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Leno4ka [110]
3 years ago
6

The period (T) of an oscillating wave is 1/5s. What happens to the frequency (f) of the wave if T increases to 1/2s

Physics
1 answer:
Anastasy [175]3 years ago
6 0
Frequency = 1/T
as the 5 is reduced, frequency is increase.
as 1 whole wave travels through a point in a lesser time now
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A greyhound's velocity changes from rest to 19 m/s in 2 seconds. What is the greyhound's average acceleration?
Andrew [12]

It should be b)9.5m/s2

6 0
3 years ago
How am I supposed to solve this?
RSB [31]

Answer:

4.02 km/hr

Explanation:

5 km/hr = 1.39 m/s

The swimmer's speed relative to the ground must have the same direction as line AC.

The vertical component of the velocity is:

uᵧ = us cos 45

uᵧ = √2/2 us

The horizontal component of the velocity is:

uₓ = 1.39 − us sin 45

uₓ = 1.39 − √2/2 us

Writing a proportion:

uₓ / uᵧ = 121 / 159

(1.39 − √2/2 us) / (√2/2 us) = 121 / 159

Cross multiply and solve:

159 (1.39 − √2/2 us) = 121 (√2/2 us)

220.8 − 79.5√2 us = 60.5√2 us

220.8 = 140√2 us

us = 1.115

The swimmer's speed is 1.115 m/s, or 4.02 km/hr.

7 0
3 years ago
Read 2 more answers
g a horizontal wheel of radius is rotating about a vertical axis. What is the magnitude of the resultant acceleration of a bug t
mihalych1998 [28]

Answer:

   a = w² r

Explanation:

In this exercise, indicate that the wheel has angular velocity w, the worm experiences the same angular velocity if it does not move, and has an acceleration towards the center of the circle, according to Newton's second law, called the centripetal acceleration.

           a = v² / r

angular and linear variables are related

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we substitute

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4 0
3 years ago
Suppose we wrap a string around the surface of a uniform cylinder of radius 38.0 cm that is supported by an axle passing through
liraira [26]

Answer:

Angular acceleration = 23.68 rad / s²

Explanation:

Given that,

acceleration = 9m/s²

Therefore acceleration of string is 9m/s²

since string is constant in length

cylinder of radius 38.0 cm = 0.38m

Angular acceleration = a / r

Angular acceleration = 9 / 0.38

                                   = 23.68 rad / s²

Angular acceleration = 23.68 rad / s²

4 0
3 years ago
At a point 1.2 m out from the hinge, 14.0 N force is exerted at an angle of 27 degrees to the moment arm in a plane which is per
geniusboy [140]

Answer:

\tau = 7.63 Nm

Explanation:

As we know that moment of force is given as

\tau = \vec r \times \vec F

now we have

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now from above formula we have

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here we know that

\theta = 27 degree

so we have

\tau = (1.2)(14) sin27

\tau = 7.63 Nm

3 0
3 years ago
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