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3 years ago

What’s the difference between red blood cells and white blood cells

Physics

2 answers:

Lerok [7]3 years ago

4 0

Red blood cells contain hermoglobin whiles white blood cells do not. Hermoglobin is what gives it its red color.

mixas84 [53]3 years ago

4 0

Red blood cells contain hermoglobin whiles white blood cells do not. Hermoglobin is what gives it its red color.

You might be interested in

How does the height from which you drop the ball relate to the height that the ball bounces back up?

Stels [109]

The higher you go the more potential energy there is, and the lower it is the more kinetic energy there is, so the more kinetic energy there is the higher the ball will bounce.

7 0

3 years ago

The unit used to measure electric current is the ampere (A). Now, assume that the current delivered at a wall socket reaches the

pav-90 [236]

Answer:

T = 0.017s

Explanation:

period is the time it takes a particle to make one oscillation

An electric current is periodic in nature

The current reaches 3.8A ten times.

So there must have been 10 cycles (10 periods) in 0.17s. let 'T' be the period:

$T=\frac{t}{n}$

t is the total time interval

n is the number of oscillations

$T=\frac{0.17}{10}$

10T = 0.17

T = 0.17/10 = 0.017s

8 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

A 5.92 g object moving to the right at 17.1 cm/s makes an elastic head-on collision with an 11.84 g object that is initially at

Lelechka [254]

Answer:

v₁f = -5.7 cm/s

Explanation:

Assuming no external forces acting during the collision, total momentum must be conserved, as follows:

$m_{1} * v_{10} = m_{1} *v_{1f} + m_{2} * v_{2f} (1)$

Rearranging terms, we have:

$m_{1} * (v_{10} - *v_{1f} ) = m_{2} * v_{2f} (2)$

We also know that the collision is elastic, so total kinetic energy must be conserved , as follows:

$\Delta K = 0 \\ \\ \frac{1}{2} * m_{1} *v_{10} ^{2} = \frac{1}{2}* m_{1} *v_{1f} ^{2} + \frac{1}{2}* m_{2} *v_{2f} ^{2} (3)$

Rearranging , and simplifying common terms, we have:

$m_{1}* (v_{10} ^{2} -v_{1f} ^{2} ) = m_{2} *v_{2f} ^{2} (4)$

Replacing by the givens, doing some algebra and dividing (4) by (2), we find the following relationship:

$v_{10} + v_{1f} = v_{2f}$

Replacing the expression above in (1), as m₂ = 2*m₁, we can find the value of v₁f, as follows:

$m_{1} * v_{10} = m_{1} * v_{1f} +2*m_{1} * (v_{10} + v_{1f})\\ \\ -(m_{1} * v_{10}) = 3* m_{1} *v_{1f} \\ \\ v_{1f} = - \frac{v_{10} }{3} = \frac{-17.1cm/s}{3} = -5.7 cm/s$

7 0

3 years ago

A 7.94-nC charge is located 1.77 m from a 4.14-nC point charge. (a) Find the magnitude of the electrostatic force that one charg

STatiana [176]

Answer:

F=94.32*10⁻⁹N , The force F is repusilve because both charges have the same sign (+)

Explanation:

Two point charges (q₁, q₂) separated by a distance (d) exert a mutual force (F) whose magnitude is determined by the following formula:

F=K*q₁*q₂/d² Formula (1)

F: Electric force in Newtons (N)

K : Coulomb constant in N*m²/C²

q₁,q₂:Charges in Coulombs (C)

d: distance between the charges in meters(m)

Equivalence

1nC= 10⁻⁹C

Data

K=8.99x10⁹N*m²/C²

q₁ = 7.94-nC= 7.94*10⁻⁹C

q₂= 4.14-nC= 4.14 *10⁻⁹C

d= 1.77 m

Magnitude of the electrostatic force that one charge exerts on the other

We apply formula (1):

$F=8.99x10^{9} *\frac{7.94*10^{-9} *4.14 *10^{-9} }{1.77^{2} }$

F=94.32*10⁻⁹N , The force F is repusilve because both charges have the same sign (+)

7 0

3 years ago