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3 years ago

13

A small 25 kilogram canoe is floating downriver at a speed of 1 m/s. What is the canoe’s kinetic energy?

1 answer:

Helen [10]3 years ago

4 0

Kinetic energy = 1/2mv^2

=1/2(25)(1^2)
= 12.5J

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If velocity is 0.2m/s, what is the kinetic energy?

Natalija [7]

$\huge\boxed{♧ \: \mathfrak{ \underline{Answer} \: ♧}}$

we know,

$\boxed{kinetic \: \: energy = \frac{1}{2} m {v}^{2} }$

So,

$\longmapsto \dfrac{1}{2} \times 10 \times 0.2 \times 0.2$

$\longmapsto0.2 \: joules$

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3 years ago

A child standing on the edge of a freely spinning merry-go-round moves quickly to the center. Which one of the following stateme

Komok [63]

Answer:

D

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2 years ago

What does energy and Newton's Laws (all three) have to do with roller coasters?

Kipish [7]

Answer:

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2 years ago

Read 2 more answers

A thin plate 6 ft long and 3 ft wide is submerged and held stationary in a stream of water (T = 60°F) that has a velocity of 17

VashaNatasha [74]

A) In the case of the Boundary Thickness Layer we use the given formula,

$\delta = \frac{4.91x}{\sqrt{Re}}$

We know as well that,

Re = Número de Reynolds = $\frac{U*x}{\upsilon}$

Where,

U = velocity

$\upsilon$ = kinematic viscosity

For water, kinematic viscosity, $\upsilon = 1.21*10^{-5} ft^2 /s$

So, $500,000 = \frac{ 17x}{(1.21*10^{-5})}$

$x = 0.355 ft$

$d = \frac{4.91*0.355}{\sqrt {500000}}$

$d = 0.002465 ft = 0.029in$

B) For flat plate boundary layer. Given the Critical Reynolds Number.= 5*10^5 we know that is equal to Re above.

Thus, $x = 0.355 ft$

C. Wall shear stress,

$\tau = \mu*\sqrt{ U^3 / (2*\nu*x) }$

For water, dynamic viscosity, $\nu$ = 2.344*10^-5 lbf-s/ft^2

$\tau = 2.344*10^-5 \sqrt {17^3 / (2*1.21*10^{-5}*0.355)}$

$\tau = 0.5605 lbf/ft^2$

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3 years ago

What is the volume of water in 150ml of the 35% w/w of sucrose solution with a specific gravity of 1.115?

pentagon [3]

Answer:108.71 mL

Explanation:

Given

Volume of sample V=150 mL

concentration of sucrose solution 35 % w/w i.e. In 100 gm of sample 35 gm is sucrose

specific gravity =1.115

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Thus

$\rho _s=1.15\times 1 gm/mL=1.115\ gm/mL$

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mass of sucrose $m_s=0.35\times 167.25=58.53\ gm$

mass of Water $m_w=108.71 gm$

Volume of water $=108.71\times 1=108.781 mL$

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2 years ago