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3 years ago

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When sugar is added to water, the sugar dissolves and the resulting liquid is clear. Have the combined sugar and water formed a

pure substance or a mixture? Explain your a answer.

2 answers:

Nuetrik [128]3 years ago

7 0

It is mixture. Mixture is a physically combined. Pure substance is a substance that made out as a atom. So you can separate it by boiling water. As a result there is a sugar left. This method is called evaporation.

Mariana [72]3 years ago

5 0

I don't know xdddddddddd

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In magazine car tests an important indicator of performance is the zero to 60 mph (0 to 96.6 km/h) acceleration time. A time bel

slavikrds [6]

Answer:

2.73414 seconds

467622.66798 J

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

g = Acceleration due to gravity = 9.81 m/s² = a

$v=60\times \dfrac{1609.34}{3600}=26.822\ m/s$

$mgh=\dfrac{1}{2}mv^2\\\Rightarrow h=\dfrac{v^2}{2g}\\\Rightarrow h=\dfrac{26.822^2}{2\times 9.81}\\\Rightarrow h=36.66766\ m$

$s=ut+\frac{1}{2}at^2\\\Rightarrow 36.66766=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{36.66766\times 2}{9.81}}\\\Rightarrow t=2.73414\ s$

or

$v=u+at\\\Rightarrow t=\dfrac{v-u}{a}\\\Rightarrow t=\dfrac{26.822-0}{9.81}\\\Rightarrow t=2.73414\ s$

The time taken is 2.73414 seconds

The potential energy is given by

$U=mgh\\\Rightarrow U=1300\times 9.81\times 36.66766\\\Rightarrow U=467622.66798\ J$

The change in potential energy is 467622.66798 J

5 0

3 years ago

A block of mass m = 1.00 kg is attached to a spring of force constant k = 500 N/m. The block is pulled to a position xi= 5.00 cm

8_murik_8 [283]

Answer:

The speed of the block is 4.96 m/s.

Explanation:

Given that.

Mass of block = 1.00 kg

Spring constant = 500 N/m

Position $x_{i}=5.00\ cm$

Coefficient of friction = 0.350

(A). We need to calculate the speed the block has as it passes through equilibrium if the horizontal surface is friction less

Using formula of kinetic energy and potential energy

$\dfrac{1}{2}mv^2=\dfrac{1}{2}kx^2-\mu mgx$

Put the value into the formula

$\dfrac{1}{2}\times1.00\times v^2=\dfrac{1}{2}\times500\times(5.00\times10^{-2})-0.350\times1.00\times9.8\times5.00\times10^{-2}$

$v^2=\dfrac{2\times12.3285}{1.00}$

$v^2=24.657$

$v=4.96\ m/s$

Hence, The speed of the block is 4.96 m/s.

6 0

2 years ago

which of the following locations has the largest electric field? A)0.4 cm from a +3.0 uC point charge B)0.2 cm frm a +1.5 uC poi

JulsSmile [24]

<span>Electric field is proportional to q/d^2, where q is the magnitude of the charge and d is the distance. Since all the given units are identical, we can just compare their relative magnitudes without calculating for the exact values.
A) 3/(0.4)^2 = 18.75
B) 1.5/(0.2)^2 = 37.5
C) 6/(0.4)^2 = 37.5
D) 3/(0.2)^2 = 75
Therefore, choice D has the largest electric field of all.
</span>

7 0

3 years ago

An observer stands on the side of the front of a stationary train. When the train starts moving with constant acceleration, the

Zarrin [17]

To solve this problem we will use the linear motion description kinematic equations. We will proceed to analyze the general case by which the analysis is taken for the second car and the tenth. So we have to:

$x = v_0 t \frac{1}{2} at^2$

Where,

x= Displacement

$v_0$ = Initial velocity

a = Acceleration

t = time

Since there is no initial velocity, the same equation can be transformed in terms of length and time as:

$L = \frac{1}{2} a t_1 ^2$

For the second cart

$2L \frac{1}{2} at_2^2$

When the tenth car is aligned the length will be 9 times the initial therefore:

$9L = \frac{1}{2} at_3^2$

When the tenth car has passed the length will be 10 times the initial therefore:

$10L = \frac{1}{2}at_4^2$

The difference in time taken from the second car to pass it is 5 seconds, therefore:

$t_2-t_1 = 5s$

From the first equation replacing it in the second one we will have that the relationship of the two times is equivalent to:

$\frac{1}{2} = (\frac{t_1}{t_2})^2$

$t_1 = \frac{t_2}{\sqrt{2}}$

From the relationship when the car has passed and the time difference we will have to:

$(t_2-\frac{t_2}{\sqrt{2}}) = 5$

$t_2 (\sqrt{2}-1) = 3\sqrt{2}$

$t_2= (\frac{5\sqrt{2}}{\sqrt{2}-1})^2$

Replacing the value found in the equation given for the second car equation we have to:

$\frac{L}{a} = \frac{1}{4} (\frac{5\sqrt{2}}{\sqrt{2}-1})^2$

Finally we will have the time when the cars are aligned is

$18 \frac{1}{4} (\frac{5\sqrt{2}}{\sqrt{2}-1})^2 = t_3^2$

$t_3 = 36.213s$

The time when you have passed it would be:

$20\frac{1}{4} (\frac{5\sqrt{2}}{\sqrt{2}-1})^2 = t_4^2$

$t_4 = 38.172$

The difference between the two times would be:

$t_4-t_3 = 38.172-36.213 \approx 2s$

Therefore the correct answer is C.

4 0

3 years ago

An electric heater containing two heating wires X and Y is connected to a power supply of electromotive force(emf) 9.0V and negl

mina [271]

Answer:

0.4 ohms.

Explanation:

From the circuit,

The voltage reading in the voltmeter = voltage drop across each of the parallel resistance.

1/R' = 1/R1+1/R2

R' = (R1×R2)/(R1+R2)

R' = (2.4×1.2)/(2.4+1.2)

R' = 2.88/3.6

R' = 0.8 ohms.

Hence the current flowing through the circuit is

I = V'/R'................ Equation 1

Where V' = voltmeter reading

I = 6/0.8

I = 7.5 A

This is the same current that flows through the variable resistor.

Voltage drop across the variable resistor = 9-6 = 3 V

Therefore, the resistance of the variable resistor = 3/7.5

Resistance = 0.4 ohms.

7 0

3 years ago