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elena55 [62]
3 years ago
8

Calculating Displacement from the Area under a Curve Try Use the graph to answer the question What is the total displacement of

the object after 5 seconds? Velocity vs. Time m 70 Velocity (m/s) 60 50 40 30 20 10 0 0 1 2 3 4 5 Time (s) Done Intro 15 of 18​

Physics
1 answer:
Agata [3.3K]3 years ago
8 0

Answer:

175 m

Explanation:

In a velocity vs time graph, displacement is the area under the curve.

We can calculate this as area of a trapezoid:

A = ½ (10 m/s + 60 m/s) (5 s)

A = 175 m

Or, we can split the area into a rectangle and a triangle.

A = (10 m/s) (5 s) + ½ (60 m/s − 10 m/s) (5 s)

A = 175 m

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You're average speed would be 6km/hour or 100m per minute.
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The temperature of the cooling water as it leaves the hot engine of an automobile is 240 °F. After it passes through the radiato
djverab [1.8K]

Answer:571.09 kJ

Explanation:

Given

Temperature of cooling water from engine exit=240^{\circ} F\approx 115.55^{\circ}C

After Passing through the radiator its temperature decreases to 175^{\circ}F\approx 79.44^{\circ}F

specific heat of water=4.184 J/g^{\circ}C

Volume of water = 1 gallon\approx 3.78 L

density of water \rho =1 gm/mL

Thus mass of water=\rho \times V=3.78\times 1=3.78 kg

Heat transferred to the surrounding is equal to heat absorbed by cooling water

Q=m\cdot c\cdot \Delta T

Q=3.78\times 4.184\times 1000\times (115.55-79.44)

Q=3.78\times 4.184\times 1000\times (36.11)

Q=571.09 kJ

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3 years ago
Where were the girls heading when the car broke down?<br> in the movie *hidden figures*
Ilia_Sergeevich [38]

Explanation:

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5 0
4 years ago
Consider a 10 gram sample of a liquid with specific heat 2 J/gK. By the addition of 400 J, the liquid increases its temperature
stich3 [128]

40 J/g is the heat of vaporization of the liquid.

Answer: Option D

<u>Explanation:</u>

Given that mass of liquid sample: m = 10 g

And, Specific heat of the liquid: S = 2 J/g K

Also, the increase in the temperature of the liquid,  \Delta T = T_{2}-T_{1} = 10 K

Therefore, the total amount of heat energy required is given by:

              q_{1} = m \times S \times\left(T_{2}-T_{1}\right) = 10 \times 2 \times 10 = 200 J

According to the given data in the question,

Total heat energy supplied, q = 400 J

Rest of heat would be q_{2}=q-q_{1}=400-200=200 \mathrm{J}

Now, 200 J vaporizes the mass, half of the liquid from full portion boiled away. So,

                 m^{\prime} = \frac{10}{2} = 5 \mathrm{g}

Latent heat of vaporization of the liquid is L_{v}. It can be calculated as below,

                       q_{2} = m^{\prime} L_{v}

                       L_{v} = \frac{q_{2}}{m^{\prime}} = \frac{200}{5} = 40 \mathrm{J} / \mathrm{g}

5 0
3 years ago
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