The graph below shows the speed of a car as it drives along a racetrack.

Physics

1 answer:

nordsb [41]2 years ago

4 0

Answer:

Average speed = distance/time

From 1 to 9 seconds:

Distance covered = 1 - 0.2 = 0.8 km

Time = 9 - 1 = 8 sec

Average speed = 0.8 km / 8 sec

Average speed = 0.1 km/s .

The average speed for the whole test is 1.6 km / 20 sec = 0.08 km/sec. A graph of speed vs time would average out as a horizontal line at 0.08 km/sec from 1 sec to 21 sec. The area under it would be (0.08 km/s) x (20 sec) = 1.6 km.

Surprise surprise ! The area under a speed/time graph is the distance covered during that time !

In closing, I want to express my gratitude for the gracious bounty of 3 points with which I have been showered. Moreover, the green breadcrust and tepid cloudy water have also been refreshing.

Explanation:

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Ket [755]

Answer:

Perfectly inelastic collision

Explanation:

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2. Inelastic collision: When the momentum the system is conserved but the kinetic energy is not conserved, the collision is said to be inelastic. For example, collision of a ball with the mud.

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Two fixed charges, q1 = +1.07µC and q2 = -3.28µC, are 61.8cm apart. Where may a third charge be located so that no net force act

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Answer:

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3 years ago

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A 200-gram liquid sample of Alcohol Y is prepared at -6°C. The sample is then added to 400 g of water at 20°C in a sealed styrof

Vinil7 [7]

The specific heat capacity of the alcohol will be 3.72 kJ/kg°C.

<h3>What is the specific heat capacity?</h3>

The amount of heat required to increase a substance's temperature by one degree Celsius is known as its "specific heat capacity."

Similarly, heat capacity is the relationship between the amount of energy delivered to a substance and the increase in temperature that results.

Given data;

Mass of liquid sample of Alcohol m₁ = 200-gram

The temperature of alcohol, T₁ = -6°C.

Mass of liquid sample of water m₂ = 400-gram

The temperature of the water, T₂= 20°C.

The specific heat capacity of the alcohol, S₁=?

The specific heat capacity of water is, S₂=4.19 kJ/kg.°C

As we know that;

<h3 />

$\rm Q_{gain}= Q{loss} \\\\ Q_{alcohol} =Q_{water} \\\\\ m_1s_1\triangle T_1 = m_2S_2 \triangle S_2 \\\\ 200 \times 10^{-3} \times S_1 [ (12-(-6) ] = 40 \times 10^{-3} \times 4.19 \times 10^{-3} \times (20-12)\\\\S_1 = 2 \times 4.19 \times 10^3 \times \frac {8}{18} \\\\ S_1 = 3.72 \ kJ /kg ^0 C$