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Nataliya [291]
3 years ago
10

An object is dropped from a platform 100 feet high. Ignoring wind resistance, what will its speed be when it reaches the ground?

Physics
1 answer:
Scorpion4ik [409]3 years ago
8 0

Answer:

80 ft/s

Explanation:

Use III equation of motion

V^2 = U^2 + 2g h

Here, U = 0, g = 32 ft/s^2, h = 100 ft

V^2 = 0 + 2 × 32 ×100

V^2 = 6400

V = 80 ft/s

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bonufazy [111]

So there is a decimal after the last zero and it looks like this 5098000. You have to move the decimal point six back to get in between the five and the zero which looks like this 5.098000 

<span>Scientific notation is the way that scientists easily handle very large numbers or very small numbers. For example, instead of writing 0.0000000056, we write 5.6 x 10^<span>9</span>.</span>

Being that we moved the decimal six places back the answer is 5.098 x 10^6

3 0
4 years ago
Acceleration is the magnitude of average velocity.​
lina2011 [118]

Answer:

false

Explanation:

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3 years ago
Bicyclists in the Tour de France do enormous amounts of work during a race. For example, the average power per kilogram generate
tigry1 [53]

Explanation:

It is given that,

Average power per unit mass generated by Lance, \dfrac{P}{m}=6.5\ W/kg

P=6.5\times 75=487.5\ W

(a) Distance to cover race, d = 160\ km =160\times 10^3\ m

Average speed of the person, v = 11 m/s

If t is the time taken to cover the race.

t=\dfrac{d}{v}

t=\dfrac{160\times 10^3\ m}{11\ m/s}

t = 14545.46 s

Let W is the work done. The relation between the work done and the power is given by :

P=\dfrac{W}{t}

W=P\times t

W=487.5\times 14545.46

W = 7090911.75 J

(b) Since, 1\ J=2.389\times 10^{-4}\ calories

So, in 7090911.75 J, W=7090911.75 \times 2.389\times 10^{-4}

W = 1694.01 J

Hence, this is the required solution.

6 0
3 years ago
I need help. Can you help me? Thank you
SashulF [63]
What the ionquestion??
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3 years ago
A box that weighs 5.00×10^2 N is sliding down a ramp at a constant speed. The angle the ramp makes with the horizontal is 25°. W
maxonik [38]

Answer:

0.466 (3 sig. fig.)

Explanation:

Frictional force acting on the box = 5.00×10^2xsin25

Normal force acting on the box = 5.00×10^2xcos25

coefficient of friction = 0.466 (3 sig. fig.)

5 0
3 years ago
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