Integrating the velocity equation, we will see that the position equation is:

<h3>How to get the position equation of the particle?</h3>
Let the velocity of the particle is:

To get the position equation we just need to integrate the above equation:


Then:


Replacing that in our integral we get:


Where C is a constant of integration.
Now we remember that 
Then we have:

To find the value of C, we use the fact that f(0) = 0.

C = -1 / 3
Then the position function is:

Integrating the velocity equation, we will see that the position equation is:

To learn more about motion equations, refer to:
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<h3><u>Answer;</u></h3>
volume = 6.3 × 10^-2 L
<h3><u>Explanation</u>;</h3>
Volume = mass/density
Mass = 0.0565 Kg,
Density = 900 kg/m³
= 0.0565 kg/ 900 kg /m³
= 6.3 × 10^-5 M³
but; 1000 L = 1 m³
Hence, <u>volume = 6.3 × 10^-2 L</u>
Answer:
Vy = V sin theta = 30 * ,574 = 17.2 m/s
t1 = 17.2 / 9.8 = 1.76 sec to reach max height
Max height = 17.2 * 1.76 - 1/2 * 4.9 * 1.76^2 = 15.1 m
H = V t - 1/2 g t^2 = 1.2 * 9.8 * 1.76^2 = 15.1 m
Time to fall from zero speed to ground = rise time = 1.76 sec
Vx = V cos 35 = 24.6 m / sec horizontal speed
Time in air = 1.76 * 2 = 3.52 sec before returning to ground
S = 24.6 * 3.52 = 86.6 m
Answer:
2.12/R mW
Explanation:
The electrical power, P generated by the rod is
P = B²L²v²/R where B = magnetic field = 0.575 T, L = length of metal rod = separation of metal rails = 20 cm = 0.2 m, v = velocity of metal rod = 40 cm/s = 0.4 m/s and R = resistance of rod = ?
So, the induced emf on the conductor is
E = BLv
= 0.575 T × 0.2 m × 0.4 m/s
= 0.046 V
= 46 mV
The electrical power, P generated by the rod is
P = B²L²v²/R
= B²L²v²/R
So, P = (0.575 T)² × (0.2 m)² × (0.4 m/s)²
= 0.002116/R W
= 2.12/R mW
Answer:
Explanation:
Magnitude of frictional force = μ mg
μ is either static or kinetic friction.
To start the crate moving , static friction is calculated .
a ) To start crate moving , force required = μ mg where μ is coefficient of static friction .
force required =.517 x 56.6 x 9.8 = 286.76 N .
b ) to slide the crate across the dock at a constant speed , force required
= μ mg where μ is coefficient of kinetic friction , where μ is kinetic friction
= .26 x 56.6 x 9.8 = 144.21 N .