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4 years ago

A taxi hurries with a costant speed of 84km/h. How far it travel in 5 hour

1 answer:

8 0

Hey there! If the taxi travels 84km in one hour (without making any changes in speed aka remaining constant), then multiply 84(5) to find how many kilometers it would drive in one hour. 84(5)=420km

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Plz download the doc and answer its due in 20 mins like bro wth

4vir4ik [10]

Answer:

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Explanation:

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4 0

3 years ago

Read 2 more answers

.00 mol of gas occupy a volume of 112 L at a pressure of 1.00 atm. What is the kelvin temperature of the gas

artcher [175]

Sorry i couldnt help

5 0

3 years ago

What is the missing number in the sequence 3,5,9,_,15,23,33

balu736 [363]

Answer:

In every sequence even numbers are added in order.

3, 5, 9, 15, 23, 33

6 0

3 years ago

Read 2 more answers

A rectangular block of solid carbon (graphite) floats at the interface of two immiscible liquids.

Sergio039 [100]

Explanation:

Let us take the volume of block is x.

Since, the block is floating this means that it is in equilibrium. Formula to calculate net force will be as follows.

$F_{net} = Buoyancy force(F_{b}) - weight force(w)$

Also, buoyancy force $(F_{b})$ = (volume submerged in water × density of water) + (volume in oil × density of oil)

$(F_{b})$ = $(0.592 V \times \rho) + (1 - 0.592)V \times 1000 g$

= $(0.592 V \times \rho + 408 V)$ g

As, W = V × density of graphite × g

It is given that density of graphite is $2.16 g/cm^{3}$ or 2160 $kg/m^{3}$ .

So, W = 2160 V g

$F_{net}$ = (0.592 V \rho + 408 V) g - 2160 V g = 0

$0.592 \rho$ = 1752

$\rho$ = 2959.46 $kg/m^{3}$ or 2.959 $g/cm^{3}$ is the density of oil.

It is given that mass of flask is 124.8 g.

Mass of 35.3 $cm^{3}$ oil = $35.3 \times 2.959$ 104.7 g

Hence, in second weighing total mass will be calculated as follows.

(124.8 + 104.7) g

= 229.27 g

Thus, we can conclude that in the second weighing mass is 229.27 g.

5 0

3 years ago

Help please. 15.0 moles of gas are in a 3.00L tank at 23.4∘C . Calculate the difference in pressure between methane and an ideal

CaHeK987 [17]

Using PV=nRT or the ideal gas equation, we substitute n= 15.0 moles of gas, V= 3.00L, R equal to 0.0821 L atm/ mol K and T= 296.55 K and get P equal to 121.73 atm. The Van der waals equation is (P + n^2a/V^2)*(V-nb) = nRT. Substituting a=2.300L2⋅atm/mol2 and b=0.0430 L/mol, P is equal to 97.57 atm. The difference is <span>121.73 atm- 97.57 atm equal to 24.16 atm.</span>

3 0

3 years ago