Answers
The position of John at a time of 4.53 s is 20.8 m.
It is essential to know that the formula for position, velocity, and acceleration is given as:
\($$x=x_0+v_0t+\frac{1}{2}at^2$$\)
\($$v=v_0+at$$\)
\($$v^2=v_0^2+2a(x-x_0)$$\)
Here, x is the position, v is the velocity, t is the time elapsed, and a is the acceleration. John's position at a time of 4.53 s is given as follows:
Given,
\($$x_0=0, v_0=4.6 m/s, t=4.53s, a=-9.8m/s^2$$\)
From the above formula, we can calculate the position of John at a time of 4.53 s.Substitute all the values in the formula for position, and we get,
\($$x=x_0+v_0t+\frac{1}{2}at^2$$\)
\($$x=0+(4.6)(4.53)+\frac{1}{2}(-9.8)(4.53)^2$$\)
\($$x=20.8 m$$\)
Therefore, the position of John at a time of 4.53 s is 20.8 m.
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Related Questions
Very far from earth (at R- oo), a spacecraft has run out of fuel and its kinetic energy is zero. If only the gravitational force of the earth were to act on it (ie., neglect the forces from the sun and other solar system objects), the spacecraft would eventually crash into the earth The mass of the earth is Me and its radius is Re. Neglect air resistance throughout this problem, since the spacecraft is primarily moving through the near vacuum of space
Find the speed s of the spacecraft when it crashes into the earth Express the speed in terms of M, Re, and the universal gravitational constant G.
Answers
Answer:
Speed of the spacecraft right before the collision: \(\displaystyle \sqrt{\frac{2\, G\cdot M_\text{e}}{R\text{e}}}\).
Assumption: the earth is exactly spherical with a uniform density.
Explanation:
This question could be solved using the conservation of energy.
The mechanical energy of this spacecraft is the sum of:
the kinetic energy of this spacecraft, andthe (gravitational) potential energy of this spacecraft.Let \(m\) denote the mass of this spacecraft. At a distance of \(R\) from the center of the earth (with mass \(M_\text{e}\)), the gravitational potential energy (\(\mathrm{GPE}\)) of this spacecraft would be:
\(\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R}\).
Initially, \(R\) (the denominator of this fraction) is infinitely large. Therefore, the initial value of \(\mathrm{GPE}\) will be infinitely close to zero.
On the other hand, the question states that the initial kinetic energy (\(\rm KE\)) of this spacecraft is also zero. Therefore, the initial mechanical energy of this spacecraft would be zero.
Right before the collision, the spacecraft would be very close to the surface of the earth. The distance \(R\) between the spacecraft and the center of the earth would be approximately equal to \(R_\text{e}\), the radius of the earth.
The \(\mathrm{GPE}\) of the spacecraft at that moment would be:
\(\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\).
Subtract this value from zero to find the loss in the \(\rm GPE\) of this spacecraft:
\(\begin{aligned}\text{GPE change} &= \text{Initial GPE} - \text{Final GPE} \\ &= 0 - \left(-\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\right) = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \end{aligned}\)
Assume that gravitational pull is the only force on the spacecraft. The size of the loss in the \(\rm GPE\) of this spacecraft would be equal to the size of the gain in its \(\rm KE\).
Therefore, right before collision, the \(\rm KE\) of this spacecraft would be:
\(\begin{aligned}& \text{Initial KE} + \text{KE change} \\ &= \text{Initial KE} + (-\text{GPE change}) \\ &= 0 + \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \\ &= \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\end{aligned}\).
On the other hand, let \(v\) denote the speed of this spacecraft. The following equation that relates \(v\!\) and \(m\) to \(\rm KE\):
\(\displaystyle \text{KE} = \frac{1}{2}\, m \cdot v^2\).
Rearrange this equation to find an equation for \(v\):
\(\displaystyle v = \sqrt{\frac{2\, \text{KE}}{m}}\).
