Answer:
As given that the car maintains a constant speed v as it traverses the hill and valley where both the valley and hill have a radius of curvature R.
(i) At point C, the normal force acting on the car is largest because the centripetal force is up. gravity is down and normal force is up. net force is up so magnitude of normal force must be greater than the car's weight.
(ii) At point A, the normal force acting on the car is smallest because the centripetal force is down. gravity is down and normal force is up. net force is up so magnitude of normal force must be less than car's weight.
(iii) At point C, the driver will feel heaviest because the driver's apparent weight is the normal force on her body.
(iv) At point A, the driver will feel the lightest.
(v)The car can go that much fast without losing contact with the road at A can be determined as follow:
Fn=0 - lose contact with road
Fg= mv²/r
mg=mv²/r
v=sqrt (gr)
Answer:
Part a)
Part b)
Explanation:
As we know that mountain climber is at rest so net force on it must be zero
So we will have force balance in X direction
now we will have force balance in Y direction
Part a)
so from above equations we have
Part b)
Now for tension in right string we will have
The main cause of this is Friction. The more oil that is laid down, the less friction there is between the ball and the lane surface. The less friction, the harder it is for the bowler to send the ball in a curved path imparted by the spin that the bowler puts on the ball at the instant of release.
Answer:
the acceleration is 130.3m/s²
Explanation:
Given data
Force F= 18.9N
Mass of ball m= 0.145kg
Acceleration a=?
Applying the Newton's second law of motion
"The rate of change of momentum of a body is proportional to the external force".
F=ma
a= F/m
a= 18.9/0.142
a= 130.3m/s²
Answer:
2.16×10⁻⁶ N
Explanation:
Applying,
F = kqq'/r² (coulomb's Law)....................... Equation 1
Where F = electrostatic force, k = coulomb's constant, q = charge on the styrofoam, q' = charge on the grain of salt, r = distance between the charges.
From the question,
Given: q = 0.002 mC = 2.0×10⁻⁶ C, q' = 0.03 nC = 3.0×10⁻¹¹ C, r = 0.5 m
Constant: k = 8.99×10⁹ Nm²/C²
Substitute these values into equation 1
F = (2.0×10⁻⁶)(3.0×10⁻¹¹)(8.99×10⁹)/0.5²
F = 2.16×10⁻⁶ N
