Which of these is an example of Newton’s first law of motion?

Electricity that comes into our homes from the grid is known as:

il63 [147K]

Power grid
All the poles and wires you see along the highway and in front of your house are called the electrical transmission and distribution system. Today, generating stations all across the country are connected to each other through the electrical system (sometimes called the "power grid").

7 0

2 years ago

(20) A rocket is launched vertically. At time t = 0 seconds, the rocket’s engine shuts down. At the time, the rocket has reached

Diano4ka-milaya [45]

Answer:

Explanation:

Given that,

h(t) = -9.8t² / 2 + 125t + 500

for t > 0.

At t = 0, the rocket is at height h = 500m, at a velocity of Vo = 125m/s.

We want to find the maximum height reached by rocket

Using mathematics maxima and minima

let find the turning point when dh/dt = 0

dh/dt = -9.8t + 125

dh / dt = 0 = -9.8t + 125

9.8t = 125

t = 125 / 9.8

t = 12.76s

Let find the turning point to know if this time t = 12.76 is maximum or minimum point

Let find d²h / dt²

d²h / dt² = -9.8

Since, d²h/dt² < 0, then, at t = 12.76s is the maximum points.

Then, the maximum height reached is

h = -9.8t² / 2 + 125t + 500

h = -9.8(12.76)² / 2 + 125(12.76) + 500

h = -797.80 + 1595 + 500

h = 1297.2 m

The maximum height reached is 1297.2 m

From the attachment, the maximum height is 1297.2m at t = 12.76sec.

Comment, the result are the same for both the analysis aspect and the graphical aspect.

3 0

3 years ago

The built in flash in a compact camera is usally capable of giving correct exsposure for distance up to how many meters?

Brut [27]

Answer:

An on-camera flash is an indispensible accessory for many photographers; it provides additional light when conditions become too dark to handhold your camera comfortably, allows you to achieve more balanced exposures in daylight conditions, permits freezing of fast-moving subjects and can also be used to control or trigger other flash light sources. Additionally, a flash can be used as a highly effective creative tool to establish an aesthetic that elevates your imagery when lighting conditions are considered less than stellar. The benefits of an external on-camera flash far outweigh those provided by a built-in camera flash, while the only drawback is keeping an additional piece of equipment.

On-Camera Flash versus Off-Camera Flash versus In-Camera Flash

The term on-camera flash simply refers to a type of strobe light (flash) that can connect directly with your camera. While it is referred to as “on-camera” this does not require the flash to be physically mounted on your camera. On-camera flashes can, and often are, used off-camera. This differs from other strobe-light sources, such as studio pack strobes and monolights in that these types of strobes are not meant to be physically connected to your camera (except under rare and unusual circumstances involving convoluted methods of adaptation). Additionally, on-camera flashes usually have a self-contained power supply, although external power sources can sometimes be used to improve performance or battery life.

On-camera external flash also refers to the type of external flash that can be used on your camera, compared to a built-in flash that is integrated into many cameras. An on-camera external flash performs better than a built-in flash in almost every regard with the one exception that it is not built into your camera. The ability to take the flash off your camera results in a significantly greater number of lighting options; far more than simply providing a blast of flat light to the scene to facilitate an adequate exposure. It is often not desirable to have your flash pointed squarely at the scene at hand; more often than not you will want to bounce the flash light off other surfaces and point in other directions to control the look of your flash. When using an in-camera flash, you are forced to use the flash at the given angle from which it extends.

Most built-in flashes are also located near the camera lens, which can often result in the red-eye effect when photographing subjects in dimly lit conditions. Red-eye occurs because pupils dilate in dim light, the built-in flash is aligned with the lens's optical axis, its beam enters the eye and reflects back at the camera from the retina at the rear of the eye, which is quite red. Being able to use an on-camera flash source off-camera, from a different angle, will help to eliminate the red-eye effect in your photographs of people.

Guide Numbers, Manual Usage, Controlling Flash Power and Sync Speeds

Before delving into the automatic technology that is contained within most contemporary flashes, it is best to understand how to manually control and grasp a flash’s power. This is directly related to having an understanding of exposure ratios—how shutter speeds and apertures affect and balance each other—even though auto-exposure metering is available and often utilized for determining the best exposure settings.

Explanation:

8 0

2 years ago

Un hamster esta sentado sobre un tocadisco cuya rapidez angular es constante si el hamster se mueve a un punto localizado al dob

kotegsom [21]

Answer:

b) se duplica

Explanation:

The disk is moving with constant angular velocity, let's call it $\omega$ .

The linear velocity of a point on the disk is given by

$v=\omega r$

where r is the distance of the point from the axis of rotation.

In this problem, the object is moved at a distance twice as far as the initial point, so

$r' = 2r$

Therefore, the new linear velocity is

$v'=\omega r' = \omega (2r) = 2 \omega r = 2 v$

So, the velocity has doubled, and the correct answer is

b) se duplica

8 0

2 years ago

A circular loop in the plane of a paper lies in a 0.45 T magnetic field pointing into the paper. The loop's diameter changes fro

Amanda [17]

Answer:

(A). The direction of the induced current will be clockwise.

(B). The magnitude of the average induced emf 16.87 mV.

(C). The induced current is 6.75 mA.

Explanation:

Given that,

Magnetic field = 0.45 T

The loop's diameter changes from 17.0 cm to 6.0 cm .

Time = 0.53 sec

(A). We need to find the direction of the induced current.

Using Lenz law

If the direction of magnetic field shows into the paper then the direction of the induced current will be clockwise.

(B). We need to calculate the magnetic flux

Using formula of flux

$\phi_{1}=BA\cos\theta$

Put the value into the formula

$\phi_{1}=0.45\times(\pi\times(8.5\times10^{-2})^2)\cos0$

$\phi_{1}=0.01021\ Wb$

We need to calculate the magnetic flux

Using formula of flux

$\phi_{2}=BA\cos\theta$

Put the value into the formula

$\phi_{2}=0.45\times(\pi\times(3\times10^{-2})^2)\cos0$

$\phi_{2}=0.00127\ Wb$

We need to calculate the magnitude of the average induced emf

Using formula of emf

$\epsilon=-N(\dfrac{\Delta \phi}{\Delta t})$

Put the value into t5he formula

$\epsilon=-1\times(\dfrac{0.00127-0.01021}{0.53})$

$\epsilon=0.016867\ V$

$\epsilon=16.87\ mV$

(C). If the coil resistance is 2.5 Ω.

We need to calculate the induced current

Using formula of current

$I=\dfrac{\epsilon}{R}$

Put the value into the formula

$I=\dfrac{0.016867}{2.5}$

$I=0.00675\ A$

$I=6.75\ mA$

Hence, (A). The direction of the induced current will be clockwise.

(B). The magnitude of the average induced emf 16.87 mV.

(C). The induced current is 6.75 mA.

5 0

3 years ago