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3 years ago

After their loss to U of A, the ASU sumdevils traveled back to Tempe which is 187

Physics

1 answer:

Dmitry_Shevchenko [17]3 years ago

5 0

The average speed is 20.8 m/s

Explanation:

The average speed for the trip is given by:

$v=\frac{d}{t}$

where

d is the distance covered

t is the time elapsed

For the trip in this problem, we have:

d = 187 km = 187,000 m is the distance travelled

The initial time is 10:00 pm while the arriving time is 12:30 am: this means that the time elapsed is 2.5 hours. Converting into seconds,

$t = 2.5 h \cdot (60)(60)=9000 s$

Therefore, the average speed for the trip is

$v=\frac{187,000}{9000}=20.8 m/s$

Learn more about average speed:

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A particular car engine operates between temperatures of 440°C (inside the cylinders of the engine) and 20°C (the temperature of

Step2247 [10]

One of the concepts to be used to solve this problem is that of thermal efficiency, that is, that coefficient or dimensionless ratio calculated as the ratio of the energy produced and the energy supplied to the machine.

From the temperature the value is given as

$\eta = 1-\frac{T_L}{T_H}$

Where,

T_L = Cold focus temperature

T_H = Hot spot temperature

Our values are given as,

T_L = 20\° C = (20+273) K = 293 K

T_H = 440\° C = (440+273) K = 713 K

Replacing we have,

$\eta = 1-\frac{T_L}{T_H}$

$\eta = 1-\frac{293}{713}$

$\eta = 0.589$

Therefore the maximum possible efficiency the car can have is 58.9%

4 0

3 years ago

The diagram shows changing wavelengths along the electromagnetic spectrum. Which labels belong in the areas marked Y and Z? Y: U

marshall27 [118]

Y: Radio
Z: Gamma
because gamma has a higher frequency then radio, therefore the higher the frequency the shorter the wavelength

8 0

3 years ago

A car goes from 40n/s to 60 m/s in 8s the acceleration of the car is

stich3 [128]

Acceleration= change in velocity/ change in time.

60-40=20 m/s (change in speed)
Time= 8 seconds

20/8= 2.5
Acceleration= 2.5 m/s^2

Hope this helps! :)

7 0

3 years ago

Read 2 more answers

Consider the power dissipated by the two circuits in the video. The ratio of power dissipation in the parallel circuit to that i

klemol [59]

Answer:

Explanation:

In a series circuit, if we assume that each bulb has a resistance R, the equivalent resistance, is the sum of all the resistances, in this case, 3*R.
The power dissipated in this resistance, can be expressed as follows:

$P_{series} = V*I = V*(\frac{V}{R_{eq}}) = \frac{V^{2}}{3*R} (1)$

In the parallel circuit, the equivalent resistance, is equal to 1/3*R.

The power dissipated in this resistance, can be expressed as follows:

$P_{//} = V*I = V*(\frac{V}{R_{eq}}) = \frac{V^{2}}{\frac{R}{3} } (2)$

In order to take the ratio of both powers, we can divide both sides of (1) and (2) , as follows:

$\frac{P_{//} }{P_{series} } =\frac{V^{2}}{\frac{R}{3}} * \frac{3*R}{V^{2}} = 9$

The ratio of power dissipation in the parallel circuit to that in the series circuit is 9 times.

5 0

4 years ago

uppose the weights of Farmer Carl's potatoes are normally distributed with a mean of 9 ounces and a standard deviation of 0.8 ou

Brrunno [24]

Answer:

x= 9.53 ounces

Explanation:

Given that

Mean ,μ= 9 ounces

Standard deviation ,σ=0.8 ounces

He wants to sell only those potatoes that are among the heaviest 25%.

P=25% = 0.25

When P= 0.25 then Z=0.674

Lest take x is the the minimum weight required to be brought to the farmer's market.

We know that

x = Z . σ + μ

x= 0.674 ₓ 0.8 + 9 ounces

x= 9.53 ounces

3 0

3 years ago