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3 years ago

Which two mixtures are homogeneous?

Physics

2 answers:

MAVERICK [17]3 years ago

7 0

Answer:

A. Gasoline

B. Apple juice

Explanation:

A homogeneous mixture can be defined as any liquid, solid or gaseous mixture which has an identical or uniform composition and properties throughout any given sample of the mixture. In Chemistry, all solutions are considered to be a homogeneous mixture.

For instance, aqueous hydrogen chloride, HCl(aq) is a homogeneous mixture. The aqueous hydrogen chloride is a homogeneous mixture of water and hydrogen chloride. This ultimately implies that, aqueous hydrogen chloride HCl(aq) is a solution of hydrogen chloride in water and it is commonly referred to as Hydrochloric acid.

Hence, the two mixtures that are homogeneous are gasoline and apple juice.

Korvikt [17]3 years ago

6 0

Answer:

A. Gasoline

B. Apple juice

A homogeneous mixture is a mixture in which the composition is uniform throughout the mixture. The salt water described above is homogeneous because the dissolved salt is evenly distributed throughout the entire salt water sample. Often it is easy to confuse a homogeneous mixture with a pure substance because they are both uniform. The difference is that the composition of the substance is always the same. The amount of salt in the salt water can vary from one sample to another. All solutions would be considered homogeneous because the dissolved material is present in the same amount throughout the solution.

One characteristic of mixtures is that they can be separated into their components. Since each part of the mixture has not reacted with another part of the mixture, the identities of the different materials is unchanged.

Hence, the two mixtures that are homogeneous are gasoline and apple juice

Explanation:

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Gantt charts are A. widely used network techniques. B. not widely used. C. planning charts used to schedule resources and alloca

kotykmax [81]

Answer:

planning charts used to schedule resources and allocate time is the correct answer

Explanation:

Gantt Charts are a project planning tool used to track the projects from the start to endpoint.

Gantt charts are used for the preparation and planning of the projects.

By using Gantt charts we can assess how much time a project would take to complete and we can also determine resources required with the help of Gantt charts.

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4 years ago

A student weighing 700 N climbs at constant speed to the top of an 8 m vertical rope in 10 s. The average power expended by the

stich3 [128]

Answer:

3. 560W

Explanation:

First we calculate the work required for the 700N weigh student to climb up by 8 meters. Work is force times distance

W = Fs = 700 * 8 = 5600J

The average power is defined as the rate of energy expended over a unit of time. In order words, work divided by time duration

P = W/t = 5600 / 10 = 560 W

3 0

3 years ago

A 4.4-µF capacitor is initially connected to a 5.1-V battery. Once the capacitor is fully charged the battery is removed and a 2

Grace [21]

Question is incomplete. Missing part:

Find the charge on the capacitor at the following times:

1) t = 0 mu S

2) t = 1 mu S

3) t = 50 mu S

1) $22.4 \mu C$

We start by calculating the initial charge on the capacitor. For this, we can use the following relationship:

$C=\frac{Q_0}{V_0}$

where

C is the capacitance

Q0 is the initial charge stored

V0 is the initial potential difference across the capacitor

When the capacitor is connected to the battery, we have:

$C=4.4\mu F = 4.4\cdot 10^{-6}F$

$V_0 = 5.1 V$

Solving for $Q_0$ ,

$Q_0 = CV_0 = (4.4\cdot 10^{-6})(5.1)=2.24 \cdot 10^{-5} C = 22.4 \mu C$

So, when the battery is disconnected, this is the charge on the capacitor at time t = 0.

2) $20.0\mu C$

To find the charge on the capacitor at any other time t, we use the equation:

$Q(t) = Q_0 e^{-\frac{t}{RC}}$

where

$Q_0 = 22.4 \mu C$

t is the time

$R=2.0 \Omega$ is the resistance

$C=4.4\mu F$ is the capacitance

Therefore, at time $t=1 \mu s$ , we have:

$Q(t) = (22.4) e^{-\frac{1}{(2.0)(4.4)}}=20.0 \mu C$

3) $0.08 \mu C$

As before, we use again the equation:

$Q(t) = Q_0 e^{-\frac{t}{RC}}$

However, here the time to consider is

$t=50 \mu C$

Substituting into the formula,

$Q(t) = (22.4) e^{-\frac{50.0}{(2.0)(4.4)}}=0.08 \mu C$

4 0

3 years ago

Please help<br> Need this for today

adoni [48]

Explanation:

The magnitude of centripetal acceleration is:

a = v² / r

The velocity is:

v = d/t

v = 2 × 2π (5 m) / 60 s

v = 1.05 m/s

a = (1.05 m/s)² / (5 m)

a = 0.219 m/s²

The direction is towards the center. So at point A, the centripetal acceleration is 0.219 m/s² downward, and at point B, the centripetal acceleration is 0.219 m/s² upward.

5 0

3 years ago

28. (a) A daredevil is attempting to jump his motorcycle over a line of buses parked end to end by driving up a 32° ramp at a sp

Ad libitum [116K]

Answer:

a) # buses = 7

Explanation:

For this exercise we use the kinematic equations, let's find the time it takes to reach the same height

y = $v_{oy}$ t - ½ g t²

Let's decompose the speed, with trigonometry

v₀ₓ = v₀ cos θ

$v_{oy}$ = v₀ sin θ

v₀ₓ = 40 cos 32

v₀ₓ = 33.9 m / s

$v_{oy}$ = 40 sin32

$v_{oy}$ = 21.2 m / s

When it arrives it is at the same initial height y = 0

0 = ( $v_{oy}$ - ½ gt) t

That has two solutions

t = 0 when it comes out

t = 2 $v_{oy}$ / g when it arrives

t = 2 21.2 /9.8

t = 4,326 s

We use the horizontal displacement equation

x = vox t

x = 33.9 4.326

x = 146.7 m

To find the number of buses we can use a direct proportions rule

# buses = 146.7 / 20

# buses = 7.3

# buses = 7

The distance of the seven buses is

L = 20 * 7 = 140 m

b) let's look for the scope for this jump

R = vo2 sin2T / g

R = 40 2 without 2 32 /9.8

R = 146.7 m

As we can see the range and distance needed to pass the seven (7) buses is different there is a margin of error of 6.7 m in favor of the jumper (security)

7 0

3 years ago