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Korvikt [17]
3 years ago
15

Two men decide to use their cars to pull a truck stuck in mud. They attach ropes and one pulls with a force of 598 N at an angle

of 29◦ with respect to the direction in which the truck is headed, while the other car pulls with a force of 941 N at an angle of 23◦ with respect to the same direction. 598 N 29 ◦ 941 N 23 ◦ What is the net forward force exerted on the truck in the direction it is headed? Answer in units of N
Physics
1 answer:
kondaur [170]3 years ago
5 0

Answer:

magnitude of the force is

F = 1537 N

direction of the force is given as

\theta = 25.3 degree

Explanation:

As we know that the first force is 598 N at 29 degree

so we have

F_1 = 598(cos29 \hat i + sin29 \hat j)

F_1 = 523 \hat i + 290 \hat j

Now another force is 941 N at 23 degree

F_2 = 941(cos23\hat i + sin23\hat j)

F_2 = 866.2 \hat i + 367.7 \hat j

so we will have

F = F_1 + F_2

F = (523 + 866.2)\hat i + (290 + 367.7)\hat j

F = 1389.2\hat i + 657.7 \hat j

magnitude of the force is

F = \sqrt{1389.2^2 + 657.7^2}

F = 1537 N

direction of the force is given as

tan\theta = \frac{F_y}{F_x}

tan\theta = \frac{657.7}{1389.2}

\theta = 25.3 degree

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Two sinusoidal waves, which are identical except for a phase shift, travel along in the same direction. The wave equation of the
Y_Kistochka [10]

Answer:

two sinusoidal waves, which are identical except for a phase shift, travel along in the same direction. The wave equation of the resultant wave is yR (x, t) = 0.70 m sin⎛ ⎝3.00 m−1 x − 6.28 s−1 t + π/16 rad⎞ ⎠ . What are the angular frequency, wave number, amplitude, and phase shift of the individual waves?

ω = 6.28 s − 1 ,

k = 3.00 m− 1 ,

φ = π rad,

A R = 2 A cos (φ 2 ) ,

A = 0.37 m

Explanation:

y1 ( x , t ) = A sin( k x − ω t +φ ) ,

y 2 ( x , t ) = A sin ( k x − ω t ) .

from the principle of superposition which states that when two or more waves combine, there resultant wave is the algebriac sum of the individual waves

y1 ( x , t ) = A sin( k x − ω t +φ ) ,   is generaL form of thw wave eqaution

A=amplitude

k=angular wave number

ω=angular frequency

φ =phase constant

k=2π/lambda

ω=2π/T

yR (x, t) = 0.70 m sin{3.00 m−1 x − 6.28 s−1 t + π/16 rad}....................*

two waves superposed to give the above, assuming they are moving in the +x direction

y1 ( x , t ) = A sin( k x − ω t +φ ) , .....................1

y 2 ( x , t ) = A sin ( k x − ω t ) ...........................2

adding the two equation will give

A sin( k x − ω t +φ )+A sin ( k x − ω t ) .................3

A( sin( k x − ω t +φ )+ sin ( k x − ω t ) ),......................4

similar to the following trigonometry identity

sina+sinb=2cos(a-b)/2sin(a+b)/2

let a= ( k x − ω t

b=k x − ω t +φ )

y(x,t)=2Acos(φ/2)sin(k x − ω t +φ/2)

k=3m^-1

lambda=2π/k=2.09m

ω=6.28= T=2π/6.28

T=1s

φ/2=π/16

φ=π/8rad

amplitude

2Acos(φ/2)=0.70 m

A=0.7/2cos(π/8)

A=0.37 m

6 0
4 years ago
N=-D(n2-n1)/(x2-x1) D is diffastion.What are the dimensions of D.
MrMuchimi

We don't have much to go on.

The dimensions of D depend on the dimensions of N, n, and x, and we don't know what any of those stand for.

It might help if we had ever heard of 'diffastion', but we're striking out there too.

8 0
3 years ago
Each wheel of a 320 kg motorcycle is 52 cm in diameter and has rotational inertia 2.1 kg m2 . The cycle and its 75 kg rider are
Amiraneli [1.4K]

Answer:

The value is  h  = 32.91 \  m

Explanation:

From the question we are told that

    The diameter of each wheel is  d =  52 \ cm  =  0.52 \  m

    The mass of the motorcycle is  m  =  320 \ kg

    The rotational kinetic inertia is  I  =  2.1 \  kg \  m^2

    The  mass of the  rider is  m_r =  75 \ kg

     The  velocity is  v  =  85 \  km/hr = 23.61 \  m/s

      Generally the radius of the wheel is mathematically represented as

      r =  \frac{d}{2}

=>     r =  \frac{0.52}{2}

=>    r =  0.26 \  m

Generally from the law of energy conservation

     Potential energy  attained  by  system(motorcycle and rider )  =  Kinetic  energy of the system  +  rotational kinetic energy of  both wheels of the motorcycle

=>  Mgh  =  \frac{1}{2}  Mv^2  +   \frac{1}{2}  Iw^2  +  \frac{1}{2}  Iw^2

=>    Mgh  =  \frac{1}{2}  *  Mv^2  +  Iw^2

Here  w is the angular velocity which is mathematically represented as

     w =  \frac{v }{r }

So

    Mgh  =  \frac{1}{2}  *  Mv^2  +  I \frac{v}{r} ^2

Here M  =  m_r +  m

         M  =  320 + 75

          M  = 395 \  kg

395 *  9.8 *  h  =   0.5    *   395 *  (23.61)^2 +  2.1  *[\frac{ 23.61}{ 0.26} ] ^2

=>   h  = 32.91 \  m

   

7 0
4 years ago
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