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3 years ago

Do Earth’s oceans gain or lose water, considering evaporation and precipitation together? How much?

1 answer:

7 0

Much of the precipitation in large bodies of water occurs at the surface. The ocean loses about 37000 km cubed considering evaporation and precipitation.

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How do iquantum tunel through a wall and how long will it take me?

Anon25 [30]

Answer:

The phenomenon known as "tunneling" is one of the best-known predictions of quantum physics, because it so dramatically confounds our classical intuition for how objects ought to behave. If you create a narrow region of space that a particle would have to have a relatively high energy to enter, classical reasoning tells us that low-energy particles heading toward that region should reflect off the boundary with 100% probability. Instead, there is a tiny chance of finding those particles on the far side of the region, with no loss of energy. It's as if they simply evaded the "barrier" region by making a "tunnel" through it.

Explanation:

5 0

2 years ago

When does a rubber band, which has been shot at a wall, have the most potential energy?

Yuri [45]

D When it is stretched ready to shoot at the wall

4 0

2 years ago

Read 2 more answers

The electronics supply company where you work has two different resistors, R1 and R2, in its inventory, and you must measure the

elixir [45]

Answer:

R₁ = 23.77 ohms and R₂ = 7.92 ohms

Explanation:

When connected in series, the equivalence resistance, R(eq) is given as

R(eq) = (R₁ + R₂)

When connected in parallel, the equivalence resistance, R(eq) is given as

[1/R(eq)] = [(1/R₁) + (1/R₂)]

R(eq) = (R₁R₂)/(R₁ + R₂)

The parallel and series combination are connected to a battery of emf 39.0 V with negligible internal resistance. And the power supplied is measured.

But power supplied is given as

P = IV = (V/R) V = (V²/R)

When connected in series, the power supplied is given as

P = 48.0 W,

V = 39.0 V,

R = R(eq) = (R₁ + R₂)

48 = (39²/R)

R = (39²/48)

R = 31.6875 ohms

R = (R₁ + R₂) = 31.6875

(R₁ + R₂) = 31.6875 (eqn 1)

When connected in series, the power supplied is given as

P = 256.0 W,

V = 39.0 V,

R = R(eq) = (R₁R₂)/(R₁ + R₂)

256 = (39²/R)

R = (39²/256)

R = 5.9414 ohms

R = R(eq) = (R₁R₂)/(R₁ + R₂) = 5.9414

(R₁R₂)/(R₁ + R₂) = 5.9414

But, recall eqn 1

(R₁ + R₂) = 31.6875

(R₁R₂)/(R₁ + R₂) = 5.9414

Substituting for (R₁ + R₂)

(R₁R₂)/(R₁ + R₂) = (R₁R₂)/31.6875 = 5.9414

(R₁R₂) = 31.6875 × 5.9414 = 188.2683

R₁ = (188.2683/R₂)

(R₁ + R₂) = 31.6875

Substituting for R₁

(188.2683/R₂) + R₂ = 31.6875

multiply through by R₂

188.2683 + R₂² = 31.6875R₂

R₂² - 31.6875R₂ + 188.2683 = 0

Solving the quadratic equation

R₂ = 23.77 ohms or 7.92 ohms

If R₂ = 23.77 ohms, R₁ = 7.92 ohms

If R₂ = 7.92 ohms, R₁ = 23.77 ohms

Since the question explains that R₁ > R₂

R₁ = 23.77 ohms and R₂ = 7.92 ohms

Hope this Helps!!!

3 0

2 years ago

You are driving a 2400.0-kg car at a constant speed of 14.0 m/s along a wet, but straight, level road. As you approach an inters

olya-2409 [2.1K]

Answer:0.43

Explanation:

Given

mass of car $m=2400 kg$

Speed of car $u=14 m/s$

Distance traveled before coming to halt $s=23.2 m$

Let $\mu$ the coefficient of friction

Maximum deceleration road can provide during motion is

$a=\mu g$

using $v^2-u^2=2 as$

$0-14^2=2\cdot (-\mu g)\cdot 23.2$

$\mu =\frac{196}{454.72}$

$\mu =0.431$

7 0

2 years ago

A small metal bead, labeled A, has a charge of 28 nC .It is touched to metal bead B, initially neutral, so that the two beads sh

Dmitry_Shevchenko [17]

Answer:

Explanation:

Let the charge on bead A be q nC and the charge on bead B be 28nC - qnC

Force F between them

4.8\times10^{-4} = $\frac{9\times10^9\times q\times(28-q)\times10^{-18}}{(5\times10^{-2})^2}$

=120 x 10⁻⁸ = 9 x q(28 - q ) x 10⁻⁹

133.33 = 28q - q²

q²- 28q +133.33 = 0

It is a quadratic equation , which has two solution

q_A = 21.91 x 10⁻⁹C or q_B = 6.09 x 10⁻⁹ C

3 0

2 years ago