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Sergeeva-Olga [200]
3 years ago
13

You drive on Interstate 10 from San Antonio to Houston, half the time at 54 km/h and the other half at 118 km/h. On the way back

you travel half the distance at 54 km/h and the other half at 118 km/h. What is your average speed?
Physics
2 answers:
Irina-Kira [14]3 years ago
7 0

Answer:

Average speed: 86 km/h

Explanation:

Driving from San Antonio to Houston:

1st. half time: 54km/h

2nd. half time: 118 km/h

Average speed = [tex] \frac{54 \frac{km}{h}+ 118 \frac{km}{h}  }{2}=86 \frac{km}{h} [\tex]

Driving way back:

1st. half time: 54km/h

2nd. half time: 118 km/h

Average speed = [tex] \frac{54 \frac{km}{h}+ 118 \frac{km}{h}  }{2}=86 \frac{km}{h} [\tex]

As in both routes we have the same average speed, then the average speed for the whole trip is 86 km/h

pogonyaev3 years ago
7 0

Answer:

v_{avg} = 79.7 km/h

Explanation:

As we move from San Antonio to Houston

let the distance is "d" from Antonio to Houston

Half the time it moves with 54 km/h and next half the time it moves 118 km/h

so we will have

54 T + 118 T = d

T = \frac{d}{172}

so total time is

2T = \frac{d}{86} = 0.0116 d

now while his return journey half the distance he move with 54 km/h and next half distance with speed 118 km/h

so we have time of return journey

T' = \frac{d/2}{54} + \frac{d/2}{118}

so now we have

T' = 0.0135 d

now for the average speed we know that

v_{avg} = \frac{distance}{time}

v_{avg} = \frac{d + d}{0.0116 d + 0.0135 d}

v_{avg} = 79.7 km/h

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Answer:

The comparisons are;

The height of the bromine in the 50 ml beaker will be twice that of the 100 ml beaker

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The actual volume of bromine in both beakers will be equivalent

Explanation:

The properties of a liquid are;

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2) A liquid will assume the shape of a container in which it is placed

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Therefore, given that the volume of the Bromine is measured in 50 ml beaker and a 100 ml beaker, there will be differences in the measured height of the same volume of bromine in each beaker.

5 0
3 years ago
A man 6 feet tall walks at a rate of 6 feet per second away from a light that is 15 feet above the ground.
Tems11 [23]

Answer:(a)10 ft/s

(b)4 ft/s

Explanation:

Given

height of light =15 feet

height of man=6 feet

\frac{\mathrm{d} x}{\mathrm{d} t}=6 ft/s

From diagram

\frac{15}{y}=\frac{6}{y-x}

5(y-x)=2y

3y=5x

differentiate both sides

3\times \frac{\mathrm{d} y}{\mathrm{d} t}=5\times \frac{\mathrm{d} x}{\mathrm{d} t}

Tip of shadow is moving at the rate of

\frac{\mathrm{d} y}{\mathrm{d} t}=\frac{5}{3}\times 6=10 ft/s

(b)rate at which length of his shadow  is changing

Length of shadow is y-x

differentiating w.r.t time

\frac{\mathrm{d} (y-x)}{\mathrm{d} t}=\frac{\mathrm{d} y}{\mathrm{d} t}-\frac{\mathrm{d} x}{\mathrm{d} t}

\frac{\mathrm{d} (y-x)}{\mathrm{d} t}=10-6=4 ft/s

7 0
3 years ago
Suppose that you are standing on a train accelerating at 0.20g (where g is the acceleration due to gravity). What minimum coeffi
prisoha [69]

Answer:

0.2

Explanation:

The given parameters are;

The acceleration of the train, a = 0.2·g

The mass of the person standing on the train = m

Let μ represent the coefficient of static friction, we have;

The force acting on the person, F = m × a = m × 0.2·g

The force of friction acting between the feet and the floor, F_f = m·g·μ

For the person not to slide we have;

The force acting on the person = The force of friction acting between the feet and the floor

F = F_f

∴ m × 0.2·g = m·g·μ

From which we get;

0.2 = μ

The coefficient of static friction that must exist between the feet and the floor if the person is not to slide, μ = 0.2.

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Answer:

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