strength alone.
A woman stands on a moving sidewalk (conveyor belt) that is moving to the right at a speed of 2 m/s relative to the ground. A dog runs on the belt toward the woman at a speed of 8 m/s relative to the belt. A conveyor belt is moving to the right at the speed v subscript belt, ground equals 2 meters per second. A girl stands stationary on the left end of the belt. A dog runs on the belt to the left toward the girl at the speed v subscript dog, belt equals 8 meters per second. What is the speed of the dog relative to the ground
Answer:
V_{2G} = - 10 m / s
Explanation:
This is an exercise on relative velocities in one dimension, let's use the subscript 1 for the girl, the 2 for the but, the index 3 for the conveyor belt and the index G for the Earth
we will consider the velocity is positive to the right.
They indicate the speed of the belt with respect to the ground = 2 m/s
the dog speed with respect to the belt = - 8 m / s
Dog speed is requested with respect to the Earth
[tex]V_{2G} = V_{23} - V_{3G}[/tex]
notice that the inner subscripts cancel out.
We calculate
= -8 - 2
V_{2G} = - 10 m / s
the negative sign indicates that the dog moves to the left
Use g = 9.80 m/s2 (down]
1. What velocity does a freely falling object reach after 4.0 s if it starts from rest?
The velocity of the freely falling object reached after 4.0 s is equal to 39.2 m/s.
What is the equation of motion?A relation can be established between the velocity of the body and acceleration of the body when the body is moving along a straight line with uniform acceleration. The distance traveled by the body in a specific time has expressed a set of equations called equations of motion.
The first equation of motion can be expressed as follows:
v = u + at
Where u is the initial velocity, v is the final velocity and a is the acceleration and t is the time of the object.
Given, the acceleration of the falling object, g = 9.8 m/s²
The initial velocity of the object, u = 0, and the time = 4 sec
The velocity of the object after time, t = 4 sec.
v = (0) + (9.80)× (4)
v = 39.2 m/s
Therefore, the velocity of the falling object will be 39.2 m/s after 4 seconds.
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define stress engineering science
Answer:
Stress, in physical sciences and engineering, force per unit area within materials that arises from externally applied forces, uneven heating, or permanent deformation and that permits an accurate description and prediction of elastic, plastic, and fluid behaviour.
I hope it's helpful!
Couldn’t you technically make infinite speed by putting a car in a vacuum chamber? Since top speed it made by the amount of force it takes until power and wind balance couldn’t you just get rid of that factor entirely?
Answer:
No
Explanation:
For infinite speed to be achevied, one must have no sink of energy to spend. The source of entropy in this example, is the tires hitting the surface, producing heat and friction. Not to mention that you'd still need fuel to start the car, and an infinite tunnel or track, which would be impossible and speed up to process of energy loss through entropy quicker.
Mae puts on her new winter scarf. Signals are sent to her brain telling her that the scarf is soft. Which sense provides her brain with this information?
Answer:
Touch
Explanation:
The resistivity of a material is 2x10-3Ωm. What is the conductivity\n
Answer:
Conductivity, [tex]\sigma=500\ (\Omega-m)^{-1}[/tex]
Explanation:
Given that,
The resistivity of a material, [tex]\rho=2\times 10^{-3}\ \Omega-m[/tex]
We need to find the conductivity of the material.
We know that the reciprocal of resistivity is called conductivity.
[tex]\sigma=\dfrac{1}{\rho}\\\\\sigma=\dfrac{1}{2\times 10^{-3}}\\\\\sigma=500\ (\Omega-m)^{-1}[/tex]
So, the conductivity of the material is [tex]500\ (\Omega-m)^{-1}[/tex].
Two insulated wires, each 2.64 m long, are taped together to form a two-wire unit that is 2.64 m long. One wire carries a current of 7.68 A; the other carries a smaller current I in the opposite direction. The two wire unit is placed at an angle of 65.0o relative to a magnetic field whose magnitude is 0.59 T. The magnitude of the net magnetic force experienced by the two-wire unit is 4.11 N. What is the current I
Answer:
[tex]4.77\ \text{A}[/tex]
Explanation:
F = Magnetic force = 4.11 N
[tex]I_n[/tex] = Net current
[tex]I_2[/tex] = Current in one of the wires = 7.68 A
B = Magnetic field = 0.59 T
[tex]\theta[/tex] = Angle between current and magnetic field = [tex]65^{\circ}[/tex]
[tex]l[/tex] = Length of wires = 2.64 m
[tex]I[/tex] = Current in the other wire
Magnetic force is given by
[tex]F=I_nlB\sin\theta\\\Rightarrow I_n=\dfrac{F}{lB\sin\theta}\\\Rightarrow I_n=\dfrac{4.11}{2.64\times 0.59 \sin65^{\circ}}\\\Rightarrow I_n=2.91\ \text{A}[/tex]
Net current is given by
[tex]I_n=I_2-I\\\Rightarrow I=I_2-I_n\\\Rightarrow I=7.68-2.91\\\Rightarrow I=4.77\ \text{A}[/tex]
The current I is [tex]4.77\ \text{A}[/tex].
