Answer:

Explanation:
Given:
- mass of vehicle,

- radius of curvature,

- coefficient of friction,

<u>During the turn to prevent the skidding of the vehicle its centripetal force must be equal to the opposite balancing frictional force:</u>

where:
coefficient of friction
normal reaction force due to weight of the car
velocity of the car

is the maximum velocity at which the vehicle can turn without skidding.
I believe that the answer is C<span />
Answer:
4.2 m/s
Explanation:
Momentum is conserved.
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
(35 g) (9 m/s) + (75 g) (-7 m/s) = (35 g) (-15 m/s) + (75 g) v
315 g m/s − 525 g m/s = -525 g m/s + (75 g) v
315 g m/s = (75 g) v
v = 4.2 m/s
Answer:
Explanation:
The equation for this, since we are talking about weight on an elevator, is Newton's 2nd Law adjusted to fit our needs:
where the Normal Force needed to lift that elevator car is the tension. So the equation then becomes
T = ma + w where T is the tension in the cable to lift the elevator, m is the mass of the elevator (which we have to solve for), a is the acceleration of the elevator (positive since it's going up), and w is the weight of the elevator (which we have as 5500 N). Solving first for mass:
w = mg and
5500 =- m(10) so
m = 550 kg. Now we have what we need to solve for the tension:
T = 550(4.0) + 5500 and
T = 2200 + 5500 so
T = 7700 N