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Whitepunk [10]
3 years ago
14

Two blocks with masses 1 and 2 are connected by a massless string that passes over a massless pulley as shown. 1 has a mass of 2

.25 kg and is on an incline of 1=43.5∘ with coefficient of kinetic friction 1=0.205 . 2 has a mass of 5.45 kg and is on an incline of 2=32.5∘ with coefficient of kinetic friction 2=0.105 . The two‑block system is in

Physics
1 answer:
Bess [88]3 years ago
3 0

Answer:

The acceleration of M_2 is  a =  0.7156 m/s^2

Explanation:

From the question we are told that

    The mass of first block is  M_1 =  2.25 \ kg

    The angle of inclination of first block is  \theta _1 =  43.5^o

    The coefficient of kinetic friction of the first block is  \mu_1  = 0.205

      The mass of the second block is  M_2 = 5.45 \ kg

     The angle of inclination of the second block is  \theta _2 =  32.5^o

      The coefficient of kinetic friction of the second block is \mu _2 = 0.105

The acceleration of M_1 \ and\  M_2 are same

The force acting on the mass M_1 is mathematically represented as

     F_1 = T -  M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1

=> M_1 a = T -  M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1

Where T is the tension on the rope

The force acting on the mass M_2 is mathematically represented as    

  F_2 =  M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2

   M_2 a =  M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2

At equilibrium

  F_1 =  F_2

So

 T -  M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1 =M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2

making a the subject of the formula

    a =  \frac{M_2 g sin \theta_2 - M_1 g sin \theta_1 - \mu_1 M_1g cos \theta - \mu_2 M_2 g cos \theta_2 }{M_1 +M_2}

substituting values a =  \frac{(5.45) (9.8) sin (32.5) - (2.25) (9.8) sin (43.5) - (0.205)*(2.25) *9.8cos (43.5) - (0.105)*(5.45) *(9.8) cos(32.5) }{2.25 +5.45}

    => a =  0.7156 m/s^2

     

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a flag of mass 2.5 kg is supported by a single rope. A strong horizontal wind exerts a force of 12 N on the flag. Calculate the
tatuchka [14]
The free-body diagram of the forces acting on the flag is in the picture in attachment.

We have: the weight, downward, with magnitude
W=mg = (2.5 kg)(9.81 m/s^2)=24.5 N
the force of the wind F, acting horizontally, with intensity
F=12 N
and the tension T of the rope. To write the conditions of equilibrium, we must decompose T on both x- and y-axis (x-axis is taken horizontally whil y-axis is taken vertically):
T \cos \alpha -F=0
T \sin \alpha -W=
By dividing the second equation by the first one, we get
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From which we find
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7 0
3 years ago
A 5 kg block is pushed across a table by a horizontal force of 40 N with an acceleration of 5 m/s^2. What is the frictional forc
julsineya [31]

Answer:

15

Explanation:

mass, M = 5Kg

horizontal force, F_h = 40N

acceleration, a =5 m/s^2

frictional force, F_f =?

net force = ma

net force = F_h - F_f = 40N - F_f

40  - F_f = 5 x 5

- F_f = 25 - 40

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photoshop1234 [79]

Answer:

a) f' = 432 Hz

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Explanation:

a) To calculate the frequency of sound waves that car receives, you take into account the Doppler effect. In this case (observer moves away of the source) you have the following formula:

f'=f(\frac{v-v_o}{v+v_s})    (1)

where

f: frequency of the source = ?

v: speed of sound = 343 m/s

vo: speed of the observer = 40.0 m/s

vs: speed of the source = 0 m/s (stationary)

You replace the values of all parameters in the equation (1):

To calculate f' you first calculate the frequency of the sound wave, by using the following formula:

v=\lambda f\\\\

v: speed of sound

λ: wavelength = 0.700 m

f=\frac{v}{\lambda}=\frac{343m/s}{0.700m}=480Hz

Next, you replace the values of all parameters in the equation (1):

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I=\frac{P}{4\pi r^2}

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r: distance to the source = 70.0 m

I=\frac{50.0 W}{4\pi(70.0m)^2}=8.12*10^{-4}\frac{W}{m^2}

hence, the intensity is 8.12*10^⁻4 W/m^2

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