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ale4655 [162]
3 years ago
7

Need Help

Physics
1 answer:
ValentinkaMS [17]3 years ago
3 0

The option that correctly compares a pure substance to a mixture is: B. A pure substance can only be separated by chemical means, whereas some mixtures can be separated by physical means.

A pure substance can be defined as a substance that has a definite and constant (fixed) structure and they are typically made up of only a kind of particles. Thus, a pure substance is a chemical substance that has a constant (fixed) chemical composition and properties.

As a result of a pure substance being a chemical substance, it cannot be separated into its constituents by a physical separation technique or method.

Instead, a pure substance being a chemical substance can only be separated by chemical separation technique or method.

In contrast, mixtures can be separated by a physical separation technique or method.

Find more information: brainly.com/question/24615100

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The device is a burner from an electric stove. It is used to transfer heat to a pot by A. Conduction B. Convection C. Radiation
Irina-Kira [14]

I’m pretty positive that it’s A. Conduction. Just refer to the definition of it.

3 0
3 years ago
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It takes you 5 min to walk with an average velocity of .75 m/s to the north from the parking lot to the entrance of the amusemen
skad [1K]
A displacement is a vector quantity that takes into account the shortest distance from the starting point to the endpoint. 

The given above gave a time interval in minutes which needs to be converted to seconds. Given that each minute is 60 seconds, 5 minutes equal 300 seconds. To determine the distance, multiply time with speed. The product is 225 m. 

Thus, the displacement is 225 m. 
3 0
3 years ago
An ice sled powered by a rocket engine starts from rest on a large frozen lake and accelerates at +44 ft/s2. After some time t1,
mash [69]

Answer:

a) t₁ = 4.76 s, t₂ = 85.2 s

b) v = 209 ft/s

Explanation:

Constant acceleration equations:

x = x₀ + v₀ t + ½ at²

v = at + v₀

where x is final position,

x₀ is initial position,

v₀ is initial velocity,

a is acceleration,

and t is time.

When the engine is on and the sled is accelerating:

x₀ = 0 ft

v₀ = 0 ft/s

a = 44 ft/s²

t = t₁

So:

x = 22 t₁²

v = 44 t₁

When the engine is off and the sled is coasting:

x = 18350 ft

x₀ = 22 t₁²

v₀ = 44 t₁

a = 0 ft/s²

t = t₂

So:

18350 = 22 t₁² + (44 t₁) t₂

Given that t₁ + t₂ = 90:

18350 = 22 t₁² + (44 t₁) (90 − t₁)

Now we can solve for t₁:

18350 = 22 t₁² + 3960 t₁ − 44 t₁²

18350 = 3960 t₁ − 22 t₁²

9175 = 1980 t₁ − 11 t₁²

11 t₁² − 1980 t₁ + 9175 = 0

Using quadratic formula:

t₁ = [ 1980 ± √(1980² - 4(11)(9175)) ] / 22

t₁ = 4.76, 175

Since t₁ can't be greater than 90, t₁ = 4.76 s.

Therefore, t₂ = 85.2 s.

And v = 44 t₁ = 209 ft/s.

3 0
3 years ago
a ball is thrown vertically upward with an initial speed of 40 m/s. how high is the ball above the ground when it stops
NISA [10]

Answer:

80m, assuming g=10m/s^2

Explanation:

40m/s will be reduced to 0m/s in 4 seconds. 4 seconds x 40m/s would be 160m up, but you will only get half of that because you decelerate linearly to 0m/s. This leaves you with 4 x 20 = 80m.

5 0
3 years ago
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If an object accelerates from rest, what will its velocity be after 1.3 s if it has a constant acceleration of 9.1 m/s^2?
HACTEHA [7]

\text{Given that,}\\\\\text{Initial velocity,} ~v_0 = 0~ \text{m~s}^{-1}\\\\\text{Time,  t = 1.3~sec}\\\\\text{Acceleration, a = 9.1 m s}^{-2}\\\\\\\\\text{Velocity,}\\\\v = v_0  +at\\\\\implies v = 0 + 9.1 \times 1.3 = 11.83~~ \text{m~s}^{-1}

5 0
2 years ago
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