An object is lifted upward at a constant speed. The gain in potential energy of the object is due to the

The convergence of two continental plates would produce

alexandr1967 [171]

That would be a subduction zone.

5 0

2 years ago

A basketball player throws a chall -1 kg up with an initial speed of his hand at shoulder height = 2.15 m Le gravitational poten

Talja [164]

Complete Question:

A basketball player tosses a basketball m=1kg straight up with an initial speed of v=7.5 m/s. He releases the ball at shoulder height h= 2.15m. Let gravitational potential energy be zero at ground level

a) Give the total mechanical energy of the ball E in terms of maximum height hn it reaches, the mass m, and the gravitational acceleration g.

b) What is the height, hn in meters?

Answer:

a) Energy = mghₙ

b) Height, hₙ = 5.02 m

Explanation:

a) Total energy in terms of maximum height

Let maximum height be hₙ

At maximum height, velocity, V=0

Total mechanical energy , E = mgh + 1/2 mV^2

Since V=0 at maximum height, the total energy in terms of maximum height becomes

Energy = mghₙ

b) Height, hₙ in meters

mghₙ = mgh + 1/2 mV^2

mghₙ = m(gh + 1/2 V^2)

Divide both sides by mg

hₙ = h + 0.5 (V^2)/g

h = 2.15m

g = 9.8 m/s^2

V = 7.5 m/s

hₙ = 2.15 + 0.5(7.5^2)/9.8

hₙ = 2.15 + 2.87

hₙ = 5.02 m

6 0

3 years ago

A fluid moves through a tube of length 1 meter and radius r=0.002±0.0002 meters under a pressure p=4⋅105±1750 pascals, at a rate

yaroslaw [1]

Answer:

The maximum error is $\Delta \eta = 2032.9$

Explanation:

From the question we are told that

The length is $l = 1\ m$

The radius is $r = 0.002 \pm 0.0002 \ m$

The pressure is $P = 4 *10^{5} \ \pm 1750$

The rate is $v = 0.5*10^{-9} \ m^3 /t$

The viscosity is $\eta = \frac{\pi}{8} * \frac{P * r^4}{v}$

The error in the viscosity is mathematically represented as

$\Delta \eta = | \frac{\delta \eta}{\delta P}| * \Delta P + |\frac{\delta \eta}{\delta r} |* \Delta r + |\frac{\delta \eta}{\delta v} |* \Delta v$

Where $\frac{\delta \eta }{\delta P} = \frac{\pi}{8} * \frac{r^4}{v}$

and $\frac{\delta \eta }{\delta r} = \frac{\pi}{8} * \frac{4* Pr^3}{v}$

and $\frac{\delta \eta }{\delta v} = - \frac{\pi}{8} * \frac{Pr^4}{v^2}$

So

$\Delta \eta = \frac{\pi}{8} [ |\frac{r^4}{v} | * \Delta P + | \frac{4 * P * r^3}{v} |* \Delta r + |-\frac{P* r^4}{v^2} |* \Delta v]$

substituting values

$\Delta \eta = \frac{\pi}{8} [ |\frac{(0.002)^4}{0.5*10^{-9}} | * 1750 + | \frac{4 * 4 *10^{5} * (0.002)^3}{0.5*10^{-9}} |* 0.0002 + |-\frac{ 4*10^{5}* (0.002)^4}{(0.5*10^{-9})^2} |* 0 ]$