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3 years ago

10

Let's examine over the next few questions an NFL kick as described in the 3rd Law video. The announcer claimed that the ball is

struck with "almost a ton of force." First of all, phrases like "almost a ton" drive physics people crazy. Give me a number! So let's guess-timate 1900 lbs of force. Convert that to Newtons. (you can always just Google this one)

1 answer:

hichkok12 [17]3 years ago

4 0

Answer:

8451.62109367671 Newtons

Explanation:

$1\ kg=1\times 9.8066500286389=9.8066500286389\ N$

$1 \lb=2.2046226218488\ kg$

It is known that

$1\ lbs=\dfrac{9.8066500286389}{2.2046226218488}=4.4482216282508\ N$

So,

$1900\ lbs=1900\times 4.4482216282508\\\Rightarrow 1900\ lbs=8451.6210936765\ N$

The force in Newtons is 8451.6210936765 Newtons

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A Person whose weight is 5.20 x 10^2 N is being pulledup

Gnoma [55]

Answer:

Explanation:

Given

Weight of Person $W=5.20\times 10^{2} N$

Cave is $h=35.1 m$ deep

Breaking stress $T=569 N$

Net Force on Person

$F_{net}=569-520=49 N$

$a_{net}=\frac{F_{net}}{\frac{W}{g}}$

$a_{net}=\frac{49}{\frac{520}{9.8}}$

$a_{net}=0.923 m/s^2$

The shortest time such that the person can be taken out of cave

$h=ut+\frac{1}{2}at^2$

where

h=distance moved

t=time

a=acceleration

$35.1=0+\frac{1}{2}(0.923)(t)^2$

$t^2=76.05$

$t=\sqrt{76.05}$

$t=8.72\ s$

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3 years ago

You drop a steel ball bearing, with a radius of 2.40 mm, into a beaker of honey. Note that honey has a viscosity of 6.00 Pa/s an

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Answer:

The “terminal speed” of the ball bearing is 5.609 m/s

Explanation:

Radius of the steel ball R = 2.40 mm

Viscosity of honey η = 6.0 Pa/s

$\text { Viscosity has Density } \sigma=1360 \mathrm{kg} / \mathrm{m}^{3}$

$\text { Steel has a density } \rho=7800 \mathrm{kg} / \mathrm{m}^{3}$

$\left.\mathrm{g}=9.8 \mathrm{m} / \mathrm{s}^{2} \text { (g is referred to as the acceleration of gravity. Its value is } 9.8 \mathrm{m} / \mathrm{s}^{2} \text { on Earth }\right)$

While calculating the terminal speed in liquids where density is high the stokes law is used for viscous force and buoyant force is taken into consideration for effective weight of the object. So the expression for terminal speed (Vt)

$V_{t}=\frac{2 \mathrm{R}^{2}(\rho-\sigma) \mathrm{g}}{9 \eta}$

Substitute the given values to find "terminal speed"

$\mathrm{V}_{\mathrm{t}}=\frac{2 \times 0.0024^{2}(7800-1360) 9.8}{9 \times 6}$

$\mathrm{V}_{\mathrm{t}}=\frac{0.0048 \times 6440 \times 9.8}{54}$

$\mathrm{V}_{\mathrm{t}}=\frac{302.9376}{54}$

$\mathrm{V}_{\mathrm{t}}=5.609 \mathrm{m} / \mathrm{s}$

The “terminal speed” of the ball bearing is 5.609 m/s

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Arte-miy333 [17]

Answer:the horizontal velocity is 40

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