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UkoKoshka [18]
3 years ago
6

HELP PLS! :/

Physics
1 answer:
deff fn [24]3 years ago
4 0

Answer:

31.28m/s

Explanation:

We can use the third key equation of accelerated motion

Δd = viΔt+1/2aΔt^2

subtitute in the values we know

41.6 = vi(1.89)+1/2(-9.8)(1.89)^2

41.6 = vi(1.89) - 17.52

59.12 = vi(1.89)

vi = 31.28

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What time of the day do you feel the most Energetic and what do you usually do in those moments?​
iren2701 [21]

Answer:

i typically feel the kost awake at night. i do my work (organize patinet info, label patient charts... etc.).

Explanation:

my circadian rythm is out of whack bc of dld

4 0
3 years ago
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What is the science behind the making of pop-up books?
IRINA_888 [86]

Answer

The moveable parts of the pop up book are of often cut out by hand and are folded and glued by hand upon the printed pages. The cover is glued or sewn to the lining. Front and backs are often made up from board, which is just a heavier gauge paper than is used for the pages.

Explanation

Hope that this helps you and have a great day:)

6 0
3 years ago
A penguin slides at a constant velocity of 3.57 m/s down an icy incline. The incline slopes above the horizontal at an angle of
igomit [66]

Answer:t=0.81 s

Explanation:

Given

Penguin slides down with constant velocity of 3.57 m/s

as the Penguin Slides with constant velocity therefore F_{net} is zero on Penguin

F_{net}=mg\sin \theta -f_r

f_r=friction Force

f_r=\mu mg

\mu =coefficient of Kinetic friction

mg\sin \theta =\mu mg

\tan \theta =\mu

\mu =\tan 9.85=0.45

after reaching on floor final velocity of penguin will be zero after time t

thus

v=u+at

here a=\mu g

a=0.45\times 9.8=4.41  (deceleration)

0=3.57-4.41\times t

t=\frac{3.57}{4.41}=0.809

t=0.81 s

8 0
3 years ago
A 14.0 m uniform ladder weighing 490 N rests against a frictionless wall. The ladder makes a 63.0°-angle with the horizontal.
crimeas [40]

Answer:

Explanation:

Given:

length of ladder r_L = 14m

weight of ladder F_L = 490N

position of firefighter r_F = 3.8m

weight of firefighter F_F = 820N

angle of ladder \alpha = 63

Unknown:

force of the wall on the ladder F_W

force of friction on base of ladder F_R

normal force on base of ladder F_N

From the free body diagram of the sketch you get 3 equations:

F_x = ma_x = F_W - F_R = 0\\ F_y = ma_y = F_N - F_F - F_L = 0\\ \tau _P = \overrightarrow{r} \times \overrightarrow{F} = r_FF_Fcos\alpha + \frac{1}{2}r_LF_Lcos\alpha - r_LF_Wsin\alpha = 0

Solving the equations gives:

F_W = F_R\\ F_N = F_F + F_L\\ F_W = \frac{r_FF_F + 0.5r_LF_L}{r_L tan\alpha}

a)

F_R = 238N\\ F_N = 1310N

b)

F_R = \mu F_N\\ \mu = \frac{F_R}{F_N} \\ \mu = 0.3

c) Using the result from b and solving for r_F

\\ \mu = 0.15\\ F_R = \mu F_N\\ r_F = 2.4m

4 0
4 years ago
A certain digital camera having a lens with focal length 7.50 cmcm focuses on an object 1.70 mm tall that is 4.70 mm from the le
Crank

Answer:

a. 7.62cm

b. Real and inverted

c. 2.76 cm

d. 3450

Explanation:

We proceed as follows;

a. the lens equation that relates the object distance to the image distance with the focal length is given as follows;

1/f = 1/p + 1/q

making q the subject of the formula;

q = pf/p-f

From the question;

p = 4.70m

f = 7.5cm = 0.075m

Substituting these, we have ;

q = (4.7)(0.075)/(4.7-0.075) = 0.3525/4.625 = 0.0762 = 7.62 cm

b. The image is real and inverted since the image distance is positive

c. We want to calculate how tall the image is

Mathematically;

h1 = (q/p)h0

h1 = (7.62/4.70)* 1.7

h1 = 2.76 cm

d. We want to calculate the number of pixels that fit into this image

Mathematically:

n = h1/8 micro meter

n = 2.76cm/8 micro meter = 2.76 * 10^-2/8 * 10^-6 = 3450

3 0
3 years ago
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