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garik1379 [7]
3 years ago
12

A truck travelling at 30m/s decelerates at 1.5m/s². How far does it travel during the 10th second after the brakes are applied?​

Physics
1 answer:
11Alexandr11 [23.1K]3 years ago
3 0

Answer

225 meters.

Explanation:

x=x0+30t-(1/2)(1.5)t^2

x=0+30(10)-(1/2)(1.5)(10)^2

x=300-75

x=225

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A water pipe is inclined 40.0° below the horizontal. The radius of the pipe at the upper end is 2.00 cm. If the gauge pressure a
aliya0001 [1]

Answer:

P_2 = -1.9 \times 10^8 Pa

Explanation:

As it is given that flow rate in the pipe is 20 cm^3/s

so we have

Q = A_1v_1 = A_2v_2

at the upper end the area is given as

A_1 = \pi r_1^2

A_1 = \pi(0.02)^2 = 1.26 \times 10^{-3} cm^2

Also at the other end

A_2 = \pi r_2^2

A_2 = \pi(0.01)^2 = 0.314 \times 10^{-3} cm^2

now the speed at two ends is given as

v_1 = \frac{20}{1.26 \times 10^{-3}}

v_1 = 159.15 m/s

v_2 = \frac{20}{0.314 \times 10^{-3}}

v_2 = 637 m/s

now by Bernoulli's theorem we have

P_1 + \frac{1}{2}\rho v_1^2 + \rho g h_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho g h_2

now we have

0.112(1.013 \times 10^5) + \frac{1}{2}1000(159.15)^2 + 1000(9.81)(2.65sin40) = P_2 + \frac{1}{2}(1000)(637)^2 + 0

Now we have

P_2 = -1.9 \times 10^8 Pa

4 0
3 years ago
Think more about the situation in the question above. If you picked a "good" thing, how might
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What falls faster? Light or heavy objects?
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Read 2 more answers
What is the pressure drop due to the bernoulli effect as water goes into a 3.00-cm-diameter nozzle from a 9.00-cm-diameter fire
Paul [167]

Answer:

\Delta P=1581357.92\ Pa

Explanation:

Given:

  • diameter of hose pipe, D=0.09\ m
  • diameter of nozzle, d=0.03\ m
  • volume flow rate, \dot{V}=40\ L.s^{-1}=0.04\ m^3.s^{-1}

<u>Now, flow velocity in hose:</u>

v_h=\frac{\dot V}{\pi.D^2\div 4}

v_h=\frac{0.04\times 4}{\pi\times 0.09^2}

v_h=6.2876\ m.s^{-1}

<u>Now, flow velocity in nozzle:</u>

v_n=\frac{\dot V}{\pi.d^2\div 4}

v_n=\frac{0.04\times 4}{\pi\times 0.03^2}

v_n=56.5884\ m.s^{-1}

We know the Bernoulli's equation:

\frac{P_1}{\rho.g}+\frac{v_1^2}{2g}+Z_1=\frac{P_2}{\rho.g}+\frac{v_2^2}{2g}+Z_2

when the two points are at same height then the eq. becomes

\frac{P_1}{\rho.g}+\frac{v_1^2}{2g}=\frac{P_2}{\rho.g}+\frac{v_2^2}{2g}

\Delta P=\frac{\rho(v_n^2-v_h^2)}{2}

\Delta P=\frac{1000(56.5884^2-6.2876^2)}{2}

\Delta P=1581357.92\ Pa

8 0
3 years ago
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velikii [3]

Answer: a) Twice Chuck's

Explanation:

If the angular speed is constant, as the merry - go -round is a rigid body, so the distance between two points must remain the same, it is needed that the points farther the center, move faster than the ones closer to it.

There exists a relationship between tangential and angular speed, as follows, that relates the definitions of linear and angular speed, and the angle definition:

angle = arc / radius ⇒ Δθ/Δt = Δs/Δt / r ⇒ ω = v/r ⇒ v = ω. r

If ω is constant, v is directly proportional to r, distance to the center (radius in a circular platform), so if r is twice for Andrea, her tangential speed must be twice Chuck's.

5 0
4 years ago
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