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3 years ago

I NEED HELP I AM SO CONFUSED, WILL GIVE BRAIN.

Physics

1 answer:

Phantasy [73]3 years ago

5 0

Answer:

$resolving \: \\ (\downarrow_{ + }) \: s = ut + \frac{1}{2} g {t}^{2} \\ u = 0 ,\: g = 9.81m. {s}^{ - 2} , \: t = 5s \\ s =h= \frac{1}{2} g {t}^{2}, = \frac{1}{2} (9.81){5}^{2}\\=4.9\times25 \\ h=122.6m$

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3 years ago

How are active and passive transport different?

sashaice [31]

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3 years ago

The driver of a pickup truck accelerates from rest to a speed of 37 mi/hr over a horizontal distance of 215 ft with constant acc

Alona [7]

Answer:

Maximum shearing force developed in each of the two pegs during acceleration is 1830 lbf

Explanation:

First we will find the acceleration of pickup truck.

As, the acceleration is uniform, therefore we can use Newton's third equation of motion:

2as = $V_{f}^{2}-V_{i}^{2}$

First convert speed into ft/sec

1 mile/hr = 1.47 ft/sec

therefore,

37 mile/hr = 37 x 1.47 ft/sec

37 mile/hr = 54.39 ft/sec

with initial speed 0 ft/sec (starting from rest), using in equation of motion:

a = [(54.39 ft/sec)² - (0 ft/sec)²]/2(215 ft)

a = 6.88 ft/sec²

Now, the total shear force will be given by Newton's second law of motion:

F = ma

F = (460 lbm +72 lbm)(6.88 ft/sec²)

F = 3660 lbf

Now for the max shear force in each of the two pegs we divide total fore by 2:

Force in each peg = F/2 = (3660 lbf)/2

<u>Force in each peg = 1830 lbf</u>

4 0

4 years ago

Jack and Nas decided to have a race from their houses to the park to know who runs the fastest. Jack whose house is 15 kilometer

Anettt [7]

Answer:

the Jack speed is greater than the Nas speed

v₁ = 2.08 m / s > v2 = 1.85 m / s

Explanation:

For this exercise we can find the average speed of each of the boys, the average speed is defined as the displacement in the time interval

v = x / t

Jack.

It tells us that it travels x = 15 km in a time of t = 2 h

let's reduce the magnitudes to the SI system

x = 15 km (1000 m / 1 km) = 15 10³ m

t = 2 h (3600 s / 1 h) = 7200 s

let's calculate

v₁ = 15 10³/7200

v₁ = 2.08 m / s

Nas travels a distance of x = 10 km in a time of t = 1.5 h

x = 10 km = 10 10³ m

t = 1.5 h (3600s / 1h) = 5400 s

let's calculate the speed

v2 = 10 10³/5400

v2 = 1.85 m / s

From these results we can see that the Jack speed is greater than the Nas speed

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3 years ago

What do lenses do?

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4 years ago

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