Could you please help me on number four

Svetlanka [38]

The 'Bulge', the 'Disk', the 'Spiral Arms', and the 'Halo' those are four parts of a spiral galaxy. Try finding out the many different types of galaxies to understand which is which.

5 0

3 years ago

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Which statement is correct about the car?

JulijaS [17]

Answer:

B

Explanation:

OOf we are doing this stuff atm

So if its faster at the front and slow at the back you can tell that its not slowing down because less of a force is there however at the front there is more of a force. Friction is low which means that its not makimg much contact so no sudden change of forces thats also why its B

6 0

3 years ago

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Janice is unsure about her future career path. She has grown up on her family farm, but she is also interested in medicine. Jani

Vika [28.1K]

Answer:

d not joining FRA and joining HOSA INSTEAD

3 0

4 years ago

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A player kicks a football from ground level with a velocity of 26.2m/s at an angle of 34.2° above the horizontal. How far back f

Amanda [17]

For the ball to go straight into the goal, the kicker needs to be no more than 6.54 meters away from the goal.

For the ball to arc into the goal, the kicker needs to be between 58.5 and 65.1 meters away from the goal.

<h3>Explanation</h3>

How long does it take for the ball to reach the goal?

Let the distance between the kicker and the goal be $x$ meters.

Horizontal velocity of the ball will always be $26.2\times\cos{34.2\textdegree}$ until it lands if there's no air resistance.

The ball will arrive at the goal in $\displaystyle \frac{x}{26.2\times\cos{34.2\textdegree}}$ seconds after it leaves the kicker.

What will be the height of the ball when it reaches the goal?

Consider the equation

$\displaystyle h(t) = -\frac{1}{2}\cdot g\cdot t^{2} + v_{0,\;\text{vertical}} \cdot t + h_0$ .

For this soccer ball:

$g = 9.81\;\text{m}\cdot\text{s}^{-2}$ ,
$v_{0,\;\text{vertical}} = 26.2\times \sin{34.2\textdegree{}}\;\text{m}\cdot\text{s}^{-2}$ ,
$h_0 = 0$ since the player kicks the ball "from ground level."

$\displaystyle t=\frac{x}{26.2\times\cos{34.2\textdegree}}$

when the ball reaches the goal.

$\displaystyle h= - 9.81 \times \frac{x^2}{(26.2\times\cos{34.2\textdegree})^2} + (26.2 \times \sin{34.2\textdegree})\times\frac{x}{26.2\times\cos{34.2\textdegree}} \\\phantom{h} = -\frac{9.81}{(26.2\times\cos{34.2\textdegree})^2}\cdot x^{2} + \frac{\sin{34.2\textdegree}}{\cos{34.2\textdegree}}\cdot x$ .

Solve this quadratic equation for $x$ , $x > 0$ .

$x = 65.1$ meters when $h = 0$ meters.
$x = 6.54$ or $58.5$ meters when $h = 4$ meters.

In other words,

For the ball to go straight into the goal, the kicker needs to be no more than 6.54 meters away from the goal.
For the ball to arc into the goal, the kicker needs to be between 58.5 and 65.1 meters away from the goal.

3 0

3 years ago

Ice has a specific heat of 2090 J/(kg C) and water has a specific heat of 4186 J/(kg C). Water has a latent heat of fusion of 3.

zepelin [54]

Answer:

3.1 × 10⁷ J

Explanation:

The total heat required is the sum of the heats required in each stage.

1) Solid: from -60°C to 0°C.

Q₁ = c(s) × m × ΔT = (2090 J/kg.°C) × 10 kg × (0°C - (-60°C)) = 1.3 × 10⁶ J

where,

c(s): specific heat of the solid

m: mass

ΔT: change in the temperature

2) Solid to liquid at 0°C

Q₂ = Qf × m = (3.3 × 10⁵ J/kg) × 10 kg = 3.3 × 10⁶ J

where,

Qf: latent heat of fusion

3) Liquid: from 0°C to 100°C

Q₃ = c(l) × m × ΔT = (4186J/kg.°C) × 10 kg × (100°C - 0°C) = 4.2 × 10⁶ J

where,

c(l): specific heat of the liquid

4) Liquid to gas at 100 °C

Q₄ = Qv × m = (2.26 × 10⁶ J/kg) × 10 kg = 2.26 × 10⁷ J

where,

Qv: latent heat of vaporization

Total heat

Q₁ + Q₂ + Q₃ + Q₄

1.3 × 10⁶ J + 3.3 × 10⁶ J + 4.2 × 10⁶ J + 2.26 × 10⁷ J = 3.1 × 10⁷ J

3 0

3 years ago