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Yakvenalex [24]
3 years ago
14

How many centimeters are in 3.50 feet?

Physics
1 answer:
Kazeer [188]3 years ago
4 0
106.68 centimetres are in 3.50 feet
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A musical chord consists of several notes, each formed by a sound wave of a different frequency. The three graphs show the patte
sweet [91]

Answer:

If a chord had notes with frequencies of 100, 1,000, and 6,000 Hz, the basilar membran would vibrate at multiple positions, with peaks at A, B, and C.

Explanation:

3 0
3 years ago
If the gravitational field strength at the top of Mount Everest is 9.772 N/kg, approximately how tall (in feet) is
Delicious77 [7]

Answer:

The height of the Everest mountain is, x =  8514.087 m

Explanation:

Given data,

The gravitational field strength at the top of mount Everest, gₓ = 9.772 m/s²

The formula for gravitational field strength is,

                               <em> gₓ = GM/(R+x)²</em>

Where, x is the height from the surface of the Earth

Therefore,

                                (R+x)² = GM/gₓ

                                     x = √(GM/gₓ) - R

Substituting the values,

                       x = √(6.67408 x 10⁻¹¹ X  5.972 x 10²⁴ / 9.772) - 6.378 x 10⁶

                       x =  8514.087 m          

Therefore, the height of the Everest mountain is, x =  8514.087 m                                                              

6 0
3 years ago
Which of the following employs the use of an external combustion engine?
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The answer should be ''all the above''
7 0
4 years ago
If you add together all of the forces exerted on a object and get a non zero value that is called
True [87]

Hello There!

<em>If you add together all of the forces exerted on a object and get a non zero value that is called the "NET FORCE" of the object</em>

3 0
3 years ago
Read 2 more answers
A box is given a push so that it slides across the floor. How far will it go, given that the coefficient of kinetic friction is
IRINA_888 [86]

Answer:

s=4.44 m

Explanation:

Given that

Coefficient of the kinetic friction ,μ = 0.13

Initial velocity ,u= 3.4 m/s

Final velocity of the box ,v= 0 m/s

The acceleration due to friction force

a= - μ g

Now by putting the values in the above equation

a= - 0.13 x 10    ( take g= 10 m/s²)

 a= - 1.3 m/s²

We know that

v²= u ² + 2 a s

s=distance

a=acceleration

v=final speed

u=initial speed

Now by putting the values in the above equation

0²= 3.4² - 2 x 1.3 x s

s=\dfrac{3.4^2}{2\times 1.3}\ m

s=4.44 m

The distance cover by box will be 4.44 m.

6 0
3 years ago
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