All categories

3 years ago

Relative densoty hydrometer puridy of milk

Physics

2 answers:

abruzzese [7]3 years ago

8 0

Answer:

A lactometer is a specialized hydrometer used measure the relative density and purity of milk. A lactometer has graduations from 1 3 (top-bottom) between the uppermost marking of W representing the density of water and the lowermost marking of M representing the density of pure milk.

Explanation:

hope it helps

good day ✨✌️

Dmitrij [34]3 years ago

7 0

Answer:

Hope this can help u

Explanation:

A lactometer is a specialized hydrometer used measure the relative density and purity of milk. ... If the milk is pure the readings will be close to 1 and if the amount of water in the milk increases then the markings will be close to 3.

You might be interested in

Two parallel-plate capacitors, identical except that one has twice the plate separation of the other, are charged by the same vo

erik [133]

Answer:

The capacitor having less distance of separation has a stronger electric field.

Explanation:

The capacitors are identical and only difference between them is that one has twice the plate separation of the other. Therefore, capacitance of the given capacitors C1 and C2 is,

C1= Aε/d and C2=Aε/2d

The charges Q1 and Q2 on the capacitors of capacitance C1 and C2 respectively, is then given by the equation,

Q1=VC1

Q1=VAε/d

Q2=VC2

Q2=VAε/2d

Therefore, the surface charge density σ1 and σ2 for the capacitors is,

σ1=Q1/A

σ1=VAε/(d*A)

σ1=Vε/d

Similarly,

σ2=Q2/A

σ2=Vε/2d

The electric field between the plates is directly proportional to the surface charge density. And so electric field is inversely proportional to the distance of separation. Therefore the capacitor whose distance of separation is less has a stronger electric field.

Check out similar questions.

brainly.com/question/14744776

#SPJ10

8 0

2 years ago

An object of mass m is dropped from height h above a planet of mass M and radius R.Find an expression for the object's speed as

Mnenie [13.5K]

Answer:

Explanation:

Mass of object = m

Height above planet = h

Mass of planet = M

Radius of planet = R

As we have to find out velocity, so let's apply the law of conservation of energies on initial( when the object was at height) and final( when object hit the surface points.

Initial energy = Final energy

$K_{i} + U_{i} = K_{f} + U_{f}$

$\frac{1}{2}mv_{i} ^{2} - \frac{GMm}{h+R} = \frac{1}{2}mv_{f} ^{2} - \frac{GMm}{R}\\v_{i} = 0\\v_{f} ^{2} = 2\frac{GM}{R} - 2 \frac{GM}{h+R}$

$v_{f} =\sqrt{\frac{2GMh}{R(R+h)} }$

4 0

4 years ago

A trebuchet was a hurling machine built to attack the walls of a castle under siege. A large stone could be hurled against a wal

Studentka2010 [4]

(a) 18.9 m/s

The motion of the stone consists of two independent motions:

- A horizontal motion at constant speed

- A vertical motion with constant acceleration ( $g=9.8 m/s^2$ ) downward

We can calculate the components of the initial velocity of the stone as it is launched from the ground:

$u_x = v_0 cos \theta = (25.0)(cos 41.0^{\circ})=18.9 m/s\\u_y = v_0 sin \theta = (25.0)(sin 41.0^{\circ})=16.4 m/s$

The horizontal velocity remains constant, while the vertical velocity changes due to the acceleration along the vertical direction.

When the stone reaches the top of its parabolic path, the vertical velocity has became zero (because it is changing direction): so the speed of the stone is simply equal to the horizontal velocity, therefore

$v=18.9 m/s$

(b) 22.2 m/s

We can solve this part by analyzing the vertical motion only first. In fact, the vertical velocity at any height h during the motion is given by

$v_y^2 - u_y^2 = 2ah$ (1)

where

$u_y = 16.4 m/s$ is the initial vertical velocity

$v_y$ is the vertical velocity at height h

$a=g=-9.8 m/s^2$ is the acceleration due to gravity (negative because it is downward)

At the top of the parabolic path, $v_y = 0$ , so we can use the equation to find the maximum height

$h_{max} = \frac{-u_y^2}{2a}=\frac{-(16.4)^2}{2(-9.8)}=13.7 m$

So, at half of the maximum height,

$h = \frac{13.7}{2}=6.9 m$

And so we can use again eq(1) to find the vertical velocity at h = 6.9 m:

$v_y = \sqrt{u_y^2 + 2ah}=\sqrt{(16.4)^2+2(-9.8)(6.9)}=11.6 m/s$

And so, the speed of the stone at half of the maximum height is

$v=\sqrt{v_x^2+v_y^2}=\sqrt{18.9^2+11.6^2}=22.2 m/s$

(c) 17.4% faster

We said that the speed at the top of the trajectory (part a) is

$v_1 = 18.9 m/s$

while the speed at half of the maximum height (part b) is

$v_2 = 22.2 m/s$

So the difference is

$\Delta v = v_2 - v_2 = 22.2 - 18.9 = 3.3 m/s$

And so, in percentage,

$\frac{\Delta v}{v_1} \cdot 100 = \frac{3.3}{18.9}\cdot 100=17.4\%$

So, the stone in part (b) is moving 17.4% faster than in part (a).

4 0

4 years ago

Speed with direction is called

Naddika [18.5K]

Velocity
if you change direction, say turn around, so does your velocity
an increase in velocity is called accelleration

7 0

3 years ago

The singularity at the center of a black hole is predicted to be a region of zero volume and infinite density that contains all

Reika [66]

Answer:

Yes

Explanation:

Science is a quest to find answers and patterns in the universe which can then be relied on for predicting future occurrence. Finding these singular points or at least to some extent, the cause and formation of this point will help with our further and deeper understanding of the universe, and we can then be able to predict when, where or when this singularities occurs.

Answer:

Explanation:

Spending a lot of time on these singularities that are hidden from view behind an event horizon that prevents any knowledge about the singularity reaching the outside universe is an absolute waste of time. This reflects no practice of science, since science is all about finding answers. If these answers we seek to find no matter how much we probe, cannot be gotten from further studies of these occurrences then it will be best to leave what can't be found. The fact that all known laws of physics breaks down at these point is a pointer that any physical means that will be employed to probe these singularities will yield no result within our known physical laws.

I will go with yes, with hope that advance in science will someday break these barriers or reconcile our known laws of physics with whatever physical law prevails within these singularities.

5 0

3 years ago

Relative densoty hydrometer puridy of milk​

Relative densoty hydrometer puridy of milk