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Anna [14]
2 years ago
11

Hdkshshgsksisbsgshsbisheghs be iwhghhfhjdjjh

Physics
1 answer:
zlopas [31]2 years ago
6 0

Answer:

dsfcwisienruyoiqnpu3p is enicuwyeiourcbt84ru

Explanation:

uqyoeyriuobyeriuov uvry84bvruoiby4roybvoiur ryvw8rvby82yrbiuo3

You might be interested in
You are in the forest with some of your friends. You’re being chased by a very angry and hungry bear.
melomori [17]

Answer:

2m head start or else you done for

Explanation:

you cant even out run a bear they run at 35mph the fastest human is 25

8 0
2 years ago
nert xenon actually forms many compounds, especially with highly electronegative fluorine. The ΔH o f values for xenon difluorid
loris [4]

Answer:

For Xenon fluoride, the average bond energy is 132kj/mol

For tetraflouride,the average bond energy is 150.5kj/mol.

For hexaflouride, the average bond energy is 146.5 kj/mol

Explanation:

For xenon fluoride

105/2 = 52.5

For F-F

159/2 = 79.5

Average bond energy of Xe-F = 79.5 + 52.5 = 132kj/mole

For tetraflouride

284/4 = 71

For F-F

159/2 = 79.5

Average bond energy = 79.5 + 71 = 150.5kj/mol

For hexaflouride

402/6 = 67

F-F = 159/2 = 79.5

Average bond energy = 67 + 79.5 = 146.5kj/ mol

3 0
3 years ago
An oil droplet is sprayed into a uniform electric field of adjustable magnitude. The 0.11 g droplet hovers
ohaa [14]

Answer:

The direction of the field is downward, and negatively charged particles will experience an upwards force due to the field.

F = N e E     where E is the value of the field and N e the charge Q

M g = N e E      and M g is the weight of the drop

N = M g / (e E)

N = 1.1E-4 * 9.8 / (1.6E-19 * 370) = 1.1 * 9.8 / (1.6 * 370) * E15 = 1.82E13

.00011 kg is a very large drop

Q = N e = M g / E = .00011 * 9.8 / 370 = 2.91E-6 Coulombs

Check:     N = Q / e = 2.91E-6 / 1.6E-19 = 1.82E13   electrons

7 0
2 years ago
A rotating fan completes 1200 revolutions every minute. Consider the tip of a blade, at a radius of 0.15 m. (a) Through what dis
olga55 [171]

Answer:

(a) 0.942 m

(b) 18.84 m/s

(c) 2366.3 m/s²

(d) 0.05 s

Explanation:

(a) In one revolution, it travels through one circumference, 2πr = 2 × 3.14 × 0.15 m = 0.942 m.

(b) Its frequency, f, is 1200 rev/min = \dfrac{1200}{60}rev/s = 20 rev/s.

Its angular frequency, ω = 2πf = 2π × 20 = 40π

The speed is given by

v = ωr = 40π × 0.15 = 6π = 18.84 m/s

(c) Its acceleration is given by, a = ω²r = (40π)² × 0.15 = 2366.3 m/s²

(d) The period is the inverse of the frequency because it is the time taken to complete one revolution.

T = \dfrac{1}{f}

T = 1/20 = 0.05 s

6 0
3 years ago
Suppose that a uniform electric field exists in a certain region of space. Now consider a mathematical plane surface of area A.
____ [38]

Answer: The normal of the plane must be parallel to the electric field vector.

Explanation:

the normal to the surface is defined as a versor that is perpendicular to the plane.

Now, if the angle between this normal and the line of the field is θ, we have that the flux can be written as:

Φ = E*A*cos(θ)

Where E is the field, A is the area and θ is the angle already defined.

Now, this maximizes when θ = 0.

This means that the normal of the surface must be parallel to the electric field

7 0
2 years ago
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