Answer:
La diferencia entre las tensiones máxima y mínima es de 19.614 newtons.
Explanation:
Puesto que la bola gira en un círculo vertical, existe claramente una diferencia entre las tensiones debido a la influencia de la gravedad y la tensión que resulta de la aceleración centrípeta experimentada por la masa. La máxima tensión ocurre cuando la bola se encuentra en el nadir (o la sima) del trayecto circular, la cual se describe por la Segunda Ley de Newton:
[tex]T_{max} - m\cdot g = m\cdot \frac{v^{2}}{L}[/tex]
En cambio, la mínima tensión aparece cuando la bola se encuentra en el cénit (o la cima) del trayecto circular, descrita por la misma ley de Newton:
[tex]T_{min} + m\cdot g = m\cdot \frac{v^{2}}{L}[/tex]
Donde:
[tex]T_{min}[/tex], [tex]T_{max}[/tex] - Tensiones mínima y máxima, medidas en newtons.
[tex]m[/tex] - Masa de la bola, medida en kilogramos.
[tex]g[/tex] - Constante gravitacional, medida en metros por segundo al cuadrado.
[tex]L[/tex] - Distancia con respecto al eje de rotación, medida en metros.
[tex]v[/tex] - Rapidez tangencial, medido en metros por segundo.
Se elimina la aceleración centrípeta de ambas expresiones por igualación:
[tex]T_{min} + m\cdot g = T_{max} - m\cdot g[/tex]
Ahora, la diferencia entre las tensiones máxima y mínima es:
[tex]T_{max} - T_{min} = 2\cdot m \cdot g[/tex]
Si [tex]m = 1\,kg[/tex] y [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], entonces:
[tex]T_{max} - T_{min} = 2\cdot (1\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)[/tex]
[tex]T_{max}-T_{min} = 19.614\,N[/tex]
La diferencia entre las tensiones máxima y mínima es de 19.614 newtons.
a Porsche 911 accelerates from rest to 27 metre per second due north in 5.8 seconds the mass of the car is 1400 kg what is the magnitude and direction of the average net force
Answer:
The average force has a magnitude 6524 N due north.
Explanation:
The average net force F = ma where m = mass of car = 1400 kg and a = acceleration.
a = (v - u)/t where u = initial velocity of car = 0 m/s (since it starts from rest)
v = final velocity of car = 27 m/s due north and t = time of motion = 5.8 s
a = (27 m/s - 0 m/s)/5.8 s = 27 m/s ÷ 5.8 s = 4.66 m/s
Since the direction of the velocity change is the direction of the acceleration, the acceleration is 4.66 m/s due north.
The average force, F = ma = 1400 kg × 4.66 m/s = 6524 N
Since the acceleration is due north, the average force takes the direction of the acceleration.
So the direction of the average force is due north
The average force has a magnitude 6524 N due north.
A ball is being rolled by a normal push of 180N. It is opposed by friction which has a force of 61N and air resistance which has a force of 23N. Calculate the resultant force.
Resultant force is basically the force left after everything is added.
if a ball is being pushed one one side with 180N, and being pushed on teh opposite side with 84N (I added friction and air resistance since they're acting on the same side), then the resultant force would be:
180N - 84N = 96N (you can determine whether it's positive or negative based on the direction of the vector)
Answer the following questions regarding the equation:
N₂ + 3H₂ → 2NH₃
1) indicates what type of reaction is
2) what represents the coefficients 3 and 2 in the previous reaction is done for
3) What would be missing in the previous equation to make it more accurate is
Explanation:
1) This is a synthesis reaction (two or more reactants combine to form a single product).
2) The coefficients are added to balance the reaction.
3) Adding the states of matter (solid, liquid, gas) will make the reaction more precise.
Mt Everest is at the height of 8850 m; its base is at an elevation of 5400m. The temperature here drops at the rate of 1 degree per 100m. If temperature at the base is -5, what will be the temperature at the top?
Answer:
The temperature at the top is -39.5 degree
Explanation:
The height of Mt Everest = 8850 m
The base is at the elevation of = 5400 m
The drop in temperature = 1 degree per 100 m
At the base, the temperature = -5
Since every 100 meter rise in height results in the one degree drop of temperature. Therefore, first find the difference between the base and height. Multiply with negative one.
Difference in heights = 8850 – 5400 = 3450
The temperature at the top = (3450 / 100) + 5 = -39.5 degree
Since there is fall in temperature so we use negative sign.
