What is the density of an object that has a mass of 30 g and a volume of 20cm cubed/ to the third power?

Consider a uniform solid sphere of radius R and mass M rolling without slipping. Which form of its kinetic energy is larger, tra

castortr0y [4]

Answer:

A. Its translational kinetic energy is larger than its rotational kinetic energy.

Explanation:

Given that

Radius = R

Mass = M

We know that mass moment of inertia for the solid sphere

$I=\dfrac{2}{5}MR^2$

Lets take angular speed =ω

Linear speed =V

Condition for pure rolling , V= ω R

Rotation energy ,RE

$RE=\dfrac{1}{2}I\omega^2$

$RE=\dfrac{1}{2}\times \dfrac{2}{5}MR^2\times \omega^2$

$RE=\dfrac{1}{5}\times MR^2\times \omega^2$

$RE=\dfrac{1}{5}\times MV^2$

RE= 0.2 MV²

The transnational kinetic energy TE

$TE=\dfrac{1}{2}MV^2$

TE= 0.5 MV²

From above we can say that transnational energy is more than rotational energy.

Therefore the answer is A.

3 0

3 years ago

how many atoms of oxygen are there in one molecule of cardon dioxide , if the chemical formula is CO2

garri49 [273]

Answer:

2 oxygen 1 carbon

Explanation:

8 0

3 years ago

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Two charged objects are 1 meter apart. Calculate the magnitude of the electric force between them if the two charges are +1.0 μC

Solnce55 [7]

Answer:

0.00899 N

Explanation:

The magnitude of the electrostatic force between two charges is given by the equation:

$F=k\frac{q_1 q_2}{r^2}$

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the charges

r is the distance between the two charges

And the force is:

- Repulsive if the two charges have same sign

- Attractive if the two charges have opposite sign

In this problem we have:

$q_1=1.0\mu C = 1.0\cdot 10^{6}C$ (charge of object 1)

$q_2=1.0\mu C = 1.0\cdot 10^{6}C$ (charge of object 2)

r = 1 m (separation between the objects)

So, the electric force is

$F=(8.99\cdot 10^9)\frac{(1.0\cdot 10^{-6})^2}{1^2}=0.00899 N$

5 0

4 years ago

______________friction is the type of kinetic friction that is difficult to overcome.

Burka [1]

Answer:

Explanation:

energy friction

8 0

3 years ago

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Two particles are moving along the x axis. Particle 1 has a mass m₁ and a velocity v₁ = +4.7 m/s. Particle 2 has a mass m₂ and a

nirvana33 [79]

Answer:

m₁ / m₂ = 1.3

Explanation:

We can work this problem with the moment, the system is formed by the two particles

The moment is conserved, to simulate the system the particles initially move with a moment and suppose a shock where the particular that, without speed, this determines that if you center, you should be stationary, which creates a moment equal to zero

p₀o = m₁ v₁ + m₂ v₂

pf = 0

m₁ v₁ + m₂ v₂ = 0

m₁ / m₂ = -v₂ / v₁

m₁ / m₂= - (-6.2) / 4.7

m₁ / m₂ = 1.3

Another way to solve this exercise is to use the mass center relationship

Xcm = 1/M (m₁ x₁ + m₂ x₂)

We derive from time

Vcm = 1/M (m₁ v₁ + m₂v₂)

As they say the velocity of the center of zero masses

0 = 1/M (m₁ v₁ + m₂v₂)

m₁ v₁ + m₂v₂ = 0

m₁ / m₂ = -v₂ / v₁

m₁ / m₂ = 1.3

4 0

3 years ago