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3 years ago

A runner begins from rest and accelerates at a rate of 0.22 meters per second' for 15

Physics

1 answer:

ziro4ka [17]3 years ago

7 0

Answer:

25 m

Explanation:

Given:

v₀ = 0 m/s

a = 0.22 m/s²

t = 15 s

Find: Δx

Δx = v₀ t + ½ at²

Δx = (0 m/s) (15 s) + ½ (0.22 m/s²) (15 s)²

Δx = 24.75 m

Rounded to two significant figures, the runner moves a distance of 25 m.

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4 years ago

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In April 1974, Steve Prefontaine completed a 10.0 km race in a time of 27 min , 43.6 s . Suppose "Pre" was at the 7.43 km mark a

Novay_Z [31]

Answer:

Acceleration, a = 0.101 m/s²

Explanation:

Average speed = total distance / total time.

At the 7.43km mark, total distance = 7.43km or 7430m

Total time = 25 * 60 s = 1500s

Average speed = 7430m/1500s = 4.95m/s

He then covers (10 - 7.43)km = 2.57 km = 2570 m

in t = 27m43.6s - 25min = 2m43.6s = 163.6 s

Then he accelerates for 60 s, and maintains this velocity V, for the remaining (163.6 - 60)s = 103.6 s.

From V = u + at; V = 4.95m/s + a *60s

Distance covered while accelerating is

s = ut + ½at² = 4.95m/s * 60s + ½ a *(60s)² = 297m + a*1800s²

Distance covered while at constant velocity, v after accelerating is

D = velocity * time

Where v = 4.95m/s + a*60s

D = (4.95m/s + a*60s) * 103.6s = 512.82m + a*6216s²

Total distance covered after initial 7.43 km, S + D = 2570 m, so

2570 m = 297m + a*1800s² + 512.82m + a*6216s²

2570 = 809.82 + a*8016

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3 years ago

A block of mass 0.221 kg is placed on top of a light, vertical spring of force constant 5365 N/m and pushed downward so that the

Anvisha [2.4K]

Answer:

The maximum height above the point of release is 11.653 m.

Explanation:

Given that,

Mass of block = 0.221 kg

Spring constant k = 5365 N/m

Distance x = 0.097 m

We need to calculate the height

Using stored energy in spring

$U=\dfrac{1}{2}kx^2$ ...(I)

Using gravitational potential energy

$U' =mgh$ ....(II)

Using energy of conservation

$E_{i}=E_{f}$

$U_{i}+U'_{i}=U_{f}+U'_{f}$

$\dfrac{1}{2}kx^2+0=0+mgh$

$h=\dfrac{kx^2}{2mg}$

Where, k = spring constant

m = mass of the block

x = distance

g = acceleration due to gravity

Put the value in the equation

$h=\dfrac{5365\times(0.097)^2}{2\times0.221\times9.8}$

$h=11.653\ m$

Hence, The maximum height above the point of release is 11.653 m.

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3 years ago

A 1/10th scale model of an airplane is tested in a wind tunnel. The reynolds number of the model is the same as that of the full

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Answer:

of the velocity of a full size plane in the air

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4 years ago

How much mili ampere is 2 ampere

Rasek [7]

Answer:

2000 mili ampere

Explanation:

1 ampere is = to 1000 miliampere so 2 x 1000 is equal to 2000 miliampere

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3 years ago