Answer:it should be Newton’s second law
Explanation: lmk if I’m wrong
Answer:
A block of mass M = 5 kg is resting on a rough horizontal surface for which the coefficient of friction is 0.2. When a force F = 40N is applied, the acceleration of the block will be then (g=10ms
2 ).
Mass of the block=5kg
Coeffecient of friction=0.2
external applied force, F=40N
The angle at which the force is applied=30degree
So the horizontal component of force=Fcos30=40×
23 =20 3 N
While the uertical component of the force acting in upward direction=Fsin30=40× 21
=20N
The normal reaction from the surface (N)=mg−Fsin30=50−20=30N
So the ualue of limiting friction=μN=0.2×30=6N
Hence the net horizontal force on the block=Fcos30=μN=20
3
N−6N=28.64N
The horizontal acceleration of the block=
m
Fcos30−μN = 528.64
=5.73m/s 2
Answer:
88 m/s
Explanation:
To solve the problem, we can use the following SUVAT equation:
where
v is the final velocity
u is the initial velocity
a is the acceleration
d is the distance covered
For the car in this problem, we have
d = 484 m is the stopping distance
v = 0 is the final velocity
is the acceleration
Solving for u, we find the initial velocity:
Answer:
m = B²qR² / 2 V
Explanation:
If v be the velocity after acceleration under potential difference of V
kinetic energy = loss of electric potential energy
1/2 m v² = Vq ,
v² = 2 Vq / m ----------------------- ( 1 )
In magnetic field , charged particle comes in circular motion in which magnetic force provides centripetal force
magnetic force = centripetal force
Bqv = mv² / R
v = BqR / m
v² = B²q²R² / m² ------------------------- (2)
from (1) and (2)
B²q²R² / m² = 2 Vq / m
m = B²q²R² / 2 Vq
m = B²qR² / 2 V
