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atroni [7]
3 years ago
11

The average speed during any time interval is equal to the total distance of travel divided by the total time. Let d represent t

he distance between A and B. Let t1 be the time for which she has the higher speed of 5.15 m/s = d/t1. Let t2 represent the longer time for the return trip at 2.80 m/s =d/t2. Then the times are t1 = d/5.15 5.15 m/s and t2 = d/2.80 2.8 m/s. The average speed vavg is given by the following equation. vavg = Total distance/Total time = d + d/t1 + t2.
Physics
1 answer:
Tanya [424]3 years ago
7 0

Answer:

Average speed = 3.63 m/s

Explanation:

The average speed during any time interval is equal to the total distance travelled divided by the total time.

That is,

Average speed = distance/ time

Let d represent the distance between A and B.

Let t1 be the time for which she has the higher speed of 5.15 m/s. Therefore,

5.15 = d/t1.

Make d the subject of formula

d = 5.15t1

Let t2 represent the longer time for the return trip at 2.80 m/s . That is,

2.80 = d/t2.

Then the times are t1 = d/5.15 5 and

t2 = d/2.80.

The average speed vavg is given by the following equation.

avg speed = Total distance/Total time

Avg speed = d + d/t1 + t2

Where

Total distance = 2d

Total time = t1 + t2

Total time = d/5.15 + d/2.80

Total time = (2.8d + 5.15d)/14.42

Total time = 7.95d/14.42

Total time = 0.55d

Substitute total distance and time into the formula above.

Avg speed = 2d / 0.55d

Avg Speed = 3.63 m/s

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What current flows through a 2.60-cm-diameter rod of pure silicon that is 20.5 cm long, when 1.00 ✕ 103 V is applied to it? (Suc
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4 years ago
A cube of ice at an initial temperature of -15.00°C weighing 12.5 g total is placed in 85.0 g of water at an initial temperature
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<u>Answer:</u> The specific heat of ice is 2.11 J/g°C

<u>Explanation:</u>

When ice is mixed with water, the amount of heat released by water will be equal to the amount of heat absorbed by ice.

Heat_{\text{absorbed}}=Heat_{\text{released}}

The equation used to calculate heat released or absorbed follows:

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]       ......(1)

where,

q = heat absorbed or released

m_1 = mass of ice = 12.5 g

m_2 = mass of water = 85.0 g

T_{final} = final temperature = 22.24°C

T_1 = initial temperature of ice = -15.00°C

T_2 = initial temperature of water = 25.00°C

c_1 = specific heat of ice = ?

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Putting values in equation 1, we get:

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c_1=2.11J/g^oC

Hence, the specific heat of ice is 2.11 J/g°C

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