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3 years ago

I got 13 min to do 20 problems helpppp

Physics

1 answer:

Alina [70]3 years ago

3 0

Answer:

I believe it's 200 miles...

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Inserting the formulas you found for Xman(t) and Xbus(t) into the conditionXman(tcatch)=Xbus(tcatch) , you obtain the following-

lara [203]

Answer:

c > √(2ab)

Explanation:

In this exercise we are asked to find the condition for c in such a way that the results have been real

The given equation is

½ a t² - c t + b = 0

we can see that this is a quadratic equation whose solution is

t = [c ±√(c² - 4 (½ a) b)] / 2

for the results to be real, the square root must be real, so the radicand must be greater than zero

c² -2a b > 0

c > √(2ab)

3 0

3 years ago

Spiderman has a mass of 80 kg. He knows his webbing will break if it is exposed to a 200 N force. What is the maximum height he

likoan [24]

Answer:

This depends on the writers

if they want they can make spiderman deny the laws of nature

8 0

3 years ago

Hydrogen gas is generated when acids come into contact with certain metals. When excess hydrochloric acid reacts with 2.2 g of (

CaHeK987 [17]

To solve the problem it is necessary to apply the concepts related to Byle's Law and Avogadro's Law.

The ideal gas equation would help us find the final solution to the problem, defined by

$PV = nRT$

Where,

T= Temperature of the gas

R = Universal as constant

n = number of moles

V = Volume

P = Pressure

For our case we have that the mass of Zn is 2.2g in moles would be

$[tex]Zn = \frac{2.2}{65}$ [/tex]

$Zn = 0.0338$

We know that 1 mole of hydrogen gas is proceed by 1 mole of zinc and the result is $Zn^{2+}$ , then Hydrogen can produce the same quantity,

$H_2 = 0.0338$

Applying the previous equation we have that

$V= \frac{nRT}{P}$

$V = \frac{0.0338*0.08206*293.15}{0.98}$

$V = 0.829L$

Therefore the volume of hydrogen gas is collected is 0.829L

6 0

3 years ago

20. A mass of gas has a volume of 4 m3, a temperature of 290 K, and an absolute pressure of 475 kPa. When the gas is allowed to

aliina [53]

$Given:\\V_1=4m^3\\T_1=290K\\p_1=475kPa\\\\V_2=6.5m^3\\T_2=277K\\\\Find:\\p_2=?\\\\Solution:\\\\ \frac{pV}{T} =const.\\\\ \frac{p_1V_1}{T_1} = \frac{p_2V_2}{T_2} \\\\\frac{p_1V_1T_2}{T_1}=p_2V_2\\\\\frac{p_1V_1T_2}{T_1V_2}=p_2\\\\p_2=p_1 \frac{V_1}{V_2} \frac{T_2}{T_1} \\\\\\p_2=475kPa\cdot \frac{4m^3}{6.5m^3} \cdot \frac{277K}{290K} \approx 279.2kPa\\\\Correct\;is\;answer\;\;C$

7 0

3 years ago

The average weather of a particular place is

aniked [119]

<span>The average weather of a particular place is "Climate"

In short, Your Answer would be Option B

Hope this helps!</span>

3 0

3 years ago

Read 2 more answers