what are the primary functions of body paragraphs in a comparative essay that focuses on genres? select four options.

What is the constant of proportionality in the proportional relationship y=45x ? 45 4 5 I don't know.

Natali5045456 [20]

Answer:

45

Explanation:

constant proportion

y = kx

5 0

3 years ago

Which change is an unlikely result of destruction of wetlands?

pantera1 [17]

Answer:

healthy fish populations

Explanation:

i just took the test and get it right

3 0

2 years ago

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

Luden [163]

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

(c) The kinetic energy given by the block = The elastic potential energy gained by the spring

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.

3 0

3 years ago

Which of the following describes how the author organizes the text?

sveticcg [70]

Answer:

Answer B

Explanation:

4 0

3 years ago

Of segments and , which of the segments has a greater area based on the given information? justify with your work. circle has ra

WITCHER [35]

Of segments and, which of the segments has a greater area based on the given information circle has a radius 12m has the greater base.

Substitute the fee of "r" into the equation for the place of a circle: place = πr^2. Note that π is the image for pi, which is about three.14. For example, a circle with a radius of three cm could yield an equation like this: place = π3^2.

Once you discover the place of the world, all you need to do is get the place of the triangle after which subtract the place of the triangle from the world and to be able to let you know the place of CFD and CED and can help you realize which one is more.
For instance, Lets discover A, area place = 90/360 x πx10 ^ 2 = 25pi
Now we could get the place of the triangle, that is place triangle = 1/2x bh xsin C
Now we could discover B of the triangle, that is place angle = 1/2x bhx sin place triangle = 1/2 x 12 x12sin 60 = 36 sqrt ( three CFD=sec place - place of gle = 24 pi-36 sqrt three)=13.04m^ .
So A has the more area base.