Ask question

Ask question

All categories

3 years ago

7

A helicopter lifts a 66 kg astronaut 15 m vertically from the ocean by means of a cable. The acceleration of the astronaut is g/

11. How much work is done on the astronaut by (a) the force from the helicopter and (b) the gravitational force on her? Just before she reaches the helicopter, what are her (c) kinetic energy and (d) speed?

1 answer:

artcher [175]3 years ago

4 0

Answer

given,

mass of the astronaut = 66 kg

height of the lift = 15 m

acceleration = g/11 = 0.89 m/s²

a) the work done by helicopter

W = F× h

= m (a+g)h

= 66 × (9.8+0.89)15

W =10583.1 J

b) work done by the gravity

W= mgh

W = - 66 × 9.8 × 15

W = -9702 J

c) Δ KE = net work done

Δ KE = 10583.1 - 9702

Δ KE = 881.1 J

d) speed

$\dfrac{1}{2}mv^2 = 881.1$

$v = \sqrt{\dfrac{2 \times 881.1}{66}}$

v =5.17 m/s

You might be interested in

What is the displacement of the runner in 4 laps? *<br> 2 point

Serhud [2]

Answer:

There is no displacement.

Explanation:

Because the runner is running laps and returning to the original place, there is no displacement as displacement is relative to the change in location from the original position.

Hope this helps. . .

ly UwU

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

3 years ago

Potassium is a crucial element for the healthy operation of the human body. Potassium occurs naturally in our environment and th

gregori [183]

Complete Question

Potassium is a crucial element for the healthy operation of the human body. Potassium occurs naturally in our environment and thus our bodies) as three isotopes: Potassium-39, Potassium-40, and Potassium-41. Their current abundances are 93.26%, 0.012% and 6.728%. A typical human body contains about 3.0 grams of Potassium per kilogram of body mass. 1. How much Potassium-40 is present in a person with a mass of 80 kg? 2. If, on average, the decay of Potassium-40 results in 1.10 MeV of energy absorbed, determine the effective dose (in Sieverts) per year due to Potassium-40 in an 80- kg body. Assume an RBE of 1.2. The half-life of Potassium-40 is $1.28 * 10^9$ years.

Answer:

The potassium-40 present in 80 kg is $Z = 0.0288 *10^{-3}\ kg$

The effective dose absorbed per year is $x = 2.06 *10^{-24}$ per year

Explanation:

From the question we are told that

The mass of potassium in 1 kg of human body is $m = 3g= \frac{3}{1000} = 3*10^{-3} \ kg$

The mass of the person is $M = 80 \ kg$

The abundance of Potassium-39 is 93.26%

The abundance of Potassium-40 is 0.012%

The abundance of Potassium-41 is 6.78 %

The energy absorbed is $E = 1.10MeV = 1.10 *10^{6} * 1.602 *10^{-19} = 1.7622*10^{-13} J$

Now 1 kg of human body contains $3.0*10^{-3}\ kg$ of Potassium

So 80 kg of human body contains k kg of Potassium

=> $k = \frac{ 80 * 3*10^{-3}}{1}$

$k = 0.240\ kg$

Now from the question potassium-40 is 0.012% of the total potassium so

Amount of potassium-40 present is mathematically represented as

$Z = \frac{0.012}{100} * 0.240$

$Z = 0.0288 *10^{-3}\ kg$

The effective dose (in Sieverts) per year due to Potassium-40 in an 80- kg body is mathematically evaluated as

$D = \frac{E}{M}$

Substituting values

$D = \frac{1.7622*10^{-13}}{80}$

$D = 2.2*10^{-15} J/kg$

Converting to Sieverts

We have

$D_s = REB * D$

$D_s = 1.2 * 2.2 *10^{-15}$

$D_s = 2.64 *10^{-15}$

So

for half-life ( $1.28 *10^9 \ years$ ) the dose is $2.64 *10^{-15}$

Then for 1 year the dose would be x

=> $x = \frac{2.64 *10^{-15}}{1.28 * 10^9}$

$x = 2.06 *10^{-24}$ per year

7 0

4 years ago

What is the acceleration if velocity increases from 10 m/s to 15m/s after travelling a distance of 5 metre

Semenov [28]

Answer:

a=1.25m/s²

Explanation:

GIVEN DATA

vi=10m/s

vf=15m/s

S=5m

TO FIND

a=?

SOLUTION

by using third equation of motion

2as=(vf)²-(vi)²

2a(5m)=(15m/s)²-(10m/s)²

10m×a=225m²/s²-100m²/s²

10m×a=125m²/s²

a= $\frac{125}{10}$

a=12.5m/s²

8 0

3 years ago

A billiard ball moving at 5 m/s strikes another ball which is initially at rest. After the collision, the first ball moves at a

ziro4ka [17]

Answer:

The velocity of the second ball is approximately 2.588 m/s

The angle direction of the second ball is 75° counterclockwise from the horizontal

Explanation:

The initial velocity of the first billiard ball = 5 m/s

The initial velocity of the billiard ball the first billiard ball strikes = 0 m/s

The final velocity of the first billiard ball = 4.35 m/s

The final direction of motion of the first billiard ball = 30° below its original motion

For perfectly elastic collision, whereby the target is at rest initially, by conservation of momentum, we have;

m₁ × $\underset{v_1}{\rightarrow}$ = m₁· $\underset{v'_1}{\rightarrow}$ + m₂· $\underset{v'_2}{\rightarrow}$

Which gives;

m₁ × 5·i = m₁·((√3)/2×5·i - 2.5·j) + m₂· $\underset{v'_2}{\rightarrow}$

∴ m₂· $\underset{v'_2}{\rightarrow}$ = m₁ × 5·i - m₁·((√3)/2×5·i - 2.5·j)

m₂· $\underset{v'_2}{\rightarrow}$ = m₁ × 5·(1 - √3/2)·i + m₁·2.5·j = m₁ × 2.5·(2 - √3)·i + m₁·2.5·j

Therefore, given that the mass of both billiard balls are equal, we have, m₁ = m₂, which gives;

m₂· $\underset{v'_2}{\rightarrow}$ = m₁· $\underset{v'_2}{\rightarrow}$ = m₁ × 2.5·(2 - √3)·i + m₁·2.5·j

∴ $\underset{v'_2}{\rightarrow}$ = 2.5·(2 - √3)·i + 2.5·j

The magnitude of the velocity of the second ball is $\underset{v'_2}{\rightarrow}$ = √((2.5·(2 - √3))² + 2.5²) ≈ 2.588 m/s

The direction of the second ball, θ = arctan(2.5/((2.5·(2 - √3))) = 75° counterclockwise from the horizontal.

3 0

3 years ago