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aliina [53]
2 years ago
14

A ball is kicked at 10.4 m/s at an angle of 32 degrees to the horizontal

Physics
1 answer:
Anarel [89]2 years ago
4 0

Answer:

3M/S

Explanation:

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Once juno reaches jupiter, what is the minimum amount of time it takes for the transmitted signals to travel from the spacecraft
Komok [63]

In very very very round figures . . .

-- Jupiter is about 5.2 times as far from the sun as the earth is.

-- So when Jupiter and the EARTH are aligned in both orbits, Jupiter is about

(4.2) x (150 million kilometers) = 630 million kilometers

Time = (distance) / (speed)

The speed of light and radio is 300,000 km/second

Time = (630 million / 300 thousand)

<em>Time = 2,100 seconds</em>

That's 35 minutes.

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3 years ago
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In the first law of Thermodynamics ΔE = Q - W, what does ΔE stand for???
Alexxx [7]
<span>Δ</span>E = q + w

q = heat (quantity of)

q and w can be positive or negative depending on if work/heat is being absorbed/done on the system or released/done by the system
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Which statement about moons is true?
kotegsom [21]

Answer:

A

Explanation:

All of the other answers don't make much sense

8 0
3 years ago
What is the frequency of a clock waveform whose period is 750 microseconds?
Allushta [10]
Use this formula to find your answer...

Determine the frequency of a clock waveform whose period is 2us or (micro) and 0.75ms

frequency (f)=1/( Time period).

Frequency of 2 us clock =1/2*10^-6 =10^6/2 =500000Hz =500 kHz.

Frequency of 0..75ms clock =1/0.75*10^-3 =10^3/0.75 =1333.33Hz =1.33kHz.

6 0
3 years ago
A parallel-plate capacitor in air has a plate separation of 1.76 cm and a plate area of
Monica [59]

Answer:

Explanation:

Plate separation, d = 1.76 cm = 0.0176 m

Area of plates, A = 25 cm^2 = 0.0025 m^2

V = 255 V

(a) Capacitance of capacitor

C = \frac{\epsilon _0A}{d}

C = \frac{8.854\times 10^{-12}\times 0.0025}{0.0176}

C = 1.258 x 10^-12 F

charge is same before and after immersion as the battery is disconnected

q = C V

q = 1.258 x 10^-12 x 255 = 3.2 x 10^-10 C

(b)

Capacitance before, C = 1.258 x 10^-12 C

capacitance after, C' = k x C = 80 x 1.258 x 10^-12 = 100.64 x 10^-12 C

Where, k is the dielectric constant of water = 80

Potential difference after immersion, V' = V / k = 255 / 80 = 3.1875 V

(c) initial energy,

U = \frac{q^{2}}{2C}

U = \frac{(3.2\times 10^{-10})^{2}}{2\times 1.258\times 10^{-12 }}=4.07\times 10^{-8}J

Final energy

U' = \frac{q^{2}}{2C'}

U' = \frac{(3.2\times 10^{-10})^{2}}{2\times 100.64\times 10^{-12}}=5.08\times 10^{-10}J

6 0
3 years ago
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