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Dovator [93]
3 years ago
15

PLEASE HELP!!!! I’ll give brainliest!

Physics
1 answer:
Jet001 [13]3 years ago
3 0
Ok well I know measure of long leg is 30 degrees and short leg is 60 degrees
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What is the distance fallen for a freely falling object 1 s after being dropped from a rest position? What is the distance for a
kondor19780726 [428]
<h2>Distance traveled in 1 second after drop is 4.9 m</h2><h2>Distance traveled in 4 seconds after drop is 78.4 m</h2>

Explanation:

We have s = ut + 0.5at²

For a free falling object initial velocity u = 0 m/s and acceleration due to gravity, g = 9.8 m/s²

Substituting

                 s = 0 x t + 0.5 x 9.8 x t²

                 s = 4.9t²

We need to find distance traveled in 1 s and 4 s

Distance traveled in 1 second

                   s = 4.9 x 1² = 4.9 m

Distance traveled in 4 seconds

                   s = 4.9 x 4² = 78.4 m

Distance traveled in 1 second after drop = 4.9 m

Distance traveled in 4 seconds after drop = 78.4 m

5 0
3 years ago
Which atoms have the highest electronegativity?
Ronch [10]

Answer:

Note that there is little variation among the transition metals. Electronegativities generally decrease from top to bottom within a group due to the larger atomic size. Of the main group elements, fluorine has the highest electronegativity (EN = 4.0) and cesium the lowest (EN = 0.79).

3 0
3 years ago
What does the pupil of the eye controls
Alja [10]
The correct answer would be D
5 0
3 years ago
If the released energy of one earthquake is 100 times that of another, how much greater is its magnitude on the richter scale?
Alekssandra [29.7K]
Greater by 2 on the Richter Scale :)
3 0
3 years ago
. A ball is thrown downward at a speed of 20 m/s. Choosing
a_sh-v [17]

The final velocity is v = 20 - gt

The distance traveled by the ball at time t is y = y_o + (-20)t - \frac{1}{2} gt^2

The maximum distance traveled by the object is 0 = (20)^2 - 2g(y-y_0)

The given parameters;

initial velocity of the ball, u = 20 m/s

acceleration due to gravity, g = 9.8 m/s²

The final velocity can be calculate as;

v = 20 - gt

The distance traveled by the ball at time t;

y = y_o + (-20)t - \frac{1}{2} gt^2

The maximum distance traveled by the object is calculated as;

v^2 = u^2 - 2g(y - y_0)\\\\0 = (20)^2 - 2g(y-y_0)

Learn more here: brainly.com/question/16878713

3 0
3 years ago
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