It is already found that right before the collision, \(\displaystyle \text{KE} = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\). Make use of this equation to find \(v\) at that moment:
\(\begin{aligned}v &= \sqrt{\frac{2\, \text{KE}}{m}} \\ &= \sqrt{\frac{2\, G\cdot M_\text{e} \cdot m}{R_\text{e}\cdot m}} = \sqrt{\frac{2\, G\cdot M_\text{e}}{R_\text{e}}}\end{aligned}\).
A 2.98-kg object oscillates on a spring with an amplitude of 8.05 cm. Its maximum acceleration is 3.55 m/s2. Calculate the total energy.
Answers
Answer:
a = ω^2 A formula for max acceleration (ignoring sign)
V = ω A formula for max velocity
V^2 = ω^2 A^2 = a A from first equation
E = 1/2 M V^2 = 1/2 * 2.98 * 3.55 * .0805 = .426 J
(kg * m/sec^2 * m = kg m^2 / sec^2 = Joule
The first P-wave of an earthquake travels 5600 kilometers from the epicenter and arrives at a seismic station at 10:05 a.m. At what time did this earthquake occur?
Ahhhhhh I have a Regent's test in 2 hours and I don't know how to solve this type of question! Any help would be appreciated.
Anyone know what the steps to do this are? I dont even need an answer, just how to get to it. Thank you!
Answers
The earthquake would occur 13 minutes before 10:05 a.m. which will be at 9.52 am.
The p-waves travel with a constant velocity of 7 km/s
The time can be calculated by using the formula
t = d / v
where
T1 = 10:05 a.m
d is the distance they take to travel from the epicenter
v is the speed of the p-waves
On average, the speed of p-waves is
v = 7 km/s
d = 5600 km (given)
Substituting the values in the formula;
t = d / v
t = 5600 ÷ 7
t = 800 seconds
Converting into minutes,
t = 800 ÷ 60
t = 13.3
≈ 13 mins
T1 - 13 mins = T2
10:05 - 13 mins = 9.52 am
It means the earthquake occurred prior 13 minutes, that is at 9.52 am.
Therefore, the earthquake occurred at 9.52 am.
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An 8.0 Kg mass is placed at = 3 where should a 10 Kg mass be placed along the − so that the center of mass will be located ay = 4.5?
Answers
Answer:
Therefore, the 10 kg mass should be placed at x = 5.7 m along the x-axis to achieve a center of mass located at y = 4.5 m.
Explanation:
To find the position along the x-axis where a 10 kg mass should be placed such that the center of mass is located at y = 4.5, we can use the formula for the center of mass:
x_cm = (m1 * x1 + m2 * x2) / (m1 + m2)
Here, m1 and x1 represent the mass and position of the 8 kg mass, respectively. m2 is the mass of the 10 kg mass, and we need to find x2, its position.
Given:
m1 = 8 kg
x1 = 3 m
x_cm = unknown (to be found)
m2 = 10 kg
y_cm = 4.5 m
Since the center of mass is at y = 4.5, we only need to consider the y-coordinate when calculating the center of mass position along the x-axis.
To solve for x2, we can rearrange the formula as follows:
x2 = (x_cm * (m1 + m2) - m1 * x1) / m2
Substituting the given values:
x2 = (x_cm * (8 kg + 10 kg) - 8 kg * 3 m) / 10 kg
Simplifying:
x2 = (x_cm * 18 kg - 24 kg*m) / 10 kg
Now, we can set the y-coordinate of the center of mass equal to 4.5 m and solve for x_cm:
4.5 m = (8 kg * 3 m + 10 kg * x2) / (8 kg + 10 kg)
Simplifying:
4.5 m = (24 kg + 10 kg * x2) / 18 kg
Multiplying both sides by 18 kg:
81 kg*m = 24 kg + 10 kg * x2
Subtracting 24 kg from both sides:
10 kg * x2 = 81 kg*m - 24 kg
Dividing both sides by 10 kg:
x2 = (81 kg*m - 24 kg) / 10 kg
Simplifying:
x2 = 8.1 m - 2.4 m
x2 = 5.7 m
(brainlest?) ples:(
Answer:
the 10 kg mass should be placed at x = -2.4 m to achieve a center of mass at y = 4.5 m.
