The gravitational force between Mars and the Sun is
Explanation:
The magnitude of the gravitational force between two objects is given by the equation:
where
is the gravitational constant
m1, m2 are the masses of the two objects
r is the separation between them
In this problem, we have:
is the mass of the Sun
is the mass of Mars
is the average distance Mars-Sun
Substituting into the equation, we find the gravitational force:
So, the closest answer is
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<h2>Diagonal of circle </h2>
Explanation:
As the skateboarder wants to cross the play ground . The surface is rough .
As we know , the force of friction is non-conservative force . Thus work is required against this force .
We have formula:
work done = Force x distance (in one direction )
Te force applied cannot be changed , so he is to decrease the distance .
In case of circle , diameter is the minimum distance . Thus he is supposed to move along it .
Responder:
20πrads ^ -1; 24πrads ^ -1; 0,1 seg; 10 Hz
Explicación:
Dado lo siguiente:
Radio (r) del círculo = 120 cm
600 revoluciones por minuto en radianes por segundo
(600 / min) * (2π rad / 1 rev) * (1min / 60seg)
(1200πrad / 60sec) = 20π rad ^ -1
Velocidad angular (w) = 20πrads ^ -1
Velocidad lineal = radio (r) * velocidad angular (w)
Velocidad lineal = (120/100) * 20πrad
Velocidad lineal = 1.2 * 20πrads ^ -1 = 24πrads ^ -1
C.) Período (T):
T = 2π / w = 2π / 20π = 0.1 seg
D.) Frecuencia (f):
f = 1 / T = 1 / 0.1
1 / 0,1 = 10 Hz
Answer:
ω = 630.2663 = 630[rad/s]
Explanation:
Solution:
- We can tackle this question by simple direct proportion relation between angular speed for the disk to rotate a cycle that constitutes 20 holes. We will use direct relation with number of holes per cycle to compute the revolution per seconds i.e frequency of speed f.
1rev(20 hole) -> 20(cycle)/rev
2006.2(cycle) -> f ?
f = 2006.2/20 = 100.31rev at second
- The relation between angular frequency and angular speed is given by:
ω = 2πf
ω = 2*3.14*100.31
ω = 630.2663 = 630[rad/s]