Ask question

Ask question

All categories

3 years ago

12

The capitary action of water is caused the cohesion of water molecules to different types of molecules and the adhesion of water

molecules with other water molecules true or false

1 answer:

s344n2d4d5 [400]3 years ago

5 0

Answer:

This is True

Explanation:

I just did this exact unit in bio last week I hope this helps ;)

You might be interested in

A stone is thrown vertically into the air at an initial velocity of 96 ft/s. On Mars, the height s (in feet) of the stone above

vladimir1956 [14]

Answer:

240 ft

Explanation:

t = Time taken

u = Initial velocity = 96 ft/s

v = Final velocity

s = Displacement

a = Acceleration = 12 m/s² on Mars 32 ft/s² on Earth negative due to upward direction

Mars

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=96\times t+\frac{1}{2}\times -12\times t^2\\\Rightarrow s=96t-6t^2\ ft$

Earth

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=96\times t+\frac{1}{2}\times -32\times t^2\\\Rightarrow s=96t-16t^2\ ft$

Differentiating the first equation with respect to time we get

$\frac{ds}{dt}=96-12t$

Equating with zero

$0=96-12t\\\Rightarrow t=\frac{96}{12}=8\ s$

Differentiating the second equation with respect to time we get

$\frac{ds}{dt}=96-32t$

Equating with zero

$0=96-32t\\\Rightarrow t=\frac{96}{32}=3\ s$

Applying the time taken to the above equations, we get

$s=96t-6t^2\ ft\\\Rightarrow s=96\times 8-6\times 8^2\\\Rightarrow s=384$

$s=96t-16t^2\\\Rightarrow s=96\times 3-16\times 3^2\\\Rightarrow s=144$

Difference in height = 384-144 = 240 ft

The stone will travel 240 ft higher on Mars

6 0

3 years ago

A 1.5kg cart moves along a track at 0.20m/s until it hits a fixed bumper at the end of the track. Find the average force exerted

DochEvi [55]

Answer:

Net force will be 4.875 N

Explanation:

We have given mass of the cart m = 1.5 kg

Initial velocity of the cart = 0.2 m/sec

And final velocity of the car = - 0.125 m/sec ( Negative direction is due to opposite direction )

Instant of time $\Delta t=0.1sec$

Change in momentum is given by

$\Delta P=1.5\times (0.2-(-0.125))=1.5\times 0.325=0.4875kgm/sec$

Now force is given by

$F=\frac{\Delta P}{\Delta t}=\frac{0.4875}{0.1}=4.875N$

Net force will be 4.875 N

6 0

2 years ago

Why do elements within a group have similar chemical properties

CaHeK987 [17]

Elements<span> in the same </span>group<span> in the periodic table </span>have similar chemical properties<span>. This is because their atoms </span>have<span> the same number of electrons in the highest occupied energy level. </span>Group<span> 1 </span>elements<span> are reactive metals called the alkali metals.</span>Group<span> 0 </span>elements<span> are unreactive non-metals called the noble gases.
</span>

4 0

3 years ago

An LC circuit is built with a 20 mH inductor and an 8.0 PF capacitor. The capacitor voltage has its maximum value of 25 V at t =

Margaret [11]

Answer:

a) the required time is 0.6283 μs

b) the inductor current is 0.5 mA

Explanation:

Given the data in the question;

The capacitor voltage has its maximum value of 25 V at t = 0

i.e V $_m$ = V₀ = 25 V

we determine the angular velocity;

ω = 1 / √( LC )

ω = 1 / √( ( 20 × 10⁻³ H ) × ( 8.0 × 10⁻¹² F) )

ω = 1 / √( 1.6 × 10⁻¹³ )

ω = 1 / 0.0000004

ω = 2.5 × 10⁶ s⁻¹

a) How much time does it take until the capacitor is fully discharged for the first time?

V $_m$ = V₀sin( ωt )

we substitute

25V = 25V × sin( 2.5 × 10⁶ s⁻¹ × t )

25V = 25V × sin( 2.5 × 10⁶ s⁻¹ × t )

divide both sides by 25 V

sin( 2.5 × 10⁶ × t ) = 1

( 2.5 × 10⁶ × t ) = π/2

t = 1.570796 / (2.5 × 10⁶)

t = 0.6283 × 10⁻⁶ s

t = 0.6283 μs

Therefore, the required time is 0.6283 μs

b) What is the inductor current at that time?

$I$ (t) = V₀√(C/L) sin(ωt)

{ sin(ωt) = 1 )

$I$ (t) = V₀√(C/L)

we substitute

$I$ (t) = 25V × √( ( 8.0 × 10⁻¹² F ) / ( 20 × 10⁻³ H ) )

$I$ (t) = 25 × 0.00002

$I$ (t) = 0.0005 A

$I$ (t) = 0.5 mA

Therefore, the inductor current is 0.5 mA

8 0

3 years ago

Calculate the work against gravity required to build the right circular cone of height 4 m and base of radius 1.2 m out of a lig

Nana76 [90]

Answer:

Work done = 35467.278 J

Explanation:

Given:

Height of the cone = 4m

radius (r) of the cone = 1.2m

Density of the cone = 600kg/m³

Acceleration due to gravity, g = 9.8 m/s²

Now,

The total mass of the cone (m) = Density of the cone × volume of the cone

Volume of the cone = $\frac{1}{3}\pi r^2 h$

thus,

volume of the cone = $\frac{1}{3}\pi 1.2^2\times 4$ = 6.03 m³

therefore, the mass of the cone = 600 Kg/m³ × 6.03 m³ = 3619.11 kg

The center of mass for the cone lies at the $\frac{1}{4}$ times the total height

thus,

center of mass lies at, h' = $\frac{1}{4}\times4=1m$

Now, the work gone (W) against gravity is given as:

W = mgh'

W = 3619.11kg × 9.8 m/s² × 1 = 35467.278 J

4 0

3 years ago