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.
A ballplayer catches the ball 4 seconds after throwing it straight up in the air. What speed did he throw it and how high did it go?
Answer:
Speed of the ball when it was thrown = 16.67 m/s
Height reached by the ball = 14.16 s
Explanation:
Given time in the ball comes to the same height from where the ball was thrown is 3.4 s.
Let us assume that the ball was thrown with a speed of u in the upward direction. Since only gravitational force acts on the particle in the downward direction, it has a constant acceleration in the downward direction.
We also know that a particle has the same magnitude of velocity when it is under a free fall at a fixed height but direction is opposite i.e., initially in the upward direction and finally in the downward direction.
https://tex.z-dn.net/?f=%5Ctherefore%20%5Cdfrac%7B-u-u%7D%7B3.4%7D%3D%20-9.8%5C%5C%5CRightarrow%20%5Cdfrac%7B2u%7D%7B3.4%7D%3D%209.8%5C%5C%5CRightarrow%202u%20%3D%209.8%5Ctimes%203.4%5C%5C%5CRightarrow%20u%20%3D%20%5Cdfrac%7B9.8%5Ctimes%203.4%7D%7B2%7D%5C%5C%5CRightarrow%20u%20%3D%2016.67%5C%20m%2Fs
Hence, the particle was thrown with a speed of 16.67 m/s.
Now, we also note that the particle has zero velocity at its maximum height.
So, from the initial position to the maximum height reached by the particle, we have
https://tex.z-dn.net/?f=0%5E2%20%3D%20u%5E2%2B2as%5C%5C%5CRightarrow%20s%20%3D%20%5Cdfrac%7Bv%5E2-u%5E2%7D%7B2a%7D%5C%5C%5CRightarrow%20s%20%3D%20%5Cdfrac%7B0%5E2-16.67%5E2%7D%7B2(-9.8)%7D%5C%5C%5CRightarrow%20s%20%3D14.16%5C%20m
Hence, the maximum height reached by the ball is 14.16 m.
How do human population growth trends differ between developed nations and developing nations?
Answer:
Replacement-Level Fertility
Another important population characteristic that differ btw develop nation and developing nations is relates to births is replacement-level fertility. Replacement-level fertility is the fertility rate that will result in the replacement of the parents in the population. Again, in an ideal world, the human replacement-level fertility rate would be exactly two. This would mean that each couple would produce two offspring that would replace them in the population. If this occurred, then the human population would stay at a stable rate
How do the displacement, velocity, and acceleration of a runner change as he races from the starting
line toward the finish line along a straight path?
Answer:
maybe line toward the finish
Explanation:that question is confusing
Velocity is the pace at which your displacement is shifting; it describes how quickly and in which direction you are travelling. And acceleration, which is measured in meters per second, is the rate at which your velocity is changing (or meters per second squared).
What factor that affect the acceleration and velocity?Net force and mass are the two primary factors that impact how quickly an item accelerates. For instance, mass has an inverse relationship with acceleration while net force has a direct relationship.
The acceleration is also influenced by other elements like friction, air or fluid resistance, and pressure.
According to the second law, the mass of the item and the net force acting on it both impact how quickly an object accelerates.
Therefore, an object's acceleration is directly proportional to the net force applied on it and inversely proportional to its mass.
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#SPJ2
As a block falls through the air by 40 meter it does work equal to -1800 joule. Determine the mass of a block.
Answer:
m = 4.5 kg
Explanation:
w = - 1800 j
Fd = - 1800 j
mgd = - 1800 j
m = - 1800 ÷(gd)
m = - 1800 ÷( 10×-40)
m = 4.5 kg
A circular ferris wheel that revolves at a constant rate once every 30 seconds. The
radius of the ferris wheel is 10 m. What is the normal force of the ferris wheel on a 10
kg toddler at the very bottom of the ferris wheel?
The normal force of the ferris wheel on a 10kg toddler at the very bottom of the ferris wheel is 102.41 N.
Centripetal force of the Ferris wheelThe normal force of the ferris wheel on a 10kg toddler at the very bottom of the ferris wheel is calculated as follows;
Fn = Fc + mg
Fn = mω²r + mg
where;
ω is the angular speed = 1 rev/30 s = 2π/30 s = 0.21 rad/sFn = 10(0.21²) 10 + 10(9.8)
Fn = 102.41 N
Thus, the normal force of the ferris wheel on a 10kg toddler at the very bottom of the ferris wheel is 102.41 N.
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If an object accelerates from rest, with a constant acceleration of
10 m/s2, what will its velocity be after 2 s?
The velocity of the body affter 2 seconds is 20 m/s
What is acceleration?Acceleration, a is the change of velocity with time. this is represented mathematically as
a = ( v - u ) / t
where:
u = intial velocity
v = final velocity
t = time
Given:
a = 10 m/s2
t = 2s
u = 0 ( from rest means initial velocity is zero )
therefore
a = v / t
v = a t
v = 10 * 2
v = 20 m/s
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How fast must a proton move so that its kinetic energy is 70% of its total
energy?
