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3 years ago

9

Which mixtures could be separated using evaporation? Check all that apply. fat in milk salt in ocean water mud in pond water but

ter in cookie dough vinegar in salad dressing

2 answers:

Charra [1.4K]3 years ago

8 0

Answer:

B. salt in ocean water

C. mud in pond water

Your welcome :D

salantis [7]3 years ago

5 0

I would go with salt in ocean water as when you heat it and the water begins to evaporate it will leave the salt behind

Also water in mud as the water would evaporate and leave the mud residue behind

GOOD LUCK
BRAINLIEST IF HELPED

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What force must the deltoid muscle provide to keep the arm in this position?

ruslelena [56]

Answer:

Deltoid Force, $F_{d} = \frac {r_{a}mgsin\alpha_{a}}{r_{d}sin\alpha_{d}}$

Additional Information:

Some numerical information are missing from the question. However, I will derive the formula to calculate the force of the deltoid muscle. All you need to do is insert the necessary information and calculate.

Explanation:

The deltoid muscle is the one keeping the hand arm in position. We have two torques that apply to the rotating of the arm.

1. The torque about the point in the shoulder for the deltoid muscle, $T_{Deltoid}$

2. The torque of the arm, $T_{arm}$

Assuming the arm is just being stretched and there is no rotation going on,

$T_{Deltoid}$ = 0

$T_{arm}$ = 0

⇒ $T_{Deltoid} = T_{arm}$

$r_{d}F_{d}sin\alpha_{d} = r_{a}F_{a}sin\alpha_{a}$

Where,

$r_{d}$ is radius of the deltoid

$F_{d}$ is the force of the deltiod

$\alpha_{d}$ is the angle of the deltiod

$r_{a}$ is the radius of the arm

$F_{a}$ is the force of the arm , $F_{a} = mg$ which is the mass of the arm and acceleration due to gravity

$\alpha_{a}$ is the angle of the arm

The force of the deltoid muscle is,

$F_{d} = \frac {r_{a}F_{a}sin\alpha_{a}}{r_{d}sin\alpha_{d}}$

but $F_{a} = mg$ ,

∴ $F_{d} = \frac {r_{a}mgsin\alpha_{a}}{r_{d}sin\alpha_{d}}$

7 0

2 years ago

A ball is thrown w a speed of 30m/s at an angle of 10. When is the vertical component of the velocity equal to zero

irga5000 [103]

Now the vertical velocity of the ball thrown at an angle 10° is given as

Voy(initial vertical velocity)= 30m/s x sin 10

Voy(initial vertical velocity)= 5.2m/s

Now the ball is decelerating with an acceleration due to gravity equivalent to 9.8m/s^2.

Let Vy be the final velocity and that is equal to zero in this case.

Now

Vy= Voy- tx9.8

Where t is the time at which the vertical velocity becomes 0.

Substituting the values we get

0= 5.2-tx9.8

9.8t=5.2

t=0.53 secs

5 0

2 years ago

Read 2 more answers

An electron starts from rest in a vacuum, in a region of strong electric field. The electron moves through a potential differenc

Semenov [28]

Answer:

The kinetic energy is $KE = 5.67*10^{-18} \ J$

Explanation:

From the question we are told that

The potential difference is $\Delta V = 36 \ volts$

The potential energy of the end is mathematically represented as

$PEs = - q * \Delta V$

q is the charge on an electron with a constant value of $q = 1.60 *10^{-19} \ C$

substituting values

$PE = - 1.60*10^{-19} * 36$

$PE = - 5.67*10^{- 18} \ J$

Now from the law of energy conservation

The $PE_e = KEe$

Where $KE _e$ is the potential energy at the end

So

$KE = 5.67*10^{-18} \ J$

The negative sign is not includes because kinetic energy can not be negative

7 0

3 years ago

A particle's trajectory is described by x = (0.5t^3-2t^2) meters and y = (0.5t^2-2t), where time is in seconds. What is the part

mestny [16]

Differentiate the components of position to get the corresponding components of velocity :

$v_x = \dfrac{\mathrm dx}{\mathrm dt} = \left(1.5\dfrac{\rm m}{\mathrm s^3}\right) t^2 - \left(4\dfrac{\rm m}{\mathrm s^2}\right)t$

$v_y = \dfrac{\mathrm dy}{\mathrm dt} = \left(1\dfrac{\rm m}{\mathrm s^2}\right)t-2\dfrac{\rm m}{\rm s}$

At <em>t</em> = 5.0 s, the particle has velocity

$v_x = \left(1.5\dfrac{\rm m}{\mathrm s^3}\right) (5.0\,\mathrm s)^2 - \left(4\dfrac{\rm m}{\mathrm s^2}\right)(5.0\,\mathrm s) = 17.5\dfrac{\rm m}{\rm s}$

$v_y = \left(1\dfrac{\rm m}{\mathrm s^2}\right)(5.0\,\mathrm s)-2\dfrac{\rm m}{\rm s} = 3.0\dfrac{\rm m}{\rm s}$

The speed at this time is the magnitude of the velocity :

$\sqrt{{v_x}^2 + {v_y}^2} \approx \boxed{17.8\dfrac{\rm m}{\rm s}}$

The direction of motion at this time is the angle $\theta$ that the velocity vector makes with the positive <em>x</em>-axis, such that

$\tan(\theta) = \dfrac{3.0\frac{\rm m}{\rm s}}{17.5\frac{\rm m}{\rm s}} \implies \theta = \tan^{-1}\left(\dfrac{3.0}{17.5}\right) \approx \boxed{9.73^\circ}$

4 0

2 years ago

When observed from earth, the wavelengths of light emitted by a star are shifted toward the red end of the electromagnetic spect

Musya8 [376]

Answer:

When observed from Earth, the wavelengths of light emitted by a star are shifted toward the red end of the electromagnetic spectrum because: the star is moving away from planet Earth.

A star is a giant astronomical or celestial object that contains a luminous sphere of plasma and bounded together by its own gravitational force.

A redshift can be defined as a displacement (shift) of the spectral lines of celestial or astronomical objects toward longer wavelengths (the red end of an electromagnetic spectrum), as a result of the Doppler effect.

Hence, a redshift is considered to be a subtle change in the color of visible electromagnetic radiation from stars (starlight), as observed from planet Earth.

In conclusion, a redshift occur when observing a star from planet Earth because the star is moving away from planet Earth.

Read more: brainly.com/question/17934476

Explanation:

have a great day :)

3 0

1 year ago