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ladessa [460]
3 years ago
10

At the periphery of a hurricane the air is ____, and several kilometers above the surface, in the eye, the air is ____.

Physics
1 answer:
tatuchka [14]3 years ago
6 0
<span>At the periphery of a hurricane the air is sinking, and several kilometers above the surface, in the eye, the air is sinking.  </span>
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A cylinder of mass 8.0 kg rolls without slipping on a horizontal surface. At a certain instant its center of mass has a speed of
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Answer:

b

Explanation:

determine the rotational kinetic energy about it's center of mass

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Three boxes in contact rest side-by-side on a smooth, horizontal floor. Their masses are 5.0-kg, 3.0-kg, and 2.0-kg, with the 3.
Ivahew [28]

Answer:

(a)Look at the attached graphic

(b)

(b)-1 Equation 1  : m1= 5kg

       50-F1= 5 *a

(b)-2 Equation 2 : m2= 3kg

        F1-F2= 3 *a

(b)-3 Equation 3 : m3= 2kg

         F2 = 2*a  

(c) F1 =25 N

(d) F2 =10 N

Explanation:

We apply Newton's second law:

∑F = m*a (Formula 1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

(a) Draw the free-body diagrams for each of the boxes

Look at the attached graphic

(b) Write Newton’s equation for each mass along the horizontal direction.

Data: m1=  5.0-kg ,m2= 3.0-kg , ,m3= 2.0-kg

<em>Look</em> <em>m1 free-body diagram:</em>

∑Fx = m1*a

50-F1= 5 *a Equation 1

<em>Look</em> <em>m2 free-body diagram:</em>

∑Fx = m2*a

F1-F2= 3 *a Equation 2

<em>Look</em> <em>m3 free-body diagram:</em>

∑Fx = m3*a

F2 = 2*a     Equation 3

(c) What magnitude force does the 3.0-kg box exert on the 5.0- kg box?

<em>Look</em> <em>Free body diagram of the mass set</em>

∑Fx = m*a   m= m1+m2+m3= 5+3+2 = 10 kg

50 = 10*a

a= 50/10 = 5 m/s²

We replace a = 5 m/s² in the equation 1:

50-F1= 5 *5

50-25= F1

F1 = 25 N

<em> (d) </em><em>What magnitude force does the 3.0-kg box exert on the 2.0kg box?</em>

We replace a= 5 m/s² in the equation 3

F2 = 2*5 = 10 N

4 0
3 years ago
In an experiment, students measure the position x of a cart as a function of time t for a cart that starts at rest and moves wit
Minchanka [31]

Given :

Initial velocity , u = 0 m/s² .

To Find :

The acceleration of the cart.

Solution :

Since, acceleration is constant.

Using equation of motion :

x = ut + \dfrac{at^2}{2}\\\\x = \dfrac{at^2}{2}

Putting, t = 1 s  and x = 4 m in above equation, we get :

4 = \dfrac{a(1)^2}{2}\\\\a = 8 \  m/s^2

Therefore, the acceleration of the cart is 8 m/s².

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3 years ago
Which of these is an advantage of parallel circuits over series circuits?
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Answer:

d

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A roundabout in a fairground requires an input power of 2.5 kW when operating at a constant angular velocity of 0.47 rad s–1 . (
natita [175]

Answer:

Explanation:

Since the roundabout is rotating with uniform velocity ,

input power = frictional power

frictional power = 2.5 kW

frictional torque x angular velocity = 2.5 kW

frictional torque x .47 = 2.5 kW

frictional torque = 2.5 / .47 kN .m

= 5.32 kN . m

= 5 kN.m

b )

When power is switched off , it will decelerate because of frictional torque .

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