Answer:
a) The period of the satellite is approximately 5,425.305 seconds (1.507 hours)
b) The orbital velocity of the satellite is approximately 7,725.8565 m/s
Explanation:
The given parameters are;
The mass of the satellite, m = 1,000 kg
The altitude at which the satellite is orbiting, h = 300 km
a) The period of the satellite is given as follows;
[tex]T = 2 \times \pi \times \sqrt{\dfrac{(R + h)^3}{g \times R^2} }[/tex]
Where;
R = The radius of the Earth = 6.371 × 10⁶ m
g = The acceleration due to gravity = 9.81 m/s²
h = The altitude of the orbit of the satellite = 300 km = 300,000 m
By substitution, we have;
[tex]T = 2 \times \pi \times \sqrt{\dfrac{((6.371 + 0.3)\times 10^6)^3}{9.81 \times (6.371 \times 10^6)^2} } \approx 5,425.305[/tex]
T ≈ 5,425.305 seconds ≈ 1.507 hours
b) The orbital velocity of the satellite is given as follows;
[tex]v_0 = \sqrt{\dfrac{G \times M_{central}}{R + h} } = \sqrt{\dfrac{g \times R^2}{R + h} }[/tex]
Where;
v₀ = The orbital velocity of the satellite
Which by substitution gives;
[tex]v_0 = \sqrt{\dfrac{9.81 \times (6.371 \times 10^6)^2}{(6.371 + 0.3) \times 10^6} } = 7,725.8565[/tex]
The orbital velocity, v₀ = 7,725.8565 m/s.
Shows a car travelling around a bend in the road. The car is travelling at a constant speed. There is a resultant force acting on the car. This resultant force is called the centripetal force. (i) In which direction, A, B, C or D, does the centripetal force act on the car? Tick ( ) one box. A B C D (1) (ii) State the name of the force that provides the centripetal force.
Answer:
This question is incomplete
Explanation:
This question is incomplete but the missing figure is in the attachment below.
When an object is travelling around a circular path, there is a force that tends to draw that object towards the center of the circular path and keep the object moving in the curved path, that force is called the centripetal force. From this description, it can be deduced that the direction of the centripetal force that acts on the car (in the attachment below) is D.
The name of the force that provides this centripetal force is frictional force. This is the force that prevents the car from slipping off the road; keeping it moving in the curved path.
Can u anser 5,6 on the picture
Answer: Number 6 is Periods
Explanation:
A pinball bangs against a bumper of a pinball machine with a speed of 0.46 m/s. If the ball has a mass of 0.058 kg, what is the ball's kinetic energy?
Answer:
either its 12 or 0.402
Explanation:
Calculate the kinetic energy of a 50 kg cart moving at a speed of 18.6 m/s.
Answer:
8649 J
Explanation:
KE = 1/2mv^2
1/2(50)(18.6)^2
1/2(50)(345.96) = 8649 J
Which of the following is NOT one of the essential components of an exercise program?
Answer:
i dont know dude but ask someone who is in your grade lol
Explanation:
how many teeth shows signs of decay?
Answer:
what do you mean
Explanation:
A 1500 kg car has an applied forward force of 5000 N and experiences an air resistance of 1250 N. What is the car's acceleration?
Answer:
[tex]2.33\ m/s^2[/tex]
Explanation:
Net Force
According to the second Newton's law, the net force exerted by an external agent on an object of mass m is:
Fn = m.a
Where a is the acceleration of the object. The net force is the sum of the individual vector forces applied to the object.
The m=1500 Kg car has two horizontal forces applied: the forward force of 5000 N that causes the movement and the air resistance force of 1250 N that opposes motion.
The net force is Fn = 5000 N - 1500 N = 3500 N
To find the acceleration, we solve the equation for a:
[tex]\displaystyle a=\frac{Fn}{m}[/tex]
[tex]\displaystyle a=\frac{3500}{1500}[/tex]
[tex]\boxed{a = 2.33\ m/s^2}[/tex]
The car's acceleration is [tex]a = 2.33\ m/s^2[/tex]
the diagram shows the apparent motion of the sun across the sky?
