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aleksley [76]
3 years ago
9

Please help Convert the following distance v time graph velocity graph

Physics
1 answer:
Reil [10]3 years ago
5 0

Explanation:

for v vs t graph for t=9 to 11,v=15kmph

similarly v2=0,v3=60kmph&v4= -40kmph

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You decide to walk part of the way around a lake. The lake is a circle with a radius of 2.0 km You start on the shore due south
AveGali [126]

Answer:

2.828 km

Explanation:

radius of lake, r = 2 km

Let O be the centre of the circular lake and you walk from A to A to D.

Displacement is the measure of shortest distance between two points.

According to the diagram , the displacement AD is

AD^{2}=AO^{2}+OD^{2} (By using the Pythagorean theorem)

AD^{2}=2^{2}+2^{2}

AD = 2.828 km

thus, the displacement is 2.828 km.

7 0
3 years ago
express the following in metres (1) 52fm (2) 26 Mm (3)12am (4) 69 pm (5) 85 mm​
AlekseyPX

Answer:........u are so sweet

7 0
3 years ago
What is the result of two displacement vectors having opposite directions?
jonny [76]

<u>Answer:</u>

<em>Resultant of two vectors having opposite direction is the difference of the two displacements having the same direction as the larger vector. </em>

<u>Explanation:</u><u> </u>

Resultant of two vectors is obtained by performing the vector addition operation. When the directions of both vectors are same the resultant’s direction will also be the same as the inputs. When two vectors have opposite directions, one direction will be taken positive making one vector positive and the other negative.

By performing addition of a positive and negative number we are actually taking the difference between both. Thus performing vector addition of two vectors with opposite directions is equivalent to finding the difference between the vectors. Consider a system consisting of a solid block, on which two forces F1 and F2 act in the opposite direction.

One force will be considered positive and the other is considered negative. The resultant is given by the difference of two force vectors. Displacement of the block will be in the direction of the greater force.  

5 0
3 years ago
Read 2 more answers
You hold glider AA of mass 0.125 kgkg and glider BB of mass 0.375 kgkg at rest on an air track with a compressed spring of negli
andrew11 [14]

Answer:

The magnitude of the velocity of glider B is 0.2m/s and the direction is the negative direction  

Explanation:

Inelastic Collision

Given data

mass of glider A m1= 0.125kg

initial velocity u1=0

final velocity v1= 0.600 m/s

mass of glider B m2= 0.375kg

initial velocity u2=0

final velocity v2=?

We know that the expression for the conservation of momentum is given as

m1u1+m2u2=m1v1+m2v2

since u1=u2=u=0m/s

u(m1+m2)=m1v1+m2v2

substituting we have

0(0.125+0.0375)=0.125*0.6+0.375*v2

0=0.075+0.375v2

0.375v2=-0.075

v2=-0.075/0.375

v2=-0.2m/s

 The magnitude of the velocity of glider B is 0.2m/s and the direction is the negative direction      

3 0
3 years ago
A transverse wave on a string is described by the following wave function.y = (0.090 m) sin (px/11 + 4pt)(a) Determine the trans
alukav5142 [94]

Explanation:

(a) It is known that equation for transverse wave is given as follows.

                 y = (0.09 m)sin(\pi \frac{x}{11} + 4 \pi t)

Now, we will compare above equation with the standard form of transeverse wave equation,

                 y = A sin(kx + \omega t)

where,    A is the amplitude = 0.09 m

              k is the wave vector = \frac{\pi}{11}

              \omega is the angular frequency = 4\pi

              x is displacement = 1.40 m

              t is the time = 0.16 s

Now, we will differentiate the equation with respect to t as follows.

The speed of the wave  will be:

                   v(t) = \frac{dy}{dt}

                v(t) = A \omega cos(kx + \omega t)

        v(t) = (0.09 m)(4\pi) cos(\frac{\pi \times 1.4}{11} + 4 \pi \times 0.16)

          v(t) = -0.84 m/s

The acceleration of the particle in the location is

            a(t) = \frac{dv}{dt}

           a(t) = -A \omega 2sin(kx + \omega t)

           a(t) = -(0.09 m)(4 \pi)2 sin(\frac{\pi \times 1.4}{11} + 4\pi \times 0.16)

           a(t) = -9.49 m/s^{2}

Hence, the value of transverse wave is 0.84 m/s and the value of acceleration is 9.49 m/s^{2} .

(b)  Wavelength of the wave is given as follows.

               \lambda = \frac{2\pi}{k}

              \lambda = (frac{2\pi}{\frac{\pi}{11})&#10;

              \lambda = 22 m

The period of the wave is

             T = \frac{2 \pi}{\omega}

             T = \frac{2 \pi}{4 \pi}

                = 0.5 sec

Now, we will calculate the speed of propagation of wave as follows.

                    v = \frac{\lambda}{T}

                       = \frac{22 m}{0.5 s}

                       = 44 m/s

therefore, we can conclude that wavelength is 22 m, period is 0.5 sec, and speed of propagation of wave is 44 m/s.

7 0
3 years ago
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