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Dmitrij [34]
3 years ago
6

A rocket is moving with constant velocity of 2000ms and accelerates uniformly at 10ms. What is the velocity after 5 minutes?

Physics
1 answer:
AnnyKZ [126]3 years ago
8 0
The velocity is 300ml in 5 mins
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B.) Carbon Dioxide because the carbon is surrounded by oxygen
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a backpack has a mass of 8 kg. it is lifted and given 54.9 J of gravitational potential energy. how high is is lifted? accelerat
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P=mgh
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The driver of a car going 86.0 km/h suddenly sees the lights of a barrier 44.0 m ahead. It takes the driver 0.75 s to apply the
Lynna [10]

Part a

Answer: NO

We need to calculate the distance traveled once the brakes are applied. Then we would compare the distance traveled and distance of the barrier.

Using the second equation of motion:

s=ut+0.5at^2

where s is the distance traveled, u is the initial velocity, t is the time taken and a is the acceleration.

It is given that, u=86.0 km/h=23.9 m/s, t=0.75 s, a=-10.0 m/s^2

\Rightarrow s=23.9 m/s \times 0.75s+0.5\times (-10.0 m/s^2)\times (0.75 s)^2=15.11 m

Since there is sufficient distance between position where car would stop and the barrier, the car would not hit it.

Part b

Answer: 29.6 m/s

The maximum distance that car can travel is s=44.0 m

The acceleration is same, a=-10.0 m/s^2

The final velocity, v=0

Using the third equation of motion, we can find the maximum initial velocity for car to not hit the barrier:

v^2-u^2=2as

0-u^2=-2 \time 10.0m/s^2 \times 44.0 m\Rightarrow u=\sqrt{880 m^2/s^2}=29.6 m/s

Hence, the maximum speed at which car can travel and not hit the barrier is 29.6 m/s.





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3 years ago
How should the student change the circuit to give negative values for current and
dmitriy555 [2]

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Flip the cell.

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what do penguins eat for lunch?

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3 0
2 years ago
Water, with a density of 1000 kg/m3, flows out of a spigot, through a hose, and out a nozzle into the air. The hose has an inner
stepan [7]

Answer:

   P₁ = 2.3506 10⁵ Pa

Explanation:

For this exercise we use Bernoulli's equation and continuity, where point 1 is in the hose and point 2 in the nozzle

          P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

          A₁ v₁ = A₂ v₂

Let's look for the areas

          r₁ = d₁ / 2 = 2.25 / 2 = 1,125 cm

          r₂ = d₂ / 2 = 0.2 / 2 = 0.100 cm

          A₁ = π r₁²

          A₁ = π 1.125²

          A₁ = 3,976 cm²

          A₂ = π r₂²

          A₂ = π 0.1²

          A₂ = 0.0452 cm²

Now with the continuity equation we can look for the speed of water inside the hose

           v₁ = v₂ A₂ / A₁

           v₁ = 11.2 0.0452 / 3.976

           v₁ = 0.1273 m / s

Now we can use Bernoulli's equation, pa pressure at the nozzle is the air pressure (P₂ = Patm) the hose must be on the floor so the height is zero (y₁ = 0)

           P₁ + ½ ρ v₁² = Patm + ½ ρ v₂² + ρ g y₂

          P₁ = Patm + ½ ρ (v₂² - v₁²) + ρ g y₂

Let's calculate

           P₁ = 1.013 10⁵ + ½ 1000 (11.2² - 0.1273²) + 1000 9.8 7.25

           P₁ = 1.013 10⁵ + 6.271 10⁴ + 7.105 10⁴

           P₁ = 2.3506 10⁵ Pa

7 0
3 years ago
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