Two teams are playing tug-of-war. Team A, on the left, is pulling on the rope with an effort of 5000 N. If the rope is moving at a constant velocity, how hard and in which direction is team B pulling?
A. 2500 N to the left
B. 5000 N to the right
C. 2500 N to the right
D. 5000 N to the left
Explanation:
If Team A is on the left, B is on the right
if the force is constant, it means that the effort applied is equal.
So Team B is pulling 5000N to the right.
6. While standing on an ice surface (assume no friction), a 60 kg skater pushes against
a wall with an average force of 240 N east for 0.360 s. Calculate the velocity
(magnitude and direction) of this student at 0.360 s.
Answer: 1.44m/s West
Explanation:
Given the following :
Mass of skater = 60kg
Average force = 240N east
Time of = 0.360 seconds
The Velocity after 0.360seconds
Recall the relation:.
Force(f) * time(t) = mass (m) * (final Velocity(V) - initial velocity(u))
Where ; final Velocity(v) - initial velocity(u) = Magnitude of velocity(V)
Ft = mV
240 × 0.360 = 60 × V
60 × V = 86.4
V = 86.4 / 60
V = 1.44m/s
According to Newton's third law, to every action, there is an equal and opposite reaction.
The action force acts in the eastern direction hitting the wall with a force of 240N, therefore direction after hitting the wall will be in the opposite (western direction) direction, which is the direction of the reaction force
Therefore, velocity is 1.44ms West
A light, rigid rod is 51.4 cm long. Its top end is pivoted on a frictionless horizontal axle. The rod hangs straight down at rest with a small, massive ball attached to its bottom end. You strike the ball, suddenly giving it a horizontal velocity so that it swings around in a full circle. What minimum speed at the bottom is required to make the ball go over the top of the circle?
Answer:
4.489 m/s
Explanation:
We are given;
Length of rod;L = 51.4 cm = 0.514 m
When the ball is at the top end, it's velocity will be zero and it's kinetic energy will be transformed to potential energy.
Thus; ½mv² = mgh
m will cancel out to give;
v² = 2gh
Where:
g is acceleration due to gravity = 9.8 m/s²
h is the maximum height of the ball which is equal to twice the length of the rod.
Thus, h = 2(0.514) = 1.028 m
v² = 2 × 9.8 × 1.028
v = √20.1488
v = 4.489 m/s
A toaster oven draws 300.0 watts of power. If it is plugged into an outlet with a voltage of 115 volts, what current is in the toaster's circuit?
OA 3.45 10% amps
ОВ.
415 amps
Ос.
285 amps
OD
2.61 amps
Answer:
I = 2.61 A
Explanation:
It is given that,
Power of a toaster is 300 watts
It is plugged into an outlet with a voltage of 115 volts
We need to find the current in the toaster's circuit.
The electrical power is given by the formula as follows :
P = V I
I is electric current
[tex]I=\dfrac{P}{V}\\\\I=\dfrac{300}{115}\\\\I=2.61\ A[/tex]
So, 2.61 A of current is in the toaster's circuit.
Answer: D.
2.61 amps
Explanation: edmentum
The magnitude of the electric potential depends on which factors?
O A. Mass and amount of charge
O B. Mass and distance
O C. Distance and amount of charge
O D. Density and amount of charge
Answer:
C
Explanation:
It is directly propotional to the amount of charge and inversly propotional to distance.
Answer:
C. Distance and amount of charge
Explanation:
The magnitude of the electric potential depends on which factors?
Distance and amount of charge
I need a story about kepler's laws for a child, something a child would understand. And i'd like it to be okay for someone if I copy anhd paste it or else my parents are gonna kill me and my teacher will be very disappointe, this is for summer school.
Answer:
(1) The orbits are ellipses, with focal points ƒ1 and ƒ2 for the first planet and ƒ1 and ƒ3 for the second planet. The Sun is placed in focal point ƒ1.
(2) The two shaded sectors A1 and A2 have the same surface area and the time for planet 1 to cover segment A1 is equal to the time to cover segment A2.
(3) The total orbit times for planet 1 and planet 2 have a ratio a13/2 : a23/2
Answer:
g the day off and I will be there at work and I will be there at work and I will
Explanation:
you are not going to be able to make it to the meeting tonight but I can tomorrow
Two projectiles are thrown from the same point with the velocity of49ms-1. First is
projected making an angle with the horizontal and the second at an angle of (90- ). The
second is found to rise 22.5m higher than the first. Find the heights to which each will rise?
