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3 years ago

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The acceleration of a particle is given by ax(t) = -(2.00 m/s2) + (2.70 m/s3)t. (a) find the initial velocity v0x, such that the

particle will have the same x-coordinate at t = 3.80 s as it had at t = 0.

1 answer:

Alexeev081 [22]3 years ago

3 0

I’ll solve them right now just wait a few

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Why does a comet's tail point away from the Sun? 1Choice 1 The solar wind blows the tail away from the Sun. 2 It is being pulled

AnnyKZ [126]

Choice 1

The Sun's radiation and solar wind cause the dust and gas around the comet (coma) to stretch the coma. The solar wind electromagnetically blows the ions in the coma away.

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3 years ago

Calculate the magnitude and the direction of the resultant forces

snow_tiger [21]

answer:

resultant = 127.65 in the positive direction

explanation:

F1 = 50N , F2 = 40N, f3 = 55N , f4 = 60N

Fy = 50 sin 50 = 50 × -0.26 = -13

Fx = 40 cos 0 = 40×1 = 40

fx = 55 cos 25 = 55×0.99 = 54.45

Fy = 60 sin 70 = 60 × 0.77 = 46.2

resultant = -13+40+54.45+46.2 = 127.65 in the positive direction

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3 years ago

Air temperature in a desert can reach 58.0°C (about 136°F). What is the speed of sound (in m/s) in air at that temperature?

Veronika [31]

The approximate speed of sound in dry (0% humidity) air, in meters per second, at temperatures near 0 °C, can be calculated from

$c_{air} = (331.3+0.606 \upsilon)$

Here

$\upsilon =$ Temperature in Celsius

Replacing with our values we have that

$\upsilon=58\° C$

$c_{air} = (331.3+0.606*58)$

$c_{air} = 366.1m/s$

Therefore the speed of sound in air at that temperature is 366.1m/s

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3 years ago

The overall length of a piccolo is 32.0 cm. The resonating air column vibrates as in a pipe that is open at both ends. (a) Find

lana66690 [7]

Answer:

lowest frequency = 535.93 Hz

distance between adjacent anti nodes is 4.25 cm

Explanation:

given data

length L = 32 cm = 0.32 m

to find out

frequency and distance between adjacent anti nodes

solution

we consider here speed of sound through air at room temperature 20 degree is approximately v = 343 m/s

so

lowest frequency will be = $\frac{v}{2L}$ ..............1

put here value in equation 1

lowest frequency will be = $\frac{343}{2(0.32)}$

lowest frequency = 535.93 Hz

and

we have given highest frequency f = 4000Hz

so

wavelength = $\frac{v}{f}$ ..............2

put here value

wavelength = $\frac{343}{4000}$

wavelength = 0.08575 m

so distance = $\frac{wavelength}{2}$ ..............3

distance = $\frac{0.08575}{2}$

distance = 0.0425 m

so distance between adjacent anti nodes is 4.25 cm

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3 years ago

Peter Piper picked a peck of pickled peppers. If Peter Piper picked a peck of pickled peppers, how many pickled peppers did Pete

maria [59]

Peter picked 8 quarts of peppers.

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2 years ago

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