Isim sehir eyşa<br> iiiiiiiimi iiiiooiiii

Rohith has been jogging to the bus stop for 2.0 minutes at 3.5 m/s when he looks at

nikdorinn [45]

Multiply these numbers and there’s your answer

6 0

3 years ago

Two children are riding on the edge of a merry-go-round that has a mass of 100.kg and radius of 1.60m and is rotating at 20.0rpm

Gre4nikov [31]

Here since both children and merry go round is our system and there is no torque acting on this system

So we will use angular momentum conservation in this

$I_1\omega_1 = I_2\omega_2$

now here we have

$I_1 = \frac{MR^2}{2} + m_1R^2 + m_2R^2$

$I_1 = \frac{100(1.60)^2}{2} + (22 + 28)(1.60)^2$

$I_1 = 256$

Now when children come to the position of half radius

then we will have

$I_2 = \frac{MR^2}{2} + m_1(\frac{R}{2})^2 + m_2(\frac{R}{2})^2$

$I_2 = \frac{100(1.6)^2}{2} + (28 + 22)(0.8)^2$

$I_2 = 160$

now from above equation we have

$256 (20.0 rpm) = 160(\omega_2)$

$\omega_2 = 32 rpm$

8 0

3 years ago

Speed can be calculated by dividing distance by what?

Nady [450]

S=d/t

speed can be calculated by dividing distance by time.

3 0

3 years ago

Read 2 more answers

The rotational kinetic energy term is often called the kinetic energy in the center of mass, while the translational kinetic ene

Nataliya [291]

Question in proper order

The rotational kinetic energy term is often called the <em>kinetic energy </em><em>in</em> the center of mass, while the translational kinetic energy term is called the <em>kinetic energy </em><em>of</em> the center of mass.

You found that the total kinetic energy is the sum of the kinetic energy in the center of mass plus the kinetic energy of the center of mass. A similar decomposition exists for angular and linear momentum. There are also related decompositions that work for systems of masses, not just rigid bodies like a dumbbell.

It is important to understand the applicability of the formula

$Ktot=Kr+Kt$

Which of the following conditions are necessary for the formula to be valid?

a. The velocity vector $v$ must be perpendicular to the axis of rotation

b.The velocity vector $v$ must be perpendicular or parallel to the axis of rotation

c. The moment of inertial must be taken about an axis through the center of mass

Answer:

Option c

Explanation:

$K_{total} = K_{rotational}+K_{translational}$

The first two conditions are untrue, this is because, you can have rotation in any direction and translation in any direction of any collection of masses. Rotational and translational velocities of masses do not depend on each other

The last statement is true because by definition, the moment of inertia, which is a measure of reluctance, is usually taken about a reference point which is the center of mass

4 0

3 years ago

Read 2 more answers

How many stars our thier in the universe

Svet_ta [14]

<h2>About 1 billion trillion stars</h2>

Since the universe is so big, we don't know for sure what the exact number is, but assuming an average of 100 billion stars per galaxy (which is smaller than a universe), we could say that there are 1 billion trillion stars in the observable universe.

8 0

3 years ago

Isim sehir eyşa iiiiiiiimi iiiiooiiii​

Isim sehir eyşa iiiiiiiimi iiiiooiiii