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4 years ago

D= ((vf+vi)/2) t Solve for vi

2 answers:

8 0

= t(Vf + Vi/2)
Vf + Vi/2 = d/t 
Vf = (d/t) - Vi/2 

Answer: Vf = (d/t) - Vi/2 OR (2dt - Vit)/2t

vlada-n [284]4 years ago

7 0

D= ((vf+vi)/2) t

D = (vf +vi)*t/2

2*D = (vf +vi)*t

2D/t = vf +vi

vf +vi = 2D/t

vi = 2D/t - vf

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Finding the work done in pulling a stranded climber to safety. In an unfortunate accident, a rock climber finds herself stuck 17

Gnesinka [82]

Answer:

-9446.22 J

Explanation:

Parameters given:

Mass of climber and harness = 56kg

Mass of cable = 0.7kg

Distance between climber and top of rock face = 17m

The work done in pulling the climber is given as:

W = F * d

F is the force applied on the rope. It is opposite the force of gravity pulling the climber, hence, it is given as:

F = -Fg

Fg = force of gravity

Fg = m * g

g = acceleration due to gravity.

The mass of the climber, harness and cable = 56 + 0.7 = 56.7kg

=> Fg = 56.7 * 9.8

Fg = 555.66 N

Therefore, the work done will be:

W = - 555.66 * 17

W = -9446.22 J

The negative value of work means that the work done is opposite the value of the force acting on the climber.

4 0

4 years ago

If the earth can travel 1800 km in 1 minute what is it’s speed in km/s

Lady bird [3.3K]

Answer:

30km/s

Explanation:

1800/60=30

4 0

4 years ago

A) A real object 4.0 cm high stands 30.0 cm in front of a converging lens of focal length 23 cm. Find the image distance, the im

vfiekz [6]

Explanation:

Given that,

Height of object = 4.0 cm

Distance of the object u= -30.0 cm

Focal length = 23 cm

We need to calculate the image distance

Using lens's formula

$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$

Where, u = object distance

v = image distance

f = focal length

Put the value into the formula

$\dfrac{1}{v}-\dfrac{1}{-30}=\dfrac{1}{23}$

$\dfrac{1}{v}=\dfrac{1}{23}-\dfrac{1}{30}$

$\dfrac{1}{v}=\dfrac{7}{690}$

$v=98.57\ cm$

We need to calculate the height of the image

Using formula of height

$\dfrac{h'}{h}=\dfrac{-v}{u}$

Where, h' = height of image

h = height of object

$\dfrac{h'}{4}=\dfrac{-98.57}{-30}$

$h'=\dfrac{98.57\times4}{30}$

$h'=13.14\ cm$

The image is real and inverted.

(b). Now, object distance u = 13.0 cm

We need to calculate the image distance

Using lens's formula

$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$

Put the value into the formula

$\dfrac{1}{v}=\dfrac{1}{23}-\dfrac{1}{13.0}$

$\dfrac{1}{v}=-\dfrac{10}{299}$

$v=-29.9\ cm$

We need to calculate the height of the image

$\dfrac{h'}{4}=\dfrac{-(-29.9)}{-13.0}$

$h=-\dfrac{29.9\times4}{13.0}$

$h=-9.2\ cm$

The image is virtual and erect.

Hence, This is required solution.

6 0

3 years ago

A racecar is equipped with a computer that records the reading on its speedometer every second during a race. If you graph this

Westkost [7]

Since the device is a speedometer, the data it read is the speed of the racecar. Data recording involving time usually uses time as the independent variable. It was also said in the problem that it records the speed every second which shows that the time interval is constant. This means that only other data, the car's speed, is the dependent variable.

6 0

3 years ago

Why does a boat that is bumped by a wave not actually move to another position in the water

cluponka [151]

Explanation:

due to its shape st the bottom

7 0

4 years ago