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sergeinik [125]
3 years ago
9

A spinning ice skater on extremely smooth ice is able to control the rate at which she rotates by pulling in her arms. Which of

the following statements are true about the skater during this process? (There could be more than one correct choice.)
Her kinetic energy remains constant.
Her moment of inertia remains constant.
Her angular momentum remains constant.
She is subject to a constant non-zero torque.
Physics
1 answer:
Korvikt [17]3 years ago
7 0

Answer:

Correct -> Her angular momentum remains constant.

Explanation:

If the skater pulls her arms, her radius changes, so her moment of inertia changes.

By definition, moment of inertia is the resistance to rotation. So, if her moment of inertia decreases, her angular velocity increases, because if there is no external torque (in this question there is none), angular momentum is conserved.

L = I\omega

K = \frac{1}{2}I\omega^2

If the moment of inertia decreases by half, the angular velocity doubles. In that case kinetic energy also increases, because the square of the angular velocity affects the kinetic energy more than the decrease of the moment of inertia.

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neptune is an average distance of 4.5×10^12m from the sun. Estimate the length of the Neptunian year.
Vikentia [17]

As per Kepler's third law we know that

\frac{T_1^2}{T_2^2} = \frac{R_1^3}{R_2^3}

now here we know that

T_1 = year of Neptune

T_2 = year of Earth

R_1 = distance of Neptune from Sun

R_2 = Distance of Earth from Sun

so now we will have

\frac{T_1^2}{1} = \frac{(4.5 \times 10^{12})^3}{(1.5 \times 10^11)^3}

T_1^2 = 27000

T_1 = 164.3 years

so length of year of Neptune is 164.3 years

6 0
3 years ago
Is an electron an antiparticle, boson, lepton, or hadron?
Masteriza [31]

Answer:

An electron is a lepton

4 0
3 years ago
Phyllis is calculating the path of a space probe that will travel to Pluto. On the way to Pluto, the probe will pass Mars, Satur
inysia [295]

Answer:

Jupiter

Explanation:

4 0
3 years ago
Read 2 more answers
Consider the following reaction proceeding at 298.15 K: Cu(s)+2Ag+(aq,0.15 M)⟶Cu2+(aq, 1.14 M)+2Ag(s) If the standard reduction
lutik1710 [3]

Answer : The cell potential for this cell 0.434 V

Solution :

The balanced cell reaction will be,  

Cu(s)+2Ag^{+}(aq)\rightarrow Cu^{2+}(aq)+2Ag(s)

Here copper (Cu) undergoes oxidation by loss of electrons, thus act as anode. silver (Ag) undergoes reduction by gain of electrons and thus act as cathode.

First we have to calculate the standard electrode potential of the cell.

E^o_{[Cu^{2+}/Cu]}=0.34V

E^o_{[Ag^{+}/Ag]}=0.80V

E^o=E^o_{[Ag^{+}/Ag]}-E^o_{[Cu^{2+}/Cu]}

E^o=0.80V-(0.34V)=0.46V

Now we have to calculate the concentration of cell potential for this cell.

Using Nernest equation :

E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Cu^{2+}][Ag]^2}{[Cu][Ag^+]^2}

where,

n = number of electrons in oxidation-reduction reaction = 2

E_{cell} = ?

Now put all the given values in the above equation, we get:

E_{cell}=0.46-\frac{0.0592}{2}\log \frac{(1.14)\times (1)^2}{(1)\times (0.15)}

E_{cell}=0.434V

Therefore, the cell potential for this cell 0.434 V

8 0
3 years ago
What will be the force if the particle's charge is tripled and the electric field strength is halved? Give your answer in terms
Alexus [3.1K]

Answer:

1.5F

Explanation:

Using

E= F/q

Where F= force

E= electric field

q=charge

F= Eq

So if qis tripled and E is halved we have

F= (E/2)3q

F= 1.5Eq=>> 1.5F

4 0
3 years ago
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