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3 years ago

Help me besties PLEASE ;-;

Physics

1 answer:

Svet_ta [14]3 years ago

6 0

Answer:

Current in resistors = 1.5 amp

Current in resistors = 0.18 amp

P = 93.75 watts

Explanation:

Given:

1. Three 5 ohm resistor(series)

Voltage = 10 volt

2. Three 5 ohm resistor(parallel)

Voltage = 9 volt

3. Three 5 ohm resistor(series)

Voltage = 6 volt

Find:

Current in resistors

Total power of circuit

Computation:

1. Total resistor (Series) R = R1 + R2 + R3

Total resistor (Series) R = 5 + 5 + 5

Total resistor (Series) R = 15 ohm

I = V/R

Current in resistors = (15)/(10)

Current in resistors = 1.5 amp

2. Total resistor (Series) R = 1/[1/R1 + 1/R2 + 1/R3

Total resistor (Series) R = 1[1/5 + 1/5 + 1/5]

Total resistor (Series) R = 1.67 ohm

I = V/R

Current in resistors = (1.67)/(9)

Current in resistors = 0.18 amp

3. Total resistor (Series) R = R1 + R2 + R3

Total resistor (Series) R = 5 + 5 + 5

Total resistor (Series) R = 15 ohm

I = V/R

I = (15)/(6)

I = 2.5 amp

P = I²(R)

P = (2.5)²(15)

P = 93.75 watts

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3 years ago

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

Help me besties PLEASE ;-;​

Help me besties PLEASE ;-;