<span>Th find the average speed of a trip we need to dived the total distance by the total time.
Let's find the total distance d.
d = (300 mi/h)(2.00 h) + 750 miles
d = 600 miles + 750 miles
d = 1350 miles
The total distance is 1350 miles
Let's find the total time t.
t = 2.00 hours + (750 mi / 250 mi/h)
t = 2.00 hours + 3.00 hours
t = 5.00 hours
The total time of the trip is 5.00 hours.
We can find the average speed.
d / t = 1350 miles / 5.00 hours
d / t = 270 miles/ hour
The average speed of the trip is 270 mi/h
(Note that the direction does not matter when we find the average speed.)</span>
Answer:
Acceleration = 5.77 m/s²
Distance cover in 13 seconds = 487.56 meter
Explanation:
Given:
Final velocity of mobile device = 75 m/s
initial velocity of mobile device = 0 m/s
Time taken = 13 seconds
Find:
Acceleration
Distance cover in 13 seconds
Computation:
v = u + at
75 = 0 + (a)(13)
13a = 75
a = 5.77
Acceleration = 5.77 m/s²
s = ut + (1/2)(a)(t²)
s = (0)(t) + (1/2)(5.77)(13²)
Distance cover in 13 seconds = 487.56 meter
Hello!
Use the equation F = m · a (Newton's Second Law) to solve. Substitute in the given values:
F = 5 · 20
F = 100N
From what I can see it's D, I did this by simply examining the other answers and seeing that they are beneficial, so, from that information, this one must not be.
Answer:
F = 4.47 10⁻⁶ N
Explanation:
The expression they give for the strength of the tide is
F = 2 G m M a / r³
Where G has a value of 6.67 10⁻¹¹ N m² / kg² and M which is the mass of the Earth is worth 5.98 10²⁴ kg
They ask us to perform the calculation
F = 2 6.67 10⁻¹¹ 135 5.98 10²⁴ 13 / (6.79 10⁶)³
F = 4.47 10⁻⁶ N
This force is directed in the single line at the astronaut's mass centers and the space station
