The percentage error in calculated value of the refractive index of acrylic block is calculated using the formula: (true value - calculated value/true value) × 100%.
What is refractive index of a refracting medium?The refractive index of a medium is the ratio of the sine of the angle of incidence on the medium and the sine of the angle of refraction.
Refractive index = sin i/sin rwhere;
i is the angle of incidencer is angle of refraction.The refractive index of a medium increases with density of the medium.
To determine the percentage error in determining the refractive index of a medium, the following formula is used:
Percent error = error/true value × 100%Therefore, the percentage error in calculated value of the refractive index of acrylic block is the ratio of the error and true value multiplied by 100%.
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Organic farming follows the
principles of health and
A. economic gain
B. wealth
C. profits
D. ecology
Answer:The Answer is economic gain
Explanation:
Given the necessary information about the two components of projectile motion, your task is to explain the difference or even describe the two components by their formulas or equations.
The two components of projectile motion include the following:
Horizontal motionVertical motion.What is a Projectile?This is defined as a missile propelled by the application of an external force and allowed to move freely under the influence of gravity and air resistance.
The equation for Horizontal motion Vx = V * cos(α)
Vertical velocity component: Vy = V * sin(α)
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Adrian wants to take a personality test to help her decide which college major to choose. How can the test help her
Answer:
It could point her in a direction that best suits her
Explanation:
Hope that helps!
Which of the following would NOT cause acceleration? A. An object falling towards the earth. B. The wind catching the sail of a sailboat. C. A car traveling at a constant velocity. D. A pitcher throwing a baseball.
A student uses a motion detector to study the kinematics of a block suspended on a vertical spring that obeys Hooke’s law. The student pulls the block a distance y from equilibrium, releases it from rest, and records the speed v of the block as it passes the equilibrium position. The student repeats this process several times for different values of y. Which variables should be plotted on the horizontal and vertical axes to yield a linear graph?
When the speed of the block is plotted on the vertical axis and displacement of the block plotted on the horizontal axis, a linear graph will be obtained showing the inverse relationship between the speed of the block and the displacement of the block.
Hooke's lawHooke's law states that force or load applied to elastic material is directly proportional to the extension produced in the material. This law can be written as;
F = kx
where;
F is the applied forcek is spring constantx is extension of the materialVelocity of the blockThe speed of the block increases with decreasing displacement of the bock. This is because the kinetic energy of the block is maximum at zero displacement and minimum at maximum displacement of the block.
Thus, when the speed of the block is plotted on the vertical axis and displacement of the block plotted on the horizontal axis, a linear graph will be obtained showing the inverse relationship between the speed of the block and the displacement of the block.
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PLEASE HELP!!!!!!! MT TIME FOR MY TEST IS ALMOST OVER!!!
An ideal spring, with a pointer attached to its end, hangs next to a scale. With a 100-N weight attached, the pointer indicates "40" on the scale as shown. Using a 200-N weight instead results in "60" on the scale. Using an unknown weight X instead results in "30" on the scale. The weight of X is:
Answer:
50 N
Explanation:
Let the natural length of the spring = L
so
100 = k(40 - L) (1)
200 = k(60 - L) (2)
(2)/(1): 2 = (60 - L)/(40 - L)
60 - L = 2(40 - L)
60 - L = 80 - 2L
2L - L = 80 - 60
L = 20
Sub it into (1):
100 = k(40 - 20) = 20k
k = 100/20 = 5 N/in
Now
X = k(30 - L) = 5(30 - 20) = 50 N
Where is the electric field between a charged balloon and a charged piece of paper 0.04 m away?
A. The electric field is in the region around the balloon and paper where there are outside charges that affect them.
B. The electric field is in the 0.04 m distance between the balloon and paper and not anywhere else around either the paper or the balloon.
C. The electric field is all around the balloon and paper including between them, where, depending on their charge, there would be a force of either attraction or repulsion.
