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DanielleElmas [232]

3 years ago

15

SpongeBob noticed that his favorite pants were not as clean as they used to be. His friend Sandy told him that he should try usi

ng Clean-O detergent, a new laundry soap she found at Sail-Mart. SpongeBob made sure to wash one pair of pants in plain water and another pair in water with the Clean-O detergent.
1. what is the independent variable?

2. what id the dependent variable?

pls help

1 answer:

Alborosie3 years ago

3 0

The pants with the detergent would be dependant

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A rocket is fired straight up. It contains two stages (Stage 1 and Stage 2) of solid rocket fuel that are designed to burn for 1

Archy [21]

Answer:

a) y= 3.5 10³ m, b) t = 64 s

Explanation:

a) For this exercise we use the vertical launch kinematics equation

Stage 1

y₁ = y₀ + v₀ t + ½ a t²

y₁ = 0 + 0 + ½ a₁ t²

Let's calculate

y₁ = ½ 16 10²

y₁ = 800 m

At the end of this stage it has a speed

v₁ = vo + a₁ t₁

v₁ = 0 + 16 10

v₁ = 160 m / s

Stage 2

y₂ = y₁ + v₁ (t-t₀) + ½ a₂ (t-t₀)²

y₂ = 800 + 150 5 + ½ 11 5²

y₂ = 1092.5 m

Speed is

v₂ = v₁ + a₂ t

v₂ = 160 + 11 5

v₂ = 215 m / s

The rocket continues to follow until the speed reaches zero (v₃ = 0)

v₃² = v₂² - 2 g y₃

0 = v₂² - 2g y₃

y₃ = v₂² / 2g

y₃ = 215²/2 9.8

y₃ = 2358.4 m

The total height is

y = y₃ + y₂

y = 2358.4 + 1092.5

y = 3450.9 m

y= 3.5 10³ m

b) Flight time is the time to go up plus the time to go down

Let's look for the time of stage 3

v₃ = v₂ - g t₃

v₃ = 0

t₃ = v₂ / g

t₃ = 215 / 9.8

t₃ = 21.94 s

The time to climb is

$t_{s}$ = t₁ + t₂ + t₃

t_{s} = 10+ 5+ 21.94

t_{s} = 36.94 s

The time to descend from the maximum height is

y = v₀ t - ½ g t²

When it starts to slow down it's zero

y = - ½ g t_{b}²

t_{b} = √-2y / g

t_{b} = √(- 2 (-3450.9) /9.8)

t_{b} = 26.54 s

Flight time is the rise time plus the descent date

t = t_{s} + t_{b}

t = 36.94 + 26.54

t =63.84 s

t = 64 s

3 0

3 years ago

The speed of light is greater in a vacuum than in air or water.<br><br> True or false

Dima020 [189]

True !! Hope I helped you out a bit!

7 0

4 years ago

Read 2 more answers

A 0.400-kg ice puck, moving east with a speed of 5.86 m/s , has a head-on collision with a 0.900-kg puck initially at rest.

andreev551 [17]

Answer:

a) The final speed of the 0.400-kg puck after the collision is 2.254 meters per second, b) The negative sign of the solution found in part a) indicates that 0.400-kg puck is moving westwards, c) The speed of the 0.900-kg puck after the collision is 3.606 meters per second eastwards.

Explanation:

a) Since collision is perfectly elastic and there are no external forces exerted on pucks system, the phenomenon must be modelled after the Principles of Momentum and Energy Conservation. Changes in gravitational potential energy can be neglected. That is:

Momentum

$m_{1}\cdot v_{1,o} + m_{2}\cdot v_{2,o} = m_{1}\cdot v_{1,f} + m_{2}\cdot v_{2,f}$

Energy

$\frac{1}{2}\cdot (m_{1}\cdot v_{1,o}^{2}+ m_{2}\cdot v_{2,o}^{2})=\frac{1}{2}\cdot (m_{1}\cdot v_{1,f}^{2}+ m_{2}\cdot v_{2,f}^{2})$

$m_{1}\cdot v_{1,o}^{2} + m_{2}\cdot v_{2,o}^{2} = m_{1}\cdot v_{1,f}^{2} + m_{2}\cdot v_{2,f}^{2}$

Where:

$m_{1}$ , $m_{2}$ - Masses of the 0.400-kg and 0.900-kg pucks, measured in kilograms.

$v_{1,o}$ , $v_{2,o}$ - Initial speeds of the 0.400-kg and 0.900-kg pucks, measured in meters per second.

$v_{1}$ , $v_{2}$ - Final speeds of the 0.400-kg and 0.900-kg pucks, measured in meters per second.