Explanation:
To find the position along the x-axis where the 10 kg mass should be placed so that the center of mass is located at y = 4.5, we can use the principle of the center of mass.
The center of mass of a system is given by the equation:
x_cm = (m1x1 + m2x2) / (m1 + m2),
where x_cm is the x-coordinate of the center of mass, m1 and m2 are the masses, and x1 and x2 are the positions along the x-axis.
Given:
m1 = 8 kg,
x1 = 3 m,
m2 = 10 kg,
y_cm = 4.5 m.
To solve for x2, we need to find the x-coordinate of the center of mass (x_cm) by using the y-coordinate:
y_cm = (m1y1 + m2y2) / (m1 + m2),
where y1 and y2 are the positions along the y-axis.
Rearranging the equation and substituting the given values:
4.5 = (83 + 10y2) / (8 + 10).
Simplifying the equation:
4.5 = (24 + 10*y2) / 18.
Multiplying both sides by 18:
81 = 24 + 10*y2.
Rearranging the equation:
10*y2 = 81 - 24,
10*y2 = 57.
Dividing both sides by 10:
y2 = 5.7.
Therefore, the y-coordinate of the 10 kg mass should be 5.7 m.
To find the x-coordinate of the 10 kg mass, we can use the equation for the center of mass:
x_cm = (m1x1 + m2x2) / (m1 + m2).
Substituting the given values:
x_cm = (83 + 10x2) / (8 + 10).
Since the center of mass is at x_cm = 0 (the origin), we can solve for x2:
0 = (83 + 10x2) / (8 + 10).
Rearranging the equation:
83 + 10x2 = 0.
24 + 10*x2 = 0.
10*x2 = -24.
Dividing both sides by 10:
x2 = -2.4.
Activities:
1. Name the instrument that is used to measure Air Pressure.
2.Explain what is Cyclone and Anticyclone
Answers
Answer: barometer.
A cyclone is a storm or system of winds that rotates around a center of low atmospheric pressure. An anticyclone is a system of winds that rotates around a center of high atmospheric pressure.
The United States is about the only country in the world that still uses the units feet, miles, and gallons. You might see some car specifications in a magazine that give fuel efficiency as 8.1 km per kilogram of fuel. A mile is 1.609 km, a gallon is 3.785 liters, and a liter of gasoline has a mass of 0.729 kg.
Answers
The United States still makes use of the form of measure that includes the units: feet, miles, and gallons while others make use of yards, kilometers and liters
What is a Scale of Measure?This refers to the classification that describes the nature of the information within the values assigned to variables.
Therefore, we can see that the United States still makes use of the form of measure that includes the units: feet, miles, and gallons while others make use of yards, kilometers and liters
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You know you can provide 600 W
of power to move large objects. You need to move a 60-kg
safe up to a storage loft, 18 m
above the floor.
Part A
With what average speed can you pull the safe straight up?
Answers
A. The average speed you can use to pull the safe is 1.02 m/s
B. The time needed to pull the safe up is 17.65 s
A. How do i determine the velocity?First, we shall obtain the force. This is shown below:
Mass of safe (m) = 60 KgAcceleration due to gravity (g) = 9.8 m/s² Force (F) =?F = mg
F = 60 × 9.8
F = 588 N
Finally, we shall obtain the average speed. Details below:
Power = 600 WForce = 588 NAverage speed =?Power = force × average speed
600 = 588 × average speed
Divide both sides by 588
Average speed = 600 / 588
Average speed = 1.02 m/s
B. How do i determine the time?The time needed to pull the safe up can be obtained as follow:
Average speed = 1.02 m/sTotal distance = 18 mTime = ?Time = Total distance / average speed
Time = 18 / 1.02
Time = 17.65 s
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Complete question:
You know you can provide 600 W
of power to move large objects. You need to move a 60-kg
safe up to a storage loft, 18 m
above the floor.
Part A
With what average speed can you pull the safe straight up?