Answer:
Height of the first projectile = 49.98 m
Height of the second projectile = 72.52 m
Explanation:
From the given information;
Two projectiles are thrown from the same point with the velocity of49m/s
First is projected making an angle θ with the horizontal
and the second at an angle of 90 - θ.
Thus; for the first height to the horizontal; we have;
[tex]H_1 = \dfrac{v^2 sin^2 \theta}{2g}[/tex] ----- (1)
the second height in the vertical direction is :
[tex]H_2 = \dfrac{v^2 cos^2 \theta}{2g}[/tex] -----(2)
However; the second is found to rise 22.5 m higher than the first; so , we have :
[tex]\dfrac{v^2 cos^2 \theta}{2g}= 22.5 + \dfrac{v^2 sin^2 \theta}{2g}[/tex]
Let's recall that :
Cos²θ = 1 - Sin²θ
Replacing it into above equation; we have:
[tex]\dfrac{v^2}{2g} - \dfrac{v^2 sin^2 \theta}{2g}= 22.5 + \dfrac{v^2 sin^2 \theta}{2g}[/tex]
[tex]\dfrac{v^2}{2g} - 22.5 = \dfrac{v^2 sin^2 \theta}{g}[/tex]
[tex]\dfrac{1}{2 } \dfrac{v^2}{g} - 22.5 = \dfrac{v^2 sin^2 \theta}{g}[/tex]
[tex]\dfrac{1}{2 } - \dfrac {9.8 \times 22.5}{(49)^2} = sin^2 \theta[/tex]
[tex]\dfrac{1}{2 } - \dfrac {220.5}{2401} = sin^2 \theta[/tex]
[tex]sin^2 \theta= 0.408[/tex]
From (1);
[tex]H_1 = \dfrac{v^2 sin^2 \theta}{2g}[/tex]
[tex]H_1 = \dfrac{49^2 \times 0.408}{2*9.8}[/tex]
[tex]H_1 = \dfrac{979.608}{19.6}[/tex]
[tex]\mathbf{H_1 =49.98 \ m }[/tex]
Height of the first projectile = 49.98 m
Similarly;
From(2)
[tex]H_2 = \dfrac{v^2 cos^2 \theta}{2g}[/tex]
[tex]H_2 = \dfrac{v^2 (1-sin^2 \theta)}{2g}[/tex]
[tex]H_2 = \dfrac{49^2 (1-0.408 )}{2 \times 9.8}[/tex]
[tex]H_2 = \dfrac{2401 (0.592 )}{19.6}[/tex]
[tex]H_2 = \dfrac{1421.392}{19.6}[/tex]
[tex]\mathbf{H_2 = 72.52 \ m}[/tex]
Height of the second projectile = 72.52 m
Show all work and answer all 4 parts. Projectile motion. 20 points. Thank you.
Answer:
i took g = 9.8m/s
A. 1.16secs
B. 2.32secs
C. 6.57m
D. 57.91m
Explanation:
A. How long does the Missile take to reach ot peak?
Time taken (t) =( U²Sin (angle) )/g
u = initial velocity = 25m/s
angle given = 30°
g = acceleration due to gravity = 9.8m/s²
t = U² x Sin (angle) / g
t = 25² x Sin(30)/9.8
t = 1.61secs
B. How long is the missile in the air in total?
T = 2t
T = 2 x 1.61 = 2.32 secs
C. what maximum Height does the missile reach?
Maximum height = U²Sin²(angle) / 2gM.H =25² x Sin(30)² / 2 x 9.8M.H=6.57mMaximum height= 6.57mD. How far does the missile travel Horizontally?
Range = U²2Sin(angle)/gRange = 25² x 2 x Sin(30) / 9.8Range = 57.91mhow does enormous energy get released from the sun
Answer:
By nuclear fission
Explanation:
The sun generates enormous energy through the process of nuclear fusion.
The core or the innermost part of the sun is characterized by high temperature and pressure. These two factors cause the separation of nuclei from electrons and the fusion of hydrogen nuclei to form a helium atom.
During the fusion process, energy is released.
Question 1 of 10
2 Points
What is the SI unit for gravitational potential energy?