Answer:
The earth rotates west to east
Explanation:
Considering that the sun doesn't really move and the earth is rotating around the sun, and it is moving from east to west with us (the earth) facing it in which means the earth rotates around the sun west to east!
a squirrel runs at a speed of 9.9 m/s with 25 J of kinetic energy
What is the squirrels mass
Answer:
yeet yeet yeet yeet
Explanation:
Kinetic energy (K.E):-
So, the Mass of the Squirrel is 0.51 Kg (or) 510 grams.
A squirrel runs at a speed of 9.9 m/s with 25 J of kinetic energy.
What is the squirrel’s mass?
Answer: 0.51 kg
Physical values in the real world have two components: magnitude and
Answer:
Dimension.
Explanation:
A 75kg bicyclist (including the bicycle), initially at rest at the top of a hill, coasts down the hill, reaching a speed of 14.6m/s at the bottom of the hill. The distance and height of the hill are shown. Neglect any friction impeding the motion and the rotational energy of the wheels. List the energy types at the initial and final time and whether work and loss (due to non-conservative forces) occur as well as the corresponding amounts of energy.
The energy type at the initial time is potential energy and the energy at the final time or position is kinetic energy.
What is the law of conservation of energy?The law of conservation of energy states that energy can neither be created nor destroyed but can be transformed from one form to another.
Based on the law of conservation of mechanical energy, the formula for the change in the kinetic energy and the potential energy of the bicyclist is given as;
K.Ei + P.Ei = K.Ef + P.Ef
where;
K.Ei is the initial kinetic energy of the bicyclistK.Ef is the final kinetic energy of the bicyclistP.Ei is the initial potential energy of the bicyclistP.Ef is the final potential energy of the bicyclistThe kinetic energy of the bicyclist increases with increase in the velocity of the bicyclist while the potential energy increases with increase in the height of the bicyclist.
At the initial position when the bicyclist is at rest, the kinetic energy is zero, so the only energy at the initial position is potential energy because the height is maximum.
In addition, at the final position, the velocity of the bicyclist is maximum and the height is zero, so the only energy at the final position is kinetic energy.
Learn more about conservation of energy here: https://brainly.com/question/166559
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If the velocity of an object changed from 30 m/s to 60 m/s over a period of 10 seconds what would the average acceleration be ?
A sound wave passes from air into water and then into steel. With each change in medium, the velocity of this wave will a(increase,then decrease b(decrease c(increase d( decrease,then increase ?
Answer:
a
Explanation:
because it can increase anything
26. For this table of data, which column is the independent variable?
radius of circle (m) area of circle (m2)
3.4
36.317
4.5
63.617
4.6
66.476
6.2
120.763
7.6
181.458
7.7
186.265
8.6
232.352
9.7
295.592
area of circle
o
radius of circle
O
None of these is correct.
Answer:
radius of a circle is measured by an instrument, so it is the independent variable
Explanation:
The independent variable corresponds to the magnitude controlled by the experimenter and in general can be measured by an instrument.
The dependent variable is the variable used to perform the calculations, obtained from a mathematical transformation of the independent variable.
The radius of a circle is measured by an instrument, so it is the independent variable and the area is calculated by
A = π r²
Therefore the area corresponds to the dependent variable
A train travels with a speed of 115km/hr. How much time does it take to cover a distance of 470km.
It would take approximately 4 hours for the train to cover a distance of 470km.
4 hours 5 minutes 13.04 seconds
A fish swimming at a rate of .6 m/s notices a huge shark. Three seconds later, the fish is swimming at a speed of 3 m/s. What is the fish's acceleration?
0.8 m/s/s
-0.8 m/s/s
12.5 m/s/s
-12.5 m/s/s
Answer:
C
Explanation:
???
i think
Alex wants to learn how to surf, but he is not a strong swimmer. He knows he needs to increase his ability to paddle out in order to catch the best waves. Which piece of advice would you give to help him get started on reaching his goal? He should swim at least three times a week at the community pool to build stamina. He should tread water every day to get more comfortable in the water. He should purchase the best surfboard he can afford because it will help him paddle faster. He should watch your friend who is a competitive surfer practice to learn her technique.
Answer:
He should swim at least three times a week at the community pool to build stamina.
Explanation:
- During a certain period, the angular position of a rotating object is given by: = − + , where is in radian and t is in seconds. Determine the angular position, angular speed, and angular acceleration of the rotating object at = Sec.
