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3 years ago

Calculate the distance travel if it accelerates from 0 to 27.8 meters per second in 2.5 seconds

Physics

1 answer:

luda_lava [24]3 years ago

8 0

The distance travel is 69.5 meters.

Explanation:

Given datas are as follows

Speed = 27.8 meters / second

Time = 2.5 seconds

The formula to calculate the speed using distance and time is

Speed = Distance ÷ Time (units)

Then Distance = Speed × Time (units)

Distance = (27.8 × 2.5) meters

Distance = 69.50 meters

Therefore the distance travelled is 69.50 meters.

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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

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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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An object floats in a beaker as shown. when it was put into the beaker, it displaced an amount of water into the graduated cylin

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Answer:

answer is C. 10 g

Explanation:

: When an object floats, it displaces an amount of water that has the same mass as itself. Therefore, the mass of the water in the graduated cylinder is equal to the mass of the object. We can see that there are 10 mL of water in the graduated cylinder. We also know that the density of water is 1 g/mL. Since each mL of water has a mass of 1 g, then 10 mL of water has a mass of 10 g. If the mass of the displaced water is 10 g, then the mass of the floating object is also 10 g.

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4.1 A steel spur pinion has a pitch of 5 teeth/in, 20 full-depth teeth, and a 20° pressure angle. The pinion runs at a speed of

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Answer:

15.07 ksi

Explanation:

Given that:

Pitch (P) = 5 teeth/in

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Pinion speed ( $n_p$ ) = 2000 rev/min

Power (H) = 30 hp

Teeth on gear ( $N_G$ ) = 50

Teeth on pinion ( $N_p$ ) = 20

Face width (F) = 1 in

Let us first determine the diameter (d) of the pinion.

Diameter (d) = $\frac{N}{P}$

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= 4 in

From the values of Lewis Form Factor Y for ( $n_p$ ) = 20 ; at 20°

Y = 0.321

To find the velocity (V); we use the formula:

$V = \frac{\pi d n_p}{12}$

$V = \frac{\pi *4*2000}{12}$

V = 2094.40 ft/min

For cut or milled profile; the velocity factor $(K_v)$ can be determined as follows:

$K_v = \frac{2000+V}{2000}$

$K_v = \frac{2000+2094.40}{2000}$

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However, there is need to get the value of the tangential load $(W^t)$ , in order to achieve that, we have the following expression

$W^t=\frac{T}{\frac{d}{2} }$

$W^t = \frac{63025*H}{\frac{n_pd}{2}}$

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$\sigma = \frac{K_vW^tp}{FY}$

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$\sigma = 15073.07 psi$

$\sigma =$ 15.07 ksi

∴ The estimate of the bending stress = 15.07 ksi

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