I thought it would be 0.7c but that is wrong. I really don't know how to do this type of problem and my text book isn't any help.
Try this solution, all the details are in the attachment. note, the answer is marked with orange colour. If it is possible, check the provided solution in other sources.
Answer: ≈0.81c.
Suppose a grower sprays (8.2x10^1) kg of water at 0 °C onto a fruit tree of mass 180 kg. How much heat is released by the water when it freezes?
Since there is no temperature change which drives heat flow, thus no heat will be released by the water.
Heat released by the water when it freezesThe heat released by the water when it freezes is calculated as follows;
Q = mcΔФ
where;
m is mass of waterc is specific heat capacity of waterΔФ is change in temperature = Фf - Фiwhen water freezes, the temperature, Фf = 0 °C
Q = 82 x 4200 x (0 - 0)
Q = 0
Since there is no temperature change which drives heat flow, thus no heat will be released by the water.
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I WILL MARK BRAINLIEST IF YOU ANSWER THIS PHYSICS QUESTION!!!
A 21 kg child is riding a 5.9 kg bike with a velocity of 4.5 m/s to the northwest.
1. What is the momentum of the child and the bike together?
2. What is the momentum of the child?
3. What is the momentum of the bike?
Answer:
1. 121.05 kg m/s NW
2. 94.5 kg m/s NW
3. 26.55 kg m/s NW
Explanation:
Define the following terms.
1. plate boundaries
2. seismic
3. seismic waves
4. body waves
5. surface waves
6. tectonic quakes
7. volcanic quakes
8. focus of earthquake
9. epicenter
10. fault
Answer:
I can only define a few of them, hope you don't mind.
Seismic- It means connected with or caused by earthquakes
Epicenter- It means the point on the earth's surface where the effects of an earthquake are most felt strongly.
Fault- A place where there is a break that is longer than usual in the layers of rock in the earth's crust.
Tectonic waves- Waves connected with the structure of the earth's surface.
Plate boundaries- The boundaries that differentiates the large sheets of rock (called PLATES) that form the earth's surface. Plate tectonics is the movement of the large sheets of rocks that form the earth's surface
[tex] \: \: \: \: \: \: \: \: \: [/tex]
As a conservation biologist for the Chesapeake Bay, you and your
colleagues have been conducting a research study that tracks the decrease
in the bald eagle population over the past few years.
What evidence can you find for the decrease in the bald eagle population?
As a conservation biologist for the Chesapeake Bay, you and your
colleagues have been conducting a research study that tracks the decrease
in the bald eagle population over the past few years.
What evidence can you find for the decrease in the bald eagle population?
9. aDefine refraction of light. Draw the ray diagram when an object is placed at C in a concave mirror Write any two natures of that image
Answer:
the process of bending of light when travling throw different mediume is called refraction of light.
A 53 g ice cube at −30◦C is dropped into a container of water at 0◦C. How much water freezes onto the ice? The specific heat of ice is 0.5 cal/g · ◦ C and its heat of fusion of is 80 cal/g. Answer in units of g.
For A 53 g ice cube at −30◦C is dropped into a container of water at 0◦C, the amount of water that freezes onto the ice? is mathematically given as
x = 9.93 g
What is the amount of water that freezes onto the ice?Where
Energy received = energy given out
Generally, the amount of water is mathematically given as
(53)(0.5)(30) = (80)(x)
Therefore
x = (49)(0.5)(16)/(80)
x = 9.93 g
In conclusion, the mass of water
x = 9.93 g
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a
Which of these is a chemical
change?
A. water boiling
B. salt disolving
C. paper burning
Answer:
Burning coal and boiling water are both chemical changes. Burning coal is a chemical change, and boiling water is a physical change. Burning coal is a physical change, and boiling water is a chemical change.
Explanation:
An unbalanced force acting on an object will cause it
A. increase in mass
B. decrease in mass
C. accelerate
D. remain at rest
The magnetic field perpendicular to a single wire loop of diameter 10.0 cm decreases from 0.50 T to zero. The wire is made of copper and has a diameter of 2.0 mm and length 1.0 cm. How much charge moves through the wire while the field is changing?
I know how to do the calculations, but can someone please explain what is the 10cm diameter and 2mm diameter? Why is there one wire and two diameters? I understand this problem mathematically but not conceptually.
Hi there!
We can begin by using Lenz's Law:
[tex]\epsilon = -N\frac{d\Phi _B}{dt}[/tex]
N = Number of Loops
Ф = Magnetic Flux (Wb)
t = time (s)
Also, we can rewrite this as:
[tex]\epsilon = -NA\frac{dB}{dt}[/tex]
A = Area (m²)
Since the area is constant, we can take it out of the derivative.
