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3 years ago

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the science building is 5.25 m tall a. How many decimeters is it? b. How many millimeter is it? show work

2 answers:

Anastasy [175]3 years ago

5 0

Answer:

5.25m to 52.5 Decimeters (10x)

5.25m to 5250 Millimeters (1000x)

12345 [234]3 years ago

4 0

If I remember correctly m means meter (hopefully) so it’s 52.5 decimeters.
And that would mean it would be 5250 millimeters. For the show your work thing, all you really had to do to solve this is move the decimal point depending on what type of measurement you were changing it into. So try and kinda write something like that down where you move the decimal point. Hope this helps

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My buddy and I are planning a shore dive. we're descending onto a very gradual slope that begins at 5 m/15 ft, so our descent an

Gnom [1K]

Answer:

C. 157 bar/2270 psi

Explanation:

Calculation to determine what we should head back when either of our SPGs read

SPGs=200 bar -[200 bar-(50 bar + 20 bar)]÷1/3]

SPGs=200 bar-[(200 bar-70 bar)÷1/3]

SPGs=200 bar-(130 bar÷1/3)

SPGs=200 bar-43 bar

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Therefore based on the above calculation we should head back when either of our SPGs read 157 bar/2270 psi

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3 years ago

A stone is thrown straight up from the edge of a roof, 925 feet above the ground, at a speed of 20 feet per second. Remembering

Black_prince [1.1K]

<h2>Answer: 469 feet</h2>

Explanation:

This problem is a good example of Vertical motion, where the main equation for this situation is:

$y=y_{o}+V_{o}t-\frac{1}{2}gt^{2}$ (1)

Where:

$y$ is the height of the stone at 6s (the value we want to find)

$y_{o}=925ft$ is the initial height of the stone

$V_{o}=20ft/s$ is the initial velocity of the stone

$t=6s$ is the time at which we need to find the height

$g=32ft/s^{2}$ is the acceleration due to gravity

Having this clear, let's find $y$ from (1):

$y=925ft+(20ft/s)(6s)-\frac{1}{2}(32ft/s^{2})(6s)^{2}$ (2)

Finally:

$y=469ft$ This is the height of the stone at t=6s

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3 years ago

What is the value of y?<br>18<br>10<br>2y + 4<br>10 + 2х

DENIUS [597]

Answer:

can u please write it correct like sorry I dont understand it

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4 years ago

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Sati [7]

Answer:

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3 0

3 years ago

A rock is thrown from the top of a 20-m building at an angle of 53Â° above the horizontal. If the horizontal range of the throw

NemiM [27]

Answer:

28.5 m/s

18.22 m/s

Explanation:

h = 20 m, R = 20 m, theta = 53 degree

Let the speed of throwing is u and the speed with which it strikes the ground is v.

Horizontal distance, R = horizontal velocity x time

Let t be the time taken

20 = u Cos 53 x t

u t = 20/0.6 = 33.33 ..... (1)

Now use second equation of motion in vertical direction

h = u Sin 53 t - 1/2 g t^2

20 = 33.33 x 0.8 - 4.9 t^2 (ut = 33.33 from equation 1)

t = 1.17 s

Put in equation (1)

u = 33.33 / 1.17 = 28.5 m/s

Let v be the velocity just before striking the ground

vx = u Cos 53 = 28.5 x 0.6 = 17.15 m/s

vy = uSin 53 - 9.8 x 1.17

vy = 28.5 x 0.8 - 16.66

vy = 6.14 m/s

v^2 = vx^2 + vy^2 = 17.15^2 + 6.14^2

v = 18.22 m/s

6 0

3 years ago