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Setler [38]
4 years ago
12

How much work will a machine with a power rating of 1.1 x 10^3 watts do in 2.0 hours?

Physics
2 answers:
eduard4 years ago
8 0

The answer to this question is 7.92 * 10^6 J

Hatshy [7]4 years ago
4 0

Power Rating = \(1.1 \times 10^3\) W

Time for which the machine is operating = 2 hours

Time in seconds  = 2 × 60 × 60

Time = 7200 seconds

Now, we need to use the formula for work done.

Work done = Power × time

Work done = \( 1.1 \times 10^3 \times 7200\)

Work done = \(7.92 \times 10^6\) Joules

Hence, work done by the machine in 2 hours is \(7.92 \times 10^6\) J.

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You are part of a design team assigned the task of making an electronic oscillator that will be the timing mechanism of a micro-
Snowcat [4.5K]

Solution :

We assume that there is a ring having a charge +Q and radius r. Electric field due to the ring at a point P on the axis is given by :

E_P=\int dE \cos

E_P=\int \frac{KdQ}{(\sqrt{r^2+x^2})^2}\times \frac{x}{\sqrt{r^2+x^2}}

\vec{E_P}=\frac{Kx}{r^2+x^2} \int dQ

\vec{E_P}=\frac{KxQ}{(r^2+x^2)^{3/2}} \hat{i}

If we put an electron on point P, then force on point e is :

\vec{F}=-|e|\vec{E_P}

F= \frac{-eKQx}{(r^2+x^2)^{3/2}}= \frac{-eKQx}{r^3[1+\frac{x^2}{r^2}]^{3/2}}

If r >> x , then    $\frac{x^2}{r^2} \approx 0$

Then,  $\frac{-eKQ}{r^3}x$

$ma =\frac{-eKQ}{r^3}x$

$a =\frac{-eKQ}{mr^3}x$

Compare, a = -ω²x

We get,

$\omega^2 = \frac{eKQ}{R^3m}$

$\omega = \sqrt{\frac{eKQ}{r^3m}}$

$2 \pi f = \sqrt{\frac{eKQ}{r^3m}}$

$f = \frac{1}{2 \pi}\sqrt{\frac{eKQ}{mr^3}}$

6 0
3 years ago
How do i find the number of leptons in an atom??
Serjik [45]

Answer:

In particle physics, a lepton is an elementary particle of half-integer spin (spin 1⁄2) that does not undergo strong interactions.[1] Two main classes of leptons exist: charged leptons (also known as the electron-like leptons or muons), and neutral leptons (better known as neutrinos). Charged leptons can combine with other particles to form various composite particles such as atoms and positronium, while neutrinos rarely interact with anything, and are consequently rarely observed. The best known of all leptons is the electron.

8 0
3 years ago
At what time of day would you be most likely to find that the air over water is significantly warmer than the air over land near
Brut [27]

This would happen later at night or early in the morning.

The reason being land becomes warm and cold quicker than the water because of the heat capacity. So during the day water warms up because of sunlight but at night the land becomes a lot cooler as compared to the water which is still war. So the air over water is significantly warmer than the air over land.

4 0
4 years ago
Read 2 more answers
If the architectural plans show the rough opening of a window to be 3'-3" x 4'-9" , the height of the opening should actually me
tatyana61 [14]

Answer:

height of the opening actually measure is 4'-9"

Explanation:

given data

window size = 3'-3" x 4'-9"

solution

height of the opening should actually measure will be 4'-9" in 3'-3" x 4'-9"

because according to architectural plan height can not be more than the opening size of window

and we can't take smaller height also

so fit in opening window we should take same height of  height of opening window and that is here 4'-9"

so here height of the opening actually measure is 4'-9"

5 0
3 years ago
In a 49 s interval, 595 hailstones strike a glass window of an area of 0.954 m at an angle of 25° to the window surface. Each ha
eduard

Average  force on the window: 0.32 N

Explanation:

The average force exerted on the window is given by Newton's second law

F=\frac{\Delta p}{\Delta t}

where

\Delta p is the net change in momentum of the hailstones in a time interval of \Delta t

In order to find the change in momentum, we have to consider only the component of the hailstone's momentum perpendicular to the window, therefore:

p_i =m u sin \theta is the initial momentum of one hailstone, with

m = 7 g = 0.007 kg is the mass

u=4.5 m/s is the initial speed

\theta=25^{\circ} is the angle with the window

The final momentum is

p_f = mv sin \theta

where

v = 4.5 m/s is the final speed (the  collision is elastic so the speed is equal, while the direction changes)

\theta=-25^{\circ} (after the rebound, the direction has changed)

So the change in momentum of 1 hailstone is

\Delta p = mv sin(-25^{\circ})-mu sin(25^{\circ})=-2mu sin(25^{\circ})=-0.0266 kg m/s

We are interested only in the magnitude, so

\Delta p = 0.0266 kg m/s

There are 595 hailstones hitting the window in 49 s, so the total change in momentum is

\Delta p = 595\cdot 0.0266 = 15.8 kg m/s

And therefore, the average force on the window is

F=\frac{\Delta p}{\Delta t}=\frac{15.8}{49}=0.32 N

Learn more about  force:

brainly.com/question/8459017

brainly.com/question/11292757

brainly.com/question/12978926

#LearnwithBrainly

3 0
3 years ago
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