Will a box sitting on an incline of 54.4 degree and a coefficient of static fric

Question

Will a box sitting on an incline of 54.4 degree and a coefficient of static friction mu_s = 0.6 start to slide on its own? Yes No A box slides from rest down an incline of 54.4 degree. Assuming no friction, how fast is the box moving after sliding 8.3 m? a. 132.3 m/s b. 16.2 m/s c. 11, 5 m/s d. 15.6 m/s Now with mu_k = 0.45, how fast would the box in the previous problem be moving after sliding the same distance? a. 5.6 m/s b. 3.3 m/s c. 12.4 m/s d. 9.5 m/s What would the gravitational attraction between two suns (M_sun = 1.989 times 10^30 kg) be if they were separated by 1.496 times 10 m? G = 6.674 times 10^-11 N middot m^2/kg^2 a. 1.765 times 10^61 N b. 1.765 times 10^39 N c. 1.180 times 10^28 N d. 1.180 times 10^171 N How fast would a satellite need to be going to orbit Earth at a height of 15000.0 m? M_Earth = 5.972 times 10^24 kg, R_Earth 6.371 times 10^5 m a. 2.656 10^10 m/s leftarrow don't choose this answer, that's faster than light speed b. 1.630 times 10^5 m/s c. 7.900 times 10^3 m/s. d. Not possible Why are geosynchronous orbits for satellites advantageous? a. They are not b. Receivers and transmitters on the ground do not have to track the satellite c. They are the most energy-efficient orbit d. They are the easiest orbit to achieve.

Explanation / Answer

14)The box will not slide only if

F(friction) >= F(gravity)

u mg cos(theta) = mg sin(theta)

But frictional force is lesser than gravity, so the answer is YES.

15)a = g sin(theta)

a = 9.81 x sin54.4 = 7.98 m/s^2

v^2 = u^2 + 2 a S

v = sqrt (2 a S) = sqrt (2 x 7.98 x 8.3) = 11.5 m/s

Hence, (c)11.5 m/s

15)a = g sin(theta) - u g cos(theta)

a = 9.81 x sin54.4 - 0.45 x 9.81 x cos54.4 = 5.41 m/s^2

v = sqrt (2 a S) = sqrt (2 x 5.41 x 8.3) = 9.5 m/s

Hence, (d)9.5 m/s

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