True or False: There is the same amount of water on earth today as there was when earth was formed.

Using the periodic table, choose the more reactive nonmetal.<br> Si or N

tatiyna

Answer:

N

Explanation:

Nitrogen

3 0

3 years ago

Suppose a cup of coï¬ee is at 100 degrees Celsius at time t = 0, it is at 70 degrees at t = 10 minutes, and it is at 50 degrees

Alexandra [31]

Answer:

T ambient = 10 degrees

Explanation:

Using Newton's Law of Cooling:

T(t) = Tamb + (Ti - Tamb)*e^(-kt) ..... Eq 1

Ti = 100

We have two points to evaluate the above equation as follows:

T = 70 @ t = 10 using Eq 1

70 = Tamb + (100 - Tamb)*e^(-10k) ... Eq 2

T = 50 @ t = 20 using Eq 1

50 = Tamb + (100 - Tamb)*e^(-20k) ... Eq 3

Solving the above Eq 2 and Eq 3 simultaneously:

Using Eq 2:

(70 - Tamb) / (100 - Tamb) = e^(-10k)

Squaring both sides we get:

((70 - Tamb) / (100 - Tamb))^2 = e^(-20k) .... Eq 4

Substitute Eq 4 into Eq 3

50 = Tamb + (100 - Tamb)*((70 - Tamb) / (100 - Tamb))^2

After simplification:

50 = (Tamb (100-Tamb) + (70-Tamb)^2) / (100 - Tamb)

5000 - 50*Tamb = 4900 - 40*Tamb

Tamb = 100 / 10 = 10 degrees

6 0

4 years ago

A 0.9 µF capacitor is charged to a potential difference of 10.0 V. The wires connecting the capacitor to the battery are then di

jeyben [28]

Answer:

3.6μF

Explanation:

The charge on the capacitor is defined by the formula

q = CV

because the charge will be conserved

q₁ = C₁V₂

q₂ = C₂V₂ where C₂ V₂ represent the charge on the newly connected capacitor and the voltage drop across the two capacitor will be the same

q = q₁ + q₂ = C₁V₂ + C₂V₂

CV = CV₂ + C₂V₂

CV - CV₂ = C₂V₂

C ( V - V₂) = C₂V₂

C ( V/ V₂ - V₂ /V₂) = C₂

C₂ = 0.9 ( 10 /2) - 1) = 0.9( 5 - 1) = 3.6μF

7 0

3 years ago

A cannon is fired from the edge of a cliff, which is 60m above the sea. The cannonball's initial velocity is 88.3m/s and it is f

wel

Answer:

a. 11.29 s b. 94.72 m/s at -39.8° c. 821.57 m

Explanation:

a. Using y - y₀ = ut - 1/2gt² where u = vertical component of velocity = v₀sinθ where v₀ = 88.3 m/s and θ = 34.5°, y₀ = + 60 m and y = water surface = 0 m, g = 9.8 m/s² and t = time it takes the cannon to reach the water surface.

So y - y₀ = ut - 1/2gt²

y - y₀ = (v₀sinθ)t - 1/2gt²

substituting the values of the variables into the equation, we have

0 - 60 = (88.3 m/s × sin34.5°)t - 1/2 × 9.8 m/s²× t²

- 60 = 50t - 4.9t²

So, 4.9t² - 50t - 60 = 0

Using the quadratic formula to find t,

$t = \frac{-(-50) +/- \sqrt{(-50)^{2} - 4 X 4.9 X -60} }{2 X 4.9} \\t = \frac{50 +/- \sqrt{2500 + 1176} }{9.8} \\t = \frac{50 +/- \sqrt{3676} }{9.8} \\t = \frac{50 +/- 60.63 }{9.8} \\t = \frac{50 + 60.63 }{9.8} or t = \frac{50 - 60.63 }{9.8} \\t = \frac{110.63 }{9.8} or t = \frac{-10.63 }{9.8} \\t = 11.29 sor -1.085$

Since t cannot be negative, t = 11.29 s

b. We first need to find the impact vertical velocity component. Using

v = u - gt where u = initial vertical velocity component = v₀sinθ and t = 11.29 s and g = 9.8 m/s². So,

v = v₀sinθ - gt

= 88.3 m/s × sin34.5° - 9.8 m/s² × 11.29 s

= 50.01 m/s - 110.64 m/s

= -60.63 m/s

Since the horizontal velocity is constant u' = v₀cosθ = 88.3 m/s × cos34.5° = 72.77 m/s.

The impact velocity is thus the resultant of the horizontal velocity and final impact velocity. So, V = √(v² + u'²)

= √((-60.63 m/s)² + (72.77 m/s)²)

= √((3676 m²/s² + 5295.48 m²/s²)

= √(8971.48 m²/s²

= 94.72 m/s

The angle θ = tan⁻¹(v/u') = tan⁻¹(-60.63 m/s ÷ 72.77 m/s) = tan⁻¹(-0.8332) = -39.8°

So the impact velocity is 94.72 m/s at -39.8°

c. The horizontal distance out from the base of the cliff that the ball strikes the water is the range, R = u't = 72.77 m/s × 11.29 s = 821.57 m

5 0

3 years ago

A sound wave, with speed of 340 m/s has a frequency of 17,000 Hz. What

Fiesta28 [93]

Wavelength = speed / frequency = 340 / 17000 = 0.02 m

6 0

3 years ago