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inysia [295]
3 years ago
5

How does the particle model indicate positive acceleration and how does it indicate negative acceleration? An example of particl

e model is attached

Physics
1 answer:
MrMuchimi3 years ago
5 0

Answer:

Explanation:

The diagram has a fairly simple explanation. In the top diagram, the space between the particle is increasing. That means that acceleration is increasing. The bottom diagram shows just the opposite. The particle starts off making large "distances" between where the particle is recorded and then the distances between recordings lessens and the particle is slowing down.

Rule: the greater the "distance" between dot positions, the greater the acceleration, because the speed is large.

Top diagram: increasing distance between dots = larger speed. The distance becomes greater as the particle moves to the right.

Bottom diagram: starts off large and decreases as we move from left to right  = - acceleration.  

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4 0
3 years ago
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A ball of mass m is thrown into the air in a 45° direction of the horizon, after 3 seconds the ball is seen in a direction 30° f
Rzqust [24]

Answer:

Velocity (magnitude) is 98.37 m/s

Explanation:

We use the vertical component of the initial velocity, which is:

v_{0y}=v_0*sin(45)=\frac{\sqrt{2} }{2}v_0

Using kinematics expression of vertical velocity (in y direction) for an accelerated motion (constant acceleration, which is gravity):

v_{y}=v_{0y}+a*t=\frac{\sqrt{2} }{2}v_0-9.8t

Now we need to find v_y as a function of v_0. We use the horizontal velocity, which is always the same as follow:

v_x=v_0cos(45\º)=\frac{\sqrt{2} }{2}v_0=v_{t=3}*cos(30\º) \\

We know the angle at 3 seconds:

v_y(t=3)=v_{t=3}*sin(30\º)\\v_{t=3}=\frac{v_y}{sin(30\º)}

Substitute  v_{t=3} in  v_x and then solve for  v_y

\frac{\sqrt{2} }{2}v_0=\frac{v_y*cos(30\º) }{sin(30\º)} \\v_y=\frac{\sqrt{6} }{6}v_0

With this expression we go back to the kinematic equation and solve it for initial speed

\frac{\sqrt{6} }{6} v_0 =\frac{\sqrt{2} }{2}v_0-29.4\\v_0(\frac{\sqrt{6}-3\sqrt{2}}{6} )=-29.4\\v_0=98.37 m/s

3 0
4 years ago
What is the magnitude of the electric field strength between them, if the potential 7.95 cm from the zero volt plate (and 2.05 c
Dmitriy789 [7]

Answer:

ΔVab = Ed

ΔVab = Va-Vb = Va-V0 = Va

E = Va/ d

   = 413V / 0.0795 m

    = 5194.97 V/M

Explanation:

the potential difference between two uniform plates is calculated by the formula of electric field.

6 0
3 years ago
A 0.46-kg cord is stretched between two supports, 7.2 m apart. When one support is struck by a hammer, a transverse wave travels
Kryger [21]

Answer:

      T = 6.0 N

Explanation:

given,

mass of the cord = 0.46 Kg

length of the supports = 7.2 m

time taken to travel = 0.74 s

tension in the chord = ?

using formula for tension calculation

T = \dfrac{v^2.m}{l}

v = \dfrac{l}{s}

v = \dfrac{7.2}{0.74}

v = 9.73 m/s

now, calculation of tension

T = \dfrac{9.73^2\times 0.46}{7.2}

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The tension in the cord is equal to 6.0 N.

7 0
3 years ago
A disk of a radius 50 cm rotates at a constant rate of 100 rpm. What distance in meters will a point on the outside rim travel d
Dmitry [639]

Answer:

the distance in meters traveled by a point outside the rim is 157.1 m

Explanation:

Given;

radius of the disk, r = 50 cm = 0.5 m

angular speed of the disk, ω = 100 rpm

time of motion, t = 30 s

The distance in meters traveled by a point outside the rim is calculated as follows;

\theta = \omega t\\\\\theta = (100 \frac{rev}{\min}  \times \frac{2\pi \ rad}{1 \ rev} \times \frac{1\min}{60 s} ) \times (30 s)\\\\\theta = 100 \pi \ rad\\\\d = \theta r\\\\d = 100\pi  \ \times \ 0.5m\\\\d = 50 \pi \ m = 157.1 \ m

Therefore, the distance in meters traveled by a point outside the rim is 157.1 m

6 0
3 years ago
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