The resulting net force of an object is represented below.

A spherical object (with non-uniform density) of mass 16 kg and radius 0.22 m rolls along a horizontal surface at a constant lin

Lapatulllka [165]

Answer:

Kr = 0.7618K

Explanation:

Suppose that the object's velocity is V, then his kinetic energy is:

K = $\frac{mv^{2} }{2}$

K = $\frac{(16)v^{2} }{2}$

K = 8 $v^{2}$

The rotational kinetic energy is

Kr = $\frac{Iw^{2} }{2}$

where I: The moment of inertia

ω: angular velocity

Kr = $\frac{(0.59)w^{2} }{2}$

Kr = 0.295 $w^{2}$

How the movement is without slipping, then

ω = $\frac{v}{r}$

ω = $\frac{v}{0.22}$

Thus

Kr = $\frac{0.295v^{2} }{0.22^{2} }$

Kr = 6.095 $v^{2}$

8 $v^{2}$ ----> 1

6.095 $v^{2}$ ----->?

Kr = 0.7618K

4 0

4 years ago

A 1.0-kg standard cart collides on a low-friction track with cart A. The standard cart has an initial x component of velocity of

Andru [333]

Answer:

Explanation:

Mathematically, linear momentum is expressed as the product of mass and velocity. Linear momentum conservation law states that a body or system of bodies retains its total momentum unless an external force is applied to the system.

In this case, the system consists of two carts.

At the start, the linear momentum (P) of the system is equal to:

$P=1.0kg*0.4m/s=0.4kg*m/s$

It's only composed of linear momentum of the standard cart because cart A doesn't have any linear momentum at that moment.

After the collision, linear momentum has to be the same

$P=0.4kg*m/s=1.0kg*0.20m/s+m_{A} *0.70m/s$

where m_A is the mass of the cart A.

Solving for m_A

$m_{A} =0.28kg$

After the cart A rebounds, the linea momentum of the system has changed (because of the force present in the rebound). The new linear momentum is:

$P=1kg*0.2m/s+0.29kg*(-0.7m/s)=-0.003kg*m/s$

Then, the lump of putty is added to the system, but the linear momentum has to be the same, because we added a mass, not a force. The mass of that putty (m_p) has to be added to the equation of the system

$-0.003kg*m/s=1.0kg*(-0.2m/s)+(0.29kg+m_{p} )(0.4m/s)$

Solving for m_p

$m_{p}=0.20kg$

6 0

3 years ago

What is mean by the value of universal gravitional constant is 6.67×10^-11

aev [14]

Answer:

According to the gravitational law of Isaac Newton, "the gravitational force between any two objects is proportional to the product of the objects’ masses and inversely proportional to the square of the separation between their centers".

Therefore gravitational constant is the proportionality constant used in Newton’s Law of Universal Gravitation, and is commonly denoted by G. It is expressed as:

F= Gm1m2/r2

Another scientist Cavendish was able to measure the gravitational force and the value of the proportionality constant. It is expressed as G = 6.673×10-11 N m2 kg-2.

8 0

3 years ago

The amount of straight pipe needed to bend a given radius. Also, the actual length of the conduit that will be bent.

slava [35]

Development length: The actual length of the bent conduit. Gain: Since a conduit bends radially rather than at an angle, the total length will not match the length required for all bends. Gain is the amount of space that is saved by a $90^{0}$ curve.

<h3>Bent Conduit</h3>

Conduit benders from Klein Tools are built to function and last longer than even the highest professional standards. To ensure a favourable experience and significantly enhance the final result of your project, it is advised that you become familiar with bending concepts, procedures, and the bender's capabilities. The benders are labelled with various alignment symbols to enable the operator make the bends required to complete any job. This aids bending while executing a ground or air bend. Arrow, teardrop, star point, and angle markings are the symbols on the Klein Tools benders. On certain bender head sides, you can see these markings.

Learn more about bent conduit here:

brainly.com/question/5023977

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8 0

2 years ago

what evidence would you expect to find on the moon if it had been subjected to plate tectonics? (select all that apply.)

irina1246 [14]

Volcanoes and mountains are the evidences that show plate tectonics.

<h3 /><h3>What evidence would you expect to find on the Moon if the Moon had been subjected to plate tectonics? </h3>

The presence of volcanos and mountains are the evidences that provides information that the moon has plate tectonic. If there is no volcanoes and mountains on the moon then we can say that there is no plate tectonic on the moon.

So we can conclude that volcanoes and mountains are the evidences that show plate tectonics.

Learn more about tectonics here: brainly.com/question/2325633

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5 0

2 years ago