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Alekssandra [29.7K]
4 years ago
7

How far will a freely falling object fall from rest in six seconds?

Physics
2 answers:
bija089 [108]4 years ago
5 0
Tabulate, then find the proper equation, then solve.

vi=0
vf=?
t=6s
a=g=9.8m/s²

vf=vi+at
vf=gt
vf= (find this yourself)
navik [9.2K]4 years ago
4 0
Here are two ways to do it:

<u>Way #1:</u>
-- You know that gravity accelerates things that are falling freely, adding 9.8 m/s to their speed every second.

-- After 6 seconds, an object that fell from rest winds up falling at (9.8 x 6) = 58.8m/s.

-- During that time, its average speed was 1/2(0 + 58.8) = 29.4 m/s .

-- In 6 seconds, at an average speed of 29.4 m/s, it covers (29.4 x 6) = <em>176.4meters</em>.

<u>Way #2:</u>
This way only works if you remember the formula for the distance covered during uniform acceleration.

             D  =  1/2 (acceleration) (time)²

Acceleration = gravity = 9.8 m/s²
Time  =  6 seconds

           D  =  1/2 (9.8) (6)²  =  <em>176.4 meters</em>.
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A mug rests on an inclined surface, as shown in (Figure 1) , θ=17∘.
Anna [14]
Refer to the figure shown below.

g =  9.8 m/s², the acceleration due to gravity.
W = mg, the weight of the mug.
θ = 17°, the angle of the ramp.

Let μ = the coefficient of static friction.

The force acting down the ramp is
F = W sin θ = W sin(17°) = 0.2924W N
The normal reaction is
N = W cosθ = W cos(17°) = 0.9563W N
The resistive force due to friction is
R = μN = 0.9563μW N

For static equilibrium,
μN = F
0.9563μW =0.2924W
μ = 0.3058

The frictional force is F = μN = 0.2924W
The minimum value of μ required to prevent the mug from sliding satisfies
 the condition
R > F
0.9563μW > 0.2924W
μ > 002924/.9563 = 0.306

Answer:
The frictional force is 0.2924mg, where m = the mass of the mug.
The minimum coefficient of static friction is 0.306

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Norma-Jean [14]

Answer:

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