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liberstina [14]

3 years ago

13

Two satellites A and B of the same mass are orbiting Earth in concentric orbits. The distance of satellite B from Earth’s center

is twice that of satellite A. What is the ratio of the tangential speed of B to that of A?

1 answer:

riadik2000 [5.3K]3 years ago

5 0

Answer:

ratio of tangential velocity of satellite b and a will be 0.707

Explanation:

We have given distance of satellite B from satellite A is twice

So $r_b=2r_a$

Tangential speed of the satellite is given by

$v=\sqrt{\frac{GM}{r}}$ , G is gravitational constant. M is mass of satellite and r is distance from the earth

We have to find the ratio of tangential velocities of b and a

From the relation we can see that tangential velocity is inversely proportional to square root of distance from earth

So $\frac{v_b}{v_a}=\sqrt{\frac{r_a}{r_b}}$

$\frac{v_b}{v_a}=\sqrt{\frac{r_a}{2r_a}}$

$\frac{v_b}{v_a}=\sqrt{\frac{1}{2}}$

$\frac{v_b}{v_a}=0.707$

So ratio of tangential velocity of satellite b and a will be 0.707

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stira [4]

Answer:

the force P required for impending motion is 132.3 N

the largest value of "h" allowed if the cabinet is not to tip over is 0.8 m

Explanation:

Given that:

mass of the cabinet m = 45 kg

coefficient of static friction μ = 0.30

A free flow body diagram illustrating what the question represents is attached in the file below;

The given condition from the question let us realize that ; the casters are locked to prevent the tires from rotating.

Thus; considering the forces along the vertical axis ; we have :

$\sum f_y =0$

The upward force and the downward force is :

$N_A+N_B = mg$

where;

$\mathbf { N_A \ and \ N_B}$ are the normal contact force at center point A and B respectively .

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$N_A+N_B = 441$ ------- equation (1)

Considering the forces on the horizontal axis:

$\sum f_x = 0$

$F_A +F_B = P$

where ;

$\mathbf{ F_A \ and \ F_B }$ are the static friction at center point A and B respectively.

which can be written also as:

$\mu_s N_A + \mu_s N_B = P$

$\mu_s( N_A + N_B) = P$

replacing our value from equation (1)

P = 0.30 ( 441)

P = 132.3 N

Thus; the force P required for impending motion is 132.3 N

b) Since the horizontal distance between the casters A and B is 480 mm; Then half the distance = 480 mm/2 = 240 mm = 0.24 cm

the largest value of "h" allowed for the cabinet is not to tip over is calculated by determining the limiting condition of the unbalanced torque whose effect is canceled by the normal reaction at $N_A$ and it is shifted to $N_B$ :

Then:

$\sum M _B = 0$

$P*h = mg*0.24$

$h =\frac{45*9.8*0.24}{132.3}$

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