Equation of velocity is given as
![V =[5.00m/s−(0.0180m/s^3)t^2]i^ + [2.00m/s+(0.550m/s^2)t]j^](https://tex.z-dn.net/?f=V%20%3D%5B5.00m%2Fs%E2%88%92%280.0180m%2Fs%5E3%29t%5E2%5Di%5E%20%2B%20%5B2.00m%2Fs%2B%280.550m%2Fs%5E2%29t%5Dj%5E)
at t = 7.93 s

so the magnitude of the velocity is given as


Part b)
the direction of the velocity is given as


part c)
for acceleration we know that


at t = 7.93 s

magnitude is given as


Part d)
for the direction of the motion


Answer:
v = 12.12 m/s
Explanation:
Given that,
Radius of the curvature, r = 30 m
To find,
The car's speed at the bottom of the dip.
Solution,
Let mg is the true weight of the passenger. When it is moving in the circular path, the centripetal force act on it. It is given by :

The normal reaction of the passenger is given by :

N = 1.5 mg
Let v is the car's speed at the bottom of the dip. It can be calculated as:



v = 12.12 m/s
So, the speed of the car at the bottom of the dip is 12.12 m/s. Hence, this is the required solution.
Answer:
dV/dt = 9 cubic inches per second
Explanation:
Let the height of the cylinder is h
Diameter of cylinder = height of the cylinder = h
Radius of cylinder, r = h/2
dh/dt = 3 inches /s
Volume of cylinder is given by

put r = h/2 so,

Differentiate both sides with respect to t.

Substitute the values, h = 2 inches, dh/dt = 3 inches / s

dV/dt = 9 cubic inches per second
Thus, the volume of cylinder increases by the rate of 9 cubic inches per second.
Answer:
The following options are true based on the properties of electric field;
a) Electric field lines near positive point charges radiate outward.
b) The electric force acting on a point charge is proportional to the magnitude of the point charge.
d) In a uniform electric field, the field lines are straight, parallel, and uniformly spaced.
Explanation:
From option b) From coulomb's law F = Kq1q2r/r2
Gravity causes the motions of planets, stars, and galaxies. It's why the Moon orbits around the Earth, and the Earth orbits around the Sun, and the solar system orbits around the galaxy.