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3 years ago

A .5kg bird is perched on its nest so that it has 50J of potential energy. how far is it off the of the ground?

Physics

1 answer:

pshichka [43]3 years ago

3 0

It is 10.20 m from the ground.

<u>Explanation:</u>

<u>Given:</u>

m = 0.5 kg

PE = 50 J

We know that the Potential energy is calculated by the formula:

$P. E = m \times g \times h$

where m is the is mass in kg; g is acceleration due to gravity which is 9.8 m/s and h is height in meters.

PE is the Potential Energy.

Potential Energy is the amount of energy stored when an object is stationary.

Here, if we substitute the values in the formula, we get

$P. E = m \times g \times h$

50 = 0.5 × 9.8 × h

50 = 4.9 × h

$h = \frac {50} {4.9}$

h = 10.20 m

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Runner A takes 45 s to run 400 m. Runner B takes 48 s to run the same distance. Which runner has the greater average speed?

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Runner A because they completed it in a less amount of time

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2 years ago

A rocket engine uses fuel and oxidizer in a reaction that produces gas particles having a velocity of 1380 ms The desired thrust

bearhunter [10]

Answer:

a. 141.3 kg/s b. 5.49 m/s² c. i. 104228.9 N ii. 8.53 m/s² d. i. 97305.2 N ii. 9.84 m/s²

Explanation:

a. What must be the fuel/oxidizer consumption rate (in kg s1)?

The thrust T = Rv where R = mass consumption rate and v = velocity of rocket. Since T = 195000 N and v = 1380 m/s,

R = T/v = 195000 N/1380 m/s = 141.3 kg/s

b. If the initial weight of the rocket is 125000 N, what is its initial acceleration?

We also know that thrust T - W = ma since the rocket has to move against gravity. where M = mass of rocket = W/g = 125000 N/9.8m/s² = 12755.1 kg, W = weight of rocket = 125000 N, a = acceleration of rocket and T = thrust = 195000 N.

So, T - W = Ma

195000 N - 125000 N = (12755.1 kg)a

70000 N = ma

a = 70000 N/12755.1 kg = 5.49 m/s²

c. What are the weight and acceleration of the rocket at t 15.0 s after ignition?

We know that the loss in mass ΔM = mass consumption rate × time = Rt. Since R = 141.3 kg/s and t = 15 s,

ΔM = 141.3 kg/s × 15 = 2119.5 kg

The new mass is thus M = M - ΔM = 12755.1 kg - 2119.5 kg = 10635.6 kg

i.The weight after 15 seconds is thus W' = M'g = 10635.6 kg × 9.8m/s² = 104228.9 N

ii. Since T - W' = M'a. where M' is our new mass and a our new acceleration,

a = (T - W')/M'

= (195000 N - 104228.9 N)/10635.6 kg

= 90771.1 N/10635.6 kg

= 8.53 m/s²

d. What are the weight and acceleration of the rocket at 20.0 s after ignition?

We know that the loss in mass ΔM" = mass consumption rate × time = Rt. Since R = 141.3 kg/s and t = 20 s,

ΔM" = 141.3 kg/s × 20 = 2826 kg

The new mass is thus M" = M - ΔM" = 12755.1 kg - 2826 kg = 9929.1 kg

i. The weight after 20 seconds is thus W" = M"g = 9929.1 kg × 9.8m/s² = 97305.2 N

ii. Since T - W" = M"a. where M" is our new mass and a our new acceleration,

a = (T - W")/M"

= (195000 N - 97305.2 N)/9929.1 kg

= 97694.8 N/9929.1 kg

= 9.84 m/s²

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3 years ago

True or false: When an object becomes polarized, it acquires a charge and becomes a charged object.

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Answer:

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Explanation:

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3 years ago

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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3 years ago

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