Part B
What is the time needed to pull the safe up?
two spheres A and B are projected off the edge of a 1.0 m high table with the same horizontal velocity . sphere A has a mass of 20.g and sphere B has a mass of 10.g.
If both spheres leave the edge of the table at the same instant, sphere A will land
a. at some time after sphere B.
b. at the same time as sphere B.
c. at some time before sphere B.
d. There is not enough information to decide.
Answers
Answer:
c. because A will land first becuase its heavier :)
Explanation:
Cassy shoots a large marble (Marble A, mass: 0.06 kg) at a smaller marble (Marble B, mass: 0.03 kg) that is sitting still. Marble A was initially moving at a velocity of 0.7 m/s, but after the collision it has a velocity of −0.2 m/s. What is the resulting velocity of marble B after the collision? Be sure to show your work for solving this problem along with the final answer.I need to use this formaulam1u1 + m2u2 = m1v1 + m2v2and replace with this letterI had a tutor that explained to me very good , but me doing it byself I got lost
Answers
Answer:
1.8 m/s
An
Explanation:
We will use the given equation:
m1u1 + m2u2 = m1v1 + m2v2
Where m1 is the mass of the large marble, m2 is the mass of the smaller marble, u1 is the velocity of the large marble before the collision, u2 is the velocity of the smaller marble before the collision, v1 is the velocity of the large marble after the collision and v2 is the velocity of the large marble after the collision.
So, replacing the values, we get:
m1 = 0.06 kg
m2 = 0.03 kg
u1 = 0.7 m/s
u2 = 0 m/s
v1 = -0.2 m/s
Therefore:
\(\begin{gathered} 0.06(0.7)+0.03(0)=0.06(-0.2)+0.03v_2 \\ 0.042=-0.012+0.03v_2 \end{gathered}\)So, solving for v2, we get:
\(\begin{gathered} 0.042+0.012=-0.012+0.03v_2+0.012 \\ 0.054=0.03v_2 \\ \frac{0.054}{0.03}=\frac{0.03v_2}{0.03} \\ 1.8m/s=v_2 \end{gathered}\)Then, the velocity of marble B is 1.8 m/s
HELP FAST PLS!!!!!!!!!!!
Answers
Answer: Picture is too small I can’t see
A van tire contacts the ground on a rectangular area of (10 cm) by (15cm). If the bus's mass is (900 kg), what pressure does the car exert on the ground as it rests on all four tires? (g= 9.8 m/s³)
3.80×10^5
14.7×10^4
6.67×10^3
58.8×10^3
Answers
Answer: Hmmm im not sure but i'd go with 3.80x10^5
Explanation: Like i mentioned im not very good at physics...sorry if its wrong
Hi please help on question! . If answer is correct I'll rate you five stars a thanks and maybe even brainliest! You will even get 54 pts!!
Here is a function machine.
Input : multiply by 6. Subtract 80: output
The input is the same as the output. Find the input.
Also can you please show me an easy to work out these type of questions
Answers
Answer:
Explanation:
Sure, I'd be happy to help you with the question!
Let's denote the input as x. According to the function machine, the input is multiplied by 6 and then 80 is subtracted from the result to obtain the output.
So, the function can be written as:
Output = (6 * x) - 80
Now, the problem states that the input is the same as the output. Therefore, we can set up the equation:
x = (6 * x) - 80
Let's solve this equation to find the value of x:
x = 6x - 80
Subtracting 6x from both sides, we get:
x - 6x = -80
Combining like terms, we have:
-5x = -80
Dividing both sides by -5, we find:
x = (-80) / (-5)
Simplifying the expression, we have:
x = 16
Therefore, the input (x) that results in the input being the same as the output is 16.
To work out these types of questions, it's important to carefully read the instructions and understand the operations being performed in the function machine. Then, you can set up an equation with the input and output, and solve for the unknown value. Always double-check your solution to ensure it satisfies the given conditions of the problem.