D. The electric field is in the region in which the forces come in contact.
The electric field is all around the balloon and paper including between them, where, depending on their charge, there would be a force of either attraction or repulsion.
What is electric field?An electric field is the physical field that surrounds electrically charged particles and exerts force on all other charged particles in the field, either attracting or repelling them.
Thus, for the charged balloon and a charged piece of paper 0.04 m away, we can conclude that, the electric field is all around the balloon and paper including between them, where, depending on their charge, there would be a force of either attraction or repulsion.
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The block falls for a time t0, but the string does not completely unwind. What is the change in angular momentum of the pulley-block system from the instant that the block is released from rest until time t0?
The change in angular momentum of the pulley-block system is given as [tex]\omega = Rm_ogt_o[/tex]
Data;
time = t0mass = MoTorqueThis is the force that makes object rotate about an axis. The formula is given as the product between the radius about the axis and the force acting upon it.
The torque about point o is given as
[tex]\tau = R * m_og[/tex]
The change in angular momentum about point 'o' is the product between the torque and time.
[tex]\omega = \tau * t_o\\\omega = Rm_ogt_o[/tex]
The change in angular momentum of the pulley-block system is given as
[tex]\omega = Rm_ogt_o[/tex]
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Why is it more difficult to roller blade across rough asphalt than it is to roller blade across smooth concrete?
A. Asphalt has more mass than concrete.
B. Friction on the roller blades is greater on asphalt than on concrete.
C. Roller blades exert a greater force on asphalt than they do on concrete.
D. The inertia of the roller blades is greater on asphalt then on concrete.
Answer:
The answer is B. Friction on the roller blades is greater on asphalt than on concrete.
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a hunter shoots a stone from his catapult with an initial velocity of 40 m per seconds add elevation of 60 degree with the aim of hitting a bird 145n away.
1. calculate the time of flight
Answer:
approximately 7 seconds
Explanation:
The given scenario is just an example of Projectile Motion.
In this question, we are given:
- Initial Velocity (u) = 40 m/s
- Angle of Elevation (θ) = 60 degrees
- Distance of Bird (s) = 145 meters [Not used for the time of flight]
Time of Flight:
[tex]T = \frac{2uSin(\theta)}{g}[/tex]
we will use g = 9.8 m/s/s since we are not told otherwise.
[tex]T = \frac{2(40)Sin(60 degrees)}{9.8}[/tex]
[tex]T = \frac{40\sqrt{3}}{9.8}[/tex]
T = 4.081(√3) seconds
rounded to 7 seconds
Why is a strong skeletal system important for leading an active lifestyle?
Answer:
Explanation:
Because if you don't have strong bones, you will have problems with your health. So, a strong skeletal system is important for leading a healthy life.
I’m sorry, but the person you are calling has a voice mailbox that has not been set up yet.
Please try again later.
Bye Bye!
Answer:
bye bye
Explanation:
Answer: Or please call again to see if they reached the phone yet.
Explanation: It always stops ringing right before someone gets to there phone so call again in 2 minutes. :3
A 1980-kg car is traveling with a speed of 15.5 m/s. What is the magnitude of the horizontal net force that is required to bring the car to a halt in a distance of 39.2 m
Answer: 6067.5 N
Explanation:
Work = Change in Energy. To start, all of the energy is kinetic energy, so find the total KE using: KE = 1/2(m)(v^2). Plug in 1980 kg for m and 15.5 m/s for v and get KE = 237847.5 J.
Now, plug this in for work: Work = Force * Distance; so, divide work by distance to get 6067.5 N.
At an altitude of 1.3x10^7 m above the surface of the earth an incoming meteor mass of 1x10^6 kg has a speed of 6.5x10^3 m/s. What would be the speed just before impact with the surface of earth?Ignore air resistance.
Show all steps.
Answer:
Approximately [tex]1.1 \times 10^{4}\; {\rm m\cdot s^{-1}}[/tex] if air friction is negligible.
