2. Because it can take one year to form just a few centimeters of new soil, 3 points

What is true of electrolytic cells but is not true of galvanic cells?

ira [324]

They are non-spontaneous

8 0

4 years ago

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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

Convert 15 joule into erg.

GuDViN [60]

Answer:

<h2><u>Joule</u><u>:</u></h2>

1 Joule of work is said to be done when a force of 1 Newton is applied to move/displace a body by 1 metre.

1 Joule= 1 Newton × 1 metre

1 Newton is the amount of force required to accelerate body of mass 1 kg by 1m/s²

So units of N is kgm/s²

So,

1 Joule

=1kgm/s² × m

=1kgm²/s²

1 erg is the amount of work done by a force of 1 dyne exerted for a distance of one centimetre.

1 Erg =1 Dyne × 1 cm

1 dyne is the force required to cause a mass of 1 gram to accelerate at a rate of 1cm/s².

1 Erg=1 gmcm/s² × cm

1 Erg=1 gmcm/s² × cm=1gmcm²/s²

this is what you need to convert 1gmcm²/s² to 1kgm²/s²

<h3><u>what you need to know for conversion</u></h3>

[1gm=0.001kg

1cm²

=1cm ×1cm

=0.01 m × 0.01 m

=0.0001m²

second remains constant

]

So,

1gmcm²/s²

=0.001kg×0.0001m²/s²

=0.001kg×0.0001m²/s² =0.0000001kgm²/s²

Hence,

<h3><u>1 Erg</u><u>=</u><u>0.0000001</u><u> </u><u>Joule</u></h3><h3><u>1</u><u> </u><u>Joule</u><u>=</u><u>1</u><u>0</u><u>,</u><u>0</u><u>0</u><u>0</u><u>,</u><u>0</u><u>0</u><u>0</u><u> </u><u>Erg</u></h3>

3 0

3 years ago

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If Superman really had x-ray vision at 0.12 nm wavelength and a 4.1 mm pupil diameter, at what maximum altitude could he disting

iVinArrow [24]

Answer:

Maximum altitude to see(L) = 1.47 × 10⁶ m (Approx)

Explanation:

Given:

wavelength (λ) = 0.12 nm = 0.12 × 10⁻⁹ m

Pupil Diameter (d) = 4.1 mm = 4 × 10⁻³ m

Separation distance (D) = 5.4 cm = 0.054 m

Find:

Maximum altitude to see(L)

Computation:

Resolving power = 1.22(λ / d)

D / L = 1.22(λ / d)

0.054 / L = 1.22 [(0.12 × 10⁻⁹) / (4 × 10⁻³ m)]

0.054 / L = 1.22 [0.03 × 10⁻⁶]

L = 0.054 / 1.22 [0.03 × 10⁻⁶]

L = 0.054 / [0.0366 × 10⁻⁶]

L = 1.47 × 10⁶

Maximum altitude to see(L) = 1.47 × 10⁶ m (Approx)

8 0

3 years ago

If the forces acting on a soccer ball are balanced , what should happen to the motion of the ball? A The ball will fall to the g

lukranit [14]

That’s what she said

5 0

3 years ago

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