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2 years ago

What is the melting point of sand

1 answer:

8 0

The kind of heat necessary to transform sand into a liquid state (eventually becoming glass) is much hotter than any sunny day. To make sand melt, you need to heat it to roughly 1700°C (3090°F), which is approximately the same temperature a space shuttle reaches as it re-enters earth's atmosphere.

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Why does refracting light bend when it enters a denser or less dense medium?

Rzqust [24]

Answer: The light bends because light travels fast but it slows down in a denser medium. For example light refracts in water or it bends after passing through air. When light passes through air ( a less dense medium ) then through water ( a more dense medium ) the beam of light bends because light travels more slowly in a denser medium then it picks up its pace again once it passes. The density of the substance determines how much the light is refracted. I hope this makes sense and I hope this answered your question!! :)

4 0

3 years ago

During an experiment of momentum, trolley, X, of mass (2.34 ± 0.01) kg is moving away from another trolley, Y, of mass (2.561 ±

Alla [95]

Answer:

P = 1 (14,045 ± 0.03 ) k gm/s

Explanation:

In this exercise we are asked about the uncertainty of the momentum of the two carriages

Δ (Pₓ / Py) =?

Let's start by finding the momentum of each vehicle

car X

Pₓ = m vₓ

Pₓ = 2.34 2.5

Pₓ = 5.85 kg m

car Y

Py = 2,561 3.2

Py = 8,195 kgm

How do we calculate the absolute uncertainty at the two moments?

ΔPₓ = m Δv + v Δm

ΔPₓ = 2.34 0.01 + 2.561 0.01

ΔPₓ = 0.05 kg m

Δ $P_{y}$ = m Δv + v Δm

ΔP_{y} = 2,561 0.01+ 3.2 0.001

ΔP_{y} = 0.03 kg m

now we have the uncertainty of each moment

P = Pₓ / $P_{y}$

ΔP = ΔPₓ/P_{y} + Pₓ ΔP_{y} / P_{y}²

ΔP = 8,195 0.05 + 5.85 0.03 / 8,195²

ΔP = 0.006 + 0.0026

ΔP = 0.009 kg m

The result is

P = 14,045 ± 0.039 = (14,045 ± 0.03 ) k gm/s

7 0

3 years ago

If 36 grams of water is to be heated from 24.0°C to 48°C to make a cup of tea, how much heat must be added? The specific heat of

Vinvika [58]

We will have the following:

$\begin{gathered} Q=mc\Delta T\Rightarrow Q=(36)(4.18)(48-24) \\ \\ \Rightarrow Q=3611.52 \end{gathered}$

So, the heat to add is 3611.52 Joules.

3 0

1 year ago

A cheap effective and simple way to help body recover is to get adequate amounts of. ?

skelet666 [1.2K]

Get adequate amounts of liquid

3 0

3 years ago

Read 2 more answers

If the coefficient of static friction is 0.357, and the same ladder makes a 58.0° angle with respect to the horizontal, how far

zavuch27 [327]

Answer: d= 0.57* l

Explanation:

We need to check that before ladder slips the length of ladder the painter can climb.

So we need to satisfy the equilibrium conditions.

So for ∑Fx=0, ∑Fy=0 and ∑M=0

We have,

At the base of ladder, two components N₁ acting vertical and f₁ acting horizontal

At the top of ladder, N₂ acting horizontal

And Between somewhere we have the weight of painter acting downward equal to= mg

So, we have N₁=mg

and also mg*d*cosФ= N₂*l*sin∅

So,

d= $\frac{N2}{mg}*l$ * tan∅

Also, we have f₁=N₂

As f₁= чN₁

So f₁= 0.357 * 69.1 * 9.8

f₁= 241.75

Putting in d equation, we have

d= $\frac{241.75}{69.1*9.8} *l$ * tan 58

d= 0.57* l

So painter can be along the 57% of length before the ladder begins to slip

3 0

3 years ago

Read 2 more answers