Explanation:
Let [tex]G[/tex] denote the gravitational cosntant. Let [tex]M[/tex] denote the mass of the earth. Lookup the value of both values: [tex]G \approx 6.67 \times 10^{-11}\; {\rm N\cdot m^{2}\cdot kg^{-2}}[/tex] while [tex]M \approx 5.697 \times 10^{24}\; {\rm kg}[/tex].
Let [tex]m[/tex] denote the mass of the meteor.
Let [tex]v_{0}[/tex] denote the initial velocity of the meteor. Let [tex]r_{0}[/tex] denote the initial distance between the meteor and the center of the earth.
Let [tex]r_{1}[/tex] denote the distance between the meteor and the center of the earth just before the meteor lands.
Let [tex]v_{1}[/tex] denote the velocity of the meteor just before landing.
The radius of planet earth is approximately [tex]6.371 \times 10^{6}\; {\rm m}[/tex]. Therefore:
At an altitude of [tex]1.3 \times 10^{7}\; {\rm m}[/tex] about the surface of the earth, the meteor would be approximately [tex]r_{0} \approx 6.371 \times 10^{6}\; {\rm m} + 1.3 \times 10^{7}\; {\rm m} \approx 1.9 \times 10^{7}\; {\rm m}[/tex] away from the surface of planet earth. The meteor would be only [tex]r_{1} \approx 6.371 \times 10^{6}\; {\rm m}[/tex] away from the center of planet earth just before landing.Note the significant difference between the two distances. Thus, the gravitational field strength (and hence acceleration of the meteor) would likely have changed significant during the descent. Thus, SUVAT equations would not be appropriate.
During the descent, gravitational potential energy ([tex]\text{GPE}[/tex]) of the meteor was turned into the kinetic energy ([tex]\text{KE}[/tex]) of the meteor. Make use of conservation of energy to find the velocity of the meteor just before landing.
Initial [tex]\text{KE}[/tex] of the meteor:
[tex]\displaystyle (\text{Initial KE}) = \frac{1}{2}\, m\, {v_{0}}^{2}[/tex].
Initial [tex]\text{GPE}[/tex] of the meteor:
[tex]\displaystyle (\text{Initial GPE}) &= -\frac{G\, M\, m}{r_{0}}[/tex].
(Note the negative sign in front of the fraction.)
Just before landing, the [tex]\text{KE}[/tex] and the [tex]\text{GPE}[/tex] of this meteor would be:
[tex]\displaystyle (\text{Final KE}) = \frac{1}{2}\, m\, {v_{1}}^{2}[/tex].
[tex]\displaystyle (\text{Final GPE}) &= -\frac{G\, M\, m}{r_{1}}[/tex].
If the air friction on this meteor is negligible, then by the conservation of mechanical energy:
[tex]\begin{aligned}& (\text{Initial KE}) + (\text{Initial GPE}) \\ =\; & (\text{Final KE}) + (\text{Final GPE})\end{aligned}[/tex].
[tex]\begin{aligned}& \frac{1}{2}\, m\, {v_{0}}^{2} - \frac{G\, M\, m}{r_{0}} \\ =\; & \frac{1}{2}\, m\, {v_{1}}^{2} - \frac{G\, M\, m}{r_{1}}\end{aligned}[/tex].
Rearrange and solve for [tex]v_{1}[/tex], the velocity of the meteor just before landing:
[tex]\begin{aligned}{v_{1}} &= \sqrt{\frac{\displaystyle \frac{1}{2}\, m\, {v_{0}}^{2} - \frac{G\, M\, m}{r_{0}} + \frac{G\, M\, m}{r_{1}}}{(1/2)\, m}} \\ &= \sqrt{{v_{0}}^{2} - \frac{G\, M}{r_{0}} + \frac{G\, M}{r_{1}}} \\ &= \sqrt{{v_{0}}^{2} - G\, M\, \left(\frac{1}{r_{1}} - \frac{1}{r_{0}}\right)}\end{aligned}[/tex].
