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2 years ago

Aye, I'm b0red so here's an easy one:

Physics

2 answers:

asambeis [7]2 years ago

8 0

1. blue red pink black

2. you can say bored look

3. why do you want to talk in brainly?

marta [7]2 years ago

6 0

Answer:

bloo B)

Explanation:

You might be interested in

The intensity I of light varies inversely as the square of the distance D from the source. If the intensity of illumination on a

emmasim [6.3K]

The intensity on a screen 20 ft from the light will be 0.125-foot candles.

<h3>What is the distance?</h3>

Distance is a numerical representation of the length between two objects or locations.

The intensity I of light varies inversely as the square of the distance D from the source;

I∝(1/D²)

The ratio of the intensity of the two cases;

$\rm \frac{I_1}{I_2} =(\frac{D_2}{D_1} )^2\\\\ \rm \frac{2}{I_2} =(\frac{20}{5} )^2\\\\ \frac{2}{I_2} =4^2 \\\\ I_2= \frac{2}{16} \\\\ I_2= 0.125 \ foot-candles$

Hence, the intensity on a screen 20 ft from the light will be 0.125 foot-candles

To learn more about the distance refer to the link;

brainly.com/question/26711747

#SPJ1

6 0

2 years ago

A car travels on a straight, level road. (a) Starting from rest, the car is going 38 ft/s (26 mi/h) at the end of 4.0 s. What is

lbvjy [14]

Answer:

$a)9.5\frac{ft}{s^2}\\ b) 12.66\frac{ft}{s^2}$

Explanation:

A body has acceleration when there is a change in the velocity vector, either in magnitude or direction. In this case we only have a change in magnitude. The average acceleration represents the speed variation that takes place in a given time interval.

$a_{avg}=\frac{\Delta v}{\Delta t}\\a_{avg}=\frac{v_{f}-v_{i}}{t_{f}- t_{i}}\\a_{avg}=\frac{38\frac{ft}{s}-0}{4 s- 0}=9.5\frac{ft}{s^2}\\$

$a_{avg}=\frac{\Delta v}{\Delta t}\\a_{avg}=\frac{v_{f}-v_{i}}{t_{f}- t_{i}}\\a_{avg}=\frac{76\frac{ft}{s}-38\frac{ft}{s}}{7 s- 4s}\\a_{avg}=\frac{38\frac{ft}{s}}{3s}=12.66\frac{ft}{s^2}$

8 0

3 years ago

Problems - Show all work.

Fittoniya [83]

Answer:

21s

Explanation:

Given parameters;

Radius = 10m

Speed or velocity = 3m/s

Unknown:

Period = ?

Solution:

To solve this problem, use the expression:

v = $\frac{2\pi r}{T}$

r is the radius

T is the unknown

Input the parameters and solve for T;

3 = $\frac{2 x \pi x 10}{T}$

62.84 = 3T

T = 21s

4 0

2 years ago

Young's Modulus refers to changes in the a Volume b- Length c- Body layers

Novosadov [1.4K]

Explanation:

Young' modulus is the ratio of normal stress to the longitudinal strain. Mathematically, it is given by :

Normal stress is given by force per unit area. Longitudinal strain is the change in length per unit original length.

The mathematical definition of Young's modulus is given by :

$Y=\dfrac{F/A}{\Delta L/L}$ ..........(1)

Where

$\Delta L$ is the change in length

F is the force

A is the area of cross section

So, the Young's modulus refers to the change in length of the object. Hence, the correct option is (b) "length".

8 0

3 years ago

Aloop of wire of area 71 cm^2 is placed with its plane parallel to a 16 mt magnetic field. the loop is then rotated so that its

kkurt [141]

Answer:

Approximately 1.62 × 10⁻⁴ V.

Explanation:

The average EMF in the coil is equal to

$\displaystyle \frac{\text{Final Magnetic Flux} - \text{Initial Magnetic Flux}}{2}$ ,

Why does this formula work?

By Faraday's Law of Induction, the EMF $\epsilon$ induced in a coil (one loop) is equal to the rate of change in the magnetic flux $\Phi$ through the coil.

$\displaystyle \epsilon(t) = \frac{d}{dt}(\Phi(t))$ .

Finding the average EMF in the coil is similar to finding the average velocity.

$\displaystyle \text{Average}\; \epsilon = \frac{1}{t}\int_0^t \epsilon(t)\cdot dt$ .

However, by the Fundamental Theorem of Calculus, integration reverts the action of differentiation. That is:

$\displaystyle \int_0^{t} \epsilon(t)\cdot dt = \int_0^{t} \frac{d}{dt}\Phi(t)\cdot dt = \Phi(t) - \Phi(0)$ .

Hence the equation

$\displaystyle \text{Average}\; \epsilon = \frac{1}{t}\int_0^t \epsilon(t)\cdot dt = \frac{\Phi(t)- \Phi(0)}{t}$ .

Note that information about the constant term in the original function will be lost. However, since this integral is a definite one, the constant term in $\Phi(t)$ won't matter.

Apply this formula to this question. Note that $\Phi$ , the magnetic flux through the coil, can be calculated with the equation

$\Phi = B \cdot A \cdot N \; \sin{\theta}$ .

For this question,

$B = \rm 16\; mT = 16\times 10^{-3}\; T$ is the strength of the magnetic field.
$A = \rm 71\; cm^{2} = 71\times \left(10^{-2}\right)^2 \; m^{2}$ is the area of the coil.
$N = 1$ is the number of loops in the coil.
$\theta$ is the angle between the field lines and the coil.
At $\rm 0\;s$ , the field lines are parallel to the coil, $\theta = 0^{\circ}$ .
At $\rm 0.7\; s$ , the field lines are perpendicular to the coil, $\displaystyle \theta = 90^{\circ}$ .