Answer:
15.8 seconds
Explanation:
Create an extended calculation to convert all the unit to what you need.
160 km 1000 m 1 hour 1 min
----------- x ------------- x -------------- x ---------- = 44.4 m/s
1 hour 1 km 60 min 60 sec
So 160km/hr is equal to 44.4m/s
Now you can figure out how many seconds it will take to go 700 meters.
44.4 m
---------- X x sec = 700 m
1 sec
Solve for x sec
x sec = 700m / 44.4 m/s
= 15.8 seconds
It’s true, because it also depends on things like mass. Higher temperature but less mass< Lower temperature but more mass.
The fact that the layers of graphite are held together by only weak Van der Walls forces implies that they can slide over each other.
<h3>Why is graphite a solid lubricant?</h3>
We know that graphite is composed of layers. These hexagonal layers are held together by weak Van Der Walls forces and as such are able to slide over each other. The carbon atom in each layer are held together by strong covalent bonds.
The fact that the layers of graphite are held together by only weak Van der Walls forces implies that they can slide over each other and as such make the graphite fluid.
Thus, the image that shows these layers of graphite is attached to this an answer
Learn more about graphite:brainly.com/question/11095487
#SPJ1
C.
Because it’s falling it has acceleration in the y direction. If you have acceleration, you usually also have velocity, and since kinetic energy is KE= Mv^2 you know you have it. It also has potential energy because it has some height to it, and PE= Mgh.
Answer:
Explanation:
Net electric field at the centre will be zero .
Since all the charges are equal and they all are symmetrically situated around the centre . So the electric field produced by each will cancel out each other and hence the resultant electric field will be zero . It happens because electric field is a vector quantity and therefore it adds up vectorially . All the four electric field will form two pairs , in each pair electric fields are acting in opposite direction . So they all cancel out to zero .
