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4 years ago

What times what equals 400

1 answer:

7 0

If you're willing to consider fractions or decimals,
then there are an infinite number of answers.
Like (2.5 x 160), and (15 x 26-2/3).

If you want to stick to only whole numbers,
then these 8 combinations do:

1, 400
2, 200
4, 100
5, 80
8, 50
10, 40
16, 25
20, 20

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On a floor directly underneath a second-floor balcony, there are several spherical drops of blood about 7 mm in diameter. Which

IrinaK [193]

The correct answer for this question is this one: "The drops dripped from a bloody knife about 2 ft above the ground."
On a floor directly underneath a second-floor balcony, there are several spherical drops of blood about 7 mm in diameter. The statement that best accounts for the drops is that the drops dripped from a bloody knife about 2 ft above the ground.

Hope this helps answer your question and have a nice day ahead.

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3 years ago

Adam is trying to identify a solid. He places a small amount of the substance in an acid solution and observes whether or not a

gayaneshka [121]

The answer for this one is chemical

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3 years ago

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A weather balloon is filled to a volume of 6000 L while it is on the ground, at a pressure of 1 atm and a temperature of 273 K.

avanturin [10]

Answer : The final volume of the balloon at this temperature and pressure is, 17582.4 L

Solution :

Using combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of gas = 1 atm

$P_2$ = final pressure of gas = 0.3 atm

$V_1$ = initial volume of gas = 6000 L

$V_2$ = final volume of gas = ?

$T_1$ = initial temperature of gas = 273 K

$T_2$ = final temperature of gas = 240 K

Now put all the given values in the above equation, we get the final pressure of gas.

$\frac{1atm\times 6000L}{273K}=\frac{0.3atm\times V_2}{240K}$

$V_2=17582.4L$

Therefore, the final volume of the balloon at this temperature and pressure is, 17582.4 L

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4 years ago

Read 2 more answers

A firefighting crew uses a water cannon that shoots water at 25.0 m/s at a fixed angle of 53.0° above the horizontal. The fire-f

zysi [14]

Answer:

8.8 m and 52.5 m

Explanation:

The vertical component and horizontal component of water velocity leaving the hose are

$v_v = vsin(\alpha) = 25sin(53^0) = 25*0.8 = 19.97 m/s$

$v_h = vcos(\alpha) = 25cos(53^0) = 25*0.6 = 15 m/s$

Neglect air resistance, vertically speaking, gravitational acceleration g = -9.8m/s2 is the only thing that affects water motion. We can find the time t that it takes to reach the blaze 10m above ground level

$s = v_vt + gt^2/2$

$10 = 19.97t - 9.8t^2/2$

$4.9t^2 - 19.97t + 10 = 0$

$t= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$t= \frac{19.9658877511823\pm \sqrt{(-19.9658877511823)^2 - 4*(4.9)*(10)}}{2*(4.9)}$

$t= \frac{19.9658877511823\pm14.24}{9.8}$

t = 3.49 or t = 0.58

We have 2 solutions for t, one is 0.58 when it first reach the blaze during the 1st shoot up, the other is 3.49s when it falls down

t is also the times it takes to travel across horizontally. We can use this to compute the horizontal distance between the fire-fighters and the building

$s_1 = v_ht_1 = 15*0.58 = 8.8 m$