Answer:
16
Explanation:
(x*6) - 80 = x
Multiply the parentheses
6x - 80 = x
Add 80 to each side to get
6x = x + 80
Subtract x from both sides to get
5x = 80
Divide both sides by 5
x = 16
A uniform wooden plank with a mass of 75kg and length of 5m is placed on top of a brick wall so that 1.5m of plank extends beyond the edge hanging freely in air. how far beyond the edge of the wall can a 100kg woman walk before the plank began to rotate about the edge of the wall
Answers
Answer:
x₂ = 1.33 m
Explanation:
For this exercise we must use the rotational equilibrium condition, where the counterclockwise rotations are positive and the zero of the reference system is placed at the turning point on the wall
Στ = 0
W₁ x₁ - W₂ x₂ = 0
where W₁ is the weight of the woman, W₂ the weight of the table.
Let's find the distances.
Since the table is homogeneous, its center of mass coincides with its geometric center, measured at zero.
x₁ = 2.5 -1.5 = 1 m
The distance of the person is x₂ measured from the turning point, at the point where the board begins to turn the girl must be on the left side so her torque must be negative
x₂ = \(\frac{M_1g }{m_2 g} \ x_1\)
let's calculate
x₂ = \(\frac{100}{75} \ 1\)
x₂ = 1.33 m
Four very long straight parallel wires, located at the corners of a square of side l, carry equal currents Io perpendicular to the page as shown in Figure. Determine the magnitude and direction of B at the center C of the square.
Answers
The magnitude and direction of B at the center C of the square is
2µNIo/πL on every corner of square.
Given that four long straight parallel wires, located at the corners of a square carry equal currents perpendicular to the page.
Side of a square = L
current in square wire = Io
Here C is located at the center then distance from any corner point to center C be r. Here r acts as the hypotenuse such that r = L/√2
We know that electric field (B) = NI/2πr and is same at every point on square. Ba = Bb = Bc = Bd
B = µNIo/2π(L/√2)
B = √2µNIo/2πL
Bnet = 4 x Bcos45 = 4 x √2µNIo/2πL x 1/√2
Bnet = 2µNIo/πL
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If a force is applied when using a lever over 2 meters to move the object 1 meter, the distance over which the force is applied is called the _____.
Answers
The distance over which the force is applied when using a lever is called the effort distance.
What is force?Force is defined as the rate of change of momentum.
Here,
The distance over which the force is applied when using a lever is called the effort distance. It refers to the distance between the point where the effort is applied and the fulcrum of the lever. In the case we described, where a force is applied when using a lever over 2 meters to move an object 1 meter, the effort distance would be 2 meters. The load distance, on the other hand, refers to the distance between the fulcrum and the point where the load is applied. Understanding the relationship between the effort and load distances is crucial to determining the mechanical advantage of a lever system.
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A runner of mass 58.7 kg starts from rest and accelerates with a constant acceleration of 1.07 m/s2 until she reaches a velocity of 7.5 m/s. She then continues running with this constant velocity.
a) How far has she run after 53.3 seconds?
b) What is the velocity of the runner at this point?
Answers
A. The distance travelled after 53.3 seconds is 1919.63 m
B. The velocity of the runner ath the given time is 64.53 m/s
A. How to determine the distance travelled after 53.3 secondsThe distance travelled can be obtained as follow:
Initial velocity (u) = 7.5 m/sAcceleration (a) = 1.07 m/s² Time (t) = 53.3 secondsDistance (s) =?s = ut + ½at²
s = (7.5 × 53.3) + (½ × 1.07 × 53.3²)
s = 399.75 + (0.535 × 2840.98)
s = 399.75 + 1519.87615
s = 1919.63 m
Thus, the distance travelled in 53.3 seconds is 1919.63 m
B. How to determine the velocityInitial velocity (u) = 7.5 m/sAcceleration (a) = 1.07 m/s² Time (t) = 53.3 secondsFinal velocity (v) =?v = u + at
v = 7.5 + (1.07 × 53.3)
v = 7.5 + 57.031
v = 64.53 m/s
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if a bust starts to move and its velocity becomes 90 km after 8 seconds . calculate its acceleration answer it quick please
Answers
Answer:
a = 3.125 [m/s^2]
Explanation:
In order to solve this problem, we must use the following equation of kinematics. But first, we have to convert the speed of 90 [km/h] to meters per second.
