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3 years ago

An acceleration of a freely falling body (9.8m/s^2) to ft/s^2

Physics

1 answer:

Ronch [10]3 years ago

5 0

9.8 m/s^2 = 32.2 ft/s^2

hope this helps :)

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the three major scales used to measure earthquakes are Mercalli scale Richard scale and magnitude scale

iVinArrow [24]

False; the three major scales used to measure earthquakes are the Mercalli Scale, the Richter Scale, and the Magnitude Scale. I hope this helps!

3 0

3 years ago

The density of an object is dependent upon the object’s mass and ---

kotegsom [21]

Answer:Volume

Explanation:

Density = mass/ Volume

7 0

3 years ago

Engineers are designing a curved section of a highway. If the radius of curvature of the curve is 194 m, at what angle should th

Brums [2.3K]

Answer:

The banking angle is 23.84 degrees.

Explanation:

Given that,

Radius of the curve, r = 194 m

Speed of the car, v = 29 m/s

On the banked curve, the centripetal force is balanced by the force of friction such that,

$mg\ tan\theta=\dfrac{mv^2}{r}$

$tan\theta=\dfrac{v^2}{rg}$

$tan\theta=\dfrac{(29)^2}{194\times 9.8}$

$\theta=23.84^{\circ}$

So, the banking angle is 23.84 degrees. Hence, this is the required solution.

5 0

3 years ago

if you run around a circle at 4.5 m/s and the circle has a radius of 7.7 m, what is your centripetal acceleration?

madreJ [45]

Answer:

Centripetal acceleration,

$a_{c} =2.63\ m/s^{2} }$

Explanation:

Centripetal acceleration:

Centripetal acceleration is the idea that any object moving in a circle, in something called circular motion, will have an acceleration vector pointed towards the center of that circle.

Centripetal means towards the center.

Examples of centripetal acceleration (acceleration pointing towards the center of rotation) include such situations as cars moving on the cicular part of the road.

An acceleration is a change in velocity.

Formula for Centripetal acceleration:

$a_{c} =\frac{(velocity)^{2} }{radius}$

Given here,

Velocity = 4.5 m/s

radius = 7.7 m

To Find :

$a_{c} = ?$

Solution:

We have,

$a_{c} =\frac{(velocity)^{2} }{radius}$

Substituting given value in it we get

$a_{c} =\frac{(4.5)^{2}}{7.7} \\\\a_{c} =\frac{20.25}{7.7}\\\\a_{c} =2.629\ m/s^{2} \\\\\therefore a_{c} =2.63\ m/s^{2$

Centripetal acceleration,

$a_{c} =2.63\ m/s^{2} }$

7 0

3 years ago

I have no idea of how to approach this problem

nataly862011 [7]

Answer:

p=1

Explanation:

Well me know that v=m/s

and that a=m/s^2

$(m/s)^{2} = (m/s^2)(x^p)\\\\(m^2/s^2)/(m/s^2)=x^p\\\\m^2s^2/ms^2=x^p\\\\(m^2/m)(s^2/s^2)=x^p\\\\m=x^p\\\\p=1$

Note: We don't take into account 2 because it's a scalar, it doesn't have units so it doesn't add anything to the equation.

4 0

3 years ago