Substitute in the values and evaluate:
[tex]\begin{aligned}v_{1} &= \sqrt{{v_{0}}^{2} - G\, M\, \left(\frac{1}{r_{1}} - \frac{1}{r_{0}}\right)} \\ &\approx \sqrt{\begin{aligned}(& 6.5 \times 10^{3}\; {\rm m \cdot s^{-1}}) \\ & - [6.67 \times 10^{-11}\; {\rm N \cdot {m}^{2}\cdot {kg}^{2} \times 5.697\; {\rm kg}}\\ &\quad\quad \times (1 / (6.371 \times 10^{6}\; {\rm m}) - 1 / (1.9371 \times 10^{7}\; {\rm m}))]\end{aligned}} \\ &\approx 1.1 \times 10^{4}\; {\rm m\cdot {s}^{-1}}\end{aligned}[/tex].
(Note that assuming a constant acceleration of [tex]g = 9.81\; {\rm m\cdot s^{-2}}[/tex] would give [tex]v_{1} \approx 1.7\times 10^{4}\; {\rm m\cdot s^{-1}}[/tex], an inaccurate approximation.
17) Which object would likely have the greatest velocity
a. a bouncy ball
b. a bowling call
a go-kart
d. a school bus
Answer: b or d
Explanation: b or d
Answer:
a
Explanation:
F= ma
interestingly
when you increase the mass the acceleration decreases while when the mass decreases the acceleration increases
(man, PHYSICS IS JUST THE BESY)
A: a
object with the smallest mass has largest acceleration
state the precautions that is taken when charging a metal objectexplain why a rubber balloon rubbed will often stick to the wall when it has been rubbed
The balloon will attach to the wall because the balloon's negative charges will drive the electrons in the wall to shift to the other side of their atoms, leaving the wall's surface positively charged.
If a rock is thrown upward on the planet Mars with a velocity of 15 m/s, its height above the ground (in meters) after t seconds is given by H = 15t − 1.86t2. (a) Find the velocity (in m/s) of the rock after 1 second. 11.5404 Incorrect: Your answer is incorrect. seenKey 11.28 m/s (b) Find the velocity (in m/s) of the rock when t = a. m/s (c) When (in seconds) will the rock hit the surface? (Round your answer to one decimal place.) t = s (d) With what velocity (in m/s) will the rock hit the surface?
(a) The velocity (in m/s) of the rock after 1 second is 11.28 m/s.
(b) The velocity of the rock after 2 seconds is 7.56 m/s.
(c) The time for the block to hit the surface is 4.03.
(d) The velocity of the block at the maximum height is 0.
Velocity of the rockThe velocity of the rock is determined as shown below;
Height of the rock after 1 second; H(t) = 15(1) - 1.86(1)² = 13.14 m
v² = u² - 2gh
where;
g is acceleration due to gravity in mars = 3.72 m/s²v² = (15)² - 2(3.72)(13.14)
v² = 127.23
v = √127.23
v = 11.28 m/s
Velocity of the rock when t = 2 secondv = dh/dt
v = 15 - 3.72t
v(2) = 15 - 3.72(2)
v(2) = 7.56 m/s
Time for the rock to reach maximum heightdh/dt = 0
15 - 3.72t = 0
t = 4.03 s
Velocity of the rock when it hits the surfacev = u - gt
v = 15 - 3.72(4.03)
v = 0
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A stage has a radius R = 2.00 m and a mass M = 100 kg. The stage, which has a circular disk
shape, rotates in a horizontal plane, without any friction, about a vertical axle. Sandra has a
mass of m = 60.0 kg and she walks from the rim of the stage toward the center. Calculate the
angular speed when she reaches a point r = 0.500 m from the center, if the angular speed of the
system is 2.0 rad/s when Sandra is at the rim
i want the answer of this exercise
A source of light emits photons with a wavelength of 8.1 x 10-8 meters. What is the frequency of this light
Answer:
Explanation:
Speed of light v = 3 x 10⁸ m/s
wavelength λ = 8.1 x 10⁻⁸ m
frquency f = v/λ = 3.7 x 10¹⁵ Hz
If a source of light emits photons with a wavelength of 8.1 x 10⁻⁸ meters, then the frequency of the light would be 3.7 × 10¹⁵ Hz, as the wavelength and the frequency of the photon are inversely proportional to each other.
