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PolarNik [594]
3 years ago
7

Based on the position vs. time graph, which velocity vs. time graph would correspond to the data? A graph with horizontal axis t

ime (seconds) and vertical axis position (meters). A line runs straight upward from 0 seconds 0 minutes. A graph labeled velocity versus time with horizontal axis time (seconds) and vertical axis velocity (meters per second). A line runs in straight segments from 0 seconds 0 meters per second to 1 second 15 meters per second to 2 seconds 20 meters per second to 4 seconds 20 meters per second to 5 seconds 0 meters per second. A graph labeled velocity versus time with horizontal axis time (seconds) and vertical axis velocity (meters per second). A straight line begins at 0 seconds 0 meters per second and runs to 5 seconds 5 meters per second. A graph horizontal axis time (seconds) and vertical axis velocity (meters per second). A straight line begins at 0 seconds 2 meters per second and runs to 5 seconds 2 meters per second. A graph labeled velocity versus time with horizontal axis time (seconds) and vertical axis velocity (meters per second). A straight line begins at 0 seconds 10 meters per second and runs to 5 seconds 2 meters per second.
Physics
2 answers:
Dmitriy789 [7]3 years ago
6 0
A graph with horizontal axis time
Anit [1.1K]3 years ago
6 0

Answer:

A,C,E

Explanation:

I shows correct

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When 1.14 g of octane (molar mass = 114 g/mol) reacts with excess oxygen in a constant volume calorimeter, the temperature of th
balandron [24]

Answer:

Q (reaction) = -69.7 kJ

Explanation:

Octane reacts with oxygen to give carbon dioxide and water.

C₈H₁₈ + 25 O₂ ---> 16 CO₂ +18 H₂O

This reaction is exothermic in nature. Therefore, the energy is released into the atmosphere. This reaction took place in a calorimeter, there the temperature (T) increases by 10 C. The heat capacity of the calorimeter is 6.97 kJ/C

The heat (q) of the reaction is calculated as follows:

Q= -cT, where c is the heat capacity of the calorimeter and T is the increase in temperature

q = -(6.97) x (10) = -69.7kJ

<em>Since the heat capacity is given in kilo -joule per degree Celsius, therefore, the mass of octane is not required </em>

7 0
2 years ago
The bright-line spectrum of an element in the gaseous phase is produced as 1. protons move from lower energy states to higher en
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Answer:

4. Electrons move from higher energy states to lower energy states.

Explanation:

When electrons fall from a higher (excited) energy state to a lower energy state, it loses/gives out energy.

This energy is given out by the emission of photons (quanta of light) by the electron.

7 0
3 years ago
A wooden block meauring 40cm x 10cm x 5cm has a mass 850gm . find the density of wood?
Lelechka [254]

Answer:

Explanation:

Density = Mass / Volume = 850 / 40*10*5 = 0.425 g /cm^3

5 0
2 years ago
An electron and a proton are held on an x axis, with the electron at x = + 1.000 m
mixas84 [53]

Answer:

  r2 = 1 m

therefore the electron that comes with velocity does not reach the origin, it stops when it reaches the position of the electron at x = 1m

Explanation:

For this exercise we must use conservation of energy

the electric potential energy is

          U = k \frac{q_1q_2}{r_{12}}

for the proton at x = -1 m

          U₁ =- k \frac{e^2 }{r+1}

for the electron at x = 1 m

          U₂ = k \frac{e^2 }{r-1}

starting point.

        Em₀ = K + U₁ + U₂

        Em₀ = \frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1}

final point

         Em_f = k e^2 ( -\frac{1}{r_2 +1} + \frac{1}{r_2 -1})

   

energy is conserved

        Em₀ = Em_f

        \frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1} = k e^2 (- \frac{1}{r_2 +1} + \frac{1}{r_2 -1})              

       

        \frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1} = k e²(  \frac{2}{(r_2+1)(r_2-1)} )

we substitute the values

½ 9.1 10⁻³¹ 450 + 9 10⁹ (1.6 10⁻¹⁹)² [ - \frac{1}{20+1} + \frac{1}{20-1} ) = 9 109 (1.6 10-19) ²( \frac{2}{r_2^2 -1} )

          2.0475 10⁻²⁸ + 2.304 10⁻³⁷ (5.0125 10⁻³) = 4.608 10⁻³⁷ ( \frac{1}{r_2^2 -1} )

          2.0475 10⁻²⁸ + 1.1549 10⁻³⁹ = 4.608 10⁻³⁷     \frac{1}{r_2^2 -1}

          \frac{2.0475 \ 10^{-28} }{1.1549 \ 10^{-37} } = \frac{1}{r_2^2 -1}

          r₂² -1 = (4.443 10⁸)⁻¹

           

          r2 = \sqrt{1 + 2.25 10^{-9}}

          r2 = 1 m

therefore the electron that comes with velocity does not reach the origin, it stops when it reaches the position of the electron at x = 1m

4 0
3 years ago
How do crystalline solids differ from amorphous solids?
LUCKY_DIMON [66]
Crystalline crystals have sharp, well-defined melting points. Amorphous Solids don't have melting points.
6 0
2 years ago
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