Answer:
Yes it is true because static friction help us to make grip between the foot and the ground surface
Explanation:
Yes, as we know that the friction between two surface is of two types
1) Static friction
2) Kinetic friction
When two surface moves relative to each other then the friction force between two surface is kinetic friction and it always opposes the relative motion of two rough surface
While when one surface has tendency to move with respect to its contact surface but if it can not move on it then in this case it is static friction
So relative motion is not possible in this type of friction
So static friction helps to prevent slipping
Answer:
vertical distance, s = 0.16 meters
Explanation:
It is given that,
An amateur player is about to throw a dart with an initial velocity of 15 meters/second onto a dartboard.
Dashboard is placed at a distance of 2.7 meters.
Calculating time from velocity and distance i.e. 
Using second equation of motion for finding vertical distance :

Here, u = 0 (for vertical velocity )
s = 0.162 meters
or s = 0.16 meters
Hence, the vertical distance by which the player will miss the target if he throws the dart horizontally, in line with the dartboard is 0.16 meters.
Work or energy spent=force times distance=25N*4m=100Nm
power=energy or work divided by time=100Nm/5s=20Nm/s
Answer:
The objects have opposite charges.
Explanation:
The object when released are moving towards each other simply because they attract each other.
- According to coulombs law of charges, like charges repel and will move away from one another.
- Unlike charges attracts and moves towards one another
- Since the objects are coming towards each other, they are of opposite charges.
6.
A. 1575 - 1265 = 310J
B. KE 1/2 MV^2
V=√2·KE/M = √2(310)/12 V = 7.2mls
C. PE = 1265 = mgh
h= 1265/mg = 1265/(12)(92) h= 10.8m
7.
A. KE = 1/2 mv^2 0.5(5)(12)^2 KE = 360J
B. PE = mgh = (5)(9.8)(2.6) PE = 127.4J
C. ME = KE + PE = 360 + 127.4 ME = 487.4J