Which statement about DDT best shows how science has been impacted by society?

What keeps the particles of a liquid from spreading out to fill an entire container in the same way that gas particles do?

Sonbull [250]

b. the forces of attraction among them limit their motion.

7 0

3 years ago

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A rotating object starts from rest at t = 0 s and has a constant angular acceleration. At a time of t = 2.5 s the object has an

Evgesh-ka [11]

Answer:

52 rad

Explanation:

Using

Ф = ω't +1/2αt²................... Equation 1

Where Ф = angular displacement of the object, t = time, ω' = initial angular velocity, α = angular acceleration.

Since the object states from rest, ω' = 0 rad/s.

Therefore,

Ф = 1/2αt²................ Equation 2

make α the subject of the equation

α = 2Ф/t².................. Equation 3

Given: Ф = 13 rad, t = 2.5 s

Substitute into equation 3

α = 2(13)/2.5²

α = 26/2.5

α = 4.16 rad/s².

using equation 2,

Ф = 1/2αt²

Given: t = 5 s, α = 4.16 rad/s²

Substitute into equation 2

Ф = 1/2(4.16)(5²)

Ф = 52 rad.

6 0

2 years ago

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The Reynolds number, rho VD/mu, is a very important parameter in fluid mechanics. Verify that the Reynolds number is dimensionle

Agata [3.3K]

Answer:

Re = 1 10⁴

Explanation:

Reynolds number is

Re = ρ v D /μ

The units of each term are

ρ = [kg / m³]

v = [m / s]

D = [m]

μ = [Pa s]

The pressure

Pa = [N / m²] = [Kg m / s²] 1 / [m²] = [kg / m s²]

μ = [Pa s] = [kg / m s²] [s] = [kg / m s]

We substitute the units in the equation

Re = [kg / m³] [m / s] [m] / [kg / m s]

Re = [kg / m s] / [m s / kg]

RE = [ ]

Reynolds number is a scalar

Let's evaluate for the given point

Where the data for methane are:

viscosity μ = 11.2 10⁻⁶ Pa s

the density ρ = 0.656 kg / m³

D = 2 in (2.54 10⁻² m / 1 in) = 5.08 10⁻² m

Re = 0.656 4 2 5.08 10⁻² /11.2 10⁻⁶

Re = 1.19 10⁴

4 0

3 years ago

A small box is held in place against a rough vertical wall by someone pushing on it with a force directed upward at 27 ∘ above t

ycow [4]

Answer:

Explanation:

Applied force, F = 18 N

Coefficient of static friction, μs = 0.4

Coefficient of kinetic friction, μs = 0.3

θ = 27°

Let N be the normal reaction of the wall acting on the block and m be the mass of block.

Resolve the components of force F.

As the block is in the horizontal equilibrium, so

F Cos 27° = N

N = 18 Cos 27° = 16.04 N

As the block does not slide so it means that the syatic friction force acting on the block balances the downwards forces acting on the block .

The force of static friction is μs x N = 0.4 x 16.04 = 6.42 N .... (1)

The vertically downward force acting on the block is mg - F Sin 27°

= mg - 18 Sin 27° = mg - 8.172 ... (2)

Now by equating the forces from equation (1) and (2), we get

mg - 8.172 = 6.42

mg = 14.592

m x 9.8 = 14.592

m = 1.49 kg

Thus, the mass of block is 1.5 kg.

6 0

3 years ago

During a baseball game, a batter hits a high pop-up. If the ball remains in the air for 6.22 s, how high does it rise? The accel

BigorU [14]

Answer:

47.4 m

Explanation:

When an object is thrown upward, it rises up, it reaches its maximum height, and then it goes down. The time at which it reaches its maximum height is half the total time of flight.

In this case, the time of flight is 6.22 s, so the time the ball takes to reach the maximum height is

$t=\frac{6.22}{2}=3.11 s$

Now we consider only the downward motion of the ball: it is a free fall motion, so we can find the vertical displacement by using the suvat equation

$s=ut+\frac{1}{2}gt^2$

where

s is the vertical displacement

u = 0 is the initial velocity

t = 3.11 s is the time

$g=9.8 m/s^2$ is the acceleration of gravity (taking downward as positive direction)

Solving the formula, we find

$s=\frac{1}{2}(9.8)(3.11)^2=47.4 m$

7 0

3 years ago