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3 years ago

What causes a satellite to move tangentially to a circular path?

Physics

2 answers:

yawa3891 [41]3 years ago

5 0

Explanation : We know that two forces acting on the satellite.

The speed of the satellite and the force of the gravity on the satellite then the result , the satellite moves around the earth in a circular motion.

A satellite to move tangentially to a circular path, while revolving around the circular path because the centripetal force is acting on it.

So, we can say gravity is responsible for the motion of the satellite in a circular path.

mezya [45]3 years ago

3 0

Answer: tangential velocity

The tangential velocity is the velocity the body presents at a given moment of time when describing a circular movement, taking into account its direction, as well as the radius by which it is traveling in a particular fraction of its trajectory.

To calculate tangential velocity, it is possible to take the <u>angular velocity</u> as a reference, nevertheless it is necessary to understand that it can be <u>constant</u>, while the tangential velocity can vary at each step, given the alterations of the path.

This is possible because movement of a satellite around the planet Earth is a good example of uniform circular motion, which is characterized by the movement of a body describing a circumference of a given radius with constant speed.

Therefore in this movement the velocity has a constant magnitude, but its direction varies continuously.

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Un bloque de 2.5kg de masa es empujado 2.2m a lo largo de una mesa horizontal sin fricción por una fuerza constante de 16.0 N di

Tanzania [10]

Answer:

Explanation:

(a) The applied force has two components Fx and Fy. The Fx component is the only one that does work

$W_{x}=F_{x}x=(16N)cos(25)(2.2m)=31.9J$

(b) There in no net force in the vertical component

$F_{N}-F_{y}-F_{g}=0\\F_{N}=F_{y}+F_{g}=(16N)sin(25)+(2.5kg)(9.8\frac{m}{s^{2}})=31.26N$

(c)

$F_{g}=Mg=(2.5kg)(9.8\frac{m}{s^{2}})=24.5N$

(d)

$F_{T}=F_{x}=(16N)cos(25)=14.5N$

I attached an scheme of the force diagram

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3 years ago

A skier has an acceleration of 2.5 m/s2. How long does it take her to come to a complete stop from a speed of 18 m/s?

aleksandr82 [10.1K]

(Assuming acceleration is -2.5m/s2) 7.2 seconds.

4 0

3 years ago

Read 2 more answers

The conservation of momentum is most closely related to

BartSMP [9]

Answer:

<h2>B) Newton's 2nd law</h2>

Explanation:

<h2>From; force= mass × acceleration </h2><h2> f= m×a </h2><h2>where a(acceleration)= velocity/time</h2><h3> force = mv/t</h3><h3>But momentum(p) = Mass × velocity </h3><h2>hence force =p/t </h2><h3>that is Momentum = force × time ( Newton's 2nd law)</h3>

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1 year ago

How many coulombs of charge do 50 * 10^31 electrons possess

Angelina_Jolie [31]

Quantity of Charge , Q = ne
Where n = number of electrons
e = charge on one electron = -1.6 * 10 ^-19 C.
n = 50 * 10^31 electrons

Q = (50 * 10^31)*( -1.6 * 10 ^-19 ) = -8 * 10^13 C.

Note that the minus sign indicates that the charge is a negative charge.

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3 years ago

Long Jump: inital center of mass height of 1.08 m, final center of mass height of 0.42 m, projection velocity of 8.7 m/s, projec

sammy [17]

Answer:

1) The maximum jump height is reached at A. $0.337s$

2) The maximum center of mass height off of the ground is B. $1.64m$

3) The time of flight is C. $0.834s$

4) The distance of jump is B. $7.49m$

Explanation:

First of all we need to decompose velocity in its rectangular components, so

$v_{xi}=8.7m/s(cos 22.3\°)=8.05m/s= constant\\v_{yi}=8.7m/s(sin 22.3\°)=3.3m/s$

1) We use, $v_{fy}=v_{iy}-gt$ , as we clear it for $t$ and using the fact that $v_{fy}=0$ at max height, we obtain $t=\frac{v_{iy}}{g} =\frac{3.3m/s}{9,8m/s^{2}} =0.337s$

2) We can use the formula $y_{max}=y_{i}+v_{iy}t-\frac{gt^{2}}{2}$ for $t=0.337s$ , so

$y_{max}=1.08m+(3.3m/s)(0.337s)-\frac{(9.8m/s^{2})(0.337)^{2}}{2}=1.64m$

3) We can use the formula $y_{f}=y_{i}+v_{iy}t-\frac{gt^{2}}{2}$ , to find total time of fligth, so $0.42=1.08+3.3t-\frac{(9.8)t^{2}}{2}\\0=-4.9t^{2}+3.3t+0.66$ , as it is a second-grade polynomial, we find that its positive root is $t=0.834s$

4) Finally, we use $x=v_{x}t=8.05m/s(0.834s)=6.71m$ , as it has an additional displacement of $0.77m$ due the leg extension we obtain,

$x=6.71m+0.77m=7.48m$ , aprox $x=7.49m$

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4 years ago