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AfilCa [17]
3 years ago
12

The filter reaches therminal velocity with zero paper clips (T or F)

Physics
1 answer:
Kaylis [27]3 years ago
6 0

Answer:t

Explanation:

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What is the wavelength of a wave if the wave speed is 24 m/s and the frequency is 48 Hz?
Mice21 [21]

Answer:

Wavelength is 0.5

Explanation:

To work it out, you divide Wave speed by the Frequency (24÷48=0.5)

4 0
3 years ago
Read 2 more answers
A ball thrown horizontally at vi = 30.0 m/s travels a horizontal distance of d = 55.0 m before hitting the ground. from what hei
tekilochka [14]
Assume no air resistance, and g = 9.8 m/s².

Let
x =  angle that the initial velocity makes with the horizontal.
u = 30 cos(x), horizontal velocity
v = 30 sin(x), vertical launch velocity

The horizontal distance traveled is 55 m, therefore the time of flight is
t = 55/[30 cos(x)] = 1.8333 sec(x)  s

With regard to the vertical velocity, and the time of flight,obtain
[30 sin(x)]*(1.8333 sec(x)) + (1/2)*(-9.8)*(1.8333 sec(x))² = 0
 55 tan(x) - 16.469 sec²x = 0
55 tan(x) - 16.469[1 + tan²x] = 0
16.469 tan²x - 55 tan(x) + 16.469 = 0
tan²x - 3.3396 tan(x) + 1 = 0

Solve with the quadratic formula.
tan(x) = 0.5[3.3396 +/- √(7.153)] = 3.007 or 0.3326
Therefore
x = 71.6° or x = 18.4°

The time of flight is
t = 1.8333 sec(x) = 5.8096 s or 1.932 s
The initial vertical velocity is
v = 30 sin(x) = 28.467 m/s or 9.468 m/s
The horizontal velocity is
u = 30 cos(x) = 9.467 m/s or 28.469 m/s

If t = 5.8096 s,
  u*t = 9.467*5.8096 = 55 m (Correct)
or
 u*t = 28.469*15.8096 = 165.4 m (Incorrect)

Therefore, reject x = 18.4°. The correct solution is
t = 5.8096 s
x = 71.6°
u = 9.467 m/s
v = 28.467 m/s

The height from which the ball was thrown is
h = 28.467*5.8096 - 0.5*9.8*5.8096² = -110.4 m
The ball was thrown from a height of 110.4 m

Answer: h = 110.4 m

7 0
3 years ago
Is the change in velocity divided by the time needed for the change to occur.
Montano1993 [528]

Answer: yes

Explanation:

6 0
2 years ago
You kick a soccer ball off the ground with a velocity of 15 m/s at an angle of 40° above the horizontal
jeyben [28]

At its maximum height, the ball will have zero vertical velocity, so the ball's velocity at this point is exactly equal to its horizontal velocity.

At any time <em>t</em>, the horizontal component of its velocity is

<em>v</em> = (15 m/s) cos(40°) ≈ 11.49 m/s

so at the highest point of its trajectory, the ball has a velocity of about 11.49 m/s pointed in the positive horizontal direction.

8 0
2 years ago
Which one would it be ?
MA_775_DIABLO [31]

the answer to you question is D.

8 0
2 years ago
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