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Nadusha1986 [10]
3 years ago
15

PLEASE HELP AND BE RIGHT

Physics
2 answers:
algol [13]3 years ago
7 0

Answer:

density

Explanation:

physical properties include:

  • Physical properties are those that can be observed without changing the identity of the substance.
  • The general properties of matter such as color, density, hardness, are examples of physical properties.
  • Flammability and corrosion/oxidation resistance are examples of chemical properties.

chemical properties include:

  • Reactivity with other chemicals.
  • Toxicity.
  • Coordination number.
  • Flammability.
  • Enthalpy of formation.
  • Heat of combustion.
  • Oxidation states.
  • Chemical stability. #answerwithquality #BAL
gregori [183]3 years ago
3 0

Answer:

The answer is A. Density

Explanation:

Took the test and got the answer right

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Answer:

vivir y servir a Dios

Explanation:

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In the 1960s what dance was being created in the west coast
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The dance that was being created was Hip Hop dance!
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Read 2 more answers
An automobile accelerates uniformly from a speed of 60 km/hr (16.67 m/s) to a speed of 80 km/hr (22.2 m/s) in 5 s. Determine the
monitta

Answer:

The acceleration is 1.106\ m/s^2 and the distance covered is 97.17 m.

Explanation:

Given that,

Initial speed of an automobile, u = 60 km/hr = 16.67 m/s

Final speed of an automobile, v = 80 km/hr = 22.2 m/s

Time, t = 5 s

We need to find the acceleration of the car and the distance traveled in this 5 sec interval.  Let a is the acceleration. Using the definition of acceleration as :

a=\dfrac{v-u}{t}\\\\a=\dfrac{22.2-16.67}{5}\\\\a=1.106\ m/s^2

Let d is the distance covered. Using the third equation of motion to find it as follows :

v^2-u^2=2as\\\\s=\dfrac{v^2-u^2}{2a}\\\\s=\dfrac{(22.2)^2-(16.67)^2}{2\times 1.106}\\\\s=97.17\ m

So, the acceleration is 1.106\ m/s^2 and the distance covered is 97.17 m.

7 0
3 years ago
Positional and non positional different between​
Sergeeva-Olga [200]

Explanation:

In positional number system,

1. each symbol represents different value depending on the position they occupy in a number.

2. In positional number system, each system has a value that relates to the number directly next to it. The total value of a positional number is the total of the resultant value of all positions.

3. Example: 12 can be 1 x 10 + 2 x 1, 10 + 2= 12

whereas in Non-Positional number

1. In non-positional number system, each symbol represents the same value regardless of its position

2. In non-positional number system each symbol represents a number with its own place value.

3. Example: Roman number system where I for 1, II for 2 etc.

~~

6 0
3 years ago
A train starts at rest in a station and accelerates at a constant 0.987 m/s2 for 182 seconds. Then the train decelerates at a co
algol [13]

Answer:

\displaystyle X_T=66.6\ km

Explanation:

<u>Accelerated Motion </u>

When a body changes its speed at a constant rate, i.e. same changes take same times, then it has a constant acceleration. The acceleration can be positive or negative. In the first case, the speed increases, and in the second time, the speed lowers until it eventually stops. The equation for the speed vf at any time t is given by

\displaystyle V_f=V_o+a\ t

where a is the acceleration, and vo is the initial speed .

The train has two different types of motion. It first starts from rest and has a constant acceleration of 0.987 m/s^2 for 182 seconds. Then it brakes with a constant acceleration of -0.321 m/s^2 until it comes to a stop. We need to find the total distance traveled.

The equation for the distance is

\displaystyle X=V_o\ t+\frac{a\ t^2}{2}

Our data is

\displaystyle V_o=0,a=0.987m/s^2,\ t=182\ sec

Let's compute the first distance X1

\displaystyle X_1=0+\frac{0.987\times 182^2}{2}

\displaystyle X_1=16,346.7\ m

Now, we find the speed at the end of the first period of time

\displaystyle V_{f1}=0+0.987\times 182

\displaystyle V_{f1}=179.6\ m/s

That is the speed the train is at the moment it starts to brake. We need to compute the time needed to stop the train, that is, to make vf=0

\displaystyle V_o=179.6,a=-0.321\ m/s^2\ ,V_f=0

\displaystyle t=\frac{v_f-v_o}{a}=\frac{0-179.6}{-0.321}

\displaystyle t=559.5\ sec

Computing the second distance

\displaystyle X_2=179.6\times559.5\ \frac{-0.321\times 559.5^2}{2}

\displaystyle X_2=50,243.2\ m

The total distance is

\displaystyle X_t=x_1+x_2=16,346.7+50,243.2

\displaystyle X_t=66,589.9\ m

\displaystyle \boxed{X_T=66.6\ km}

8 0
4 years ago
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