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DerKrebs [107]
3 years ago
14

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
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