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2 years ago

WILL GIVE BRAINLIEST!!! NO LINKS PLZ!!!!

Physics

1 answer:

Nesterboy [21]2 years ago

3 0

Answer:

Explanation:

a) the acceleration of the puck on the rough ice.

a = μg = 0.550(9.81) = 5.3955 = 5.40 m/s²

(comes from μ = F/N = ma/mg = a/g)

b) the distance from the end boards the puck is when it comes to a stop.

v² = u² + 2as

0² = 12.0² + 2(-5.40)s

s = 13.3 ft

so distance from the boards is

15.7 - 13.3 = 2.4 m

by the way...that's some VERY rough ice...more like sand.

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The pressure difference, , across a partial blockage in an artery (called a stenosis) is approximated by the equation where is t

Strike441 [17]

The question is incomplete. The complete question is :

The pressure difference, Δp, ac $K_u$ ross a partial blockage in an artery (called a stenosis) is approximated by the equation :

$\Delta p=K_v\frac{\mu V}{D}+K_u\left(\frac{A_0}{A_1}-1\right)^2 \rho V^2$

Where V is the blood velocity, μ the blood viscosity {FT/L2}, ρ the blood density {M/L3}, D the artery diameter, $A_0$ the area of the unobstructed artery, and A1 the area of the stenosis. Determine the dimensions of the constants $K_v$ and $K_u$ . Would this equation be valid in any system of units?

Solution :

From the dimension homogeneity, we require :

$\Delta p=K_v\frac{\mu V}{D}+K_u\left(\frac{A_0}{A_1}-1\right)^2 \rho V^2$

Here, x means dimension of x. i.e.

$[ML^{-1}T^{-2}]=\frac{[K_v][ML^{-1}T^{-1}][LT^{-1}]}{[L]}+[K_u][1][ML^{-3}][L^2T^{-2}]$

$=[K_v][ML^{-1}T^{-2}]+[K_u][ML^{-1}T^{-2}]$

So, $[K_u]=[K_v]=[1 ]=$ dimensionless

So, $K_u$ and $K_v$ are dimensionless constants.

This equation will be working in any system of units. The constants $K_u$ and $K_v$ will be different for different system of units.

5 0

2 years ago

A basketball has a coefficient of restitution of 0.821 in collisions with the wood floor of a basketball court. The ball is drop

Tanya [424]

Answer: The height of its fourth bounce = 0.43m

Explanation:

The coefficient of restitution denoted by (e), is the ratio that shows the final velocity to initial relative velocity between two objects after collision

IT is given by the formula in terms of height as

Coefficient of Restitution, e = √(2gh))/√(2gH) = √(h/H)

Where

Coefficient of Restitution, e= 0.821

H = 2.07 m

At fourth bounce , we have that

Coefficient of Restitution, e⁴ =√(h₄/H)

Putting the given values and solving , we have,

e⁴ =√(h₄/H)

= 0.821⁴ = √(h₄/2.07)

(0.821⁴ )² =h₄/2.07

0.2064 x 2.07 = 0.427 = 0.43

At fourth bounce, h₄ height = 0.43m

7 0

3 years ago

You can see lightning before you can hear the thunder. Which is the best explanation for this?

liubo4ka [24]

Salutations!

<span>You can see lightning before you can hear the thunder. Which is the best explanation for this?

The best explanation for this statement is the transmission of light in air is faster than the transmission of sound in the air. Light is faster than sound, thus you can see the thunder then you can hear the sound.

Hope I helped :D</span>

8 0

2 years ago

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I WILL MARK BRAINLIEST IF CORRECT!!!

julsineya [31]

0.05*0.2=0.05*0.15+0.015*m2',

m2'=1/6 m/s,

m2'=0.17 m/s

N I C E - D A Y!

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2 years ago

A boat is traveling with a velocity of 18 meters/second relative to water and the river is flowing at a velocity of 2.5 meters/s