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goblinko [34]
2 years ago
10

What type of system would allow light and air to enter and exit?

Physics
2 answers:
BartSMP [9]2 years ago
8 0

Answer:

Closed

Explanation:

Gekata [30.6K]2 years ago
6 0

Answer:

closed..........................

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Two charges of magnitude 5nC and -2nC are placed at points (2cm,0,0) and
scZoUnD [109]

Answer:

20 cm

Explanation:

Te electric potential enery U = kq₁q₂/r  were q₁ = 5 nC = 5 × 10⁻⁹ C and q₂ = -2 nC = -2 × 10⁻⁹ C and r = √(x - 2)² + (0 - 0)² +(0 - 0)² = x - 2. U =  -0.5 µJ = -0.5 × 10⁻⁶ J, k = 9 × 10⁹ Nm²/C².

So r = kq₁q₂/U

x - 2 = kq₁q₂/U

x = 0.02 + kq₁q₂/U m

x = 0.02 + 9 × 10⁹ Nm²/C² × 5 × 10⁻⁹ C × -2 × 10⁻⁹ C/-0.5 × 10⁻⁶ J

x = 0.02 - 90 × 10⁻⁹ Nm²/-0.5 × 10⁻⁶ J

x = 0.02 + 0.18 = 0.2 m = 20 cm

7 0
2 years ago
A curtain hangs straight down in front of an open window. A sudden gust of wind blows past the window; and the curtain is pulled
VikaD [51]

Answer:

option B.

Explanation:

The correct answer is option B.

The phenomenon of the curtains to pull out of the window can be explained using Bernoulli's equation.

According to Bernoulli's Principle when the speed of the moving fluid increases the pressure within the fluid decrease.

When wind flows in the outside window the pressure outside window decreases and pressure inside the room is more so, the curtain moves outside because of low pressure.

3 0
2 years ago
Which statements describe the applications of nuclear medicine scans? Check all that apply.
coldgirl [10]

Answer:

- asses disease progression and tissue function

- utilize a biologically active molecule

- utilize a radionuclide tracer

Explanation:

4 0
2 years ago
Read 2 more answers
A point charge Q = -400 nC and two unknown point charges, q1 and q2, are placed as shown. Point charge q1 is located 1.3 meters
nalin [4]
120 nC is the answer




Sorry if I’m wrong
6 0
2 years ago
Two cyclists start on a race between points A and D on two different routes. Cyclist X takes the route passing through the equid
ollegr [7]
The displacement is the shortest distance between two points, which is 546.41. The displacement for both is 546.41 meters

Average velocity of X = (200 + 200 + 200) / 30
Average velocity of X = 20 m/s

Average velocity of Y = 546.41 / 30 = 18.2 m/s
8 0
2 years ago
Read 2 more answers
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