If $m_{1} = 0.400\,kg$ , $m_{2} = 0.900\,kg$ , $v_{1,o} = +5.86\,\frac{m}{s}$ , $v_{2,o} = 0\,\frac{m}{s}$ , the system of equation is simplified as follows:

$2.344\,\frac{kg\cdot m}{s} = 0.4\cdot v_{1,f} + 0.9\cdot v_{2,f}$

$13.736\,J = 0.4\cdot v_{1,f}^{2}+0.9\cdot v_{2,f}^{2}$

Let is clear $v_{1,f}$ in first equation:

$0.4\cdot v_{1,f} = 2.344 - 0.9\cdot v_{2,f}$

$v_{1,f} = 5.86-2.25\cdot v_{2,f}$

Now, the same variable is substituted in second equation and resulting expression is simplified and solved afterwards:

$13.736 = 0.4\cdot (5.86-2.25\cdot v_{2,f})^{2}+0.9\cdot v_{2,f}^{2}$

$13.736 = 0.4\cdot (34.340-26.37\cdot v_{2,f}+5.063\cdot v_{2,f}^{2})+0.9\cdot v_{2,f}^{2}$

$13.736 = 13.736-10.548\cdot v_{2,f} +2.925\cdot v_{2,f}^{2}$

$2.925\cdot v_{2,f}^{2}-10.548\cdot v_{2,f} = 0$

$2.925\cdot v_{2,f}\cdot (v_{2,f}-3.606) = 0$

There are two solutions:

$v_{2,f} = 0\,\frac{m}{s}$ or $v_{2,f} = 3.606\,\frac{m}{s}$

The first root coincides with the conditions before collision and the second one represents a physically reasonable solution.

Now, the final speed of the 0.400-kg puck is: ( $v_{2,f} = 3.606\,\frac{m}{s}$ )

$v_{1,f} = 5.86-2.25\cdot (3.606)$

$v_{1,f} = -2.254\,\frac{m}{s}$

The final speed of the 0.400-kg puck after the collision is 2.254 meters per second.

b) The negative sign of the solution found in part a) indicates that 0.400-kg puck is moving westwards.

c) The speed of the 0.900-kg puck after the collision is 3.606 meters per second eastwards.

3 0

3 years ago

A concave mirror always forms a virtual image of a real object. O True O False

kifflom [539]

Answer:

The given statement is false.

Explanation:

The spherical mirrors are the mirror that are a part of a sphere. Concave and convex mirrors are two types of spherical mirrors.

A concave mirror always forms real and inverted image. A convex mirror forms real and virtual images.

For concave mirror, the value of magnification is less that 1. Also, the focal length is negative for concave mirrors.

So, the given statement is false as a concave mirror always forms a real and inverted image. Hence, this is the required solution.

4 0

4 years ago

A box of mass 60 kg is at rest on a horizontal floor that has a static coefficient of friction of 0.6 and a kinetic coefficient

gavmur [86]

Answer:

a) The minimum force required to start moving the box is 352.86 N

b) i) The friction force for the box in motion is 147.025 N

ii) The acceleration of box is 4.21625 m/s²

Explanation:

The parameters of the box at rest and the floor are;

The mass of the box = 60 kg

The static coefficient friction of the floor = 0.6

The kinetic coefficient friction of the floor = 0.25

Frictional force = Normal force × Friction coefficient

For an horizontal floor and the box laying on the floor, we have;

The normal force = The weight of the box = Mass of the box × Acceleration due to gravity, g

The acceleration due to gravity, g = 9.81 m/s²

The weight of the box = 60 × 9.81 = 588.6 N

a) The static coefficient gives the frictional force observed by the box and which must be surpassed to bring about motion

Therefore;

The minimum force required to start moving the box = The static frictional force = Weight of the box × The static coefficient of friction

The minimum force required to start moving the box = 588.1 × 0.6 = 352.86 N

The minimum force required to start moving the box = 352.86 N

b) i) When an horizontal force of 400 N is applied, the applied force is larger than the static friction force, and the box will be in motion with the kinetic coefficient of friction being the source of friction

The friction force for the box in motion = 588.1 × 0.25 = 147.025 N

ii) The force, F with the box is in motion, is given as follows;

F = Mass of box × Acceleration of box, a = Applied force - Kinematic friction force

F = 60 × a = 400 - 147.025 = 252.975 N

60 × a = 252.975 N

a = 252.975 N/(60 kg) = 4.21625 m/s²

Acceleration of box, a = 4.21625 m/s².

6 0

3 years ago