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Anit [1.1K]
3 years ago
6

When air mass is caught between two cold fronts the result is a _______ front.

Physics
1 answer:
Gre4nikov [31]3 years ago
3 0
A. Occluded
Explanation- At an occluded front, the cold air mass from the cold front meets the cool air that was ahead of the warm front.
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A 16.9 kg monkey is swinging on a 5.32 m long vine. It starts at rest, with the vine at a 43.0° angle. How fast is the monkey mo
NemiM [27]

Answer:

v = 5.7554 m/s

Explanation:

First of all we need to know if the angle of the vine is measured in the horizontal or vertical.

To do this easier, let's assume the angle is measured with the horizontal. In this case, the innitial height of the monkey will be:

h₀ = h sinα

h₀ = 5.32 sin43° = 3.6282 m

As the monkey is dropping from the innitial point which is the suspension point, is also dropping from 5.32. Then the actual height of the monkey will be:

Δh = 5.32 - 3.63 = 1.69 m

In order to calculate the speed of the monkey we need to understand that the monkey has a potential energy. This energy, because of the gravity, is converted in kinetic energy, and the value will be the same. Therefore we can say that:

Ep = Ek

From here, we can calculate the speed of the monkey.

Ep = mgΔH

Ek = 1/2 mv²

The potential energy is:

Ep = 16.9 * 9.8 * 1.69 = 279.9

Now with the kinetic energy:

1/2 * (16.9) * v² = 279.9

v² = (279.9) * 2 / 16.9

v² = 33.12

v = √33.12

<h2>v = 5.7554 m/s</h2>

Hope this helps

3 0
3 years ago
contractor will use a package emulsion having a specific gravity of 1.25 and relative bulk strength of 130 to open an excavation
tester [92]

Answer:

The burden distance is 7 ft

Solution:

As per the question:

Specific gravity of package emulsion, SG_{E} = 1.25

Specific gravity of diabase rock, SG_{R} = 2.76

Diameter of the packaged sticks, d = 3 in

Now,

To calculate the first trail shot burden distance, B:

B = [\frac{2SG_{E}}{SG_{R}} + 1.5]\times d

B = [\frac{2\times 1.25}{2.76} + 1.5]\times 3 = 7.22

B = 7 ft

5 0
3 years ago
What is the potential difference per unit length between two infinitely long concentric cylindrical shells with inner radius 1.5
Vinil7 [7]

Answer:

165.8 V/m

Explanation:

The capacitance of a long concentric cylindrical shell of length, L and inner radius, a and outer radius, b is C = 2πε₀L/㏑(b/a)

Since the charge on the cylindrical shells, Q = CV where V = the potential difference across the capacitor(which is the potential difference between the concentric cylindrical shells)

V = Q/C

V = Q ÷ 2πε₀L/㏑(b/a)

V = Q㏑(b/a)/2πε₀L

So, the potential difference per unit length V' is

V' = V/L = Q㏑(b/a)/2πε₀

Given that a = inner radius = 1.5 cm, b = outer radius = 5.6 cm and Q = 7.0 nC = 7.0 × 10⁻⁹ C and ε₀ = 8.854 × 10⁻¹² F/m substituting the values of the variables into the equation, we have

V' = Q㏑(b/a)/2πε₀

V' = 7.0 × 10⁻⁹ C㏑(5.6 cm/1.5 cm)/(2π × 8.854 × 10⁻¹² F/m)

V' = 7.0 × 10⁻⁹ C㏑(3.733)/(55.631 × 10⁻¹² F/m)

V' = 7.0 × 10⁻⁹ C × 1.3173/(55.631 × 10⁻¹² F/m)

V' = 9.2211 × 10⁻⁹ C/(55.631 × 10⁻¹² F/m)

V' = 0.16575 × 10³ V/m

V' = 165.75 V/m

V' ≅ 165.8 V/m

6 0
3 years ago
Two identical closely spaced circular disks form a parallel-plate capacitor. Transferring 2.1×109 electrons from one disk to the
Bad White [126]

Answer:

d = 0.018 m

Explanation:

Charge on the plates of the capacitor due to transfer of electrons is given as

Q = Ne

here we know that

N = 2.1 \times 10^9

so we have

Q = (2.1 \times 10^9)(1.6 \times 10^{-19})

Q = 3.36 \times 10^{-10} C

now we have electric field between the plates is given as

E = \frac{Q}{A\epsilon_0}

here we have

1.5 \times 10^5 = \frac{3.36 \times 10^{-10}}{A(8.85 \times 10^{-12})}

A = 2.53 \times 10^{-4} m^2

now we have

A = \frac{\pi d^2}{4}

2.53 \times 10^{-4} = \frac{\pi d^2}{4}

d = 0.018 m

3 0
3 years ago
If objects A and B are separately in thermal equilibrium with a third object C, then A and B are in thermal equilibrium with eac
Monica [59]
Your answer is c your welcome
8 0
3 years ago
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