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3 years ago

Why are generic pharmaceuticals significantly cheaper than name brand ones? NEED HELP ASAP

SAT

1 answer:

RSB [31]3 years ago

3 0

Answer:

Generic drugs tend to cost less than their brand-name counterparts because generic drug applicants do not have to repeat animal and clinical (human) studies that were required of the brand-name medicines to demonstrate safety and effectiveness.

Generic pharmaceuticals are significantly cheaper than name brand ones because generic pharmaceuticals are not protected by patent law, so the lack of barriers to entry and increased competition keep prices down.

Also, the makers of generic drugs do not have to cover the cost of developing and marketing a new product.

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a 15.0 kg block is attached to a very light horizontal spring of force constant 575 n/m and is resting on a smooth horizontal ta

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The instantaneous velocity of the 15 kg. mass just after collision can be found by the principle of linear momentum.

a) The speed of the 15 kg. just after collision is 2 m/s.

b) The type of collision is inelastic collision

c) The compression of the spring is approximately 0.323 m.

Reasons:

The given parameters are;

Mass of the block attached to the spring, m₁ = 15.0 kg

Force constant of the spring, K = 575 N/m

Mass of the stone that strikes the block, m₂ = 3.00 kg

Speed of the stone, v₂ = 8.00 m/s

Speed with which the stone rebounds, v₃ = 2.00 m/s

a) The total initial momentum = 3 kg. × 8 m/s = 24 kg·m/s

The final momentum, just after collision = 3 × (-2) kg·m/s + 15 kg ×v₁

By conservation of momentum, we have;

24 kg·m/s = 3 × (-2) kg·m/s + 15 kg ×v₁

$v_1 = \dfrac{24 \, kg \cdot m/s + 6 \, kg \cdot m/s}{15 \, kg} = 2 \, m/s$

The speed of the 15 kg. just after collision, v₁ = 2 m/s.

b) A collision is elastic when the kinetic energy of the collision is conserved

The initial kinetic energy, K.E.₁ = 0.5 × 3 kg. ×(8 m/s)² = 96 J

The sum of the final kinetic energy are;

0.5 × 3 kg. × (2 m/s)² + 0.5 × 15 kg × (2 m/s)² = 36 J

The initial kinetic energy ≠ The final kinetic energy

Therefore, the collision is not elastic

Kinetic energy of the block, K.E. = 0.5 × 15 kg × (2 m/s)² = 30 J

Elastic energy gained by the block = 0.5 × K × x² = 0.5 × 575 N/m × x²

Therefore;

0.5 × 575 N/m × x² = 30 J

$x^2 = \dfrac{30 \, J}{0.5 \times 575 \, N/m} = \dfrac{12}{115} \, m^2$

$x = 2 \cdot \sqrt{\dfrac{3}{115} } \approx 0.323$

The compression of the spring, x ≈ 0.323 m.

Learn more here:

brainly.com/question/7694106

Questions;

(a) The speed of the 15 kg mass immediately after the collision.

(b) Determine the type of collision; Elastic or inelastic collision.

(c) The distance to which the spring is compressed by the block.

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