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3 years ago

A water balloon is dropped off the top of a building and bursts when it hits

Physics

1 answer:

Dennis_Churaev [7]3 years ago

5 0

Answer:

The building is 176.4 meters high.

Explanation:

We use the formula for kinematics of an accelerated object (in this case acceleration of gravity):

Distance travelled = 1/2 g t^2

Distance travelled = 1/2 9.8 (6)^2 = 176.4 meters

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frictonal force due to the surface of irregularities

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3 years ago

A researcher measures the thickness of a layer of benzene (nn = 1.50) floating on water by shining monochromatic light onto the

baherus [9]

Answer:

The minimum thickness is $t= 8.75*10^{-8} m$

Explanation:

generally the equation for thin film interference is mathematically represented as

$2nt = (m + \frac{1}{2} ) \lambda$

Where t the thickness

m is any integer

n is the refractive index of the film

$\lambda$ is the wavelength of light

Since we are looking for the thickness we make t the subject of the formula

$t = \frac{(m+ \frac{1}{2} ) \lambda}{2n}$

m= 0 cause the thickness is minimum at m=0

Substituting values

$t = \frac{(0 +\frac{1}{2}) 8525*10^{-9} }{2 *1.5}$

$t= 8.75*10^{-8} m$

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3 years ago

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You're driving down the highway late one night at 20 m/s when a deer steps onto the road 39 m in front of you. Your reaction tim

miss Akunina [59]

Answer:

a) 10.8 m

b) 24.3 m/s

Explanation:

In order to get the total distance traveled since you see the deer till the car comes to an stop, we need to take into account that this distance will be composed of two parts.
The first one, is the distance traveled at a constant speed, before stepping on the brakes, which lasted the same time that the reaction time, i.e., 0.5 sec.
We can find this distance simply applying the definition of average velocity, as follows:

$\Delta x_{1} = v_{1o} * t_{react} = 20 m/s * 0.5 s = 10 m (1)$

The second part, is the distance traveled while decelerating at -11 m/s2, from 20 m/s to 0.
We can find this part using the following kinematic equation (assuming that the deceleration keeps the same all time):

$v_{1f} ^{2} - v_{1o} ^{2} = 2* a* \Delta x (2)$

where v₁f = 0, v₁₀ = 20 m/s, a = -11 m/s².
Solving for Δx, we get:

$\Delta x_{2} = \frac{-(20m/s)^{2}}{2*(-11m/s)} = 18.2 m (3)$

So, the total distance traveled was the sum of (1) and (3):
Δx = Δx₁ + Δx₂ = 10 m + 18.2 m = 28.2 m (4)
Since the initial distance between the car and the deer was 39 m, after travelling 28.2 m, the car was at 10.8 m from the deer when it came to a complete stop.

We need to find the maximum speed, taking into account, that in the same way that in a) we will have some distance traveled at a constant speed, and another distance traveled while decelerating.
The difference, in this case, is that the total distance must be the same initial distance between the car and the deer, 39 m.
⇒Δx = Δx₁ + Δx₂ = 39 m. (5)
Δx₁, is the distance traveled at a constant speed during the reaction time, so we can express it as follows:

$\Delta x_{1} = v_{omax} * t_{react} = 0.5* v_{omax} (6)$

Δx₂, is the distance traveled while decelerating, and can be obtained using (2):

$v_{omax} ^{2} = 2* a* \Delta x_{2} (7)$

Solving for Δx₂, we get:

$\Delta x_{2} = \frac{-v_{omax} ^{2} x}{2*a} = \frac{-v_{omax} ^{2}}{(-22m/s2)} (8)$

Replacing (6) and (8) in (5), we get a quadratic equation with v₀max as the unknown.
Taking the positive root in the quadratic formula, we get the following value for vomax:
v₀max = 24.3 m/s.

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3 years ago