\(90\frac{km}{h}*\frac{1000m}{1km}*\frac{1h}{3600s} \\= 25 \frac{m}{s}\)
\(v_{f} =v_{i} + (a*t)\)
where:
Vf = final velocity = 25 [m/s]
Vi = initial velocity = 0
a = acceleration [m/s^2]
t = time = 8 [s]
The initial speed is zero as the bus starts to koverse from rest. The positive sign of the equation means that the bus increases its speed.
25 = 0 + a*8
a = 3.125 [m/s^2]
Vocabulary Matching
The specialized equipment used to conduct research and repair
damaged equipment
Instruments
Space Station
Space Suit
Accomodations
Answers
Answer:
instruments
Explanation:
What three things do weather satellites best help accomplish?
Answers
Answer:Weather satellites provide vital information about cloud patterns and ground and sea temperatures. They collect and share information with remote data collecting stations such as weather buoys, and observatories from around the world.
Explanation:
object P1 and P2 are in a straight line with the normal to a plane mirror.If P1 and P2 are 18m and 21m away from the mirror. Calculate 1)distance between P1 and it's image I1
Answers
The distance between object P1 and its image formed is determined as 36 m.
Distance of the image
The distance of the image formed by object P1 is calculated as follows;
In a plane mirror; object distance = image distance
image distance of P1 = 18 m
distance between object and image = 18m + 18 m = 36 m
Thus, the distance between object P1 and its image formed is determined as 36 m.
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A light ray is incident on a smooth surface at an angle of 25 degrees with the surface. What is the angle of incidence?
25 degrees
65 degrees
45 degrees
90 degrees
Answers
The angle will have 45 degree ments
An object of height 5cm is placed 20cm in front of pin hole camera from a cubic box of side 6cm . Determine the height of the image formed
Answers
which chareged particles contribute to electricity, magnetism, and light
Answers
Answer:
Explanation:charged particles create an electric force field. Moving charged particles create a magnetic force field. Accelerating charged particles produce changing electric and magnetic force fields which propagate as EM waves.
20) A 5 Kg watermelon is dropped off a 10 meter balcony. What will the watermelon's velocity be right before it hits the ground?
Answers
Answer:
14m/s
Explanation:
Air enters into the hollow propeller tube at A with a mass flow of 4 kg/s and exits at the ends B and C with a velocity of 400 m/s, measured relative to the tube. If the tube rotates at 1500 rev/min, determine the frictional torque M on the tube.
Answers
Answer:
643N.m
Explanation:
From this question we have:
Mass flow = 4kg/s
Velocity V = 400m/s
Rotation N = 1500rev/min
We get the relative velocity at exit to be:
V2 = V - r2w
400-0.5x [(2*π*1500)/60]
= 400-78.5
= 321.5m/s
Then we have to calculate the frictional torque My
Mt = Mr2 x V2
= 4x0.5x321.5
= 643Nm
From the calculations above, we get the frictional torque M on the tube to be 643Nm.
An object is placed at the position x1 = 73 cm and a second mass that is 4/3 times as large is placed at x2 = 247 cm. Find the location of the center of mass of the system.
Answers
If An object is placed at the position x1 = 73 cm and a second mass that is 4/3 times as large is placed at x2 = 247 cm, the location is at 172.44cm
How to find the locationThe location of he center of the mass of the object would be
73 cm + (4/3 * 247) / 1 + 4/3
= 73 CM + 329.3 / 2.333
= 402.3 /2.33
= 172.44 CM
Hence we can conclude that the location is at If An object is placed at the position x1 = 73 cm and a second mass that is 4/3 times as large is placed at x2 = 247 cm, the location is at 172.44cm
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If a resistance is added in serles to a circult, the circult resistance is then:
O less
O greater
o the same
Answers
Answer:
the overall resistance increases
Explanation:
so, if a resistance is added in series to a circuit, the circuit resistance is greater
what is the state of y when y = k/x, x is halved.
