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4 years ago

When is it easiest for a person to build up static electricity?

Physics

1 answer:

Elena-2011 [213]4 years ago

3 0

Idkk look it up on quizlet, they flash cards to tell you lots of stuff

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An underwater scuba diver sees the Sun at an apparent angle of 43.0° above the horizontal. What is the actual elevation angle of

Inessa05 [86]

Answer:

The actual elevation angle is 12.87 degrees

Explanation:

In the attachment you can clearly see the situation. The angle of elevation as seen for the scuba diver is shown in magenta, we conclude that $\theta_2=90-43=47$ .

Using Snell's Law we can write:

$n_1\sin(\theta_1)=n_2\sin(\theta_2)$

$\implies \sin(\theta_1)=\frac{n_2}{n_1}\sin(\theta_2)$ ,

Let's approximate the index of refraction of the air (medium 1 in the picture) to 1.

We thus have:

$\sin(\theta_1)=n_2\sin(\theta_2)=1.333\sin(47)$

$\implies\theta_1=\arcsin[n_2\sin(\theta_2)]=\arcsin[1.333\sin(47)]\approx 77.13$ . Calling $\alpha$ the actual angle of elevation, we get from the picture that $\alpha=90-77.13=12.97$

7 0

3 years ago

Zad. 2 W marszobiegu chłopiec przebiegł 900 m w ciągu 205 min, a następnie przeszedł 300m w tym danym czasie. Jaka była średnia

USPshnik [31]

Odpowiedź:

0,049 m / s

Wyjaśnienie:

Biorąc pod uwagę, że:

Dystans biegu = 900m

Czas trwania = 205 minut

Długość przejścia = 300 m

Zajęty czas = 205 minut

Średnia prędkość :

(Przebieg + pokonany dystans) / całkowity czas

Średnia prędkość :

(900 m +. 300 m) / 205 + 205

1200 m / 410 minut

Minuty do sekund

1200 / (410 * 60)

1200/24600

= 0,0487804

= 0,049 m / s

8 0

3 years ago

What information should be posted on or near a dishwasher?

Murljashka [212]

A.. in case of any problems that may occur you would know what company to call

7 0

3 years ago

Electrical systems are governed by Ohm’s law, which states that V = IR, where V is the voltage, I is the current, and R is the r

ella [17]

Answer:

$\frac{dR(t)}{dt}=0.06\Omega$

Explanation:

Since $R(t)=\frac{V}{I(t)}$ , we calculate the resistance rate by deriving this formula with respect to time:

$\frac{dR(t)}{dt}=\frac{d}{dt}(\frac{V}{I(t)})=V\frac{d}{dt}(\frac{1}{I(t)})$

Deriving what is left (remember that $(\frac{1}{f(x)})'=-\frac{1}{f(x)^2}f'(x)$ ):

$\frac{d}{dt}(\frac{1}{I(t)})=-\frac{1}{I(t)^2}\frac{dI(t)}{dt}$

So we have:

$\frac{dR(t)}{dt}=-\frac{V}{I(t)^2}\frac{dI(t)}{dt}$

Which for our values is (the rate of <em>I(t)</em> is decreasing so we put a negative sign):

$\frac{dR(t)}{dt}=-\frac{24V}{(56A)^2}(-8A/s)=0.06\Omega$

8 0

3 years ago

The height of a typical playground slide is about 1.8 m and it rises at an angle of 30 ∘ above the horizontal.

Salsk061 [2.6K]

Answer:

$5.94\ \text{m/s}$

$1.7$

$0.577$

Explanation:

g = Acceleration due to gravity = $9.81\ \text{m/s}^2$

$\theta$ = Angle of slope = $30^{\circ}$

v = Velocity of child at the bottom of the slide

$\mu_k$ = Coefficient of kinetic friction

$\mu_s$ = Coefficient of static friction

h = Height of slope = 1.8 m

The energy balance of the system is given by

$mgh=\dfrac{1}{2}mv^2\\\Rightarrow v=\sqrt{2gh}\\\Rightarrow v=\sqrt{2\times 9.81\times 1.8}\\\Rightarrow v=5.94\ \text{m/s}$

The speed of the child at the bottom of the slide is $5.94\ \text{m/s}$

Length of the slide is given by

$l=h\sin\theta\\\Rightarrow l=1.8\sin30^{\circ}\\\Rightarrow l=0.9\ \text{m}$

$v=\dfrac{1}{2}\times5.94\\\Rightarrow v=2.97\ \text{m/s}$

The force energy balance of the system is given by

$mgh=\dfrac{1}{2}mv^2+\mu_kmg\cos\theta l\\\Rightarrow \mu_k=\dfrac{gh-\dfrac{1}{2}v^2}{gl\cos\theta}\\\Rightarrow \mu_k=\dfrac{9.81\times 1.8-\dfrac{1}{2}\times 2.97^2}{9.81\times 0.9\cos30^{\circ}}\\\Rightarrow \mu_k=1.73$

The coefficient of kinetic friction is $1.7$ .

For static friction

$\mu_s\geq\tan30^{\circ}\\\Rightarrow \mu_s\geq0.577$

So, the minimum possible value for the coefficient of static friction is $0.577$ .

8 0

3 years ago