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denis23 [38]
3 years ago
7

If you answer it I’ll love you forever!!!

Physics
2 answers:
Ilia_Sergeevich [38]3 years ago
6 0

Answer:

Performance tests can be used to see if an implemented training program is working for the athlete or if the program needs alterations. They can also assess current abilities in specific athletic areas to help the athlete choose what to focus their energy on improving.

Explanation:

seraphim [82]3 years ago
5 0

Answer:

ok my answer was They measure an athlete's performance in that area being tested. This is done to compare to past performance evaluation tests and to measure if the athlete is getting better or worse. It also determines if the athlete should stop or keep doing his/her sport.

Explanation:

is that good :)

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A rat runs 2m right, turns around and runs 3m left. Then goes 2m right. What is its displacement?
geniusboy [140]

Answer:

Explanation:

I think this answer would be 1m to the left.

7 0
3 years ago
A,b, e are complete. Help on the others would be so appreciated!!
bixtya [17]

Answer:

serie Ceq=0.678 10⁻⁶ F  and the charge Q = 9.49 10⁻⁶ C

Explanation:

Let's calculate all capacity values

a) The equivalent capacitance of series capacitors

    1 / Ceq = 1 / C1 + 1 / C2 + 1 / C3 + 1 / C4 + 1 / C5

    1 / Ceq = 1 / 1.5 + 1 / 3.3 + 1 / 5.5 + 1 / 6.2 + 1 / 6.2

    1 / Ceq = 1 / 1.5 + 1 / 3.3 + 1 / 5.5 + 2 / 6.2

    1 / Ceq = 0.666 + 0.3030 +0.1818 +0.3225

    1 / Ceq = 1,147

    Ceq = 0.678 10⁻⁶ F

b) Let's calculate the total system load

   Dv = Q / Ceq

   Q = DV Ceq

   Q = 14 0.678 10⁻⁶

   Q = 9.49 10⁻⁶ C

In a series system the load is constant in all capacitors, therefore, the load in capacitor 5.5 is Q = 9.49 10⁻⁶ C

c) The potential difference

   ΔV = Q / C5

   ΔV = 9.49 10⁻⁶ / 5.5 10⁻⁶

   ΔV = 1,725 ​​V

d) The energy stores is

    U = ½ C V²

    U = ½ 0.678 10-6 14²

    U = 66.4 10⁻⁶ J

e) Parallel system

   Ceq = C1 + C2 + C3 + C4 + C5

   Ceq = (1.5 +3.3 +5.5 +6.2 +6.2) 10⁻⁶

   Ceq = 22.7 10⁻⁶ F

f) In the parallel system the voltage is maintained

   Q5 = C5 V

   Q5 = 5.5 10⁻⁶ 14

   Q5 = 77 10⁻⁶ C

g) The voltage is constant V5 = 14 V

h) Energy stores

   U = ½ C V²

   U = ½ 22.7 10-6 14²

   U = 2.2 10⁻³ J

8 0
3 years ago
Two wires of the same metal having the same cross-
USPshnik [31]
The answer is true

Step by step explanation:
3 0
3 years ago
Read 2 more answers
Due to the wave nature of light, light shined on a single slit will produce a diffraction pattern? Green light (520 nm) is shine
TiliK225 [7]

Answer:

Yes, it will produce a diffraction pattern.

a. 3.9 mm b. 1.95 mm

Explanation:

The light shined from a single slit will produce a diffraction pattern because,  the wavefront act as wavelets which generates its own wave according to Huygens principle. This therefore causes the diffraction pattern.

Given

wavelength of green light, λ = 520 nm = 520 × 10⁻⁹ m = 5.20 × 10⁻⁷ m

width of slit, d = 0.440 mm = 0.44 × 10⁻³ m = 4.4 × 10⁻⁴ m

Distance of slit from central maximum , D = 1.65 m

Distance of first minimum from central maximum, y = ?

a. The relationship between the slit width and wavelength is given by [tex} dsinθ = mλ [/tex]where d = slit width, θ = angular distance from central maximum, λ = wavelength of light and m = ±1, ±2, ±3...

The relationship between y and D is given by tanθ = y/D

Since θ is small, sinθ ≈ θ ≈ tanθ

so, dθ = mλ ⇒ θ = mλ/d = y/D

Therefore, y = mλD/d

Now, for the first minimum above the slit, m = +1 and for the first minimum below the slit, m = -1. So, y₁ =  λD/d and y₋₁ =  -λD/d. So, the width of the central maximum Δy is the difference between the first minima below and above the central maximum. So, Δy = y₁ - y₋₁ = λD/d -(-λD/d) = 2λD/d

Substituting the values from above, Δy= 2 × 5.20 × 10⁻⁷ × 1.65/4.4 × 10⁻⁴ =  3900 × 10⁻⁶ m = 3.9 × 10⁻³ m = 3.9 mm

b. The first order fringe is the fringe located between the first minimum and the second minimum. From dsinθ = mλ and tanθ = y/D when θ is small, sinθ ≈ θ ≈ tanθ. So, y = mλD/d. Let m= 1 and m=2 be the first and second minima respectively. So,y₁ =  λD/d and y₂ =  2λD/d. The difference Δy₁ = y₂ - y₁ is the width of the first order fringe. Therefore, Δy₁ = 2λD/d - λD/d= λD/d. Substituting the values from above, we have

λD/d= 5.20 × 10⁻⁷ × 1.65/4.4 × 10⁻⁴= 1.95 × 10⁻³ m = 1.95 mm

7 0
3 years ago
4. Interference is an example of which aspect of electromagnetic radiation?
Lisa [10]

Answer:

D is the answer wave behavior

6 0
3 years ago
Read 2 more answers
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