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3 years ago

How can you determine that a coin

Physics

1 answer:

OleMash [197]3 years ago

8 0

Answer:

by checking if its malleble with a hammer

Explanation:

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HI, I really need help with this assignment for Physics:

Elena L [17]

Answer:

yes

Explanation:

3 0

3 years ago

the face of a cube towards A is brightly and shiny and the face towards V is full black.State with reason the adjustments that s

MariettaO [177]

Explanation:

increase the distance of cube from black and dull substance

5 0

3 years ago

A man stands on a platform that is rotating (without friction) with an angular speed of 1.0 rev/s; his arms are outstretched and

Airida [17]

Answer:

Explanation:

Given

$N_1=1 rev/s$

angular velocity $\omega =2\pi N_1=6.284 rad/s$

Combined moment of inertia of stool,student and bricks $=6\ kg.m^2$

Now student pull off his hands so as to increase its speed to suppose $N_2$ rev/s

$\omega _2=2\pi N_2$

After Pulling off hands so final moment of inertia is

$I_2=2\ kg-m^2$

Conserving angular momentum as no external torque is applied

$I_1\omega _1=I_2\omega _2$

$6\times 6.284=2\times \omega _2$

$\omega _2=18.85\ rad/s$

$N_2=3 rev/s$

7 0

3 years ago

What is the name of the force that acts between any two objects because of<br>their masses?

svetoff [14.1K]

Answer:

Explanation:

The force of attraction between 2 masses.

4 0

3 years ago

3. The center of mass (or center of gravity) of a two-particle system is at the origin. One particle

nika2105 [10]

Answer:

B) (-2.0 m, 0.0 m)

Explanation:

Given:

Mass of particle 1 is, $m_1=2.0\ kg$

Mass of particle 2 is, $m_2=3.0\ kg$

Position of center of mass is, $(x_{cm},y_{cm})=(0,0)$

Position of particle 1 is, $(x_1,y_1)=(3.0\ m,0.0\ m)$

Position of particle 2 is, $(x_2,y_2)=(?\ m,?\ m)$

We know that, the x-coordinate of center of mass of two particles is given as:

$x_{cm}=\frac{m_1x_1+m_2x_2}{m_1+m_2}$

Plug in the values given.

$0=\frac{(2.0\times 3)+(3.0\times x_2)}{2.0+3.0}\\\\0=6+3x_2\\\\3x_2=-6\\\\x_2=\frac{-6}{3}=-2.0\ m$

We know that, the y-coordinate of center of mass of two particles is given as:

$y_{cm}=\frac{m_1y_1+m_2y_2}{m_1+m_2}$

Plug in the values given.

$0=\frac{(2.0\times 0)+(3.0\times y_2)}{2.0+3.0}\\\\0=0+3y_2\\\\3y_2=0\\\\y_2=\frac{0}{3}=0.0\ m$

Therefore, the position of particle 2 of mass 3.0 kg is (-2.0 m, 0.0 m).

So, option (B) is correct.

8 0

3 years ago