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NeTakaya
3 years ago
11

For the simple harmonic motion equation

Physics
1 answer:
My name is Ann [436]3 years ago
3 0
Given that : d = 5sin(pi t/4), So, maximum displacement, d = 5*(+1) = 5 Also, maximum displacement, d = 5*(-1) = -5 
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5. A sled with no initial velocity accelerates at a rate of 5.0 m/s2 down a hill. How
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Answer:

60 m/s

Explanation:

a =  \frac{vf - vi}{t}

5.0 =  \frac{vf - 0}{12}

5.0 \times 12 = vf

60 = vf

6 0
3 years ago
A ball rolls down a hill and hits a box. The momentum of the ball decreases. Which happens to the momentum? Select one of the op
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3 years ago
Two cars are traveling along a straight road. Car A maintains a constant speed of 95 km/h and car B maintains a constant speed o
natulia [17]

Answer

given,

Speed of car A = 95 Km/h

                         = 95 x 0.278 = 26.41 m/s

Speed of Car B = 121 Km/h

                         = 121 x 0.278 = 33.64 m/s

Distance between Car A and B at t=0 = 41 Km

a) Distance travel by car B

   d = 26.41 t + 41000

speed of the car A = 33.64 m/s

distance = s x t

26.41 t + 41000 = 33.64 x t

7.23 t = 41000

t = 5670.82 s

time taken by Car B to cross Car A is equal to t = 5670.82 s

distance traveled by car A

D = s x t  = 26.41 x 5670.82 = 149766.25 m = 149.76 Km

b) distance travel by the car B in 30 s after overtaking car A

   D' = s x t = 33.64 x 30 = 1009.2 m = 1 Km  

8 0
2 years ago
1. A 180 kg motorcycle travels in a straight line on a horizontal road. The relationship between
Sergio039 [100]
Eh not really sure bout this one
5 0
2 years ago
What is the force per unit area at this point acting normal to the surface with unit nor- Side View √√ mal vector n = (1/ 2)ex +
Mumz [18]

Complete Question:

Given \sigma = \left[\begin{array}{ccc}10&12&13\\12&11&15\\13&15&20\end{array}\right] at a point. What is the force per unit area at this point acting normal to the surface with\b n = (1/ \sqrt{2} ) \b e_x + (1/ \sqrt{2}) \b e_z   ? Are there any shear stresses acting on this surface?

Answer:

Force per unit area, \sigma_n = 28 MPa

There are shear stresses acting on the surface since \tau \neq 0

Explanation:

\sigma = \left[\begin{array}{ccc}10&12&13\\12&11&15\\13&15&20\end{array}\right]

equation of the normal, \b n = (1/ \sqrt{2} ) \b e_x + (1/ \sqrt{2}) \b e_z

\b n = \left[\begin{array}{ccc}\frac{1}{\sqrt{2} }\\0\\\frac{1}{\sqrt{2} }\end{array}\right]

Traction vector on n, T_n = \sigma \b n

T_n =  \left[\begin{array}{ccc}10&12&13\\12&11&15\\13&15&20\end{array}\right] \left[\begin{array}{ccc}\frac{1}{\sqrt{2} }\\0\\\frac{1}{\sqrt{2} }\end{array}\right]

T_n = \left[\begin{array}{ccc}\frac{23}{\sqrt{2} }\\0\\\frac{27}{\sqrt{33} }\end{array}\right]

T_n = \frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z

To get the Force per unit area acting normal to the surface, find the dot product of the traction vector and the normal.

\sigma_n = T_n . \b n

\sigma \b n = (\frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z) . ((1/ \sqrt{2} ) \b e_x + 0 \b  e_y +(1/ \sqrt{2}) \b e_z)\\\\\sigma \b n = 28 MPa

If the shear stress, \tau, is calculated and it is not equal to zero, this means there are shear stresses.

\tau = T_n  - \sigma_n \b n

\tau =  [\frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z] - 28( (1/ \sqrt{2} ) \b e_x + (1/ \sqrt{2}) \b e_z)\\\\\tau =  [\frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z] - [ (28/ \sqrt{2} ) \b e_x + (28/ \sqrt{2}) \b e_z]\\\\\tau =  \frac{-5}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{5}{\sqrt{2} } \b e_z

\tau = \sqrt{(-5/\sqrt{2})^2  + (27/\sqrt{2})^2 + (5/\sqrt{2})^2} \\\\ \tau = 19.74 MPa

Since \tau \neq 0, there are shear stresses acting on the surface.

3 0
3 years ago
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