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konstantin123 [22]

3 years ago

7

About how long does a cumulus cloud last?

1 answer:

Tcecarenko [31]3 years ago

8 0

On what i know, is that they last about 5 to 45 minutes.

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Josh and Jake are both helping to build a brick wall which is 6 meters in height. They each lay 250 bricks, but Josh finishes th

Molodets [167]

<span>Each laid 250 bricks but while Jake was still working, Josh was lounging in the shade. Josh has more power but that power was only on for 3 hours out of 4.5. Obviously Josh could get more done is less time as long as he keeps working. Jake will get the hang of it soon.</span>

5 0

3 years ago

What is the mass of a box if it accelerated 5 m/s from a force of 12 N?

LuckyWell [14K]

Answer:

60

Explanation:

You multiply the acceleration and the force and you get your answer

4 0

2 years ago

mass is 5 lb attached to a rope wound around a pulley. The radius of the pulley is 3 in. If the mass falls at a constant velocit

suter [353]

Answer:

The power transmitted to the pulley is 0.0455 hp.

Explanation:

Given;

mass attached to the rope, m = 5 lb

radius of the pulley, r = 3 in

constant rate of fall of the mass, v = 5 ft/s

acceleration due to gravity, g = 32.2 ft/s²

1 lbf = 32.2 lb.ft/s²

The power transmitted to the pulley is calculated as;

P = Fv

P = (mg)v

$P = 5 \ lb \ \times \ 32.2 \ \frac{ft}{s^2} \ \times \ 5 \ \frac{ft}{s} \ \times \ \frac{1 \ lbf}{32.2 \ lb.ft/s^2} \ \ = 25 \ \frac{ft.lbf}{s} \\\\$

in horse power, the power transmitted is calculated as;

$P = \frac{25 \ ft.lbf}{s} \ \times \ \frac{1 \ hp}{550 \ ft.lbf/s} \ \ = 0.0455 \ hp$

Therefore, the power transmitted to the pulley is 0.0455 hp.

3 0

3 years ago

A 4.4-µF capacitor is initially connected to a 5.1-V battery. Once the capacitor is fully charged the battery is removed and a 2

Grace [21]

Question is incomplete. Missing part:

Find the charge on the capacitor at the following times:

1) t = 0 mu S

2) t = 1 mu S

3) t = 50 mu S

1) $22.4 \mu C$

We start by calculating the initial charge on the capacitor. For this, we can use the following relationship:

$C=\frac{Q_0}{V_0}$

where

C is the capacitance

Q0 is the initial charge stored

V0 is the initial potential difference across the capacitor

When the capacitor is connected to the battery, we have:

$C=4.4\mu F = 4.4\cdot 10^{-6}F$

$V_0 = 5.1 V$

Solving for $Q_0$ ,

$Q_0 = CV_0 = (4.4\cdot 10^{-6})(5.1)=2.24 \cdot 10^{-5} C = 22.4 \mu C$

So, when the battery is disconnected, this is the charge on the capacitor at time t = 0.

2) $20.0\mu C$

To find the charge on the capacitor at any other time t, we use the equation:

$Q(t) = Q_0 e^{-\frac{t}{RC}}$

where

$Q_0 = 22.4 \mu C$

t is the time

$R=2.0 \Omega$ is the resistance

$C=4.4\mu F$ is the capacitance

Therefore, at time $t=1 \mu s$ , we have:

$Q(t) = (22.4) e^{-\frac{1}{(2.0)(4.4)}}=20.0 \mu C$

3) $0.08 \mu C$

As before, we use again the equation:

$Q(t) = Q_0 e^{-\frac{t}{RC}}$

However, here the time to consider is

$t=50 \mu C$

Substituting into the formula,

$Q(t) = (22.4) e^{-\frac{50.0}{(2.0)(4.4)}}=0.08 \mu C$

4 0

3 years ago

Is the distance a baseball travels in the air after being hit distance a baseball travels in the air after being hit a discrete

Nookie1986 [14]

Answer:

Continuous random variable

Explanation:

The distance that baseball travels after being hit is a random variable and it assume any real value defined on the sample space.

The distance is measurable and thus is continuous random variable because continuous variable cannot be counted but could be measured.

3 0

3 years ago