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sergeinik [125]
3 years ago
5

Light-rail passenger trains that provide transportation within and between cities speed up and slow down with a nearly constant

(and quite modest) acceleration. A train travels through a congested part of town at 5.0 m/s. Once free of this area, it speeds up to 12 m/s in 8.0 s. At the edge of town, the driver again accelerates, with the same acceleration, for another 16 s to reach a higher cruising speed. What is the final speed?
Physics
1 answer:
snow_lady [41]3 years ago
3 0

Answer:

26m/s

Explanation:

Assuming that the acceleration is constant, we can start by calculating the train speed when it's free of the congested area:

a = \frac{\deltav}{\deltat} = \frac{12 - 5}{8} = \frac{7}{8} = 0.875 m/s^2

Then with the same acceleration we can find out the final speed:

v = v_0 + at = 12 + 0.875*16 = 26m/s

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A barrel 1 m tall and 60 cm in diameter is filled to the top with water. What is the pressure it exerts on the floor beneath it?
allsm [11]

Answer:

The pressure on the ground is about 9779.5 Pascal.

The pressure can be reduced by distributing the weight over a larger area using, for example, a thin plate with an area larger than the circular area of the barrel's bottom side.  See more details further below.

Explanation:

Start with the formula for pressure

(pressure P) = (Force F) / (Area A)

In order to determine the pressure the barrel exerts on the floor area, we need the calculate the its weight first

F_g = m \cdot g

where m is the mass of the barrel and g the gravitational acceleration. We can estimate this mass using the volume of a cylinder with radius 30 cm and height 1m, the density of the water, and the assumption that the container mass is negligible:

V = h\pi r^2=1m \cdot \pi\cdot 0.3^2 m^2\approx 0.283m^3

The density of water is 997 kg/m^3, so the mass of the barrel is:

m = V\cdot \rho = 0.283 m^3 \cdot 997 \frac{kg}{m^3}= 282.151kg

and so the weight is

F_g = 282.151kg\cdot 9.8\frac{m}{s^2}=2765.08N

and so the pressure is

P = \frac{F}{A} = \frac{F}{\pi r^2}= \frac{2765.08N}{\pi \cdot 0.3^2 m^2}\approx 9779.5 Pa

This answers the first part of the question.

The second part of the question asks for ways to reduce the above pressure without changing the amount of water. Since the pressure is directly proportional to the weight (determined by the water) and indirectly proportional to the area, changing the area offers itself here. Specifically, we could insert a thin plate (of negligible additional weight) to spread the weight of the barrel over a larger area. Alternatively, the barrel could be reshaped (if this is allowed) into one with a larger diameter (and smaller height), which would achieve a reduction of the pressure.  

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3 years ago
What is the kinetic energy of a toy truck with a mass of 0.75 kg and a velocity of 4 m/s? (Formula: )
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<span>The answer is 6 Joules. The kinetic energy (KE) of an object is the product of half of the mass (m) of the object and the square of its velocity (v²): KE = 1/2m*v². We know that m = 0.75 kg and that v = 4 m/s. Therefore, KE = 1/2 * 0.75 kg * (4 m/s)² = 1/2 * 0.75 kg * 16 m²/s² = 6 kg*m²/s² = 6 J.</span>
6 0
3 years ago
Read 2 more answers
Simple harmonic motion: the total mechanical energy of a simple harmonic oscillating system is:___.
nataly862011 [7]

The answer is zero.

A simple harmonic motion has zero total mechanical energy as it moves beyond the equilibrium point, when it achieves the maximum displacement, when it moves past the equilibrium point, and when it moves past the equilibrium point and a non-zero constant.

What is simple harmonic motion?

  • Simple Harmonic Motion (SHM) is a periodic, to and fro shifting rotation about the mean position of the body. The restoring force on an oscillating body is geared toward its mean direction since it is exactly proportional to its displacement.
  • The sum of the potential energy and the kinetic energy constitutes the total energy in brainly.com/question/17315536.
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To learn more about simple harmonic motion visit: brainly.com/question/25380728?

#SPJ4

3 0
2 years ago
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