This is a single wire loop, so N = 1.
Now, we can develop an expression for the induced emf.
We can begin by solving for the area:
[tex]A = \pi r^2 \\\\d = r/2 r = 0.05cm \\\\A = \pi (0.05^2) = 0.007854 m^2[/tex]
We can also express dB/dt as:
[tex]\frac{dB}{dt} = \frac{\Delta B}{t} = \frac{0-0.5}{t} = \frac{-0.5}{t}[/tex]
Now, we can create an equation.
[tex]\epsilon = -(1)(0.007854)\frac{-0.5}{t} = \frac{0.003927}{t}[/tex]
To solve the system, we must now develop an expression for current given an emf and resistance.
Begin by calculating the resistance of the copper wire:
[tex]R = \frac{\rho L}{A}[/tex]
ρ = Resistivity of copper (1.72 * 10⁻⁸ Ωm)
L = Length of wire (0.01 m)
A = cross section area (m²)
Solve:
[tex]R = \frac{(1.72*10^{-8})(0.01)}{\pi (0.001^2)} = 5.475 * 10^{-5} \Omega m[/tex]
Now, we can use the following relation (Ohm's Law):
[tex]\epsilon = iR\\\\\epsilon = \frac{Q}{t}R[/tex]
*Since current is equivalent to Q/t.
Plug in the value of R and set the two equations equal to each other.
[tex]\frac{Q}{t}(5.475 * 10^{-5}) = \frac{0.003927}{t}[/tex]
Cancel out 't'.
[tex]Q (5.475 * 10^{-5}) = 0.003927 \\\\Q = \frac{0.003927}{5.475*10^{-5}} = \boxed{71.73 C}[/tex]
A quantity of 1.922 g of methanol (CH3OH) was burned in a constant-volume calorimeter. Consequently,
the temperature of the water rose by 4.20 ºC. If the heat capacity of the bomb plus water was 10.4 kJ/ºC,
calculate the molar heat of combustion of methanol.
You see a car that appears very small, so you assume that it must be far from you. You are using the monocular cue of
The monocular cue of relative size
1. What is the distance covered by a T-Rex that goes from 0 m/s to 9 m/s in 6.78 seconds? (10
points)
With the use of first and third equation of motion, the distance covered by a T-Rex is 30.51 m
Linear MotionWhen a body is in linear motion, the body is moving in a straight line. some of the parameters to consider are:
Distance coveredSpeedVelocityAccelerationE.T.CGiven that a T-Rex move from 0 m/s to 9 m/s in 6.78 seconds, the distance covered can be found by calculating the acceleration.
Let us use equation 1
V = U + at
9 = 0 + 6.78a
a = 9 / 6.78
a = 1.33 m/[tex]s^{2}[/tex]
Now let us use equation 3
[tex]v^{2}[/tex] = [tex]u^{2}[/tex] + 2as
[tex]9^{2}[/tex] = 2 x 1.33 x S
81 = 2.655S
S = 81/2.655
S = 30.51 m
Therefore, the distance covered by a T-Rex is 30.51 m.
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a 1. You found that the MCB was tied with a thread and the thread was fixed with a nail on the wall in your friend's house. i. Is it good idea to do this? ii. What could be the possible hazard of this? iii. What should have done to keep the circuit safe?
The miniature circuit breaker should rather be fastned to a wall using nails and other neccessary tools.
What is a miniature circuit breaker?A miniature circuit breaker is a circuit breaker that is used in homes as a means of guarding against damage to appliances due to a very high current.
This miniature circuit breaker is also harzardous in the sense that it could lead to an electrical fault related fire outbreak especially when it is being blown freely by wind as you tie it with a thread. Doing this a very bad idea because of the risk of a fire hazard.
The miniature circuit breaker should rather be fastned to a wall using nails and other neccessary tools.
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Two bodies of specific heats S1 and S2 having the same heat capacities are combined to form a single composite body. What is the specific heat of the composite body?
