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Dennis_Churaev [7]

3 years ago

11

Which equation could be rearranged to calculate the frequency of a wave?

2 answers:

Flauer [41]3 years ago

8 0

Answer:

C. wavelength = speed/frequency

Explanation: I got it right

34kurt3 years ago

8 0

Answer:

Which equation could be rearranged to calculate the frequency of a wave?

1) wavelength = frequency/speed <<<---CORRECT

2) frequency = wavelength/speed

3) wavelength = speed/frequency

4) frequency = speed x wavelength

Explanation:

2021 EDGE 100% test score

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A certain corner of a room is selected as the origin of a rectangular coordinate system. If a fly is crawling on an adjacent wal

lisov135 [29]

Answer:

S=2.693m

Explanation:

Let point $O$ be the origin of the room and the fly is crawling at point $P$ as shown in the attached photo:

#Distance of the fly from the corner of the room is defined as:

$OP=S=\sqrt{{(x_1-x_0)^2+(y_1+y_0)^2}$

#Substitute the coordinate values in the above equation:

$S=\sqrt{{(2.6m-0m)^2+(0.7m-0m)^2}}\\=\sqrt{7.25}\\=2.693m$

Hence, the distance of the fly from the corner of the room is 2.693m

6 0

3 years ago

An astronaut standing on a platform on the moon drops a hammer. if the hammer falls 6.0 meters vertically in 2.7 seconds, what i

Amanda [17]

The distance that is traveled by the astronaut given that the motion is free-fall can be calculated through the equation,

d = Vot + 0.5at²

where d is the distance, Vo is the initial velocity, t is the time, and a is the acceleration. Substituting the known,

6 = (0 m/s)(2.7 s) + 0.5(a)(2.7 s)²

Determining the value of a,
a = 1.646 m/s²

ANSWER: 1.646 m/s²

4 0

4 years ago

The rms (root-mean-square) speed of a diatomic hydrogen molecule at 50∘C is 2000 m/s. Note that 1.0 mol of diatomic hydrogen at

denis-greek [22]

Answer:

A) d. (1/4)(2000m/s) = 500 m/s

B) c. 4000 J

C) f. None of the above (2149.24 m/s)

Explanation:

A)

The translational kinetic energy of a gas molecule is given as:

K.E = (3/2)KT

where,

K = Boltzman's Constant = 1.38 x 1^-23 J/K

T = Absolute Temperature

but,

K.E = (1/2) mv²

where,

v = root mean square velocity

m = mass of one mole of a gas

Comparing both equations:

(3/2)KT = (1/2) mv²

v = √(3KT)/m _____ eqn (1)

<u>FOR HYDROGEN:</u>

v = √(3KT)/m = 2000 m/s _____ eqn (2)

<u>FOR OXYGEN:</u>

velocity of oxygen = √(3KT)/(mass of oxygen)

Here,

mass of 1 mole of oxygen = 16 m

velocity of oxygen = √(3KT)/(16 m)

velocity of oxygen = (1/4) √(3KT)/m

using eqn (2)

<u>velocity of oxygen = (1/4)(2000 m/s) = 500 m/s</u>

B)

K.E = (3/2)KT

Since, the temperature is constant for both gases and K is also a constant. Therefore, the K.E of both the gases will remain same.

K.E of Oxygen = K.E of Hydrogen

<u>K.E of Oxygen = 4000 J</u>

C)

using eqn (2)

At, T = 50°C = 323 k

v = √(3KT)/m = 2000 m/s

m = 3(1.38^-23 J/k)(323 k)/(2000 m/s)²

m = 3.343 x 10^-27 kg

So, now for this value of m and T = 100°C = 373 k

v = √(3)(1.38^-23 J/k)(373 k)/(3.343 x 10^-27 kg)

<u>v = 2149.24 m/s</u>

<u></u>

8 0

4 years ago

PLZ HELP ON #22-26!!!! <br><br>Please explain why and how you got your answer.

AleksAgata [21]

22. a - (vf^2 - vi^2)/(2d)
a = (0 - 23^2)/(170)
a = -3.1 m/s^2

23. Find the time (t) to reach 33 m/s at 3 m/s^2
33-0/t = 3
33 = 3t
t = 11 sec to reach 33 m/s^2
Find the av velocuty: 33+0/2 = 16.5 m/s
Dist = 16.5 * 11 = 181.5 meters to each 33m/s speed. Runway has to be at least this long.

24. The sprinter starts from rest. The average acceleration is found from:
(Vf)^2 = (Vi)^2 = 2as ---> a = (Vf)^2 - (Vi)^2/2s = (11.5m/s)^2-0/2(15.0m) = 4.408m/s^2 estimated: 4.41m/s^2
The elapsed time is found by solving
Vf = Vi + at ----> t = vf-vi/a = 11.5m/s-0/4.408m/s^2 = 2.61s

25. Acceleration of car = v-u/t = 0ms^-1-21.0ms^-1/6.00s = -3.50ms^-2
S = v^2 - u^2/2a = (0ms^-1)^2-(21.0ms^-1)^2/2*-3.50ms^-2 = 63.0m

26. Assuming a constant deceleration of 7.00 m/s^2
final velocity, v = 0m/s
acceleration, a = -7.00m/s^2
displacement, s - 92m
Using v^2 = u^2 - 2as
0^2 - u^2 + 2 (-7.00) (92)
initial velocity, u = sqrt (1288) = 35.9 m/s
This is the speed pf the car just bore braking.

I hope this helps!!

5 0

3 years ago

Atoms with the same number of protons but different numbers of neutrons are called isomers

Ludmilka [50]

False, atoms with the the same number of protons but a differing number of neutrons is called an isotope.

An isomer is a compound that has the same atomic makeup as another, but in a different arrangement.

8 0

3 years ago