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frez [133]
4 years ago
7

The word inverse means ____ and is an alternative word for __________ relationship.

Physics
1 answer:
AleksandrR [38]4 years ago
4 0

Answer:

B - Opposite, and direct

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A hula hoop is rolling along the ground with a translational speed of 26 ft/s. It rolls up a hill that is 16 ft high. Determine
nikklg [1K]

Answer:12.8 ft/s

Explanation:

Given

Speed of hoop v=26\ ft/s

height of top h=16\ ft

Initial energy at bottom is

E_b=\frac{1}{2}mv^2+\frac{1}{2}I\omega ^2

Where m=mass of hoop

I=moment of inertia of hoop

\omega=angular velocity

for pure rolling v=\omega R

I=mR^2

E_b=\frac{1}{2}mv^2+\frac{1}{2}mR^2\times (\frac{v}{R})^2

E_b=mv^2=m(26)^2=676m

Energy required to reach at top

E_T=mgh=m\times 32.2\times 16

E_T=512.2m

Thus 512.2 m is converted energy is spent to raise the potential energy of hoop and remaining is in the form of kinetic and rotational energy

\Delta E=676m-512.2m=163.8m

Therefore

163.8 m=mv^2

v=\sqrt{163.8}

v=12.798\approx 12.8\ ft/s

7 0
3 years ago
an athlete running 2.6m/ increases his velocity to 3.7m/s in 2.7s. what is the acceleration of the runner
iVinArrow [24]
Acceleration= velocity/time

1.1/2.7=0.41

The acceleration is .41 m/s^2
4 0
3 years ago
A physics major is cooking breakfast when he notices that the frictional force between
White raven [17]
Let's list the given information. The frictional force, denoted as Ff, is equal to 0.200 N. We have to find the normal force, denoted as Fn. The relationship between Ff and Fn is written as:

Ff = μFn
where μ is the coefficient of friction

If there is no given data for μ, we can't solve this problem. Suppose μ = 0.5, then the normal force would be:
Fn = Ff/μ = 0.2/0.5 = 0.4 N
8 0
3 years ago
A heavy piece of hanging sculpture is suspended by a90-cm-long, 5.0 g steel wire. When the wind
Yuliya22 [10]

As we know that fundamental frequency is given as

f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}

here we know that

\mu = \frac{m}{l}

here we have

m = mass of wire = 5 g

l = length of wire = 90 cm

\mu = \frac{0.005}{0.90} kg/m

\mu = 5.56 \times 10^{-3} kg/m

from above formula now

80 = \frac{1}{2(0.90)}\sqrt{\frac{T}{5.56\times 10^{-3}}}

144 = \sqrt{180 T}

T = 115.2 N

now we know that tension is due to weight of the sculpture so we will have

Mg = 115.2 N

M = 11.76 kg

so its mass will be 11.76 kg

6 0
4 years ago
What is the magnitude of the total acceleration of point A after 2 seconds? The bar starts from rest and has a constant angular
monitta

Answer:

a_total = 2 √ (α² + w⁴) ,   a_total = 2,236 m

Explanation:

The total acceleration of a body, if we use the Pythagorean theorem is

          a_total² = a_T²2 + a_{c}²

where

the centripetal acceleration is

  a_{c} = v² / r = w r²

tangential acceleration

   a_T = dv / dt

angular and linear acceleration are related

         a_T = α  r

we substitute in the first equation

       a_total = √ [(α r)² + (w r² )²]

       a_total = 2 √ (α² + w⁴)

Let's find the angular velocity for t = 2 s if we start from rest wo = 0

        w = w₀ + α t

        w = 0 + 1.0 2

        w = 2.0rad / s

       

we substitute

        a_total = r √(1² + 2²) = r √5

        a_total = r 2,236

In order to finish the calculation we need the radius to point A, suppose that this point is at a distance of r = 1 m

         a_total = 2,236 m

7 0
3 years ago
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