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3 years ago

A ball rolls off of a table (take v0 as 0m/s) that is 1.25m tall. How long does it take for the ball to hit the floor?

1 answer:

7 0

Let
h=height of table= 1.25m
g=acceleration due to gravity= 9.8m/s^2
t=time taken=?

Solution
h =(vi)(t)+1/2(g)(t^2)
1.25=(0)(t)+1/2(9.8)(t^2)
t^2 = 0.255
t. = 0.5s

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For when white light is passed through a diffraction grating, the color closest to the center bright spot corresponds toa. Yello

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A ship maneuvers to within 2500 m of an island's 1800 m high mountain peak and fires a projectile at an enemy ship 610 m on the

Ne4ueva [31]

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.

t=(0-(250sin75)^2)/-9.8
the distance one is (2500+610)- (250m/s*cos75)*t=Dh Dh=horizontal distance 

the max height one is d=0.5*9.8*t^2 
d= max height subtract 1800-d

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3 years ago

Read 2 more answers

A ball is thrown vertically upwards with a velocity

zhuklara [117]

Answer:

Explanation:

The acceleration of gravity is 9.8m/s^2.

So to calculate the time it will take to make the ball stop(which btw means the ball now reach its greatest height), use the formula V1=V0+at. V1 is the final velocity(which is 0), V0 is the starting velocity(which is 30m/s), and the a(cceleration) is 9.8m/s^2.

(You can ignore the fact "at" is -30 instead 30, it's because the directions two velocity travel are opposite. )

We can now know the time it takes to make the ball stop just by the gravitational force is about 3 sec.

Use another formula S=1/2at^2, to find out the S(height) is 1/2*9.8*3^2=44.1, which is approximately D.45m .

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3 years ago

Suppose you want to determine the resistance of a resistor that is nominally 100 . You should be able to apply 10 V across the r

Butoxors [25]

Answer:

a) For y = 102 mA, R = 98.039 ohms

For y = 97 mA, R = 103.09 ohms

b) Check explanatios for b

Explanation:

Applied voltage, V = 10 V

For the first measurement, current $y_{1} = 102 mA = 0.102 A$

According to ohm's law, V = IR

R = V/I

Here, $I = y_{1}$

$R = \frac{V}{y_{1} } \\R = \frac{10}{0.102} \\R = 98.039 ohms$

For the second measurement, current $y_{2} = 97 mA = 0.097 A$

$R = \frac{V}{y_{2} }$

$R = \frac{10}{0.097} \\R = 103 .09 ohms$

b) $y = \left[\begin{array}{ccc}y_{1} &y_{2} \end{array}\right] ^{T}$

$y = \left[\begin{array}{ccc}y_{1} \\y_{2} \end{array}\right]$

$y = \left[\begin{array}{ccc}102*10^{-3} \\97*10^{-3} \end{array}\right]$

A linear equation is of the form y = Gx

The nominal value of the resistance = 100 ohms

$x = \left[\begin{array}{ccc}100\end{array}\right]$

$\left[\begin{array}{ccc}102*10^{-3} \\97*10^{-3} \end{array}\right] = \left[\begin{array}{ccc}G_{1} \\G_{2} \end{array}\right] \left[\begin{array}{ccc}100\end{array}\right]\\\left[\begin{array}{ccc}G_{1} \\G_{2} \end{array}\right] = \left[\begin{array}{ccc}102*10^{-5} \\97*10^{-5} \end{array}\right]$

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