Whats the question?
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Answer:
The car manufacturers could increase bore of the cylinders, place the engine in the center or back of the car, add 1 to 2 turbochargers, and lower the center of gravity of the vehicle to increase traction.
Explanation:
Turbochargers would be recommended because they significantly increase both the torque of the engine as well as the amount of horses powering the car while also increasing original efficiency both with and without the additional power. Weight adjustment allows for lightweight vehicles with good traction. This is important to both keep control of the car under acceleration, but it also makes the vehicle more efficient due to the now sheddable unnecessary weight. A more obvious approach would be to increase the base horsepower and torque of the engine by increasing the bore of the cylinders and the weight of the pistons. This acts as an inertial lever, because the extra piston weight will drag the crankshaft faster. This could also be achieved by taking away piston weight, but this could be catastrophic should a piston slip.
Answer:
The reading of the scale during the acceleration is 446.94 N
Explanation:
Given;
the reading of the scale when the elevator is at rest = your weight, w = 600 N
downward acceleration the elevator, a = 2.5 m/s²
The reading of the scale can be found by applying Newton's second law of motion;
the reading of the scale = net force acting on your body
R = mg + m(-a)
The negative sign indicates downward acceleration
R = m(g - a)
where;
R is the reading of the scale which is your apparent weight
m is the mass of your body
g is acceleration due to gravity, = 9.8 m/s²
m = w/g
m = 600 / 9.8
m = 61.225 kg
The reading of the scale is now calculated as;
R = m(g-a)
R = 61.225(9.8 - 2.5)
R = 446.94 N
Therefore, the reading of the scale during the acceleration is 446.94 N
Answer:
option A
Explanation:
given,
For exerted by the worker = 245 N
angle made with horizontal = 55°
we need to calculate Force which is not used to move the crate = ?
Movement of crate is due to the horizontal component of the force.
Crate will not move due to vertical force acting on the it.
hence, worker's force not used to move the crate is equal to 200.69
The correct answer is option A
Answer:
26 m/s
69 m
Explanation:
Given:
v₀ = 20 m/s
a = 2 m/s²
t = 3 s
Find: v and Δx
v = at + v₀
v = (2 m/s²) (3 s) + 20 m/s
v = 26 m/s
Δx = v₀ t + ½ at²
Δx = (20 m/s) (3 s) + ½ (2 m/s²) (3 s)²
Δx = 69 m
