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Masteriza [31]
2 years ago
6

What is one property of a suspension that is different from that of a solution or a colloid?

Physics
1 answer:
Step2247 [10]2 years ago
3 0

Answer:

A

Explanation:

If left to rest it will separate.  

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HELP ASAP!!!
miss Akunina [59]

Write each force in component form:

<em>v </em>₁ : 50 N due east   →   (50 N) <em>i</em>

<em>v</em> ₂ : 80 N at N 45° E   →   (80 N) (cos(45°) <em>i</em> + sin(45°) <em>j</em> ) ≈ (56.5 N) (<em>i</em> + <em>j</em> )

The resultant force is the sum of these two vectors:

<em>r</em> = <em>v </em>₁ + <em>v</em> ₂ ≈ (106.5 N) <em>i</em> + (56.5 N) <em>j</em>

Its magnitude is

|| <em>r</em> || = √[(106.5 N)² + (56.5 N)²] ≈ 121 N

and has direction <em>θ</em> such that

tan(<em>θ</em>) = (56.5 N) / (106.5 N)   →   <em>θ</em> ≈ 28.0°

i.e. a direction of about E 28.0° N. (Just to clear up any confusion, I mean 28.0° north of east, or 28.0° relative to the positive <em>x</em>-axis.)

4 0
3 years ago
Two students are watching a person riding a skateboard up and down a ramp. Each student shares what they think about the energy
avanturin [10]

Answer:

Explanation:

We know that , If the frictional force on a system is zero , then the total energy of a system will be conserved.

By using energy conservation

KE₁ +  U₁ = KE₂ + U₂

KE₁=Kinetic energy at location 1

U₁ =Potential energy at location 1

KE₂=Kinetic energy at location 2

U₂=Potential energy at location 2

Therefore, Raymond is thinking in a right way.

7 0
3 years ago
Salmon often jump waterfalls to reach their breeding grounds. Starting downstream, 3.04 m away from a waterfall 0.585 m in heigh
S_A_V [24]

Answer:

V₀ = 5.47 m/s

Explanation:

The jumping motion of the Salmon can be modelled as the projectile motion. So, we use the formula for the range of projectile motion here:

R = V₀² Sin 2θ/g

where,

R = Range of Projectile = 3.04 m

θ = Launch Angle = 41.7°

V₀ = Minimum Launch Speed = ?

g = 9.81 m/s²

Therefore,

3.04 m = V₀² [Sin2(41.7°)]/(9.81 m/s²)

V₀² = 3.04 m/(0.10126 s²/m)

V₀ = √30.02 m²/s²

<u>V₀ = 5.47 m/s</u>

6 0
3 years ago
An ambulance is traveling north at 55.9 m/s, approaching a car that is also traveling north at 28.4 m/s. The ambulance driver he
wlad13 [49]

Answer:

915 Hz

Explanation:

The observed frequency from a sound source is given as

f₀ = f [(v + v₀)/(v+vₛ)]

where

f₀ = observed frequency of the sound by the observer = ?

f = actual frequency of the sound wave = 983 Hz

v = actual velocity of the sound waves = 343 m/s

vₛ = velocity of the source of the sound waves = 55.9 m/s

v₀ = velocity of the observer = 28.4 m/s

f₀ = 983 [(343+28.4)/(343+55.9)]

f₀ = 915.2 Hz = 915 Hz

6 0
3 years ago
A car traveling on a flat (unbanked), circular track accelerates uniformly from rest with a tangential acceleration of 1.85 m/s2
Tasya [4]

Answer:

Coefficient of friction = 0.836

Explanation:

If v be the speed after one quarter of the circular path

v² = 2as = 2 x 1.85 x 2πr/4 ; v²/r = 1.85 x 3.14 = 5.8

tangential acceleration = 5.8 m/s²

radial acceleration = v² /r = 5.8

total acceleration = √2 x 5.8

m x√2 x 5.8 = m x g xμ

μ = √2 x 5.8 / 9.8 = 0.836

7 0
3 years ago
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