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Bingel [31]
3 years ago
8

A car starts from rest and after 10 seconds is traveling at 20 m/s. Assuming that it continues to accelerate at the same rate it

will take another ________ seconds to reach 40 m/s.
Physics
1 answer:
Alenkinab [10]3 years ago
5 0

Assuming that it continues to accelerate at the same rate it will take another 10 seconds to reach 40 m/s.

Answer:

Explanation:

Since the first question states that there is a change in the velocity from rest to 20 m/s in 10 seconds time interval. So the acceleration experienced by the car during this 10 seconds should be determined first as follows:

Acceleration = (final velocity-initial velocity)/Time

Acceleration = (20-0)/10 = 2 m/s².

So this means the car is traveling with an acceleration of 2 m/s².

As it is stated that the car continues to move with same acceleration, then in the second case, the acceleration is fixed as 2 m/s², initial velocity as 20 m/s and final velocity as 40 m/s. So the time taken for the car to reach this velocity with the constant acceleration value will be as follows:

Time = Change in velocity/Acceleration

Time = (40-20)/2 = 20/2=10 s

So again in another 10 seconds by the car to reach 40 m/s from 20 m/s. Similarly the car will take a total of 20 seconds to reach from rest to 40 m/s value for velocity.

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A solution is prepared by dissolving 17.75 g sulfuric acid, h2so4, in enough water to make 100.0 ml of solution. if the density
Yuliya22 [10]

The solution of Sulfuric Acid (H2SO4) has the following mole fractions:

  • mole fraction (H2SO4)= 0.034
  • mole fraction (H2O)= 0.966

To solve this problem the formula and the procedure that we have to use is:

  • n = m / MW
  • = ∑ AWT
  • mole fraction = moles of A component / total moles of solution
  • ρ = m /v

Where:

  • m = mass
  • n = moles
  • MW = molecular weight
  • AWT = atomic weight
  • ρ = density
  • v = volume

Information about the problem:

  • m solute (H2SO4) = 17.75 g
  • v(solution) = 100 ml
  • ρ (solution)= 1.094 g/ml
  • AWT (H)= 1 g/mol
  • AWT (S) = 32 g/mol
  • AWT (O)= 16 g/mol
  • mole fraction(H2SO4) = ?
  • mole fraction(H2O) = ?

We calculate the moles of the H2SO4 and of the H2O from the Pm:

MW = ∑ AWT

MW (H2SO4)= AWT (H) * 2 + AWT (S) + AWT (O) * 4

MW (H2SO4)= (1 g/mol * 2) + (32,064 g/mol) + (16 g/mol * 4)

MW (H2SO4)= 2 g/mol + 32 g/mol + 64 g/mol

MW (H2SO4)=  98 g/mol

MW (H2O)= AWT (H) * 2 + AWT (O)

MW (H2O)= (1 g/mol * 2) + (16 g/mol)

MW (H2O)= 2 g/mol + 16 g/mol

MW (H2O)=  18 g/mol

Having the Pm we calculate the moles of H2SO4:

n = m / MW

n(H2SO4) = m(H2SO4) / MW (H2SO4)

n(H2SO4) = 17.75 g / 98 g/mol

n(H2SO4) = 0.1811 mol

With the density and the volume of the solution we get the mass:

ρ(solution)= m(solution) /v(solution)

m(solution) = v(solution) * ρ(solution)

m(solution) = 100 ml * 1.094 g/ml

m(solution) = 109.4 g

Having the mass of the solution we calculate the mass of the water in the solution:

m(H2O) = m(solution) - m solute (H2SO4)

m(H2O) = 109.4 g - 17.75 g

m(H2O) = 91.65 g

We calculate the moles of H2O:

n = m / MW

n(H2O) = m(H2O) / MW (H2O)

n(H2O) = 91.65 g / 18 g/mol

n(H2O) = 5.092  mol

We calculate the total moles of solution:

total moles of solution = n(H2SO4) + n(H2O)

total moles of solution = 0.1811 mol + 5.092  mol

total moles of solution = 5.2731 mol

With the moles of solution we can calculate the mole fraction of each component:

mole fraction (H2SO4)= moles of (H2SO4) / total moles of solution

mole fraction (H2SO4)= 0.1811 mol / 5.2731 mol

mole fraction (H2SO4)= 0.034

mole fraction (H2O)= moles of (H2O) / total moles of solution

mole fraction (H2O)= 5.092  mol / 5.2731 mol

mole fraction (H2O)= 0.966

<h3>What is a solution?</h3>

In chemistry a solution is known as a homogeneous mixture of two or more components called:

  • Solvent
  • Solute

Learn more about chemical solution at: brainly.com/question/13182946 and brainly.com/question/25326161

#SPJ4

8 0
2 years ago
If you have a 1.0 m aqueous solution of NaCl, by how much will it increase the water’s boiling point, if KB = 0.512 °C/m? In oth
fgiga [73]

<u>Answer:</u> The elevation in boiling point is 1.024°C.

<u>Explanation:</u>

To calculate the elevation in boiling point, we use the equation:

\Delta T_b=ik_b\times m

where,

i = Van't Hoff factor = 2 (for NaCl)

\Delta T_b = change in boiling point  = ?

k_b = boiling point constant = 0.512^oC/m

m = molality = 1.0 m

Putting values in above equation, we get:

\Delta Tb=2\times 0.512^oC/m\times 1.0m\\\\\Delta Tb=1.024^oC

Hence, the elevation in boiling point is 1.024°C.

5 0
4 years ago
Which term is a scalar quantity?
balu736 [363]
Vector quantities have both magnitude and direction. Distance is a scalar quantity. It refers only to how far an object has traveled. For example, 4 feet is a distance; it gives no information about direction.
7 0
4 years ago
Read 2 more answers
(4.3 x 10-2 )4.950 x 105)<br> O 21.285 x 103<br> O 2.13 x 104<br> O 21 x 103<br> O 2.1 x 104
Natalka [10]

Answer:

The answer is:

It's 21.285 × 103

7 0
3 years ago
A child of mass 40.0 kg is in a roller coaster car that travels in a loop of radius 7.00 m. at point a the speed of the car is 1
pav-90 [236]
I attached the missing picture.
The force of seat acting on the child is a reaction the force of child pressing down on the seat. This is the third Newton's law. The force of a child pressing down the seat and the force of the seat pushing up on the child are the same.
There two forces acting on the child. The first one is the gravitational force and the second one is centrifugal force. In this example, the force of gravity is always pulling down, but centrifugal force always acts away from the center of circular motion.
Part A
For point A we have:
F_a=F_cf-F_g
In this case, the forces are aligned, centrifugal is pointing up and gravitational is pulling down.
F_a=m\frac{v^2}{r}-mg=179 $N
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At the point, B situation is a bit more complicated. In this case force of gravity and centrifugal force are not aligned. We have to look at y components of this forces, y-axis, in this case, is just pointing upward.
F=F_{cf}\cos(30)-mg=m\frac{v^2}{r}\cos(30)-mg=153.2$N
Part C
The child will stay in place at point A when centrifugal force and force of gravity are in balance:
F_g=F_{cf}\\&#10;mg=m\frac{v^2}{r}\\&#10;gr=v^2\\&#10;v=\sqrt{gr}=8.29\frac{m}{s}

6 0
3 years ago
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