A. Newton
B. Joule
C. Meter
D. Watt
Answer:
Joule
Explanation:
As any unit of energy (kinetic, potential, etc.), the gravitational potential energy has SI units of Joule. This s the work done by a force of 1 Newton to displace an object 1 meter in the direction of application of the force.
Oil at 150 C flows slowly through a long, thin-walled pipe of 30-mm inner diameter. The pipe is suspended in a room for which the air temperature is 20 C and the convection coefficient at the outer tube surface is 11 W/m2 K. Estimate the heat loss per unit length of tube.
Answer:
1.01 W/m
Explanation:
diameter of the pipe d = 30 mm = 0.03 m
radius of the pipe r = d/2 = 0.015 m
external air temperature Ta = 20 °C
temperature of pipe wall Tw = 150 °C
convection coefficient at outer tube surface h = 11 W/m^2-K
From the above, we assumed that the pipe wall and the oil are in thermal equilibrium.
area of the pipe per unit length A = [tex]\pi r ^{2}[/tex] = [tex]7.069*10^{-4}[/tex] m^2/m
convectional heat loss Q = Ah(Tw - Ta)
Q = 7.069 x 10^-4 x 11 x (150 - 20)
Q = 7.069 x 10^-4 x 11 x 130 = 1.01 W/m
The heat loss per unit length of tube should be considered as the 1.01 W/m.
Calculation of the heat loss:Since
diameter of the pipe d = 30 mm = 0.03 m
radius of the pipe r = d/2 = 0.015 m
external air temperature Ta = 20 °C
temperature of pipe wall Tw = 150 °C
convection coefficient at outer tube surface h = 11 W/m^2-K
Now
area of the pipe per unit length A should be
= πr^2
= 7.069*10^-4 m^2/m
Now
convectional heat loss Q = Ah(Tw - Ta)
Q = 7.069 x 10^-4 x 11 x (150 - 20)
Q = 7.069 x 10^-4 x 11 x 130
= 1.01 W/m
hence, The heat loss per unit length of tube should be considered as the 1.01 W/m.
Learn more about heat here: https://brainly.com/question/15170783
An airplane on a runway has an initial speed of 15 meters/second. What is its final velocity when it travels 600 meters on the runway with a constant acceleration of 6 meters/second2?
Answer:
86.17m/s
Explanation:
Since the motion is of constant acceleration, we can use one of the equations of motion as follows;
v² = u² + 2as
Where;
v = final velocity of the airplane
u = initial velocity of the airplane
s = distance covered by the plane
a = acceleration
From the question,
u = 15m/s
s = 600m
a = 6m/s²
Substitute these values into the equation above as follows;
v² = 15² + 2(6)(600)
v² = 225 + 7200
v² = 7425
v = √7425
v = 86.17m/s
Therefore, the final velocity of the airplane is 86.17m/s
An object of mass 800g hangs on a spring. Calculate the force exerted by the object if acceleration due to gravity is 10/s2
Answer:
8 N
Explanation:
Using the equation F=ma (F: force/ m: mass in kg/ a: acceleration),
F = (800/1000)(10)
F = 8 N
a) Under what circumstances would a constant force result in increasing acceleration on a body? b) Under what circumstances would a constant force result in zero acceleration on a body?
Answer:
Remember the equation:
F = m*a
where F is force, m is mass and a is acceleration.
If we have F constant. and we want that increases, then we can have the case where m decreases.
The mass can decrease in cases like a rocket, where as the fuel of the rocket consumes, the mass of the rocket decreases and the acceleration increases.
b) The cases where a constant force results in a constant acceleration of zero, is when the force is canceled, an example of this is the constant force of the gravity in all the objects. The objects that are in the ground are being affected by this force, but the gravitational force is canceled with the normal force of the ground. Then we have a constant force that does not cause any acceleration.
A ball is moving up an inclined plane, when it's velocity changes from 15m/s to 7m/s in 3 seconds . What is the acceleration.
