A red car passes a blue car. Which is true?

Describe the relationship of attractive forces between molecules

Ede4ka [16]

<span>Describe the relationship of attractive forces between molecules and the ability of a solvent to dissolve a substance. Solvents can dissolve a substance only if the attraction of the solvent molecules is greater than the attraction between the molecules of the substance.</span>

4 0

3 years ago

two students are on a balcony 19.6 m above the street. one student throws a ball vertically downward at 14.7 m:ds. at the same i

NARA [144]

A. The difference in the two ball's time in the air is 3 seconds

B. The velocity of each ball as it strikes the ground is 24.5 m/s

C. The balls 0.500 s after they are thrown are 14.7 m apart

<h3>Further explanation</h3>

Acceleration is rate of change of velocity.

$\large {\boxed {a = \frac{v - u}{t} } }$

$\large {\boxed {d = \frac{v + u}{2}~t } }$

<em>a = acceleration ( m/s² )</em>

<em>v = final velocity ( m/s )</em>

<em>u = initial velocity ( m/s )</em>

<em>t = time taken ( s )</em>

<em>d = distance ( m )</em>

Let us now tackle the problem!

<u>Given:</u>

Initial Height = H = 19.6 m

Initial Velocity = u = 14.7 m/s

<u>Unknown:</u>

A. Δt = ?

B. v = ?

C. Δh = ?

<u>Solution:</u>

<h2>Question A:</h2><h3>First Ball</h3>

$h = H - ut - \frac{1}{2}gt^2$

$0 = 19.6 - 14.7t - \frac{1}{2}(9.8)t^2$

$0 = 19.6 - 14.7t - 4.9t^2$

$4.9t^2 + 14.7t - 19.6 = 0$

$t^2 + 3t - 4 = 0$

$(t + 4)(t - 1) = 0$

$(t - 1) = 0$

$\boxed {t = 1 ~ second}$

<h3>Second Ball</h3>

$h = H + ut - \frac{1}{2}gt^2$

$0 = 19.6 + 14.7t - \frac{1}{2}(9.8)t^2$

$0 = 19.6 + 14.7t - 4.9t^2$

$4.9t^2 - 14.7t - 19.6 = 0$

$t^2 - 3t - 4 = 0$

$(t - 4)(t + 1) = 0$

$(t - 4) = 0$

$\boxed {t = 4 ~ seconds}$

The difference in the two ball's time in the air is:

$\Delta t = 4 ~ seconds - 1 ~ second$

$\large {\boxed {\Delta t = 3 ~ seconds} }$

<h2>Question B:</h2><h3>First Ball</h3>

$v^2 = u^2 - 2gH$

$v^2 = (-14.7)^2 + 2(-9.8)(-19.6)$

$v^2 = 600.25$

$v = \sqrt {600.25}$

$\boxed {v = 24.5 ~ m/s}$

<h3>Second Ball</h3>

$v^2 = u^2 - 2gH$

$v^2 = (14.7)^2 + 2(-9.8)(-19.6)$

$v^2 = 600.25$

$v = \sqrt {600.25}$

$\boxed {v = 24.5 ~ m/s}$

The velocity of each ball as it strikes the ground is 24.5 m/s

<h2>Question C:</h2><h3>First Ball</h3>

$h = H - ut - \frac{1}{2}gt^2$

$h = 19.6 - 14.7(0.5) - \frac{1}{2}(9.8)(0.5)^2$

$\boxed {h = 11.025 ~ m}$

<h3>Second Ball</h3>

$h = H + ut - \frac{1}{2}gt^2$

$h = 19.6 + 14.7(0.5) - \frac{1}{2}(9.8)(0.5)^2$

$\boxed {h = 25.725 ~ m}$

The difference in the two ball's height after 0.500 s is:

$\Delta h = 25.725 ~ m - 11.025 ~ m$

$\large {\boxed {\Delta h = 14.7 ~ m} }$

<h3>Learn more</h3>

Velocity of Runner : brainly.com/question/3813437
Kinetic Energy : brainly.com/question/692781
Acceleration : brainly.com/question/2283922
The Speed of Car : brainly.com/question/568302

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle

6 0

2 years ago

Do radio waves travel at the speed of sound

adell [148]

Sound travels at approximately 1,100 feet per second (766 miles per hour). Radio waves travel at the speed of light, which is approximately 186,000 miles per second. This means that in the time radio waves travel the length of a football field, light can travel further than all the way around the world.

8 0

2 years ago

Read 2 more answers

As 390 g of hot milk cools in a mug it transfers 30 000 j of heat to the environment. whats is the temperature change of the mil

Fed [463]

You have to use the specific heat equation.

Q = cmΔT where Q is the energy, c is specific heat, m is mass, and ΔT is change in temp.

So we can substitute our variables into the equation.

30000J = (390g)(3.9J*g/C)ΔT

Solving for ΔT, we get:

30000J/[(390g)*(3.9J*g/C) = ΔT

ΔT = 19.72386588C

I'm assuming the temperature is C, since it was not specified.

Hope this helps!

8 0

3 years ago

There is a seasaw that's holding two men. The seesaw has a length of 18m that can pivot from a point at its center. Man 1 has a

Alex

Answer:

Distance= 2.3864m

Explanation:

So that the balance is in equilibrium parallel to the floor, we must match the moment each man makes with respect to the pivot point.

In many cases the point of application of force does not coincide with the point of application in the body. In this case the force acts on the object and its structure at a certain distance, by means of an element that transfers that action of this force to the object.

This combination of force applied by the distance to the point of the structure where it is applied is called the moment of force F with respect to the point. The moment will attempt a rotation shift or rotation of the object. The distance from the force to the point of application is called the arm.

Mathematically it is calculated by expression:

M= F×d

The moment caused by the first man is:

M1= 75kg × (9.81m/s²) × 1.75m= 1287.5625 N×m

The moment caused by the second man must be equal to that caused by the first by which:

M2= 1287.5625 N×m= 55kg × (9.81m/s²) × distance ⇒

⇒distance= (1287.5625 N×m)/( (55kg × (9.81m/s²) )= 2.3864m

At this distance from the pivot point, the second should sit down so that the balance is balanced parallel to the ground.

3 0

2 years ago