What is Wavelength?It can be understood in terms of the distance between any two similar successive points across any wave for example wavelength can be calculated by measuring the distance between any two successive crests.
C = λν
As given in the problem if a source of light emits photons with a wavelength of 8.1 x 10⁻⁸ meters, then we have to find out the frequency of the light,
The frequency of the light = 3 × 10⁸ / 8.1 x 10⁻⁸
=3.7 × 10¹⁵ Hz
Thus, the frequency of the light would be 3.7 × 10¹⁵ Hz
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A women runs a kilometer using a of 210N and a pwower output of 500W how long does it take this woman to complete 1 kilometer
Answer:
7 minutes
Explanation:
Explanation:
Power = work / time
Power = force × distance / time
500 W = 210 N × 1000 m / t
t = 420 s
t = 7 min
An electron with an initial speed of 700,000 m/s is brought to rest by an electric field. What was the potential difference that stopped the electron? What was the initial kinetic energy of the electron, in electron volts?
Answer:
See below.
Explanation:
According to the question, we know that,
work done is given by, [tex]W=qV[/tex]
and change in kinetic energy is, Δ [tex]KE=W=1=1/2[mv^{2} ][/tex]
therefore equating both the equations we get,
[tex]qV=1/2[mv^{2} ][/tex] ⇒ [tex]V=\frac{mv^{2} }{2q}[/tex]
m= mass of electron = [tex]9.1*10^{-31} kg[/tex]
q= charge on an electron = [tex]1.6*10^{-19} C[/tex]
v= speed of electron= 700000m/s
substituting the values in the above equation, we get
[tex]V=\frac{9.1*10^{-31} *(700000)^{2} }{2*1.6*10^{-19} } =1.39V[/tex]
(1). the potential difference that stopped the electron is 1.39 volts.
now the kinetic energy equation is : 2 ways[tex]KE=1/2[mv^{2} ]=\frac{9.1*10^{-31} *700000^{2} }{2} =2.22*10^{-19} J\\[/tex]
or [tex]KE=\frac{2.22*10^{-19} }{1.6*10^{-19} } =1.39eV[/tex]
(2). the initial kinetic energy of the electron is 1.39eV.
can somebody please help
The amplitude of the wave on the given sinusoidal wave graph is 10 cm.
What is amplitude of wave?
The amplitude of a wave is the maximum displacement of a wave. This is the highest vertical position of the wave from the origin.
Amplitude of the wave is calculated as follows;
From the graph, the amplitude of the wave or maximum displacement of the wave is 10 cm.
Thus, the amplitude of the wave on the given sinusoidal wave graph is 10 cm.
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Express this as y=mx + c
1.
[tex] \frac{1}{y} = \alpha ( {x - 6)}^{ - b} [/tex]
Answer:
let alpha be y,
1/2=y(x-6)^(-b)
1/2=y(1/x-6)
1/2=y/x-6
x-6=2y
x-6/2=y
(1/2)x-(6/2)=y
y=(1/2)x-3
The weight of your science book presses on the desk.
A. Gravity pulls on the book.
B. Gravity pulls on the desk.
C. The desk pushes on the book.
Answer:
A. Gravity pulls on the book :)
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A crate of oranges is shoved across a grocery store floor with a force of 100 N for a distance of 1 meter. The crate travels an additional 1 meter after the shove and stops just short of the display. You can conclude: the magnitude of work done by frictional forces on the crate is 100 J the magnitude of work done by the applied force on the crate is 100 J the magnitude of the force of friction on the crate is 50 N all of the above conclusions are correct none of the above conclusions are correct
The statements that are true are;
the magnitude of work done by frictional forces on the crate is 100 Jthe magnitude of work done by the applied force on the crate is 100 JWhat is work done?Work is said to be done when te force applied travels a distance in the direction of the force. Now we can see that when the force is applied by shoving the crate, work is done, an additional work is done by friction to bring the crate to a stop.