Explanation:
[tex]a = \frac{change \: in \: velocity}{change \: in \: time} [/tex]
[tex]a = \frac{v - u}{t} [/tex]
[tex]v = 7 {ms}^{ - 1} [/tex]
[tex]u = 15 {ms}^{ - 1} [/tex]
[tex]t = 3 \: seconds[/tex]
[tex]a = \frac{7 - 15}{3} [/tex]
[tex]a = \frac{ - 8}{3} [/tex]
[tex]a = - 2.67 \: {ms}^{ - 2} [/tex]
Using the formula W = mg, how many milliliters of water with a density of 1g/mL are required to weigh 0.75 newtons and g = 9.81 m/s2? Round to the nearest tenth. (Note: The mass will be in kg in your answer, thus convert to g and then to mL.) Answer: mL
Given that,
Weight = 0.75 N
Acceleration due to gravity = 9.81 m/s²
Density of water = 1 g/ml
We need to calculate the volume of water
Using formula of weight
[tex]W=mg[/tex]
[tex]W=\rho Vg[/tex]
[tex]mg=\rho\times V\times g[/tex]
Where, V = volume
g = acceleration
m = mass
Put the value into the formula
[tex]0.076\times 9.8\times10^3=1\times V\times9.8[/tex]
[tex]V= \dfrac{0.076\times 9.8\times10^3}{1\times9.8}[/tex]
[tex]V=76\ ml[/tex]
Hence, The volume of water is 76 mL.
Answer:
76.5mL
Explanation:
w = mg
0.75N = m * 9.81
m = 0.75 / 9.81 = 0.0765 kg
The mass in grams is 0.0765 * 1000 = 76.5g
At a density of 1 g/mL,
76.5g ÷ 1g/ml = 76.5mL
Which of the following is an example of the Doppler effect? A water bug on the surface of a pond is producing small ripples in the water. The bug is not moving. A police car with its siren on is driving towards you, and you perceive the pitch of the siren to increase. Scientists measure the speed of light and find that it is constant. The gravitational pull of two objects on each other
Answer:
A police car with its siren on is driving towards you, and you perceive the pitch of the siren to increase.
Explanation:
In Physics, Doppler effect can be defined as the change in frequency of a wave with respect to an observer in motion and moving relative to the source of the wave.
Simply stated, Doppler effect is the change in wave frequency as a result of the relative motion existing between a wave source and its observer.
The term "Doppler effect" was named after an Austrian mathematician and physicist known as Christian Johann Doppler while studying the starlight in relation to the movement of stars.
The phenomenon of Doppler effects is generally applicable to both sound and light.
An example of the Doppler effect is a police car with its siren on is driving towards you, and you perceive the pitch of the siren to increase. This is so because when a sound object moves towards you, its sound waves frequency increases, thereby causing a higher pitch. However, if the sound object is moving away from the observer, it's sound waves frequency decreases and thus resulting in a lower pitch.
Other fields were the Doppler effects are applied are; astronomy, flow management, vibration measurement, radars, satellite communications etc.
Answer:
A police car with its siren on is driving towards you, and you perceive the pitch of the siren to increase.
Explanation:
This should be the correct answer.
NEED HELP!! WILL GIVE BRAINIEST!! When a driver presses the brake pedal, his car can stop with an acceleration of -5.4 meters per second squared. how far will the car travel while coming to a complete stop if its initial speed was 25 meters per second?
Answer:
d = 57.87 m
Explanation:
It is given that,
Acceleration of car due to applied brake is [tex]-5.4\ m/s^2[/tex]
Its initial speed was 25 m/s
Its final speed was 0 as its comes to rest.
Leat d is the distance covered by the car. It can be calculated using third equation of kinematics as :
[tex]v^2-u^2=2ad\\\\d=\dfrac{v^2-u^2}{2a}\\\\d=\dfrac{0^2-(25)^2}{2\times (-5.4)}\\\\d=57.87\ m[/tex]
So, the distance covered by the car is 57.87 m.
Answer:
d = 57.87 m
Explanation: have a nice day :)
Describe the method you would use to measure the cicrumfrence of a cylinder using a thread and a meter
rule(4mk)
0)
Answer:
I take the thread and wrap it round the cylinder until it goes round the circumference of the cylinder.
Explanation:
I take the thread and wrap it round the cylinder until it goes round the circumference of the cylinder.
I then cut the thread at this length and extend it out on a table or floor.
I then place the length of the thread along the meter rule and mark of the position of its tip and bottom along the meter rule.
The difference between the bottom and top points is the circumference of the cylinder.
If the thread is longer than the meter rule, i mark off where the the length of the meter rule stops along the thread and then place the meter rule at that point and then mark off where the length of the thread stops.
The circumference of the cylinder is then 1 meter plus the extra measurement along the meter rule.
