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3 years ago

A motor cycle travelling at 10 m/s accelerates at 4m/s(squared) for 8s.

1 answer:

8 0

10 x 4^2 = 160 / 8..

V = 20m/s...

...x 8 = 100 miles,meters, metric what ever m stands for after 8 seconds.

This is my guess since the problem says 4m/s^2

V= distance/ ST (traveled/used)

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1 year ago

Describe a solar eclipse. Be sure to include the positions of the sun, moon, and earth.

Alex777 [14]

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7 0

3 years ago

The sound level at a distance of 2.30 m from a source is 115 dB. At what distance will the sound level have the following values

Aleksandr [31]

Answer:

distance is 13 m for 100 dB

distance is 409 km for 10 dB

Explanation:

Given data

distance r = 2.30 m

source β = 115 dB

to find out

distance at sound level 100 dB and 10 dB

solution

first we calculate here power and intensity and with this power and intensity we will find distance

we know sound level β = 10 log(I/ $I_{0}$ ) ......................a

put here value (I/ $I_{0}$ ) = 10^−12 W/m² and β = 115

115 = 10 log(I/10^−12)

I = 0.316228 W/m²

and we know power = intensity × 4π r² ...............b

power = 0.316228 × 4π (2.30)²

power = 21.021604 W

we know at 100 dB intensity is 0.01 W/m²

so by equation b

power = intensity × 4π r²

21.021604 = 0.01 × 4π r²

so by solving r

r = 12.933855 m = 13 m

distance is 13 m

and

at 10 dB intensity is 1 × 10^–11 W/m²

so by equation b

power = intensity × 4π r²

21.021604 = 1 × 10^–11 × 4π r²

by solving r we get

r = 409004.412465 m = 409 km

5 0

4 years ago

Air is cooled in a process with constant pressure of 150 kPa. Before the process begins, air has a specific volume of 0.062 m^3/

Mama L [17]

Answer:

The pressure is constant, and it is P = 150kpa.

the specific volumes are:

initial = 0.062 m^3/kg

final = 0.027 m^3/kg.

Then, the specific work can be written as:

$W = \int\limits^{vf}_{vi} {Pdv} \, = P(vf - vi) = 150kPa*(0.0027 - 0.062)m^3/kg = -5.25 kPa*m^3/kg.$

The fact that the work is negative, means that we need to apply work to the air in order to compress it.

Now, to write it in more common units we have that:

1 kPa*m^3 = 1000J.

-5.25 kPa*m^3/kg = -5250 J/kg.

7 0

3 years ago

A car is pushed with a force of 450 N for 19.4 seconds. What impulse was applied to the car?

harina [27]

Answer:

impulse = 8820 kg· $\frac{m}{s}$ or 8820 N·s

Explanation:

Impulse J is equal to the average force $F_{av}$ multiplied by the elapsed time Δt or in equation form, J = $F_{av}$ Δt

As long as your force of 450 N is constant then that value is your average force $F_{av}$ and your elapsed time is 19.4 seconds.

Multiply these values.

You will get an impulse of 8820 kg· $\frac{m}{s}$ or 8820 N·s.

6 0

3 years ago