If a rock is thrown upward on the planet Mars with a velocity of 13 m/s, its height in meters t
seconds later is given by y =( 13t - 1.86t^2)... (Round your answers to two decimal places.)
(a) Find the average velocity over the given time intervals.
(i) [1, 2]
______ m/s
(ii) [1, 1.5]
______ m/s
(iii) [1, 1.1]
______ m/s
(iv) [1, 1.01]
______ m/s
(v) [1, 1.001]
______ m/s
Solution
To find average velocity, you simply have to find the total distance divided by the total time for
each interval:
(i) Time = 2 - 1 = 1s
Distance = (13(2) - 1.86(2)2) - (13(1) - 1.86(1)2) = 7.42 m.
Average velocity = 7.42m / 1s = 7.42 m/s.
(ii) T = .5s
D = (13(1.5) - 1.86(1.5)2) - (13(1) - 1.86(1)2) = 4.175 m
V = 4.175m / .5s = 8.35 m/s.
(iii) T = .1s
D = (13(1.1) - 1.86(1.1)2) - (13(1) - 1.86(1)2) = .9094 m
V = .9094m / .1s = 9.094 m/s.
(iv) T = .01s
D = (13(1.01) - 1.86(1.01)2) - (13(1) - 1.86(1)2) = .092614 m
V = .092614m / .01s = 9.2614 m/s.
(v) T = .001s
D = (13(1.001) - 1.86(1.001)2) - (13(1) - 1.86(1)2) = .00927814 m
V = .00927814m / .001s = 9.27814 m/s..
GPS satellites are usually in the Medium Earth Orbit with a typical a.pdffootstatus
GPS satellites are usually in the Medium Earth Orbit with a typical altitude of 20,200 km. In
such an orbit, objects move with a speed of 3072 m/s relative to an observer standing on the
ground, (a) A clock on the satellite is ticking once every second. To the observer on the ground,
what is the time interval between the ticks? (b) Suppose the satellite is sending you radio signals
(traveling at speed of light) and you use it to calculate the distance (d = vt). If you don\'t take into
account the time dilation, how long will it take for your distance calculations to be off by 100 m?
Solution
speed of the satellite v = 3872 m/s
Relativistic time dilation t = t’/sqrt(1-v2/c2)
v/c = 3872/3e+8 = 1.29e-5
v2/c2 = 1.67e-10
the above term being very small we can approximate
t = t’(1+(1.67e-10)/2)
= t’ (1+0.835e-10)
For the ground observer the time interval between the ticks is
1+0.835e-10 s
i.e. 83.5 ps more
d=vt here v= c= 3e+10 m/s
for the distance to be off by 100m
100 = 3e+10t
t = 33.3e-10 s.
A Martian leaves Mars in a spaceship that is heading to Venus- On the.docxkaseya1
A Martian leaves Mars in a spaceship that is heading to Venus. On the way, the spaceship passes earth with a speed v = 0.77c relative to it. Assume that the three planets do not move relative to each other during the trip. The distance between Mars and Venus is 1.20 × 1011 m, as measured by a person on earth. (a) What does the Martian measure for the distance between Mars and Venus? (b) What is the time of the trip (in seconds) as measured by the Martian? Limit your answer to 2 significant digits.
Solution
the three planets can be thought to be stationary.
distance between mass and venus=d1=1.2*10^11 m
part a:
using length contraction, distance measured by the martian=1.2*10^11*sqrt(1-(v/c)^2)
=7.6565*10^10 m
part b:
time of the trip by martian=distance measured by martian/speed
=7.6565*10^10/(0.77*3*10^8)
=331.45 seconds
.
GPS satellites are usually in the Medium Earth Orbit with a typical a.pdffootstatus
GPS satellites are usually in the Medium Earth Orbit with a typical altitude of 20,200 km. In
such an orbit, objects move with a speed of 3072 m/s relative to an observer standing on the
ground, (a) A clock on the satellite is ticking once every second. To the observer on the ground,
what is the time interval between the ticks? (b) Suppose the satellite is sending you radio signals
(traveling at speed of light) and you use it to calculate the distance (d = vt). If you don\'t take into
account the time dilation, how long will it take for your distance calculations to be off by 100 m?
Solution
speed of the satellite v = 3872 m/s
Relativistic time dilation t = t’/sqrt(1-v2/c2)
v/c = 3872/3e+8 = 1.29e-5
v2/c2 = 1.67e-10
the above term being very small we can approximate
t = t’(1+(1.67e-10)/2)
= t’ (1+0.835e-10)
For the ground observer the time interval between the ticks is
1+0.835e-10 s
i.e. 83.5 ps more
d=vt here v= c= 3e+10 m/s
for the distance to be off by 100m
100 = 3e+10t
t = 33.3e-10 s.
