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3 years ago

What is word meaning "heated to a glow"?

Physics

1 answer:

adell [148]3 years ago

5 0

I think the word you want is "incandescent".

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2nd grade work. Anyone?

Temka [501]

Answer:

A. shadow

Explanation:

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3 years ago

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What does the dashed line represent in this diagram?

lys-0071 [83]

Motor branch circuit

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3 years ago

Lolz plz help me with this it’s due soon

White raven [17]

Answer:

Explanation:

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7 0

3 years ago

Water drips from the nozzle of a shower onto the floor 189 cm below. The drops fall at regular (equal) intervals of time, the fi

laiz [17]

Answer:

0.83999 m

0.20999 m

Explanation:

g = Acceleration due to gravity = 9.81 m/s² = a

s = 189 cm

$s=ut+\frac{1}{2}at^2\\\Rightarrow 1.89=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{1.89\times 2}{9.81}}\\\Rightarrow t=0.62074\ s$

When the time intervals are equal, if four drops are falling then we have 3 time intervals.

So, the time interval is

$t'=\dfrac{t}{3}\\\Rightarrow t'=\dfrac{0.62074}{3}\\\Rightarrow t'=0.206913\ s$

For second drop time is given by

$t''=2t'\\\Rightarrow t''=2\times 0.2069133\\\Rightarrow t''=0.4138266\ s$

Distance from second drop

$s=ut+\dfrac{1}{2}at^2\\\Rightarrow y''=ut''+\dfrac{1}{2}at''^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 9.81\times 0.4138266^2\\\Rightarrow s=0.839993\ m$

Distance from second drop is 0.83999 m

Distance from third drop

$s=ut+\dfrac{1}{2}at^2\\\Rightarrow y''=ut'+\dfrac{1}{2}at'^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 9.81\times 0.206913^2\\\Rightarrow s=0.20999\ m$

Distance from third drop is 0.20999 m

6 0

4 years ago

An object at rest on a flat, horizontal surface explodes into two fragments, one seven times as massive as the other. The heavie

IgorLugansk [536]

Answer:

The distance the lighter fragments slides is 343 m

Explanation:

Here, we have

Let the mass of the heavier fragments be m₁

Let the distance the heavy object slide be d₁

Let the mass of the lighter fragments be m₂

Let the final velocity of the heavier fragments be v₁

Let the final velocity of the lighter fragments be v₂

Let the distance the light object slide be d₂

The distance traveled by the heavy fragment = 7.00 m

Therefore since m₁ = 7 × m₂ we have

Initial total momentum = final total momentum

Since the initial total momentum = 0 we have

m₁·v₁ + m₂·v₂ = 0 or

7·m₂·v₁ = -m₂·v₂

∴ v₂ = 7·v₁

The net work done by the heavier block is

W $_{net, 7m}$ = μk × m₁ × g × d₁ = 1/2×m₂×v₂²

Also v₁² = 2μk·g·d₁ and

v₂² = 2μk·g·d₂ so that since v₂ = 7·v₁ or v₁ = v₂/7 we get

(v₂/7)² = 2μk·g·d₁

So v₂²/49 = 2μk·g·d₁

or (2μk·g·d₂)/49 = 2μk·g·d₁

∴ d₂/49 = d₁

d₂ = d₁×49 = 7 × 49 = 343 m

The distance the lighter fragments slides = 343 m.

7 0

3 years ago