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2 years ago

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100 POINTS! I will mark brainliest! Record your hypothesis as an “if, then” statement for the rate of dissolving the compounds:

Record your hypothesis as an “if, then” statement for the boiling point of the compounds:

1 answer:

love history [14]2 years ago

6 0

Answer:

<u><em>Rate of dissolving compounds:</em></u>

If we increase the temperature of the solution, then the dissolving compound would dissolve more easily.

<u><em>Boiling Point of Compounds:</em></u>

If the inter-molecular forces of any compound is really strong, then the boiling point of the compound would be really high.

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89. An electron is moving in a straight line with a velocity of 4.0×105 m/s. It enters a region 5.0 cm long where it undergoes a

ddd [48]

Explanation:

Given that,

Initial speed of the electron, $u=4\times 10^5\ m/s$

Distance, s = 5 cm = 0.05 cm

Acceleration of the electron, $a=6\times 10^{12}\ m/s^2$

(a) Let v is the electron's velocity when it emerges from this region. It can be calculated as :

$v^2=u^2+2as$

$v^2=(4\times 10^5)^2+2\times 6\times 10^{12}\times 0.05$

v = 871779.788 m/s

or

$v=8.71\times 10^5\ m/s$

(b) Let t is the time for which the electron take to cross the region. It can be calculated as:

$t=\dfrac{v-u}{a}$

$t=\dfrac{8.71\times 10^5-4\times 10^5}{6\times 10^{12}}$

$t=7.85\times 10^{-8}\ s$

Hence, this is the required solution.

4 0

2 years ago

Maria throws an apple vertically upward from a height of 1.3 m with an initial velocity of +2.8 m/s. Will the apple reach a frie

evablogger [386]

Answer:

No, the apple will reach 4.20041 m below the tree house.

Explanation:

t = Time taken

u = Initial velocity = 2.8 m/s

v = Final velocity = 0

s = Displacement

g = Acceleration due to gravity = -9.81 m/s² = a (negative as it is going up)

Equation of motion

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-2.8^2}{2\times -9.81}\\\Rightarrow s=0.39959\ m$

The height to which the apple above the point of release will reach is 0.39959 m

From the ground the distance will be 1.3+0.39959 = 1.69959 m

Distance from the tree house = 5.9-1.69959 = 4.20041 m

No, the apple will reach 4.20041 m below the tree house.

The values in the option do not reflect the answer.

5 0

2 years ago

on the surface of planet x a body with a mass of 10 kilograms weighs 40 newtons. The magnitude of the acceleration due to gravit

tamaranim1 [39]

Based on the Newton's second law of motion, the value of the net force acting on the object is equal to the product of the mass and the acceleration due to gravity. If we let a be the acceleration due to gravity, the equation that would allow us to calculate it's value is,
W = m x a
where W is weight, m is mass, and a is acceleration. Substituting the known values,
40 kg m/s² = (10 kg) x a
Calculating for the value of a from the equation will give us an answer equal to 4.
ANSWER: 4 m/s².

6 0

2 years ago

Read 2 more answers

A box rests on the (horizontal) back of a truck. The coefficient of static friction between the box and the surface on which it

vredina [299]

Answer:

The distance is 11 m.

Explanation:

Given that,

Friction coefficient = 0.24

Time = 3.0 s

Initial velocity = 0

We need to calculate the acceleration

Using newton's second law

$F = ma$ ...(I)

Using formula of friction force

$F= \mu m g$ ....(II)

Put the value of F in the equation (II) from equation (I)

$ma=\mu mg$ ....(III)

$a = \mu g$

Put the value in the equation (III)

$a=0.24\times9.8$

$a=2.352\ m/s^2$

We need to calculate the distance,

Using equation of motion

$s = ut+\dfrac{1}{2}at^2$

$s=0+\dfrac{1}{2}2.352\times(3.0)^2$

$s=10.584\ m\ approx\ 11\ m$

Hence, The distance is 11 m.

3 0

2 years ago

An engineer wants to design a circular racetrack of radius R such that cars of mass m can go around the track at speed without t

gtnhenbr [62]

1. $tan \theta = \frac{v^2}{Rg}$

For the first part, we just need to write the equation of the forces along two perpendicular directions.

We have actually only two forces acting on the car, if we want it to go around the track without friction:

- The weight of the car, mg, downward

- The normal reaction of the track on the car, N, which is perpendicular to the track itself (see free-body diagram attached)

By resolving the normal reaction along the horizontal and vertical direction, we find the following equations:

$N cos \theta = mg$ (1)

$N sin \theta = m \frac{v^2}{R}$ (2)

where in the second equation, the term $m\frac{v^2}{R}$ represents the centripetal force, with v being the speed of the car and R the radius of the track.

Dividing eq.(2) by eq.(1), we get the following expression:

$tan \theta = \frac{v^2}{Rg}$

2. $F=\frac{m}{R}(w^2-v^2)$

In this second situation, the cars moves around the track at a speed

$w>v$

This means that the centripetal force term

$m\frac{v^2}{R}$

is now larger than before, and therefore, the horizontal component of the normal reaction, $N sin \theta$ , is no longer enough to keep the car in circular motion.

This means, therefore, that an additional radial force F is required to keep the car round the track in circular motion, and therefore the equation becomes

$N sin \theta + F = m\frac{w^2}{R}$

And re-arranging for F,

$F=m\frac{w^2}{R}-N sin \theta$ (3)

But from eq.(2) in the previous part we know that

$N sin \theta = m \frac{v^2}{R}$

So, susbtituting into eq.(3),

$F=m\frac{w^2}{R}-m\frac{v^2}{R}=\frac{m}{R}(w^2-v^2)$

4 0

2 years ago