Help me out!! please!

Glycerin is poured into an open U-shaped tube until the height in both sides is 20 cm. Ethyl alcohol is then poured into one arm

lina2011 [118]

Answer:

Difference in height = 7.5 cm

Explanation:

We are given;.

Height of ethyl alcohol;h2 = 20 cm = 0.2 m

Density of glycerin: ρ1 = 1260 kg/m³

Density of ethyl alcohol; ρ2 = 790 kg/m³

To get the difference in height, the pressure at the top of the open end must be equal to the pressure at the point where the liquids do not mix since both points will be at different levels after the pouring.

Thus;

P1 = P2

Formula for pressure is; P = ρgh

Thus;

ρ1 × g × h1 = ρ2 × g × h2

g will cancel out to give;

ρ1 × h1 = ρ2× h2

Making h1 the subject, we have;

h1 = (ρ2× h2)/ρ1

h1 = (790 × 0.2)/1260

h1 = 0.125 m

Difference in height will be;

Δh = h2 - h1

Δh = 0.2 - 0.125

Δh = 0.075 m = 7.5 cm

4 0

3 years ago

You are trying to find out how high you have to pitch a water balloon in order for it to burst when it hits the ground. You disc

FrozenT [24]

Answer:

The balloon hit the ground with velocity -15.34 m/s

Explanation:

Lets explain how to solve the problem

You found that the best height to pitch a water balloon in order for it to

burst when it hits the ground is 12 meters.

We consider that the 12 meters is the maximum height, so the velocity

at this height is zero.

To find the velocity when the balloon hits the ground lets use the rule

v² = u² + 2gh, where v is the final velocity, u is the initial velocity, g is

the acceleration of gravity and h is the height.

u = 0 , h = 12 m , g = 9.8 m/s²

Substitute these values in the equation above

v² = 0 + 2(9.8)(12)

v² = 235.2

Take square root for both sides

v = ± $\sqrt{235.2}$

The velocity is downward, then it's a negative value

Then v = -15.34 m/s

The balloon hit the ground with velocity -15.34 m/s

6 0

3 years ago

Mohs scale is used in describing which characteristic? luster toughness hardness crystal form

Lady_Fox [76]

The answer is hardness not luster

3 0

3 years ago

Read 2 more answers

A 20-kg child running at 1.4 m/s jumps onto a playground merry-go-round that has mass 180 kg and radius 1.6m. She is moving tang

Dominik [7]

Answer:

ωf = 0.16 rad/s

Explanation:

Moment of inertia of the child = mr² = 20(1.6²) = 51.2 kg•m²

Moment of Inertia of the MGR = ½mr² = ½(180)1.6² = 230.4 kg•m²

(ASSUMING it is a uniform disk)

Initial angular momentum of the child = Iω = I(v/r) = 51.2(1.4/1.6) = 44.8 kg•m²/s

Conservation of angular momentum

44.8 = (51.2 + 230.4)ωf

ωf = 0.15909090...

4 0

3 years ago

An open rectangular tank 1 m wide and 2 m long contains gasoline to a depth of 1 m. If the height of the tank sides is 1.5 m, wh

lara [203]

Answer:

$a_y = 4.9\ m/s^2$

Explanation:

Given,

Width of rectangular tank, b = 1 m

Length of the tank, l = 2 m

height of the tank, d = 1.5 m

Depth of gasoline on the tank, h = 1 m

$\dfrac{dz}{dy}=-\dfrac{1.5-1}{1}$

$\dfrac{dz}{dy}=-0.5$

The differential form with the acceleration

$\dfrac{dz}{dy}=\dfrac{-a_y}{a_z + g}$

$-0.5=-\dfrac{a_y}{a_z + g}$

acceleration in z-direction = 0 m/s²

g = 9.8 m/s²

a_y is the horizontal acceleration of the gasoline.

$0.5=\dfrac{a_y}{0 + 9.8}$

$a_y = 9.8\times 0.5$

$a_y = 4.9\ m/s^2$

Hence, Horizontal acceleration of the gasoline before gasoline would spill is equal to 4.9 m/s²

3 0

3 years ago