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3 years ago

14

5 x 10 to the power electron cross through the cross section of a wire per second. What is the magnitude of electric current pas

sing through it?

1 answer:

Rom4ik [11]3 years ago

8 0

. She replaces a straight electrical

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It's nighttime, and you've dropped your goggles into a 3.2-mm-deep swimming pool. If you hold a laser pointer 1.1 mm above the e

Savatey [412]

Answer: 5.30m

Explanation:

depth of pool = 3.2 m

i = 67.75°

Using snell's law, we have,

n₁ × sin(i) = n₂ × 2 × sin(r)

n₁ = 1, n₂ =1.33, r= 44.09°

Hence,

Distance of Google from edge if pool is:

2.2 + d×tan(r) = 2.2 + (3.2 × tan(44.09°) =5.30m

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3 years ago

What describes the movement of a fluid during convection?

damaskus [11]

Convection is the movement<span> of groups of molecules within </span>fluids<span> such as gases and liquids, including molten rock (rheid).</span>

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2 years ago

Which of the following is most useful to determine how much energy is being used by a circuit in a given amount of time?

user100 [1]

Answer:

The answer is A.

Explanation:

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2 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 $0.577$ .

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2 years ago

About 10-15% of all galaxies are which shape?

myrzilka [38]

The answer is; Irregular.

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3 years ago