It has been suggested, and not facetiously, that life might have originated on Mars and been carried to Earth when a meteor hit Mars and blasted pieces of rock (perhaps containing primitive life) free of the surface. Astronomers know that many Martian rocks have come to Earth this way. One objection to this idea is that microbes would have to undergo an enormous, lethal acceleration during the impact. Let us investigate how large such an acceleration might be. To escape Mars, rock fragments would have to reach its escape velocity of 5.0 km/s, and this would most likely happen over a distance of about 4.0m during the impact.
1) What would be the acceleration, in m/s, of such a rock fragment?
2) What would be the acceleration, in g's, of such a rock fragment?
3) How long would this acceleration last?
4) In tests, scientists have found that over 40% of Bacillius subtilis bacteria survived after an acceleration of 450000g. In light of your answer to part A, can we rule out the hypothesis that life might have been blasted from Mars to Earth?
Answer and Explanation:
Given that
v_f = 5 km/s = 5,000 m/s
d = 4 m
v_i = 0 m/s
The computation is shown below:
1. The acceleration in m/s is
Here we use the motio third equation which is
[tex]v_f^2 = v_i^2 + 2ad[/tex]
5000^2 = 0^2 + 2 (a) (4)
So
[tex]a = 3.125 \times 10^{6} m/s^2[/tex]
2. Now acceleration in g is
[tex]= \frac{3.125 \times 10^{6} m/s^2}{9.81}[/tex]
[tex]= 3.18 \times 10^{5}g[/tex]
3. The long of acceleration last is
[tex]t = \frac{v-u}{a}[/tex]
[tex]= \frac{5000 - 0}{3.125 \times 10^{6}}[/tex]
[tex]= 1.6 \times 10^{-3}s[/tex]
4.As we can see that
[tex]3.18 \times 10^{5}[/tex] is smaller than the [tex]4.5 \times 10^{5}g[/tex]
So, it should not be ruled out
Distinguish between homogeneous and heterogeneous mixture with 2 example each. Do not use the examples give above
whats the texture of copper wire
Copper is a reddish orange, soft metal with a bright metallic luster. It is an excellent conductor of heat and electricity. Copper surfaces when exposed to air gradually turns to a dull, brownish color.
Choose all correct statements: Refractive index of medium A with respect to medium B is 4/3 and refractive index of medium A with respect to medium C is 3/2. Medium B is denser than medium C When light travels from B to C light bends away from the normal Refractive index of C with respect to B is 8/9 Ratio of speed of light in medium B to the speed of light in medium A = 4/3
Answer:
Medium B is denser than medium C
When light travels from B to C light bends away from the normal .
Refractive index of C with respect to B is 8/9
Ratio of speed of light in medium B to the speed of light in medium A = 4/3
Explanation:
Refractive index of medium A / Refractive index of medium B = 4/3
Refractive index of medium A / Refractive index of medium C = 3/2
Dividing ,
Refractive index of medium B / Refractive index of medium C
= 3/2 x 3/4
= 9 / 8
Refractive index of medium B > Refractive index of medium C
Refractive index of C with respect to B is 8/9
speed of light in medium B / the speed of light in medium A
= refractive index of A / refractive index of B = 4 / 3
helppppp me it's urgent please
the last one is the light will move in some manner not determined by the information here.
Light bends away from the normal, because it's moving from higher to lower refractive index.
Same bend-direction as when it goes from water into air.
as the mass of the cart increases, the acceleration of the cart
Answer:
decrease
Explanation:
as the mass of the cart increase the acceleration of the cart decrease
Consider a satellite in a circular orbit around the earth. Why is it important to give a satellite a horizontal speed when placing it in orbit? What will happen if the horizontal speed is too small? What will happen if the speed is too large?
Answer:
In this case, the horizontal velocity of the rocket starts from the acceleration, so if its velocity drops to zero,
Explanation:
When a satellite is in orbit the most important force is the docking of gravity with the Earth
F = m a
where the acceleration is centripetal and F is the force of universal attraction
centripetal acceleration is
a = v² / r
F = m v² / r
In this case, the horizontal velocity of the rocket starts from the acceleration, so if its velocity drops to zero, the force also drops to serious and the satellite steels to Earth.
The speed of the satellite is provides the speed, by local for smaller speeds in satellite, it descends in its orbits and when the speed is amate you have the energy to stop an orb to go to a higher orbit.