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Elanso [62]
3 years ago
10

9. How does the length of the hypotenuse in a right triangle relate to the lengths of the legs? (2 points)

Physics
2 answers:
konstantin123 [22]3 years ago
6 0
<span>The pythagorean theorem addresses the length of the hypotenuse in relation to the length of the legs. The square root of the length of the hypotenuse is equal to the sum of one leg squared plus the other leg squared. In other words, A squared plus B squared equals C squared where A and B are the lengths of the legs of the triangle and C is the length of the hypotenuse.</span>
seropon [69]3 years ago
4 0

Answer:

hypotenuse² = base² +height²

Explanation:

Let us suppose a right angled triangle having b as base, h as hypotenuse and p as perpendicular distance.

The relation between the length of legs, the base and the hypotenuse is called Pythagoras theorem. It is given by :

h^2=b^2+p^2

or

hypotenuse² = base² +height²

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A motorcycle starts from rest and has a constant acceleration. In a time interval t, it undergoes a displacement x and attains a
iren [92.7K]

Answer:

√(6ax)

Explanation:

Hi!

The question states that during a time t the motorcyle underwent a displacement x at constant acceleration a starting from rest, mathematically we can express it as:

x=(1/2)at^2

Then the we need to find the time t' for which the displacement is 3x

3x=(1/2)a(t')^2

Solving for t':

t'=√(6x/a)

Now, the velocity of the motorcycle as a function of time is:

v(t)=a*t

Evaluating at t=t'

v(t')=a*√(6x/a)=√(6*x*a)

Which is the final velocity

Have a nice day!

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3 years ago
Why are you more likely to get a concussion when hit by a field hockey ball than a
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because it can be hard

Explanation:

I said that because they be on bed rest

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Estimate how many electrons there are in the human's body. asssume the mass is 70 kg. (hint: most of the atoms in your body have
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How about 8 X 10 to the 100th power.
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PLEASE HELPPP Count the number of atoms of each element in this molecule: 2H2K(OH2)2​
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Read 2 more answers
A car is traveling at a constant speed of 33 m/s on a highway. At the instant this car passes an entrance ramp, a second car ent
Paha777 [63]

Answer:

0.8712 m/s²

Explanation:

We are given;

Velocity of first car; v1 = 33 m/s

Distance; d = 2.5 km = 2500 m

Acceleration of first car; a1 = 0 m/s² (constant acceleration)

Velocity of second car; v2 = 0 m/s (since the second car starts from rest)

From Newton's equation of motion, we know that;

d = ut + ½at²

Thus,for first car, we have;

d = v1•t + ½(a1)t²

Plugging in the relevant values, we have;

d = 33t + 0

d = 33t

For second car, we have;

d = v2•t + ½(a2)•t²

Plugging in the relevant values, we have;

d = 0 + ½(a2)t²

d = ½(a2)t²

Since they meet at the next exit, then;

33t = ½(a2)t²

simplifying to get;

33 = ½(a2)t

Now, we also know that;

t = distance/speed = d/v1 = 2500/33

Thus;

33 = ½ × (a2) × (2500/33)

Rearranging, we have;

a2 = (33 × 33 × 2)/2500

a2 = 0.8712 m/s²

3 0
3 years ago
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