Answer:
Velocity = v = 2.24m/s
Acceleration = a = 0.20m/s²
Explanation:
Please see attachment below.
Given
z=(−8 cosθ) and θ = 0.3t
z = -8Cos (0.3t)
V = dz/dt
a = v²/R.
Please see full solution below.
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Answer:
26.82m/s
Explanation:
Given
Mass = m= 0.4kg
Initial Velocity = u = 0
Charge = 4.0E-5C
Distance= d = 0.5m
Object Charge = 2E-4C
First, we'll calculate the initial energy (E)
E = Potential Energy
PE = kQq / d
Where k = coulomb constant = 8.99E9Nm²/C²
Energy is then calculated by;
PE = 8.99E9 * 4E-5 * 2E-4 / 0.5
PE = 143.84J
Energy = Potential Energy = Kinetic Energy
K.E = ½mv² = 143.84J
½mv² = ½ * 0.40 * v² = 143.85
0.2v² = 143.85
v² = 143.85/0.2
v² = 719.25
v = √719.25
v = 26.81883666380777
v = 26.82m/s
Hence, the object is 26.82m/s fast when the cart moving is very far (infinity) from the fixed charge
Answer:
Explanation:
As per the Kepler's law of planetary motions :
1- The orbits are of elliptical shape having two foci and Sun is located on one foci.
2- The line segment that joins a planet and the Sun sweeps out equal are at equal interval of time.
3- The orbital period square is directly proportional to the cube of semi major axis of its orbit.
Kepler's law are applied on each of the planets of our solar system as the distance of the Sun from the planet is calculated through this.
For example : from Kepler's first law we can see the eccentricity of the Earth's orbit is 0.0167.
Answer:
10.93m/s with the assumption that the water in the lake is still (the water has a speed of zero)
Explanation:
The velocity of the fish relative to the water when it hits the water surface is equal to the resultant velocity between the fish and the water when it hits it.
The fish drops on the water surface vertically with a vertical velocity v. Nothing was said about the velocity of the water, hence we can safely assume that the velocity if the water in the lake is zero, meaning that it is still. Therefore the relative velocity becomes equal to the velocity v with which the fish strikes the water surface.
We use the first equation of motion for a free-falling body to obtain v as follows;
v = u + gt....................(1)
where g is acceleration due to gravity taken as 9.8m/s/s
It should also be noted that the horizontal and vertical components of the motion are independent of each other, hence we take u = 0 as the fish falls vertically.
To obtain t, we use the second equation of motion as stated;
![h=ut+gt^2/2.................(2)](https://tex.z-dn.net/?f=h%3Dut%2Bgt%5E2%2F2.................%282%29)
Given; h = 6.10m.
since u = 0 for the vertical motion; equation (2) can be written as follows;
![h=\frac{1}{2}gt^2............(3)](https://tex.z-dn.net/?f=h%3D%5Cfrac%7B1%7D%7B2%7Dgt%5E2............%283%29)
substituting;
![6.1=\frac{1}{2}*9.8*t^2\\6.1=4.9t^2\\hence\\t^2=6.1/4.9\\t^2=1.24\\t=\sqrt{1.24}=1.12s](https://tex.z-dn.net/?f=6.1%3D%5Cfrac%7B1%7D%7B2%7D%2A9.8%2At%5E2%5C%5C6.1%3D4.9t%5E2%5C%5Chence%5C%5Ct%5E2%3D6.1%2F4.9%5C%5Ct%5E2%3D1.24%5C%5Ct%3D%5Csqrt%7B1.24%7D%3D1.12s)
Putting this value of t in equation (1) we obtain the following;
v = 0 + 9.8*1.12
v = 10.93m/s