[tex]\qquad\qquad\huge\underline{{\sf Answer}}♨[/tex]
Heat capacity of body 1 :
[tex]\qquad \sf \dashrightarrow \:m_1s_1[/tex]
Heat capacity of body 2 :
[tex]\qquad \sf \dashrightarrow \:m_2s_2[/tex]
it's given that, the the head capacities of both the objects are equal. I.e
[tex]\qquad \sf \dashrightarrow \:m_1s_1 = m_2s_2[/tex]
[tex]\qquad \sf \dashrightarrow \:m_1 = \dfrac{m_2s_2}{s_1} [/tex]
Now, consider specific heat of composite body be s'
According to given relation :
[tex]\qquad \sf \dashrightarrow \:(m_1 + m_2) s' = m_1s_1 + m_2s_2[/tex]
[tex]\qquad \sf \dashrightarrow \:s' = \dfrac{ m_1s_1 + m_2s_2}{m_1 + m_2}[/tex]
[tex]\qquad \sf \dashrightarrow \:s' = \dfrac{ m_2s_2+ m_2s_2}{ \frac{m_2s_2}{s_1} + m_2 }[/tex]
[ since, [tex] m_2s_2 = m_1s_1 [/tex] ]
[tex]\qquad \sf \dashrightarrow \:s' = \dfrac{ 2m_2s_2}{ m_2(\frac{s_2}{s_1} + 1)}[/tex]
[tex]\qquad \sf \dashrightarrow \:s' = \dfrac{ 2 \cancel{m_2}s_2}{ \cancel{m_2}(\frac{s_2}{s_1} + 1)}[/tex]
[tex]\qquad \sf \dashrightarrow \:s' = \dfrac{ 2 s_2}{ (\frac{s_2 + s_1}{s_1} )}[/tex]
[tex]\qquad \sf \dashrightarrow \: s' = \dfrac{2s_1s_2}{s_1 + s_2} [/tex]
➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖
Two identical spheres, A and B, carry charges of
+6 microcoulombs and -2 microcoulombs,
respectively. If these spheres touch, what will be
the resulting charge on sphere A?
The resulting charge in sphere A from the two given different charges is determined as +2 μC.
Resulting of the two speheres
The resulting charge of the two spheres is determined by suming the two individual charges together as shown below;
Resulting charge = +6 μC - 2 μC
Resulting charge = +4 μC
Charge in sphere AThe resulting charge will be distributed uniformly between the two spheres and charge in sphere A and B will be equal.
Charge in sphere A = charge in sphere B = ¹/₂ x +4 μC = + 2 μC
Thus, the resulting charge in sphere A from the two given different charges is determined as +2 μC.
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If interstellar gas has a density of 1 atom/cm3, how big a volume of material must be used to make a star with the mass of the Sun
The volume of the material that must be used to make a star with the mass of the sun is 1.2×10⁵¹ m³.
What is volume?Volume is the amount of space occupied by an object or a plane figure.
To calculate the volume of the material that must be used to make a star with the mass of the sun, we use the formula below.
Formula:
D = m/V............ Equation 1Where:
D = Density of the interstellar gasm = mass of the sunV = Volume of the materialMake V the subject of the equation
V = m/D........... Equation 2From the question,
Given:
m = mass of the sun = 1.9891×10³⁰ kgD = 1 atom/cm³ = 1.66×10⁻²¹ kg/m³Substitute these values into equation 2
V = ( 1.9891×10³⁰)/(1.66×10⁻²¹)V = 1.2×10⁵¹ m³Hence, The volume of the material that must be used to make a star with the mass of the sun is 1.2×10⁵¹ m³.
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Explain the differences between horticulture and agriculture.
[tex] \tt{Difference \: Between \: \underline\red {Horticulture} \: And \: \underline\green{Agriculture}}[/tex]
[tex] \text{\underline\red {Horticulture}}[/tex]
The Cultivation of Fruits, Vegetables and Flowers for domestic and international markets are called Horticulture.
[tex] \text{\underline\green{Agriculture}}[/tex]
It is science, art and occupation of cultivating the soil, producing crops and livestock. It is systematic and controlled use of living organisms and the environment to improve the human condition.
Hope This HelpsWhile cruising on the high sea at speed of v = 22 km/hr, a cargo ship suddenly encounters a problem with its propeller shaft leading to the main engine being shut down. This causes the vessel to slow down to 12 km/hr over the next 14 minutes. Assuming that its deceleration profile is following a correlation of as a = -0.85ct², where c is the acceleration constant for any time t, determine; i) The magnitude for constant c from the two boundary limits of deceleration profile given above, and ii) Plot against the timescale for same interval above, a total distance covered by the vessel as it's slowing down from 22 km/hr to 12 km/hr. Also plot the speed profile for the vessel during that 14 minutes' span.
Answer:
how can you support to leprosy victims?
If an object accelerates from rest, with a constant acceleration of
10 m/s2, what will its velocity be after 2 s?
Answer: 20 m/s
Explanation:
Acceleration = change in velocity / time → 10 m/s^2 = final velocity - 0 (original velocity) / 2s → final velocity = 20 m/s
Please help solve. Image down below
Answer:
thats is so hard
Explanation:
i cant answer that
Which of the following particles is similar to a He nucleus?
alpha
beta
gamma
neutrino
Alpha
I hope this helps you
:)
how r u
________________.
What is an analogy of two different roads or rivers to compare a series and parallel circuit?
Answer:
In a series circuit, the same amount of current flows through all the components placed in it. On the other hand, in parallel circuits, the components are placed in parallel with each other due to which the circuit splits the current flow.