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3 years ago

15

Suppose you were coasting on a level surface on a bicycle, and there was no friction. Ignoring air resistance, what would happen

to your speed?

1 answer:

Alika [10]3 years ago

8 0

Nothing should happen to your speed

You might be interested in

how do the boiling point and freezing point of a solution of water and calcium chloride at standard pressure compare to the boil

Snezhnost [94]

Both the increase in the boling point and the depression on the freezing point are colliative properties.

This is, they are proportional to the number of particles dissolved in the solvent, which is measured by the molality of the solution and the factor i (Van'f Hoff).

The answer to the question is that 1) the boling point of a solution of water and calcium chloride at standard pressure will be higher than the normal boiling point of pure water, and 2) the freezing point of a solution of water and calcium chloride at standard pressure will be lower than the normal freezing point of pure water.

3 0

3 years ago

What is the value of work done on an object when a 50–newton force moves it 15 meters in the same direction as the force?

victus00 [196]

Answer:

work done = 750 J

Explanation:

Given data:

Force on object = 50 N

Distance covered = 15 m

Work done = ?

Solution:

W = F. d

W = work done

F = force

d = distance

Now we will put the values in formula.

W = 50 N × 15 m

W = 750 N. m = 750 J

7 0

3 years ago

Every Sunday, Sarah prepares blue fruit punch to

MrRissso [65]

Answer:

Fruit punch with a very dark blue color

Explanation:

8 0

3 years ago

Read 2 more answers

Starting with 2.50 mol of N2 gas (assumed to be ideal) in a cylinder at 1.00 atm and 20.0C, a chemist first heats the gas at con

arsen [322]

Answer:

a) $T_b=590.775k$

b) $W_t=1.08*10^4J$

d) $Q=3.778*10^4J$

d) $\triangle V=4.058*10^4J$

Explanation:

From the question we are told that:

Moles of N2 $n=2.50$

Atmospheric pressure $P=100atm$

Temperature $t=20 \textdegree C$

$t = 20+273$

$t = 293k$

Initial heat $Q=1.36 * 10^4 J$

a)

Generally the equation for change in temperature is mathematically given by

$\triangle T=\frac{Q}{N*C_v}$

Where

$C_v=Heat\ Capacity \approx 20.76 J/mol/K$

$T_b-T_a=\frac{1.36 * 10^4 J}{2.5*20.76 }$

$T_b-293k=297.775$

$T_b=590.775k$

b)

Generally the equation for ideal gas is mathematically given by

$PV=nRT$

For v double

$T_c=2*590.775k$

$T_c=1181.55k$

Therefore

$PV=Wbc$

$Wbc=(2.20)(8.314)(1181_590.778)$

$Wbc=10805.7J$

Total Work-done $W_t$

$W_t=Wab+Wbc$

$W_t=0+1.08*10^4$

$W_t=1.08*10^4J$

c)

Generally the equation for amount of heat added is mathematically given by

$Q=nC_p\triangle T$

$Q=2.20*2907*(1181.55-590.775)\\$

$Q=3.778*10^4J$

d)

Generally the equation for change in internal energy of the gas is mathematically given by

$\triangle V=nC_v \triangle T$

$\triangle V=2.20*20.76*(1181.55-293)k$

$\triangle V=4.058*10^4J$

3 0

3 years ago

Glycerol boils at a high temperature than water. What does this indicate about the attractive forces of glycerol?

forsale [732]

Glycerol attractive forces are great than water. The harder to break, the more energy is needed.

5 0

3 years ago