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1 year ago

Dashun Mack: Attempt

Physics

1 answer:

Montano1993 [528]1 year ago

4 0

The time elapsed when the ball reaches the window is 1.55s

As per the question:

Final velocity, v = 11 m/s

Height, h = 29m

Acceleration, a = g = -9.8 m/s

From the third equation of motion:

v² = u² +2as

u² = v² - 2as

u² = (11)² - 2 × (-9.8) × (29)

u² = 689.4

u = √689.4

u = 26.25 m/s

Now, from the first equation of motion:

v = u + at

$t=\frac{v-u}{-g}$

$t = \frac{11 -26.25}{-9.8}$

t = 1.55 s

Therefore, the time elapsed when the ball reaches the window is 1.55s.

Read more about Equations of Motion:

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Bill and Ted are standing on a bridge 40 ft above a river. Bill drops a stone, while Ted decides to throw a stone downward at 10

USPshnik [31]

Answer:

Explanation:

In order to know how long after Bill released his rock should Ted throw his if they want the stones to hit the water simultanously, we need to calculate the time needed to hit the water to both rocks independent each other, and just take the difference.

For the rock dropped by Bill, as the only influence on it is gravity (accelerating it downwards with an acceleration equal to g), and v₀ =0, we can use the following kinematic equation:

$y = \frac{1}{2} * g * t^{2}$

where y = height = 40 ft.

As all the parameters are given in SI units, it is advisable to convert this value to m, as follows:

$y = 40 ft*\frac{0.3048m}{1 ft} = 12.2 m$

Now, we can solve for t, as follows:

$t = \sqrt{\frac{2*y}{g}} = \sqrt{\frac{2*12.2m}{9.8m/s2}} = 1.58 s$

For the rock thrown down at 10 m/s, the kinematic equation we just have used becomes:

$y = v0*t +\frac{1}{2} * g * t^{2}$

This leaves us a quadratic equation on t, as follows:

$t = \frac{-10m/s}{9.8m/s2} +/- \sqrt{10m/s^{2} -4*4.9m/s2*(-12.2m)} = -1.02 s +/- 1.88s$

Taking the positive root, we have:

t = -1.02 s + 1.88 s = 0.86 s

So, in order to get that both rocks hit the water at the same time, Ted will need to wait the difference between both times:

Δt = 0.86 s - 1.58s = -0.72 s

5 0

3 years ago

E) VAVC

gogolik [260]

Answer:

b) tan a = 3

Explanation:

Draw a free body diagram (see attached).

There are three forces acting on the insect. Weight downwards, normal force towards the center of the hemisphere, and friction tangent to the surface.

Sum of the forces in the radial direction:

∑F = ma

N − mg cos α = 0

N = mg cos α

Sum of the forces in the tangential direction:

∑F = ma

μN − mg sin α = 0

Substituting:

μ(mg cos α) − mg sin α = 0

μ cos α − sin α = 0

μ cos α = sin α

μ = tan α