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Ann [662]
3 years ago
15

(kinematics) need help ASAP. please give me a step by step explanation

Physics
1 answer:
harkovskaia [24]3 years ago
6 0

-- 4.5 minutes lasts 270 seconds

-- the case covers 15m in each sec

-- its total distance is

(15 m/s) x (270 sec) = 4,050 meters

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Jake drove 160 kilometres in 2 hours. What was his speed, in kmh-1
ruslelena [56]

Answer:

80 kmh

Explanation:

IDK lol i just divided it by 2 because he drove 80 kilometres in one hour

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A proton with a speed of 3.2 x 106 m/s is shot into a region between two plates that are separated by a distance of 0.23 m. As t
harina [27]

Answer:

<h2>The magnitude of the magnetic is 0.145 T</h2>

Explanation:

Given :

Speed of proton v = 3.2 \times 10^{6}\frac{m}{s}

Mass of proton m = 1.67 \times 10^{-27} Kg

The force on the proton in magnetic field is given by,

  F = q (v \times B)

  F = qvB \sin \theta

But \sin 90 = 1    (∵ Force is perpendicular to the velocity so \theta = 90)

  F = qvB

When particle enter in magnetic field at the angle of 90° so particle moves in circle

So force is given by,

  F = \frac{mv^{2} }{r}

Where r = radius but in our case 0.23 m, q = 1.6 \times 10^{-19} C

By comparing above two equation,

  B = \frac{mv}{qr}

  B = \frac{1.67 \times 10^{-27} \times 3.2 \times 10^{6}  }{1.6 \times 10^{-19} \times 0.23 }

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5 0
3 years ago
A segment of wire carries a current of 25 A along the x axis from x = −2.0 m to x = 0 and then along the y axis from y = 0 to y
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Answer: F = 2N

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The aqueduct passes under Johnson Road in Lancaster through a siphon. The maximum capacity of the aqueduct is 350 m3/s. The heig
Mariulka [41]

Answer:

D ≈ 8.45 m

L ≈ 100.02 m

Explanation:

Given

Q = 350 m³/s (volumetric water flow rate passing through the stretch of channel, maximum capacity of the aqueduct)

y₁ - y₂ = h = 2.00 m (the height difference from the upper to the lower channels)

x = 100.00 m (distance between the upper and the lower channels)

We assume that:

  • the upper and the lower channels are at the same pressure (the atmospheric pressure).
  • the velocity of water in the upper channel is zero (v₁ = 0 m/s).
  • y₁ = 2.00 m  (height of the upper channel)
  • y₂ = 0.00 m  (height of the lower channel)
  • g = 9.81 m/s²
  • ρ = 1000 Kg/m³ (density of water)

We apply Bernoulli's equation as follows between the point 1 (the upper channel) and the point 2 (the lower channel):

P₁ + (ρ*v₁²/2) + ρ*g*y₁ = P₂ + (ρ*v₂²/2) + ρ*g*y₂

Plugging the known values into the equation and simplifying we get

Patm + (1000 Kg/m³*(0 m/s)²/2) + (1000 Kg/m³)*(9.81 m/s²)*(2 m) = Patm + (1000 Kg/m³*v₂²/2) + (1000 Kg/m³)*(9.81 m/s²)*(0 m)

⇒ v₂ = 6.264 m/s

then we apply the formula

Q = v*A  ⇒   A = Q/v ⇒   A = Q/v₂

⇒   A = (350 m³/s)/(6.264 m/s)

⇒   A = 55.873 m²

then, we get the diameter of the pipe as follows

A = π*D²/4   ⇒   D = 2*√(A/π)

⇒   D = 2*√(55.873 m²/π)

⇒   D = 8.434 m ≈ 8.45 m

Now, the length of the pipe can be obtained as follows

L² = x² + h²

⇒ L² = (100.00 m)² + (2.00 m)²

⇒ L ≈ 100.02 m

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