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lyudmila [28]
3 years ago
15

A train is uniformly accelerated and passes

Physics
1 answer:
cupoosta [38]3 years ago
6 0

Answer:

Explanation:

18 km/hr(1000 m/km) / (3600 s/hr) = 5 m/s

36 km/hr = 10 m/s

v² = u² + 2as

a = (v² - u²)/2s

a = (10² - 5²)/2(1000) = 0.0375 m/s²

v² = 10² + 2(0.0375)(1000)

v² = 175

v = 13.22875... ≈ 13 m/s ≈ 48 km/hr

v = u + at

t = (v - u) / a

t = (10 - 5)/0.0375 = 133.333... 133 s

t = (13.22875 - 10) / 0.0375 = 86.1 s

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Your friend tells you “I no the moon does not rotate because we always see the same side.” do you agree or disagree with your fr
nydimaria [60]

Answer:

no the moon does not rotate it only goes in circle just like the sun so I disagree with your friend

3 0
3 years ago
When an object oscillating in simple harmonic motion is at its maximum displacement from the equilibrium position, which of the
ziro4ka [17]

Answer:

E. Zero Maximum

Explanation:

At the point of maximum displacement, the speed is zero while the restoring force is maximum. In fact:

- The restoring force is given by F=kx, where k is the spring constant and x is the displacement - at the point of maximum displacement, x is maximum, so F is maximum as well

- the total energy of the system is sum of kinetic energy and elastic potential energy:

E=K+U=\frac{1}{2}mv^2+\frac{1}{2}kx^2

where m is the mass of the system and v is the speed. Since E (the total energy) is constant due to the law of conservation of energy, we have that when K increases, U decreases, and viceversa. As a result, when x increases, v decreases, and viceversa. At the point of maximum displacement, x is maximum, so v will have its minimum value (which is zero, since the system is changing direction of motion).

4 0
3 years ago
Question 29 of 43
liraira [26]
I agree with the first responses
3 0
2 years ago
A centrifugal pump is operating at a flow rate of 1 m3/s and a head of 20 m. If the specific weight of water is 9800 N/m3 and th
tamaranim1 [39]

Answer:

<em>The power required by the pump is nearly 230.588 kW</em>

Explanation:

Flow rate of the pump Q = 1 m^3/s

the head flow H = 20 m

specific weight of water γ = 9800 N/m^3

efficiency of the pump η = 85%

First note that specific gravity of water is the product of the density of water and acceleration due to gravity.

γ = ρg

where ρ is density. For water its value is 1000 kg/m^3

g is the acceleration due to gravity = 9.81 m/s^2

The power to lift this water at this rate will be gotten from the equation

P = ρgQH

but ρg = γ

therefore,

P = γQH

imputing values, we'll have

P = 9800 x 1 x 20 = 196000 W

But the centrifugal pump that will be used will only be able to lift this amount of water after the efficiency factor has been considered. The power of pump needed must be greater than this power.

we can say that

196000 W is 85% of the power of the pump power needed, therefore

196000 = 85% of P_{p}

where P_{p} is the power of the pump needed

85% = 0.85

196000 = 0.85P_{p}

P_{p} = 196000/0.85 = 230588.24 W

<em>Pump power = 230.588 kW</em>

3 0
3 years ago
Resistors and reactors, for use over 600 volts, shall not be installed in close enough proximity to combustible materials to con
suter [353]

Resistors and reactors, for use over 600 volts, shall not be installed in close enough proximity to combustible materials to constitute a fire hazard and shall have a clearance of not less than<u> 300 mm </u>from combustible materials.

Explanation:

  • The hazards associated with high power industrial resistors are primarily due to their open construction, which is necessary for cooling.
  • The exposed conductors which make up the resistors can be not only a shock hazard but also a thermal burn hazard.
  • When a resistor fails, it either goes open or the resistance increases. When the resistance increases, it can burn the board, or burn itself up.
  • Avoid touching non-flammable resistors in operation; the surface temperature ranges from approximately 350 °C to 400°C when utilized at the full rated value. Maintaining a surface temperature of 200°C or less will extend resistors service life.
  • Do not apply power to a circuit while measuring resistance. When you are finished using an ohmmeter, switch it to the OFF position if one is provided and remove the leads from the meter.
  • Always adjust the ohmmeter for 0 (or in shunt ohmmeter) after you change ranges before making the resistance measurement.

4 0
3 years ago
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