The area of a rectangular park is 3/5 square mile.

Join my zoom meeting plz id: 368-807-1425 pass:4pWadQ

irga5000 [103]

Okay i will! hehe! lollllll

8 0

4 years ago

Christopher has at most \$18.50$18.50dollar sign, 18, point, 50 to spend at a convenience store. He buys one bag of bananas for

Juli2301 [7.4K]

Answer:

Amount to spend = $18.50

Purchased banana =$1.50

Apple Juice =$1.55

Gasoline $2.35

$1.50 + $1.55 = $3.05

18.50 - 3.05 = 15.45

1 Gallon of Gasoline = 2.35

Therefore,

2.35 x 6 = 14.1

15.35 - 14.1 = 1.25

To avoid exceeding budget, he can only buy

6 gallons of gasoline

4 0

4 years ago

There are 28 seats in the front row of a theater each successive row contains two more seats than th

Vinvika [58]

Answer:

74

Explanation:

28+2*(24-1)= 74

Given:- There are 28 seats in the front row of theater

Number of rows=24

To find:- The number of seats in the last row.

Use the formula to find the 24th term of this sequence.

Also each successive row has two more seats. This difference is same for the whole theater.

So, this forms an arithmetic sequence.

a1=21

D=2

We need to find number of seats in 24th row; which is given by a24

We use the formula for it

8 0

3 years ago

Study the table and give solution to the problem using the six steps.

faltersainse [42]

What table? you didn’t attach anything?

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