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Irina18 [472]
3 years ago
8

Most street lights are made from one or two elements mercury or sodium. Explain why astronomers preferred that cities you sodium

streetlights instead of mercury streetlights
Physics
1 answer:
lora16 [44]3 years ago
4 0
Mercury is very harmful to the average human being. the mercury can easily be released from the lamp if the lamp is knocked over and broken. mercury is also harmful if inhaled. sodium on the other hand is not harmful in any way.
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As a 5.0 x 102 N basketball player jumps from the floor up toward the basket, the magnitude of the force of her feet on the floo
s2008m [1.1K]
Here is your answer:

First find the notations:

5.0×10^2=500.
1.0×10^3=1,000

Then you multiply:

1,000×500=500,000

Your answer is...
=500,000 or 5.0×10^4
8 0
4 years ago
A truck, initially at rest, rolls down a frictionless hill and attains a speed of 20 m/s at the bottom. To achieve a speed of 40
Crank

Answer:

To achieve the velocity of 40 m/sec height will become 4 times  

Explanation:

We have given initially truck is at rest and attains a speed of 20 m/sec

Let the mass of the truck is m

At the top of the hill potential energy is mgh and kinetic energy is \frac{1}{2}mv^2

So total energy at the top of the hill =mgh+0=mgh

At the bottom of the hill kinetic energy is equal to \frac{1}{2}mv^2 and potential energy will be 0

So total energy at the bottom of the hill is equal to 0+\frac{1}{2}mv^2

Form energy conservation mgh=\frac{1}{2}mv^2

v=\sqrt{2gh}, for v = 20 m/sec

20=\sqrt{2\times 9.8\times h}

Squaring both side

19.6h=400

h = 20.408 m

Now if velocity is 0 m/sec

40=\sqrt{2gh}

19.6h=1600

h = 81.63 m

So we can see that to achieve the velocity of 40 m/sec height will become 4 times

5 0
3 years ago
You hear a thunder 3 seconds after you see lightning. How far away is the storm?
kondor19780726 [428]
3 seconds (330 m) / second =
990 meter
3 0
3 years ago
2. Air at a temperature of 20 ºC passes through a pipe with a constant velocity of 40 m/s. The pipe goes through a heat exchange
frozen [14]

Answer:

a) Q = 1436 kW

b) P ≈ 776 kW

Explanation:

Let's begin by listing out the given parameters:

T1 = 20 °C, u = 40 m/s, T2 = 820 °C, P = 4.3 kW, m = 2.5 kg/s, T3 = 510 °C, V1 = 40 m/s,

V2 = 40 m/s, V3 = 55 m/s, ṁ = 2.5 kg/s

To solve the question, we make this assumption that the size of the pipe is constant

a) No change in velocity implies that heat added is isochoric

Q = m * C * ΔT

Cv of air at 300 K(≈20 °C) = 0.718

Q = 2.5 * 0.718 * (820 − 20)

Q = 1436 kW

b) P = ṁ * Cp * ΔT + ṁ * (V2² - V3²) ÷ 2000] - Ql

V2² - V3² = 55² - 40² = 1425

ΔT = T2 - T3 = 820 - 510 = 310 °C

Cp of air at 300 K(≠20 °C) = 1.005 kJ/kgK

Ql = 4.3 kW

P = 2.5 * (1.005 * 310) + 2.5 * (1425 ÷ 2000) - 4.3

P = 778.875 + 1.78125 - 4.3 = 776.35625

P ≈ 776 kW

7 0
4 years ago
Liquids with convex meniscus​
kobusy [5.1K]

Mercury in a glass tube.

7 0
3 years ago
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