Answers
Answer:
y becomes doubled.
Explanation:
If;
y = \(\frac{k}{x}\)
what is the state of y when x is halved;
the given expression is an inverse relationship. When y increases, x is supposed to decrease and vice versa.
if x is halved; x = \(\frac{x}{2}\)
\(\frac{k}{\frac{x}{2} }\) = \(\frac{2k}{x}\)
Now compare :
\(\frac{k}{x}\) : \(\frac{2k}{x}\)
we see that y becomes doubled
A simple model of the human eye ignores its lens entirely. Most of what the eye does to light happens at the outer surface of the transparent cornea. Assume that this surface has a radius of curvature of 6.50 mm and that the eyeball contains just one fluid, with a refractive index of 1.41. Determine the distance from the cornea where a very distant object will be imaged.
Answers
Answer:
the distance from the cornea where a very distant object will be imaged is 23.35 mm
Explanation:
Given the data in the question;
For a spherical refracting surface;
\(n_i\)/\(d_0\) + \(n_t\)/\(d_i\) = ( \(n_t\) - \(n_i\) )/R
where \(n_i\) is the index of refraction of the light of ray in the incident medium
\(d_0\) is the object distance
\(n_t\) is the index of refraction of light ray in the refracted medium
\(d_i\) is the image distance
R is the radius of curvature
Now, let \(d_0\) = ∞, such that;
\(n_i\)/∞ + \(n_t\)/\(d_i\) = ( \(n_t\) - \(n_i\) )/R
0 + \(n_t\)/\(d_i\) = ( \(n_t\) - \(n_i\) )/R
we make \(d_i\) subject of the formula
\(n_t\)R = \(d_i\)( \(n_t\) - \(n_i\) )
\(d_i\) = ( \(n_t\) × R ) / ( \(n_t\) - \(n_i\) )
given that; R = 6.50 mm, \(n_t\) = 1.41, we know that \(n_i\) = 1.00
so we substitute
\(d_i\) = (1.41 × 6.50 mm ) / ( 1.41 - 1.00 )
\(d_i\) = 9.165 / 0.41
\(d_i\) = 23.35 mm
Therefore, the distance from the cornea where a very distant object will be imaged is 23.35 mm
In the event of a car collision, car 1 (m=1300 kg, v=12.5 m/s) collides with a stationary car (m=1500 kg). The two cars slide off with a velocity of 6.3 m/s, how much energy is lost to friction, sound, etc., during the collision?
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Answers
The amount of energy lost to friction, sound, and other factors during the collision is 48620.15 J.
In the event of a car collision, the momentum of the two vehicles changes as a result of the impact, and energy is transferred from one vehicle to the other.
The impact will result in a loss of energy due to friction, sound, and other factors. Therefore, to calculate the amount of energy lost during the collision, we need to determine the total kinetic energy of the system before and after the collision, and the difference between the two is the energy lost.
To calculate the initial kinetic energy, we need to use the formula 1/2mv², where m is the mass of the object and v is its velocity. Therefore, the initial kinetic energy of car 1 is:KE1 = 1/2 x 1300 kg x (12.5 m/s)² = 101562.5 J The stationary car has no initial kinetic energy since it is at rest.
The total initial kinetic energy of the system is therefore: KE initial = KE1 + KE2 = 101562.5 J To calculate the final kinetic energy, we need to use the velocity of the two cars after the collision.
The problem states that the two cars slide off with a velocity of 6.3 m/s. Therefore, the final kinetic energy of the system is: KE final = 1/2 x (1300 kg + 1500 kg) x (6.3 m/s)² = 52942.35 J.
The energy lost during the collision is the difference between the initial and final kinetic energy: Energy lost = KE initial - KE final = 48620.15 J.
Therefore, the amount of energy lost to friction, sound, and other factors during the collision is 48620.15 J.
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