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Norma-Jean [14]
3 years ago
12

All of the following statements are correct except ________ .

Physics
2 answers:
Korvikt [17]3 years ago
8 0

Answer:

d

Explanation:

a) Is a correct statement as belts are reinforcing materials that encircle the tire under the thread. it acts as a protecting layer.

b) This is also a correct statement as the carcass of tire is actually made of plies, layers of clothes and rubber.

c) Again true as the pattern of most of the thread are made and designed in such a ways that vehicle can drive on any road whether wet or dry.

d) This is an incorrect statement as beads are not the decorative pattern at the outer edge of the tread.

nasty-shy [4]3 years ago
3 0

Answer:

d

Explanation:

I took the test and got it correct

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A 1.00 kg object is attached to a horizontal spring. the spring is initially stretched by 0.500 m, and the object is released fr
valina [46]
The  spring is initially stretched, and the mass released from rest (v=0). The next time the speed becomes zero again is when the spring is fully compressed, and the mass is on the opposite side of the spring with respect to its equilibrium position, after a time t=0.100 s. This corresponds to half oscillation of the system. Therefore, the period of a full oscillation of the system is
T=2 t = 2 \cdot 0.100 s = 0.200 s
Which means that the frequency is
f= \frac{1}{T}= \frac{1}{0.200 s}=5 Hz
and the angular frequency is
\omega=2 \pi f = 2 \pi (5 Hz)=31.4 rad/s

In a spring-mass system, the maximum velocity of the object is given by
v_{max} = A \omega
where A is the amplitude of the oscillation. In our problem, the amplitude of the motion corresponds to the initial displacement of the object (A=0.500 m), therefore the maximum velocity is
v_{max} = A \omega = (0.500 m)(31.4 rad/s)= 15.7 m/s
6 0
3 years ago
Why are not all<br>materials conductive<br>to electricity​
Ilia_Sergeevich [38]

Answer:

In conductive materials, the outer electrons in each atom can easily come or go, and are called free electrons. In insulating materials, the outer electrons are not so free to move. All metals are electrically conductive.

5 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
3 years ago
Which of the following statements explains how total time spent in the air is affected as a projectile's angle of launch is incr
charle [14.2K]

Answer:

Therefore letter <u>C is the correct answer.</u>

Explanation:

In a projectile motion the total time in the air can be calculated using the following equation:

We analyze the y-component motion.

v_{fy}=v_{iy}-gt

When the final velocity (v(f)) is equal to zero we calculate the upward time and multiplying it by 2 we find the total time in the air. So we will have:

t_{tot}=2\frac{v_{iy}}{g}

t_{tot}=2\frac{v_{i}sin(\theta)}{g}

We can see that the <u>total time is directly proportional to the angle</u>, then when <u>θ increase t increase.</u>

Therefore letter C is the correct answer.

I hope it helps you!  

3 0
3 years ago
A loaf of bread is removed from an oven at 350° F and cooled in a room whose temperature is 70° F. If the bread cools to 210° F
jasenka [17]

Answer:

The bread will take 216.40 minutes to cool to 185°F

Explanation: Newton law of cooling

states that the rate of cooling of an object is inversely proportional to the difference of temperatures between the object and its surroundings i.e. dTdt=−kT, where t is the time taken and T is the difference of the temperatures between the object and its surroundings.

This gives us T as a function of t and is given by T(t)=ce−kt.

T(0)= Ce^-kt

C=35₩F- 185F= 165F

T(20) = 165×e^-20k=(350-70)

T(20)= e^-20k =280/165

-20k= ln(280/165)

-20k= ln(1.697)

-20k= 0.5289

K= 0.5289/20

K= 0.0264

If it cools at 185F in t minutes

165e^-0.0264×t=0.5

e^-0.0264t= 0.5/165

-0.0264t= on 3.03×10^-3

-0.0264t= -5.7129

t= 5.7129/0.0264

t=216.40minutes

7 0
3 years ago
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