Distance = 170 km Displacement = 30 km East Speed = 56-2/3 km/hr Velocity = 10 km/hr East
Answer:
The average force has a magnitude 6524 N due north.
Explanation:
The average net force F = ma where m = mass of car = 1400 kg and a = acceleration.
a = (v - u)/t where u = initial velocity of car = 0 m/s (since it starts from rest)
v = final velocity of car = 27 m/s due north and t = time of motion = 5.8 s
a = (27 m/s - 0 m/s)/5.8 s = 27 m/s ÷ 5.8 s = 4.66 m/s
Since the direction of the velocity change is the direction of the acceleration, the acceleration is 4.66 m/s due north.
The average force, F = ma = 1400 kg × 4.66 m/s = 6524 N
Since the acceleration is due north, the average force takes the direction of the acceleration.
So the direction of the average force is due north
The average force has a magnitude 6524 N due north.
Answer:
solid, liquid, gas
Explanation:
The three phases of water are the three states of matter water can be found in, and that is:
solid (as in ice)
liquid (as in water)
gas (as in water vapor)
Answer:
x=2.4t+4.9t^2
Explanation:
This equation is one of the kinematic equations to solve for distance. The original equation is as follows:
X=Xo+Vt+1/2at^2
We know that the ball starts at rest meaning that its initial velocity and position is zero.
X=0+Vt+1/2at^2
Since it is going down the ramp, you can use the acceleration of gravity constant. (9.81 m/s^2) and simplify that with the 1/2.
X=Vt+4.9t^2
Note: Since the positive direction in this problem is down, you are adding the 4.9t^2, but if a question says that the downward direction is negative, you would subtract those values.
Now, substitute in your velocity value.
X=2.4t+4.9t^2
here we have to calculate the net positive charge present on he surface of the conducting sphere.
as the sphere is a conducting one the,the charge will be stored on its surface.
though the charge is present at the surface,still it will behave just like the total charge is concentrated at the centre of the sphere.
the electric field at outside of the sphere is E=![\frac{1}{4\pi\epsilon} \frac{q}{r^2}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B4%5Cpi%5Cepsilon%7D%20%5Cfrac%7Bq%7D%7Br%5E2%7D)
as E= ![\frac{-dv}{dr}](https://tex.z-dn.net/?f=%5Cfrac%7B-dv%7D%7Bdr%7D)
so V=![\frac{1}{4\pi\epsilon} \frac{q}{r}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B4%5Cpi%5Cepsilon%7D%20%5Cfrac%7Bq%7D%7Br%7D)
here V=27 volt,radius of sphere is 0.800 m and the point at which the potential is considered is at 1.20 m from th centre of the sphere.hence the point is at the outside of the sphere .
hence we have 27=![\frac{1}{4\pi\epsilon} \frac{q}{1.20}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B4%5Cpi%5Cepsilon%7D%20%5Cfrac%7Bq%7D%7B1.20%7D)
q=27×
coulomb
=3.6×
coulomb [ans]