What is diffrence between ammeter and voltmeter

Your in an escape room. Only its real and your oxygen runs out in 3 minutes unless you solve this problem, The problem :a ball i

alex41 [277]

The sum of the series allows to find the result for the total distance that the ball bounces is:

total distance = 59.52 in

A series is a set of things or numbers related by a specific operation.

They indicate that the ball falls from an initial height y₀ = 30 in. and it bounces 50% of the height and the process is repeated until it stops, see attached.

Let's build a table to observe the sequence.

drop height rebound

1 30 15

2 15 7.5

3 7.5 3.75

If we call the first term y₀

The first bounce can be found.

y₁ = $\frac{y_o}{2}$

The second bounces.

$y_2 = \frac{y_1}{2} \\y_2 = \frac{y_o}{4}$

The third bounce.

$y_3 = \frac{y_2}{2} \\y_3 = \frac{y_0}{ 8}$

By observing this table we can construct a series of the form

Total distance = $y_o \ ( 1 + \frac{1}{2} + \frac{1}{4}+ \frac{1}{8} + ... +\frac{1}{2n} )$

The sum of the serie has a result of

sum = 127/64 = 1,984

Let's calculate

distance total = 30 1,984

Distance total = 59.52 cm

In conclusion, using the sum of the series we can find the result for the total distance that the ball bounces is:

total distance = 59.52 in

Learn more here: brainly.com/question/8879163

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3 years ago

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A 2.4-m high 200-m2 house is maintained at 22Â°C by an air-conditioning system whose COP, is 3.2. It is estimated that the kitch

kotegsom [21]

Answer:

The amount that is "vented" out by "the fans" is <u>$0.50</u> for 10 hours.

Option: a

<u>Explanation</u>:

"Energy discharged by air in every hour" can be determined by,

$\mathrm{Q}=\mathrm{m}_{\mathrm{air}} \mathrm{C}_{\mathrm{p}} \Delta \mathrm{T}$

Q = heat energy (Joules, J)

m = mass of a substance (kg)

c = specific heat (units J/kg∙K)

$\mathrm{m}_{\mathrm{air}}=\rho \mathrm{v}$

$\text { Density of air } \rho=1.20 \mathrm{kg} / \mathrm{m}^{3}$

$\text { Density of air } \rho=1.20 \times 200 \times 2.4$

$\text { Density of air } \rho=576 \mathrm{kg}$

∆T = 10 hours

$\text { Specific Heat Capacities of Air. The nominal values used for air at } 300 \mathrm{K} \text { are } \mathrm{C_P}=1.00 \mathrm{kJ} / \mathrm{kg} . \mathrm{K}$

Q = 576 × 1.00 × 10

Q = 5760 kJ/hours

W = 1.6 kwh

We know that, “Coefficient of performance” (COP)

$\mathrm{Cop}=\frac{Q}{w}$

$\mathrm{W}=\frac{Q}{\mathrm{cop}}$

Given that, COP = 3.2

$\mathrm{W}=\frac{1.6}{3.2}$

W = 0.5 kwh

The unit cost of electricity is $0.10/kWh

Unit electricity cost for 10 hours = 0.5 × 10 × 0.1$

Unit electricity cost for 10 hours = $0.5

The amount that is "vented out" by "the fans" is $0.50 for 10 hours.

6 0

3 years ago

Water drips from the nozzle of a shower onto the floor 200 cm below. The drops fall at regular (equal) intervals of time, the fi

Delicious77 [7]

Answer:

(A) 88.92 cm

(B) 22.22 cm

Explanation:

distance (s) = 200 cm = 0.2 m

initial velocity (v) = 0 m/s

acceleration due to gravity (g) = 9.8 m/s^{2}

lets first find the time (T) it takes for the first drop to strike the floor

from s = ut + $0.5at^{2}$

200 = 0 + $0.5 x 9.8 x T^{2}$

200 = $4.9 x T^{2}$

200 / 4.9 = $T^{2}$

T = 6.4

(A) When the first drop strikes the floor, how far below the nozzle is the second drop.

we can find how far the second drop was when the first drop hits the ground from the formula s = ut + $0.5at^{2}$

where

s = distance
u = initial velocity = 0
t = time, since the drops fall at regular (equal) intervals of time, the first drop striking the floor at the instant the fourth drop begins to fall there wold be 3 time intervals and this can be seen illustrated in the attached diagram. therefore the time of the second drop = 2/3 of the time it takes the first drop to strike the ground ( $\frac{2}{3}.T$ )
a = acceleration due to gravity = 9.8 m/s^{2}

substituting all required values we have

s = 0 + ( $0.5 x 9.8 x (\frac{2}{3}T)^{2}$ )

s = 0 + ( $0.5 x 9.8 x (\frac{2}{3} x 6.4)^{2}$ )

s = 88.92 cm

(B) When the first drop strikes the floor, how far below the nozzle is the third drop.

we can find how far the third drop was when the first drop hits the ground from the formula s = ut + $0.5at^{2}$

where

s = distance
u = initial velocity = 0
t = time, since the drops fall at regular (equal) intervals of time, the first drop striking the floor at the instant the fourth drop begins to fall there wold be 3 time intervals and this can be seen illustrated in the attached diagram. therefore the time of the third drop = 1/3 of the time it takes the first drop to strike the ground ( $\frac{2}{3}.T$ )
a = acceleration due to gravity = 9.8 m/s^{2}

substituting all required values we have

s = 0 + ( $0.5 x 9.8 x (\frac{1}{3}T)^{2}$ )