A Martian leaves Mars in a spaceship that is heading to Venus- On the.docxkaseya1
A Martian leaves Mars in a spaceship that is heading to Venus. On the way, the spaceship passes earth with a speed v = 0.77c relative to it. Assume that the three planets do not move relative to each other during the trip. The distance between Mars and Venus is 1.20 × 1011 m, as measured by a person on earth. (a) What does the Martian measure for the distance between Mars and Venus? (b) What is the time of the trip (in seconds) as measured by the Martian? Limit your answer to 2 significant digits.
Solution
the three planets can be thought to be stationary.
distance between mass and venus=d1=1.2*10^11 m
part a:
using length contraction, distance measured by the martian=1.2*10^11*sqrt(1-(v/c)^2)
=7.6565*10^10 m
part b:
time of the trip by martian=distance measured by martian/speed
=7.6565*10^10/(0.77*3*10^8)
=331.45 seconds
.
Planet Velocity Is Defined As A Function Of Its Diameter (II)Gerges francis
Paper hypothesis
The Earth moon is created by effect of Uranus and Pluto motions
The hypothesis explanation
- Although the Earth moon is created by Planets collisions But
- The Earth moon data is created by an effect of light motion calculations
- That supports the conclusion that (light motion must be a companying with planet motion an effect on it)
Paper questions
- The paper tries to answer the following question
o Why Mercury Day needs 5040 seconds to be 176 solar days?
o Why Uranus motion distance during (5040 seconds)= Pluto motion distance during 7511 seconds.
Paper conclusion
- Light Motion Is Accompanying With Planet Motion.
Gerges Francis Tawdrous +201022532292
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If a population of rabbits is increasing by 14 rabbits every month , is the population increasing
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2nd month = 14 + 14 = 28 rabit
3rd month = 14 + 14 + 14 = 42 rabbits
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So, the relation will be
no. of rabbits = 14 * m
m= no. of months
m= 1, 2, 3.....
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individual values between 450 and 550.
a. Be less than
b. Be greater than
c. Be the same as
d. Not be comparable to.
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If a rock is thrown upward on the planet Mars with a velocity of 13 .pdf
1. If a rock is thrown upward on the planet Mars with a velocity of 13 m/s, its height in meters t
seconds later is given by y =( 13t - 1.86t^2)... (Round your answers to two decimal places.)
(a) Find the average velocity over the given time intervals.
(i) [1, 2]
______ m/s
(ii) [1, 1.5]
______ m/s
(iii) [1, 1.1]
______ m/s
(iv) [1, 1.01]
______ m/s
(v) [1, 1.001]
______ m/s
Solution
To find average velocity, you simply have to find the total distance divided by the total time for
each interval:
(i) Time = 2 - 1 = 1s
Distance = (13(2) - 1.86(2)2) - (13(1) - 1.86(1)2) = 7.42 m.
Average velocity = 7.42m / 1s = 7.42 m/s.
(ii) T = .5s
D = (13(1.5) - 1.86(1.5)2) - (13(1) - 1.86(1)2) = 4.175 m
V = 4.175m / .5s = 8.35 m/s.
(iii) T = .1s
D = (13(1.1) - 1.86(1.1)2) - (13(1) - 1.86(1)2) = .9094 m
V = .9094m / .1s = 9.094 m/s.
(iv) T = .01s
D = (13(1.01) - 1.86(1.01)2) - (13(1) - 1.86(1)2) = .092614 m
2. V = .092614m / .01s = 9.2614 m/s.
(v) T = .001s
D = (13(1.001) - 1.86(1.001)2) - (13(1) - 1.86(1)2) = .00927814 m
V = .00927814m / .001s = 9.27814 m/s.
![If a rock is thrown upward on the planet Mars with a velocity of 13 m/s, its height in meters t
seconds later is given by y =( 13t - 1.86t^2)... (Round your answers to two decimal places.)
(a) Find the average velocity over the given time intervals.
(i) [1, 2]
______ m/s
(ii) [1, 1.5]
______ m/s
(iii) [1, 1.1]
______ m/s
(iv) [1, 1.01]
______ m/s
(v) [1, 1.001]
______ m/s
Solution
To find average velocity, you simply have to find the total distance divided by the total time for
each interval:
(i) Time = 2 - 1 = 1s
Distance = (13(2) - 1.86(2)2) - (13(1) - 1.86(1)2) = 7.42 m.
Average velocity = 7.42m / 1s = 7.42 m/s.
(ii) T = .5s
D = (13(1.5) - 1.86(1.5)2) - (13(1) - 1.86(1)2) = 4.175 m
V = 4.175m / .5s = 8.35 m/s.
(iii) T = .1s
D = (13(1.1) - 1.86(1.1)2) - (13(1) - 1.86(1)2) = .9094 m
V = .9094m / .1s = 9.094 m/s.
(iv) T = .01s
D = (13(1.01) - 1.86(1.01)2) - (13(1) - 1.86(1)2) = .092614 m](https://image.slidesharecdn.com/ifarockisthrownupwardontheplanetmarswithavelocityof13-230403034505-de9a0b3c/85/If-a-rock-is-thrown-upward-on-the-planet-Mars-with-a-velocity-of-13-pdf-1-320.jpg)
