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postnew [5]
2 years ago
13

Suppose you have a container filled with iron and sand. You can separate the iron from the sand if you ____________ so this is a

_________.
A) use a magnet; mixture
B) melt the sand; mixture
C) add baking soda; chemical compound
D) dissolve the sand in water; solution
Physics
2 answers:
My name is Ann [436]2 years ago
5 0
A) use a magnet; mixture
koban [17]2 years ago
4 0

Answer: Use a magnet; mixture

Explanation:

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A wooden disk of mass m and radius r has a string of negligible mass is wrapped around it. If the disk is allowed to fall and th
Tju [1.3M]

Answer:

a = \frac{2}{3}g

T = \frac{mg}{3}

Explanation:

As the disc is unrolling from the thread then at any moment of the time

We have force equation as

mg - T = ma

also by torque equation we can say

TR = I\alpha

TR = \frac{1}{2}mR^2(\frac{a}{R})

T = \frac{1}{2}ma

Now we have

mg - \frac{1}{2}ma = ma

mg = \frac{3}{2}ma

a = \frac{2}{3}g

Also from above equation the tension force in the string is

T = \frac{1}{2}ma

T = \frac{mg}{3}

7 0
2 years ago
Two long conducting cylindrical shells are coaxial and have radii of 20 mm and 80 mm. The electric potential of the inner conduc
xxMikexx [17]

Answer: 14.52*10^6 m/s

Explanation: In order to explain this problem we have to consider the energy conservation for the electron within the coaxial cylidrical wire.

the change in potential energy for the electron; e*ΔV is  equal to energy kinetic gained for the electron so:

e*ΔV=1/2*m*v^2  v^=(2*e*ΔV/m)^1/2= (2*1.6*10^-19*600/9.1*10^-31)^1/2=14.52 *10^6 m/s

3 0
2 years ago
Imagine a small child whose legs are half as long as her parent’s legs. If her parent can walk at maximum speed V, at what maxim
AnnZ [28]

Answer:

\boxed{v=\frac {V}{\sqrt {2}}}

Explanation:

We know that speed is given by dividing distance by time or multiplying length and frequency. The speed of the father will be given by Lf where L is the length of the father’s leg ad f is the frequency.

We know that frequency of simple pendulum follows that f=\frac {1}{2\pi} \sqrt {\frac {g}{l}}

Now, the speed of the father will be V=Lf= L\times (\frac {1}{2\pi} \sqrt {\frac {g}{l}}) while for the child the speed will be v=\frac {L}{2}\times (\frac {1}{2\pi} \sqrt {\frac {g}{0.5l}})

The ratio of the father’s speed to the child’s speed will be

\frac {V}{v}=\frac {\frac {L}{2}\times (\frac {1}{2\pi} \sqrt {\frac {g}{0.5l}})}{ L\times (\frac {1}{2\pi} \sqrt {\frac {g}{l}})}\\\frac {V}{v}=\frac {\sqrt {2}}{2}\\\boxed{v=\frac {V}{\sqrt {2}}}

8 0
2 years ago
A particle is in uniform circular motion. Assume a standard rtz coordinate system. If you deconstruct the net force acting on th
Kitty [74]

Answer:

a) One

Explanation:

In a uniform circular motion there must be a force acting to keep it in the circular track. This force can either be centripetal or a centrifugal force.

According to the Newton's first law of motion a particle continues to be in state of rest or in uniform motion until acted upon by an external force.

Here the term uniform motion need to be understood which refers to the uniform velocity of the particle in accordance to the vector laws.

3 0
2 years ago
A particle in the first excited state of a one-dimensional infinite potential energy well (with U = 0 inside the well) has an en
nataly862011 [7]

Answer:

The energy of this particle in the ground state is E₁=1.5 eV.

Explanation:

The energy E_{n} of a particle of mass <em>m</em> in the <em>n</em>th energy state of an infinite square well potential with width <em>L </em>is:

                                                    E_{n}=\frac{n^{2}h^{2}}{8mL^{2}}

In the ground state (n=1). In the first excited state (n=2) we are told the energy is E₂= 6.0 eV. If we replace in the above equation we get that:

                                                    E_{1}=\frac{h^{2}}{8mL^{2}}            

                                                    E_{2}=\frac{h^{2}}{2mL^{2}}

So we can rewrite the energy in the ground state as:

                                                   E_{1}=\frac{1}{4}(\frac{h^{2}}{2mL^{2}})

                                                      E_{1}=\frac{1}{4} E_{2}

                                                   E_{1}=\frac{1}{4} ( 6.0\ eV)

Finally

                                                    E_{1}=1.5\ eV

                                                   

                                                   

6 0
2 years ago
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