Answer:
Friction force always acts tangent to the surface at points of contact. Friction force acts opposite to the direction of motion. There are 2 types of friction: Static friction: If the two surfaces in contact do not move relative to each other, one has static friction.
Answer: 12Mg/h
Explanation:
Let the spring is compressed by a distance x,before the lift stops,then
Mg(h+x)= 1/2 kx^2 ............... 1
Kx - Mg = M ( 5g ) ............ 2
Make x the subject in equation 2
Kx = 5Mg + Mg
Kx = 6Mg
x = 6Mg/k ............ 3
Put equation 3 into 1
Mg ( h + x ) = 1/2 kx^2
Mgh + Mgx = 1/2kx^2
Mgh + Mg × 6Mg/k = 1/2k × ( 6Mg/k )^2
Mgh + Mg× 6Mg/k = 1/2k 36M^2g^2/ k^2
h =18Mg/k - 6Mg/h
K = 12Mg/h
Answer:
The maximum height reached by the water is 117.55 m.
Explanation:
Given;
initial velocity of the water, u = 48 m/s
at maximum height the final velocity will be zero, v = 0
the water is going upwards, i.e in the negative direction of gravity, g = -9.8 m/s².
The maximum height reached by the water is calculated as follows;
v² = u² + 2gh
where;
h is the maximum height reached by the water
0 = u² + 2gh
0 = (48)² + ( 2 x -9.8 x h)
0 = 2304 - 19.6h
19.6h = 2304
h = 2304 / 19.6
h = 117.55 m
Therefore, the maximum height reached by the water is 117.55 m.
Answer:
6.1 × 10^9 Nm-1
Explanation:
The electric field is given by
E= Kq/d^2
Where;
K= Coulombs constant = 9.0 × 10^9 C
q = magnitude of charge = 1.62×10−6 C
d = distance of separation = 1.53 mm = 1.55 × 10^-3 m
E= 9.0 × 10^9 × 1.62×10−6/(1.55 × 10^-3 )^2
E= 14.58 × 10^3/2.4 × 10^-6
E= 6.1 × 10^9 Nm-1
