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3 years ago

What is the relationshop between a pure substance and a mixture

1 answer:

7 0

The difference between a pure substance and a mixture is that mixtures are combinations of two or more pure substances, while pure substances cannot be separated into any other kinds of matter.

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Write a collision scenario here. If you choose your own collision, you can have neither, one, or both of the objects break. Be s

OleMash [197]

Answer:

My scenario would be A Car vs. a guard rail on a road. You have a car that is coming down a Highway at a speed of 43 Mph Miles per hour (69.2018 Kmh)

And it hits a steel guardrail and the car smashes in at the front and the guardrail is only bent while the car has the bumper and the hood along with the headlights and windshield along with the passenger side window break.

Explanation:

This is caused by so much force reacting from one object to another but also depends on molecular density.

5 0

3 years ago

Write the untis of following physical quantities electric current , force, frequency, density

Shtirlitz [24]

Answer:

1) electric current - ampere

2) force - joule

3 frequency - hertz

4) density - kilogram per cubic metre

pls follow me friends ..

4 0

3 years ago

Read 2 more answers

I am really struggling with this question because I can't find anything on aphelion and perihelion, it's not a topic we went ove

Hoochie [10]

I have a strange hunch that there's some more material or previous work
that goes along with this question, which you haven't included here.

I can't easily find the dates of Mercury's extremes, but here's some of the
other data you're looking for:

Distance at Aphelion (point in it's orbit that's farthest from the sun):
69,816,900 km
0. 466 697 AU

 Distance at Perihelion (point in it's orbit that's closest to the sun):
46,001,200 km
0.307 499 AU 

Perihelion and aphelion are always directly opposite each other in
the orbit, so the time between them is 1/2 of the orbital period.

Mercury's Orbital period = 87.9691 Earth days

1/2 (50%) of that is 43.9845 Earth days

The average of the aphelion and perihelion distances is

1/2 ( 69,816,900 + 46,001,200 ) = 57,909,050 km
or
1/2 ( 0.466697 + 0.307499) = 0.387 098 AU

This also happens to be 1/2 of the major axis of the elliptical orbit.

3 0

4 years ago

A stone is tied to a string and whirled at constant speed in a horizontal circle. The speed is then doubled without changing the

Tanya [424]

Answer:

Afterward the magnitude of the acceleration of the stone is four times as great

Explanation:

A stone tied to a string and whirled at a constant speed in a horizontal circle will have a centripetal acceleration given by

$a = \frac{v^{2} }{r}$

Where $a$ is the centripetal acceleration

$v$ is the speed

and $r$ is the radius of the circle

Here, the radius of the circle is the length of the string.

Now, from the question

The speed is then doubled without changing the length of the string,

Let the new speed be $v_{2}$ , that is

$v_{2} = 2v$

and without changing the length of the string means radius $r$ is constant.

To determine the magnitude of the acceleration of the stone afterwards,

Let the new acceleration be $a_{2}$ .

Then we can write that

$a_{2} = \frac{v_{2}^{2} }{r}$

From

$a = \frac{v^{2} }{r}$

$v = \sqrt{ar}$

Recall that

$v_{2} = 2v$

∴ $v_{2} = 2\sqrt{ar}$

Now, we will put the value of $v_{2}$ into

$a_{2} = \frac{v_{2}^{2} }{r}$

Then,

$a_{2} = \frac{(2\sqrt{ar}) ^{2} }{r}$

$a_{2} = \frac{4ar }{r}$

$a_{2} = 4a$

The new magnitude of the acceleration of the stone is four times the initial acceleration.

Hence,

Afterward the magnitude of the acceleration of the stone is four times as great

3 0

3 years ago

An 80 kg astronaut has gone outside his space capsule to do some repair work. Unfortunately, he forgot to lock his safety tether

Anna [14]

Answer:

the time taken by the astronaut to reach safety = 9.8 hr

Explanation:

The equation for intensity can be written as :

$I = \frac{P}{A}$

where :

$\frac{I}{c}= \frac{F}{A}$

Replacing that into the above previous equation; we have:

$\frac{P}{Ac}=\frac{F}{A}$

$F = \frac{P}{c}$

However ; the force needed to push the astronaut is as follows:

F = ma

where ;

m = mass of the astronaut and a = its acceleration

we as well say;

$\frac{P}{c} = ma$

$a = \frac{P}{mc}$

Replacing P with 1000 W ; m with 80 kg and $3*10^{8} \ m/s$ for c

Then; a = $\frac{1000 \ W}{(80)(3.0*10^8)}$

a = $4.2*10^{-8} \ m/s$

It is also known that the battery will run for one hour and after which the battery on the laser will run out

Then to determine the change in the position after the first hour ; we have:

$\Delta x_1 = \frac{1}{2}*4.2*10^{-8} \ m/s^2 ) (1.0 \ h)^2$

$\Delta x_1 = \frac{1}{2}*4.2*10^{-8} \ m/s^2 ) (1.0 *3600 s)^2$

= 0.27 m

Furthermore, the final velocity of the astronaut is determined as:

$v_1 = at_1$

where ;

$v_1 = final \ velocity$

replacing $t_1 = 1.0 \ h$ and a = $4.2*10^{-8} \ m/s$ ; Then:

$v_1 = (4.2*10^8 \ m/s * 1.0 \ h * \frac{ 3600\ s}{1.0 \ h})$

$v_1 = 1.51 *10^{-4} \ m/s$

Also; when he drifted 5.0 m away from the capsule; the distance is far short of the 5 m but he still have 9 hours left of oxygen . In addition to that, he acceleration is also zero and the final velocity remains the same, so:

To find the final distance traveled by the astronaut ;we have:

$\Delta x_2 = d - \Delta x_1$

where;

$\Delta x_2$ = the final distance

d = total distance

So;

$\Delta x_2 = 5 m - 0.27 m \\ \\ \Delta x_2 = 4.73 \ m$

The time taken to reach the final distance can be calculated as:

$t_2 = \frac{\Delta x_2 }{v_1}$

where;

$t_2$ = is the time to reach the final distance

Replacing 4.73 for ${\Delta x_2 }$ and $1.51*10^{-4}$ m/s for $v_1$

$t_2 = \frac{4.73 \ m }{1.51*10^{-4} \ m/s}$

$t_2 = 31500 \ s (\frac{1.0 \ h}{3600 \ s} )$

$t_2 = 8.8 \ h$

We knew the laser was operated for 1 hour; thus the total time taken by the astronaut to reach the final distance is the sum of the time taken to reach the final distance and the operated time of the laser.

Hence ; the time taken by the astronaut to reach safety = 9.8 hr

8 0

3 years ago