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Tresset [83]
3 years ago
15

A particle moves in the xy plane with constant acceleration. At time zero, the particle is at x = 6 m, y = 8.5 m, and has veloci

ty ~vo = (9 m/s) ˆı + (−2.5 m/s) ˆ . The acceleration is given by ~a = (4.5 m/s 2 ) ˆı + (3 m/s 2 ) ˆ . What is the x component of velocity after 3.5 s? Answer in units of m/s.
Physics
1 answer:
andreev551 [17]3 years ago
8 0

Answer:

24.75 m/s

Explanation:

Since it's only asking for the component we can use v_final=v_initial+at which then gives v_final=9+3.5*4.5=24.75 m/s

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A 40 kg dog is sitting on top of a hillside and has a potential energy of 1,568 J. What is the height of the hillside? (Formula:
igomit [66]

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B) 4.0

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A guitar string with a length of 80.0 cm is plucked. The speed of a wave in the string is 400 m/sec. Calculate the frequency of
Sholpan [36]

Answer:

The first harmonic is: 250Hz, second harmonic 500Hz, third harmonic 750Hz.

Explanation:

Use the frequency f, speed v, and wavelentgh L relationship:

v = f\cdot L\implies f = \frac{v}{L}

We are given the speed v=400 m/s. The base wavelength on a string of length 80cm is twice the length of the string (a "half wave" along the full length of the string), so:

f = \frac{400\frac{m}{s}}{2\cdot0.8 m}= 250\frac{1}{s}=250 Hz

The fundamental frequency (first harmonic) is 250 Hz

The second harmonic is produced by one full wave across the string (adding one node in the middle), so L=80cm in this case, therefore the second harmonic frequency is: f2 = 2*250=500Hz

the third harmonic add another node (and a half wave) to the pattern and the wavelength will be 2/3 of 80cm, so f3=3*250Hz = 750Hz


8 0
2 years ago
A 3.45-kg centrifuge takes 100 s to spin up from rest to its final angular speed with constant angular acceleration. A point loc
Dafna11 [192]

Answer:

(a) 18.75 rad/s²

(b) 14920.78 rev

Explanation:

(a)

First we find the acceleration of the centrifuge using,

a = (v-u)/t......................... Equation 1

Where v = final velocity, u = initial velocity, t = time.

Given: v = 150 m/s,  u = 0 m/s ( from rest), t = 100 s

Substitute into equation 1

a = (150-0)/100

a = 1.5 m/s²

Secondly we calculate for the angular acceleration using

α = a/r..................... Equation 2

Where α = angular acceleration, r = radius of the centrifuge

Given: a = 1.5 m/s², r = 8 cm = 0.08 m

substitute into equation 2

α = 1.5/0.08

α = 18.75 rad/s²

(b)

Using,

Ф = (ω'+ω).t/2........................... Equation 3

Where Ф = number of revolution of the centrifuge, ω' = initial angular velocity, ω = Final angular velocity.

But,

ω = v/r and ω' = u/r

therefore,

Ф = (u/r+v/r).t/2

where u = 0 m/s (at rest),  = 150 m/s, r = 0.08 m, t = 100 s

Ф = [(0/0.08)+(150/0.08)].100/2

Ф = 93750 rad

If,

1 rad = 0.159155 rev,

Ф = (93750×0.159155) rev

Ф = 14920.78 rev

6 0
3 years ago
Plz urgent give me about that question which are below
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B

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