A 2.00-m rod of negligible mass connects two very small objects at its ends. The mass of one object is 1.00 kg and the mass of t
1 answer:
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We know that acceleration is change in velocity by time taken for that change.
In this case velocity change is 3.7 m/s
Time taken for this change = 60 ms =
So acceleration of frog =
= 61.66 m/
So acceleration of frog is 61.66 m/
o it is evident that frog is capable of remarkable accelerations.
Answer:
51.2 mi/h
Explanation:
Total distance, d = 100 miles
First 60 miles with speed 55 mi/h
Next 40 miles with speed 75 mi/h
Time taken for first 60 miles, t1 = 60 / 55 = 1.09 h
Time taken for 40 miles, t2 = 40 / 75 = 0.533 h
Time spent to get stuck, t3 = 20 min = 0.33 h
Total time, t = t1 + t2 + t3 = 1.09 + 0.533 + 0.33 = 1.953 h
The average speed is defined as the ratio of total distance traveled to the total time taken.
Average speed =
Thus, the average speed of the journey is 51.2 mi/h.
Answer:
a) the values of the angle α is 45.5°
b) the required magnitude of the vertical force, F is 41 lb
Explanation:
Applying the free equilibrium equation along x-direction
from the diagram
we say
∑Fₓ = 0
Pcosα - 425cos30° = 0
525cosα - 368.06 = 0
cosα = 368.06/525
cosα = 0.701
α = cos⁻¹ (0.701)
α = 45.5°
Also Applying the force equation of motion along y-direction
∑Fₓ = ma
Psinα + F + 425sin30° - 600 = (600/32.2)(1.5)
525sin45.5° + F + 212.5 - 600 = 27.95
374.46 + F + 212.5 - 600 = 27.95
F - 13.04 = 27.95
F = 27.95 + 13.04
F = 40.99 ≈ 41 lb
