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Inga [223]
3 years ago
11

A car slid off an icy 10m bridge and landed 12m away from the bridge. How much time was the car in the air? (Hunt: Projectile)

Physics
2 answers:
Contact [7]3 years ago
8 0

You will use the height of the bridge from the ground.

Solution:

Formula to be used is y=Viy(t)+g(t^2)/2

Where:

Vi=initial velocity which is 0 m/s

 y=10 m

Gravitational acceleration or g =9.8m/s^2

T= time you need

Substitute all the given to the formula

10m=(0m/s)(t)+(9.8m/s^2)(t^2)/2

10mx2=9.8m/s^2(t^2)

Now isolate the variable you want to find which is T or time

10mx2/9.8m/s^2=t^2

20m/9.8m/s^2=t^2

Square root of 2.04= square root of t^2

T=1.43 secs

The answer is 1.43 seconds

STatiana [176]3 years ago
3 0
You just need to use the height of the bridge from the ground.

The formula to be used is y = Voy*t + g *(t^2) / 2, where the Voy is the intitial vertical velocity, which is zero.

Then, y = 10 m = g * (t^2) / 2 => t^2 = 2 * 10 m / g

=> t^2 = 2 * 10 m / 9.8 m/s^2 = 2.04 s^2

=> t = 1.43 s <------- answer


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How far can a person run in 13.22 minutes if he or she runs at an average speed of 12.33 km/hr?
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Answer:

2.72 km

Explanation:

(12.33 km)/ 1 hr * (1 hr)/ 60 min

0.2055 km/ min

distance=rate * time (assuming v is constant,

a=0)

(0.2055 km/ min)*(13.22 min)

2.72 km OR 2716.71 m

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Which of the following forces can change velocity?
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3 0
3 years ago
A person hums into the top of a well (tube open at only one end) and finds that standing waves are established at frequencies of
S_A_V [24]

Answer:

Explanation:

The resonance in a tube that is open at one end and closed at the other, we can find it because in the closed part you have a node and in the open part you have a belly, so for the fundamental frequency you have ¼ Lam, the different resonances are:

Fundamental              λ = 4L

3 harmonica               λ  = 4L / 3

5 Harmonica               λ = 4L / 5

General harmonica     λ = 4L / (2n-1)                n = 1, 2, 3

Let's apply this equation to our case

The speed of sound is given by

          v =  λ  f

          λ = v / f

Let's look for wavelengths

         λ₁ = 40.6 / 343 = 0.1184 m

         λ₂ = 67.7 / 343 = 0.1974 m

         λ₃ = 94.7 / 343 = 0.2761 m

Since the wavelengths are close we can assume that it corresponds to consecutive integers, where for the first one it corresponds to the integer n

          λ ₁ = 4L / (2n-1)

          λ₂ = 4L / (2 (n + 1) -1) = 4L / (2n +1)

          λ₃ = 4L / (2 (n + 2) -1) = 4L / (2n + 3)

Let's clear in the first and second equations

          2n-1 = 4L / λ₁

          2n +1 = 4L / λ₂

Let's solve the system of equations

         4L / λ₁ + 1 = 4L / λ₂ -1

         4L / λ₂ – 4L / λ₁ = 2

         2L (1 / λ₂ - 1 / λ₁) = 1

         1 / L = 2 (1 / λ₂ -1 / λ₁)

         1 / L = 2 (1 / 0.1974 - 1 / 0.1184)

         1 / L = 2 (5,066 - 8,446) = -6.76

         L = 0.1479 m

3 0
3 years ago
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