All categories

3 years ago

Explain the main difference between mixtures and compounds

1 answer:

8 0

The different substances are not chemically joined together. Each substance in themixture, keeps its own properties. The compound has properties different from the elements it contains. Each substance is easily separated from the mixture.

You might be interested in

The period of a simple pendulum in a grandfather clock on another planet is 1.80 s. What is the acceleration due to gravity (in

weeeeeb [17]

Answer:

12.17 m/s²

Explanation:

The formula of period of a simple pendulum is given as,

T = 2π√(L/g)........................ Equation 1

Where T = period of the simple pendulum, L = length of the simple pendulum, g = acceleration due to gravity of the planet. π = pie

making g the subject of the equation,

g = 4π²L/T²................... Equation 2

Given: T = 1.8 s, l = 1.00 m

Constant: π = 3.14

Substitute into equation 2

g = (4×3.14²×1)/1.8²

g = 12.17 m/s²

Hence the acceleration due to gravity of the planet = 12.17 m/s²

4 0

3 years ago

Air pressure is 1.0 · 105 N/m2, air density is 1.3 kg/m3, and the density of soft drinks is 1.0 · 103 kg/m3. If one blows carefu

natita [175]

Answer:

v = 27.456 m/s

Explanation:

The support pressure needed of the water in the straw can be calculated by the formula

Given that,

P = r*g*h

= 1000*9.8*0.05 Pa.= 490 Pa

This pressure is compensated by 0.5*r*v^2 of the air,

Hence,

0.5*1.3*v^2 = 490

velocity of air blown into the straw =

v = 27.456 m/s

8 0

3 years ago

A jogger runs at a constant rate of 10.0 m every 2.0 seconds. The jogger starts at the origin and runs in the positive direction

Elis [28]

Answer:

(a) 25 m

(b) 75 m

Explanation:

Given that the jogger runs at a constant rate of 10.0 m every 2.0 seconds.

So, the speed of the jogger,

$v=\frac{10}{2}=5m/s\;\cdots(i)$

Let d be the distance covered by him in time, t s.

As distance=(speed) x (time)

So, $d=vt$

From equation (i)

$\Rightarrow d=5t\;\cdots(ii)$

As the jogger starts from origin, so, the distance, $d$ , also represents the position of the jogger at the time $t$ s.

The position-time graph has been shown.

(a) From equation (ii), for t=5.0 s

$d=5\times 5=25 m$

So, the jogger is at a distance of 25 m from the origin.

(b) Similarly, for t=15.0 s

$d=5\times 15=75 m$

So, the jogger is at a distance of 75 m from the origin.

8 0

3 years ago

The water in a river flows uniformly at a constant speed of 2.50 m/s between parallel banks 80.0 m apart. You are to deliver a p

NISA [10]

Answer:

a) The swimmer should travel perpendicular to the bank to minimize the spent in getting to the other side.

b) 133.33 m

c) 53.13°

d) 106.67 m

Explanation:

a) The swimmer should travel perpendicular to the bank to minimize the spent in getting to the other side.

b) velocity = distance * time

Let the velocity of the swimmer be $v_{s}$ = 1.5 m/s

The separation of the two sides of the river, d = 80 m

The time taken by the swimmer to get to the other end of the river bank,

$t = \frac{d}{v_{s} }$

t = 80/1.5

t = 53.33 s

The swimmer will be carried downstream by the river through a distance, s

Let the velocity of the river be $v_{r}$ = 2.5 m/s

$S = v_{r} t$

S = 53.33 * 2.5

S = 133.33 m

c) To minimize the distance traveled by the swimmer, his resultant velocity must be perpendicular to the velocity of the swimmer relative to water

That is ,

$cos \theta = \frac{v_{s} }{v_{r} } \\cos \theta = 1.5/2.5\\cos \theta = 0.6\\\theta = cos^{-1} 0.6\\\theta = 53.13^{0}$

d) Downstream velocity of the swimmer, $v_{y} = v_{s} sin \theta\\$

$v_{y} = 1.5 sin 53.13\\v_{y} = 1.2 m/s$

The vertical displacement is given by, $y = v_{y} t$

80 = 1.2 t

t = 80/1.2

t = 66.67 s

the horizontal speed,

$v_{x} = 2.5 - 1.5cos53.13\\v_{x} = 1.6 m/s$

The downstream horizontal distance of the swimmer, $x = v_{x} t$

x = 1.6 * 66.67

x = 106.67 m

7 0

3 years ago

Hii! help asap. i’ll give brainliest thanks!

o-na [289]

I believe it’s the mass of the box but I don’t no if I’m right

Hope this helped

5 0

3 years ago