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3 years ago

From readong these two paragraph only the best description of Austin is that he is a ___ character

Physics

2 answers:

Andrei [34K]3 years ago

5 0

Explanation:

may you send the paragraph pic please im curious as to the answer too

Yuliya22 [10]3 years ago

3 0

Answer:

can I get the passage please

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In a simple electric circuit, ohm's law states that v=irv=ir, where vv is the voltage in volts, ii is the current in amperes, an

Tju [1.3M]

We take the derivative of Ohm's law with respect to time: V = IR
Using the product rule:
dV/dt = I(dR/dt) + R(dI/dt)
We are given that voltage is decreasing at 0.03 V/s, resistance is increasing at 0.04 ohm/s, resistance itself is 200 ohms, and current is 0.04 A. Substituting:
-0.03 V/s = (0.04 A)(0.04 ohm/s) + (200 ohms)(dI/dt)
dI/dt = -0.000158 = -1.58 x 10^-4 A/s

8 0

3 years ago

A car is traveling up a hill that is inclined at an angle θ above the horizontal. Determine the ratio of the magnitude of the no

padilas [110]

Explanation:

The weight of the car is equal to, $W_c=m\times g$ ...........(1)

Where

m is the mass of car

g is the acceleration due to gravity

The normal or vertical component of the force is, $F_N=mg\ cos\theta$

$F_N=mg\ cos(13)$ .............(2)

The horizontal component of the force is, $F_H=mg\ sin\theta$

Taking ratio of equation (1) and (2) as :

$\dfrac{F_N}{W_c}=\dfrac{mg\ cos\theta}{mg}$

$\dfrac{F_N}{W_c}=cos(13)$

$\dfrac{F_N}{W_c}=0.97$

$\dfrac{F_N}{W_c}=\dfrac{97}{100}$

Hence, this is the required solution.

5 0

3 years ago

in a solar system far, far away the sun's intensity is 200 w/m2 for an inner planet located a distance r away. what is the sun's

GarryVolchara [31]

The sun's intensity for an outer planet located at a distance 6r from the sun is 5.55 W/m². The result is obtained by using the inverse square law formula.

<h3>What is the Inverse Square Law formula?</h3>

The Inverse Square Law formula describes the intensity of light is inversely proportional to the square of the distance. It can be expressed as

$\frac{I_{1} }{I_{2} } = \frac{d_{2}^{2} }{d_{1}^{2}}$

Where

I₁ = Intensity at distance 1 (W/m²)
I₂ = Intensity at distance 2 (W/m²)
d₁ = distance 1 from a light source (m)
d₂ = distance 2 from a light source (m)

Given the case the sun's intensity is 200 W/m² for an inner planet at the distance r. If an outer planet is at a distance 6r, what is the sun's intensity?

By using the inverse square law formula, the sun's intensity for an outer planet is

$\frac{I_{1} }{I_{2} } = \frac{d_{2}^{2} }{d_{1}^{2}}$

$\frac{200 }{I_{2} } = \frac{(6r)^{2} }{r^{2}}$

$\frac{200 }{I_{2} } = \frac{36r^{2} }{r^{2}}$

$I_{2} = \frac{200} {36}$

I₂ = 5.55 W/m²

Hence, the sun's intensity for a planet at a distance 6r from the sun is 5.55 W/m².

Learn more about intensity of light here:

brainly.com/question/13155277

#SPJ4

3 0

1 year ago

A roller coaster car is traveling at a constant 3 m/s when it reaches a downward slope. On the slope, the car accelerates at a c

kirill [66]

Answer:

Velocity of the car at the bottom of the slope: approximately $20.3\; \rm m \cdot s^{-2}$ .

It would take approximately $3.9\; \rm s$ for the car to travel from the top of the slope to the bottom.

Explanation:

The time of the travel needs to be found. Hence, make use of the SUVAT equation that does not include time.

Let $v$ denote the final velocity of the car.
Let $u$ denote the initial velocity of the car.
Let $a$ denote the acceleration of the car.
Let $x$ denote the distance that this car travelled.

$v^2 - u^2 = 2\, a\cdot x$ .

Given:

$u = 3\; \rm m \cdot s^{-1}$ .
$a = 4.5\; \rm m \cdot s^{-2}$ .
$x = 45\; \rm m$ .

Rearrange the equation $v^2 - u^2 = 2\, a\cdot x$ and solve for $v$ :

$\begin{aligned}v &= \sqrt{2\, a \cdot x + u^2} \\ &= \sqrt{2 \times 4.5\; \rm m \cdot s^{-2} \times 45\; \rm m + \left(3\; \rm m \cdot s^{-1}\right)^{2}} \\ &\approx 20.3\; \rm m \cdot s^{-1}\end{aligned}$ .

Calculate the time required for reaching this speed from $u = 3\; \rm m \cdot s^{-1}$ at $a = 4.5\; \rm m \cdot s^{-2}$ :

$\begin{aligned}t &= \frac{v - u}{a} \\ &\approx \frac{20.3\; \rm m \cdot s^{-1} - 3\; \rm m \cdot s^{-1}}{4.5\; \rm m \cdot s^{-2}} \approx 3.9\; \rm m \cdot s^{-1}\end{aligned}$ .

3 0

3 years ago

Trees in the mid-west change colors and lose their leaves in the cooler parts of the year.

wariber [46]

Deciduous
GIVE ME BRAINLIEST NOW

3 0

3 years ago

Read 2 more answers