Hence, the following are true;
the magnitude of work done by frictional forces on the crate is 100 Jthe magnitude of work done by the applied force on the crate is 100 JLearn more about work: https://brainly.com/question/18094932?
What potential difference is required across an 32 -Ω resistor to cause 33.72 A to flow through it?
[tex]\text{Voltage,}~ V = IR =33.72\times 32 = 1079.04~ \text{volt}[/tex]
What is the approximate electrostatic force between two protons separated by a distance of 1.0 x 10-6 meter?
1) 2.3 x 10-16 N and repulsive
2) 9.0 x 1021 N and repulsive
3) 9.0 x 1021 N and attractive
4) 2.3 10-16 N and attractive
Answer:
2.3×10^-16
Explanation:
Charge of proton = 1.6×10^-19
let q(a) and q(b) be the two protons.
According to the Coulumbs Laws ;
F = kq(a)×q(b) / r^2
where k = 9.0×10^9 and r is the distance of seperation.
So having a distance of 1.0×10^-6 in our case ;
F = (9.0×10^9×1.6×10^-19×1.6×10^-19)/(1.0××10^-6)^2
F = (2.304×10^-28)/(1.0×10^-12)
F = 2.304×10^-16N
Since the both charges are positively charged they will repel .
. Radiation travels at the speed of light
T or F?
Answer:
electromagnetic radiation moves at the speed of light
The two graphs shown represent the motion of two blocks with different masses, m1 and m2. The blocks are oscillating on identical springs. Which of the following statements correctly describes the relationship between m1 and m2 and provides evidence from the graphs?
Answer:
M1 would seem to be slower because of a larger mass
x1 = A1 sin ω1 t1 describes the displacement
ω1 / ω2 = ((k1 / k2) / (m1 / m2))^1/2 = (m2 / m1)^1/2 since k's are equal
ω1 / ω2 = 1/2 from graph (frequency of 2 is greater)
(m1 / m2)^1/2 = ω2 / ω1 from above
m1 / m2 = 2^2 = 4 so m1 would have 4 times the mass of m2
M1 would seem to be slower because of a larger mass
x1 = A1 sin ω1 t1
ω1 / ω2 = ((k1 / k2) / (m1 / m2))^1/2 = (m2 / m1)^1/2 since k's are equal
ω1 / ω2 = 1/2 from graph (frequency of 2 is greater)
(m1 / m2)^1/2 = ω2 / ω1 from above
m1 / m2 = 2^2 = 4 so m1 would have 4 times the mass of m2.
What is the graph represents?The two graphs shown represent the motion of two blocks with different masses, m1 and m2. The blocks are oscillating on identical springs. For the system consisting of the two blocks, the change in the kinetic energy of the system is equal to work done by gravity on the system. For the system consisting of the two blocks, the pulley and the Earth, the change in the total mechanical energy of the system is zero.
The two graphs shown represent the motion of two blocks with different masses, m1 and m2. The blocks are oscillating on identical springs. For the system consisting of the two blocks, the change in the kinetic energy of the system is equal to work done by gravity on the system.
Therefore, M1 would seem to be slower because of a larger mass
x1 = A1 sin ω1 t1
ω1 / ω2 = ((k1 / k2) / (m1 / m2))^1/2 = (m2 / m1)^1/2 since k's are equal
ω1 / ω2 = 1/2 from graph (frequency of 2 is greater)
(m1 / m2)^1/2 = ω2 / ω1 from above
m1 / m2 = 2^2 = 4 so m1 would have 4 times the mass of m2.
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Q. 1 MWH is equal to ------- joules
a.3.6*10^10
b.3.6*10^6
c.3.6*10^9
d.3.6