The question is not complete. The complete question is :
During a certain period of time, the angular position of a rotating object is given by [tex]$\theta =2t^2 +10t+5$[/tex], where θ is in radians and t is in seconds. Determine the angular position, angular speed, and angular acceleration of the door (a) at t = 0.00 seconds, (b) at t = 3.00 seconds.
Solution :
Given :
Displacement or angular position of the object, [tex]$\theta =2t^2 +10t+5$[/tex]
∴ Angular speed is [tex]$\omega = \frac{d \theta}{dt}$[/tex]
ω = 10 + 4t
And angular acceleration is [tex]$\alpha = \frac{d \omega}{dt}$[/tex]
α = 4
a). At time, t = 0.00 seconds :
Angular displacement is [tex]$\theta =2t^2 +10t+5$[/tex]
[tex]$\theta =2(0)^2 +10(0)+5$[/tex]
= 5 rad
Angular speed is ω = 10 + 4t
ω = 10 + 4(0)
= 10 rad/s
Angular acceleration is α = 4 [tex]$rad/s^2$[/tex]
b). At time, t = 3.00 seconds :
Angular displacement is [tex]$\theta =2t^2 +10t+5$[/tex]
[tex]$\theta =2(3)^2 +10(3)+5$[/tex]
= 53 rad
Angular speed is ω = 10 + 4t
ω = 10 + 4(3)
= 22 rad/s
Angular acceleration is α = 4 [tex]$rad/s^2$[/tex]
Using Velocity vs Time Graphs to Find Acceleration
A graph titled velocity versus time has horizontal axis time (seconds) and vertical axis velocity (meters per second). A line has 4 straight segments. Line segment A runs from 0 seconds 0 meters per second to 1 seconds 15 meters per second. Then segment B runs to 2 seconds 20 meters per second. Then segment C runs to 4 seconds 20 meters per second. Then segment D runs to 5 seconds 0 meters per second.
The acceleration of segment D is m/s2.
Rank segments A, B, and C from least acceleration to greatest acceleration.
Least:
Greatest:
The acceleration at segment D is -20m/s²
The rank of the acceleration from the least to the greatest is -20m/s² < 0m/s² < 5m/s² < 10m/s² (D<C<B<A)
Acceleration is the change in velocity with respect to time.
a = v-u/t
Acceleration at segment A:
Aa = 15-0/1-0
Aa = 15m/s²
Acceleration at segment B:
Ab = 20-15/2-1
Ab = 5m/s²
Acceleration at segment C:
Aa = 0-0/4-2
Aa = 0m/s²
Acceleration at segment D:
Ac = 0-20/5-4
Ac = -20m/s²
Hence the acceleration at segment D is -20m/s²
The rank of the acceleration from the least to the greatest is -20m/s² < 0m/s² < 5m/s² < 10m/s² (D<C<B<A)
Learn more here: https://brainly.com/question/18398656
A merry go round exerts a force of 1000 N on a rider on the
outer ring of animals when it takes 15 seconds to make a
complete revolution. If the person weighs 750 N, the radius
of the circle he is making is m. Round your answer to
the nearest tenth.
Answer:
The radius of the circle made by the person on the merry go round is 74.55 meters
Explanation:
The given parameters are;
The force the merry go round exerts on the rider = 1000 N
The time it takes the merry go round to make one complete revolution = 15 seconds
The weight of the person = 750 N
The radius of the circle made by the person on the merry go round = r
We have;
[tex]F_c = \dfrac{m \cdot v^2}{r} = m \cdot \omega ^2 \cdot r[/tex]
Where;
m = The mass of the person
v = The velocity of the person
[tex]F_c[/tex] = The centrifugal force acting on the person = 1,000 N
r = The radius of the circle made by the person on the merry go round
ω = Angular velocity = 2·π/15 rad/s
We have;
The mass of the person = The weight/(The acceleration due to gravity, g)
∴ The mass of the person = 750/9.81 ≈ 76.45 kg
By substituting the calculated and known values into the equation for the centripetal force, we have;
[tex]F_c[/tex] = m × ω² × r
1000 = 76.45 × (2·π/15)² × r
r = 1000/(76.45 × (2·π/15)²) = 74.55 m
The radius of the circle made by the person on the merry go round = r = 74.55 m.