Ask question

Ask question

All categories

3 years ago

11

6. Sandy adds 3 sugar packets to her tea in order to make it sweeter. She stirs in the sugar and then takes a sip. She notices t

hat not all the sugar was dissolved. What type of solution did she make?
A. Saturated B.Unsaturated C.Supersaturated D.Mildly saturated

1 answer:

irina1246 [14]3 years ago

7 0

Answer:

C.Supersaturated

Explanation:

There are three types of solution:

<u>SATURATED SOLUTION</u>:

It is the solution that contains maximum amount of solute dissolved in a solution in normal conditions.

<u>UNSATURATED SOLUTION</u>:

It is the solution that contains less than the maximum amount of solute dissolved in a solution in normal conditions. It has space for more solute to be dissolved in it.

<u>SUPERSATURATED SOLUTION:</u>

It contains more than the maximum amount of solute dissolved in it. Such a solution has no capacity to dissolve any more solute under any conditions.

Since the sugar is no more dissolving in the tea and has settled down. Therefore, the solution is:

<u>C.Supersaturated</u>

You might be interested in

What is light energy

Oliga [24]

Light energy is defined as how nature moves energy at an extremely rapid rate, and it makes up about 99% of the body's atoms and cells, and signal all body parts to carry out their respective tasks. An example of light energy is the movement of a radio signal.

8 0

3 years ago

A 97.6-kg baseball player slides into second base. The coefficient of kinetic friction between the player and the ground is μk =

Mila [183]

Answer:

$v=6.65m/sec$

Explanation:

From the Question we are told that:

Mass $m=97.6$

Coefficient of kinetic friction $\mu k=0.555$

Generally the equation for Frictional force is mathematically given by

$F=\mu mg$

$F=0.555*97.6*9.8$

$F=531.388N$

Generally the Newton's equation for Acceleration due to Friction force is mathematically given by

$a_f=-\mu g$

$a_f=-0.555 *9.81$

$a_f=-54455m/sec^2$

Therefore

$v=u-at$

$v=0+5.45*1.22$

$v=6.65m/sec$

4 0

3 years ago

a ship travels a port p and travels 30 km due north. then it changes course and travels 20 km in a direction 30° east of north

liq [111]

When we represent what is given to us on a coordinate plane, we have a figure as shown in the attachment.

To find the distance between P and R, we have to find the Net Displacement of the ship (brown arrow in the figure).

For that, we use the rules for Vector addition.

We see that the first displacement $D_{1}$ = 30 km (blue arrow) is along the y-axis, but the second part of the ship's journey $D_{2}$ = 20 km (red arrow) is at an angle with reference to y-axis.

So, we first find the components of the red arrow along X and Y.

Component of $D_{2}$ along X-axis is given by $D_{2x} = D_{2} Sin 30$ = 10 km

Component of $D_{2}$ along Y-axis is given by $D_{2y} = D_{2} Cos 30$ = 17.32 km

We now add all the vectors along X and along Y separately.

Net Displacement along X $D_{netX} = 10 km$

Net Displacement along Y $D_{netY} = 30 + 17.32$ = 47.32 km

Now that we have the components of the net displacement along X and Y, we make use of Pythagorean Theorem to calculate the $D_{net}$

$D_{net} = \sqrt{D_{netX} ^{2} + D_{netY} ^{2}}$

Therefore, [tex]D_{net} = 48.37 km.

Hence, the distance between the ports P and R is 48 km.

6 0

4 years ago

The Hawaiian hot spot sits below the Pacific plate. As the plate moves over the hot spot, a chain of volcanoes is formed. The Ea

Viktor [21]

Answer:

$3.2048179721\times 10^{-9}\ m/s$

Explanation:

Assuming that the pacific plate moved 178 km in 1.76 million years.

s = Distance = 178 km

t = Time taken = 1.76 million years

Speed is given by

$v=\dfrac{s}{t}\\\Rightarrow v=\dfrac{178000}{1.76\times 10^6\times 365.25\times 24\times 3600}\\\Rightarrow v=3.2048179721\times 10^{-9}\ m/s$

The speed of the plate is $3.2048179721\times 10^{-9}\ m/s$

6 0

3 years ago

A mass of 1000 kg drops from a height of 10 m on a platform of negligible mass. It is desired to design a spring and dashpot on

Juliette [100K]

Answer:

k = $5\times 10^{4}\ N/m$

b = $0.707\times 10^{3}$

t = $7.1\times 10^{- 5}\ s$

Solution:

As per the question:

Mass of the block, m = 1000 kg

Height, h = 10 m

Equilibrium position, x = 0.2 m

Now,

The velocity when the mass falls from a height of 10 m is given by the third eqn of motion:

$v^{2} = u^{2} + 2gh$

where

u = initial velocity = 0

g = 10 $m/s^{2}$

Thus

$v = \sqrt{2\times 10\times 10} = 10\sqrt{2}\ m/s$

Force on the mass is given by:

F = mg = $1000\times 10 = 10000 N = 10\ kN$

Also, we know that the spring force is given by:

F = - kx

Thus

$k = \frac{F}{x} = \frac{10000}{0.2} = 5\times 10^{4}\ N/m$

Now, to find the damping constant b, we know that:

F = - bv

$b = \frac{F}{v} = \frac{10000}{10\sqrt{2}} = 0.707\times 10^{3}$

Now,

Time required for the platform to get settled to 1 mm or 0.001 m is given by:

$t = \frac{0.001}{v} = \frac{0.001}{10\sqrt{2}} = 7.1\times 10^{- 5}\ s$

4 0

3 years ago