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3 years ago

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1 answer:

larisa [96]3 years ago

5 0

Answer:

wow i have know idea

Explanation:

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4. A driver travels 135 km, east in 1.5 h, stops for 45 minutes for lunch,

stira [4]

Answer:

The driver's average velocity is 82.35 km/h.

Explanation:

Given:

The motion of the driver can be divided into 3 parts:

i. Displacement of the driver in 1.5 hours = 135 km

ii. Rest for 45 minutes.

iii. Displacement in next 2 hours = 215 km

The direction of motion remains same (east).

Now, total displacement of the driver is, $D_{Total}=135+215=350$ km.

Rest time is 45 minutes. Converting it to hours, we need to use the conversion factor $1\textrm{ min} = \frac{1}{60}$ hour.

So, 45 minutes in hours is equal to $\frac{45}{60}=0.75$ hours.

Now, total time taken for the complete journey is, $\Delta t=1.5+\frac{45}{60}+2=1.5+0.75+2=4.25\textrm{ h}$

Average velocity is given as:

$v_{avg}=\frac{\textrm{Total displacement}}{Total time}=\frac{350}{4.25}=82.35\textrm{ km/h}$

Therefore, the driver's average velocity is 82.35 km/h

4 0

3 years ago

Use technology and the given confidence level and sample data to find the confidence interval for the population mean muμ. Assum

Ronch [10]

Answer:

a. μ $_{95%} =$ 3 ± 1.8 = [1.2,4.8]

b. The correct answer is option D. No, because the sample size is large enough.

Explanation:

a. The population mean can be determined using a confidence interval which is made up of a point estimate from a given sample and the calculation error margin. Thus:

μ $_{95%} = x_$ ±(t*s)/sqrt(n)

where:

μ $_{95%} =$ = is the 95% confidence interval estimate

x_ = mean of the sample = 3

s = standard deviation of the sample = 5.8

n = size of the sample = 41

t = the t statistic for 95% confidence and 40 (n-1) degrees of freedom = 2.021

substituting all the variable, we have:

μ $_{95%} =$ 3 ± (2.021*5.8)/sqrt(41) = 3 ± 1.8 = [1.2,4.8]

b. The correct answer is option D. No, because the sample size is large enough.

Using the the Central Limit Theorem which states that regardless of the distribution shape of the underlying population, a sampling distribution of size which is ≥ 30 is normally distributed.

4 0

3 years ago

A bird flies at a speed of 2.3 m/s if it has 14 j of kinetic energy what is the mass

Sidana [21]

Kinetic Energy = 1/2 * mv²

Kinetic Energy = 14 J, v = 2.3 m/s , m = ?

14 = 1/2 * m* 2.3²

14 = 0.5*m*2.3*2.3

m = 14 / (0.5*2.3*2.3)

m = 5.29 kg.

Mass = 5.29 kg.

7 0

3 years ago

Read 2 more answers

A train traveled from Station A to Station B at an average speed of 80 kilometers per hour and then from Station B to Station C

Vinil7 [7]

Answer:

1)

75 kmh⁻¹

2)

75 kmh⁻¹

Explanation:

1)

$v_{ab}$ = Speed of train from station A to station B = 80 kmh⁻¹

$d_{ab}$ = distance traveled from station A to station B

$t_{ab}$ = time of travel between station A to station B

we know that

$Time = \frac{distance}{speed}$

$t_{ab} = \frac{d_{ab}}{v_{ab}} = \frac{d_{ab}}{80}$

$d_{bc}$ = distance traveled from station B to station C

$v_{bc}$ = Speed of train from station B to station C = 60 kmh⁻¹

$t_{bc} = \frac{d_{bc}}{v_{bc}} = \frac{d_{bc}}{60}$

Total distance traveled is given as

$d = d_{ab} + d_{bc}$

Total time of travel is given as

$t = t_{ab} + t_{bc}$

Average speed is given as

$v_{avg} = \frac{d}{t} \\v_{avg} = \frac{d_{ab} + d_{bc}}{t_{ab} + t_{bc}}\\v_{avg} = \frac{d_{ab} + d_{bc}}{(\frac{d_{ab}}{80} ) + (\frac{d_{bc}}{60} ) }$

Given that :

$d_{ab} = 4 d_{bc}$

So

$v_{avg} = \frac{4 d_{bc} + d_{bc}}{(\frac{4 d_{bc}}{80} ) + (\frac{d_{bc}}{60} ) }\\v_{avg} = \frac{4 + 1}{(\frac{4 }{80} ) + (\frac{1}{60} ) }\\v_{avg} = 75 kmh^{-1}$

2)

$v_{ab}$ = Speed of train from station A to station B = 80 kmh⁻¹

$t_{ab}$ = time of travel between station A to station B

$d_{ab}$ = distance traveled from station A to station B

we know that

$distance = (speed) (time)$

$d_{ab} = v_{ab} t_{ab}\\d_{ab} = 80 t_{ab}$

$d_{bc}$ = distance traveled from station B to station C

$v_{bc}$ = Speed of train from station B to station C = 60 kmh⁻¹

$t_{bc}$ = time of travel for train from station B to station C

we know that

$distance = (speed) (time)$

$d_{bc} = v_{bc} t_{bc}\\d_{bc} = 60 t_{bc}$

Total distance traveled is given as

$d = d_{ab} + d_{bc}\\d = 80 t_{ab} + 60 t_{bc}$

Total time of travel is given as

$t = t_{ab} + t_{bc}$

Average speed is given as

$v_{avg} = \frac{d}{t} \\v_{avg} = \frac{d_{ab} + d_{bc}}{t_{ab} + t_{bc}}\\v_{avg} = \frac{80 t_{ab} + 60 t_{bc}}{t_{ab} + t_{bc}}$

Given that :

$t_{ab} = 3 t_{bc}$

So

$v_{avg} = \frac{80 t_{ab} + 60 t_{bc}}{t_{ab} + t_{bc}}\\v_{avg} = \frac{80 (3) t_{bc} + 60 t_{bc}}{(3) t_{bc} + t_{bc}}\\v_{avg} = \frac{(300) t_{bc}}{(4) t_{bc}}\\v_{avg} = 75 kmh^{-1}$

4 0

3 years ago

What is relation of acceleration of a body to its mass and applied force

irinina [24]

The law states that external forces cause objects to accelerate, and the amount of acceleration is directly proportional to the net force and inversely proportional to the mass of the object.

If you can’t copy and paste it just reword it

3 0

4 years ago