Consider a collection of charges in a given region and suppose all other charges are distant and have a negligible effect. Further, the electric potential is taken to be zero at infinity. If the electric potential at a given point in the region is zero, which of the following statements must be true?
A. The electric field is zero at that point.
B. The charges have the same sign and are symmetrically arranged around the given point.
C. The electric potential energy is a minimum at that point.
D. There is no net charge in the region.
E. Some charges in the region are positive, and some are negative.
Answer:
E. Some charges in the region are positive, and some are negative.
Explanation:
Electric potential is given as;
[tex]V = \frac{W}{Q}[/tex]
where;
W is the work done in moving a charge between two points which have a difference in potential
Q is quantity of charge in the given region
If the electric potential at a given point in the region is zero, then sum of the charges in the given region must be equal to zero. For the charges to sum to zero, some will be positive while some will be negative,.
Therefore, the correct statement in the given options is "E"
E. Some charges in the region are positive, and some are negative.
How high will a 1-kilogram ball go if it is thrown straight up into the air with an initial velocity of 30 m/sec?
h max = 45.92 m
Further explanationGiven
1 kg ball
vo=initial velocity = 30 m/s
Required
Height
Solution
We can use the principle of energy conservation or parabolic motion
For parabolic motion
h max = vo²sin²θ/2g(θ= 90° for straight up)
For conservation energy :
KE₁=PE₂
1/2.m.vo² = m.g.h
We choose parabolic motion
h max = vo²sin²θ/2g
h max = 30²sin²90/2 x 9.8
h max = 45.92 m
A ship sets out to sail to a point 141 km due north. An unexpected storm blows the ship to a point 102 km due east of its starting point. (a) How far and (b) in what direction (as an angle from due east, where north of east is a positive angle) must it now sail to reach its original destination
Answer:
Explanation:
The point of destination is 141 north and 102 km east . Vectorially it is represented by unit vector as follows .
D = 102 i + 141 j
magnitude of D = √ ( 102² + 141² )
= √ ( 10404 + 19881)
= √ 30285
= 174 km
To go back to original position , ship should move on the following vector
D = -102 i - 141 j
Tan Ф = 141 / 102 = 1.38
Ф = 54⁰ , direction will be south of west . From north east , angle will be
180 + 54 = 234⁰
Fred is doing an experiment with basketballs. He drops one basketball while standing on a chair, and drops another basketball while standing on the roof. Which basketball will probably bounce the highest?
A. The basketball from the chair
B. The basketball from the roof
C. They will bounce the same
D. Neither ball will bounce
A car travels 100 km due East in 2 hours. It then travels 50 km South in 1 hour. What is its average velocity?
The average velocity of the car is 37.27 km/h.
The given parameters;
Initial displacement of the car, x = 100 kmTime of motion, t = 2 hoursFinal displacement of the car, y = 50 kmtime of motion, t = 1 hourThe average velocity of the car is calculated as follows;
[tex]average \ velocity = \frac{Total \ displacement}{Total \ time}[/tex]
The total displacement of the car is calculated as follows;
[tex]d = \sqrt{x^2 + y^2} \\\\d = \sqrt{100^2 \ + \ 50^2} \\\\d = 111.803 \ km[/tex]
The average velocity of the car is calculated as follows;
[tex]v = \frac{111.803}{3} \\\\v = 37.27 \ km/h[/tex]
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Newtons third law is called law of action and reaction?
Answer:
His third law states that for every action (force) in nature there is an equal and opposite reaction. In other words, if object A exerts a force on object B, then object B also exerts an equal and opposite force on object A. ... In reaction, a thrusting force is produced in the opposite direction.
Explanation:
Answer:
These two forces are called action and reaction forces and are the subject of Newton's third law of motion Formally stated Newton's third law is For every action
Explanation:
In the Dorben Company and industrial engineer designed a workstation where the seeing task was difficult because of the size of the components going into the assembly. The desired brightness was 100 fL and the workstation was painted a medium green with a reflectance of 50 percent. What illumination in foot-candles would be required at this workstation to provide the desired brightness? Estimated the required illumination if you repainted the workstation with a light cream paint. what is the luminance of a surface having a 50 % reflectance and a 4 fc illumination?
Answer:
Illumination = 200 fc
luminance = 2FL
Explanation:
given data
desired brightness = 100 fL
painted medium green reflectance = 50 percent
illumination = 4c
solution
we get here Illumination by using the formula that is
Illumination = luminance ÷ reflectance ......................1
Illumination = 100 fL ÷ 0.50
Illumination = 200 fc
With light colored cream paint with a reflectance of 75% the required illumination is 133 fc.
and
luminance will be
luminance = Illumination × reflectance = 4fc × 0.5 = 2FL