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bagirrra123 [75]

2 years ago

7

A turntable with a moment of inertia of 7.2 × − ⋅ rotates freely with an angular speed of 6.5 ⁄ . Riding on the rim of the turnt

able, 2 from the center, is a hamster. When the hamster walks to the center of the turntable, the angular speed of the turntable becomes ⁄. What is the mass of hamster?

1 answer:

irina1246 [14]2 years ago

6 0

Answer:

turntable with a moment of inertia of 7.2 × − ⋅ rotates freely with an angular speed of 6.5 ⁄ . Riding on the rim of the turntable, 2 from the center, is a hamster. When the hamster walks to the center of the turntable, the angular speed of the turntable becomes ⁄. What is the mass of hamster?

Explanation:

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An air bubble underwater has the same pressure as that of the surrounding water. As the air bubble rises toward the surface (and

goldenfox [79]

Answer:

D Remains constant

Explanation:

3 0

3 years ago

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1

notka56 [123]

Complete Question

Part of the question is shown on the first uploaded image

The rest of the question

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1 and q2 at x3 = -1.220 m ? Your answer may be positive or negative, depending on the direction of the force. Express your answer numerically in newtons to three significant figures.

Answer:

The net force exerted on the third charge is $F_{net}= 3.22*10^{-5} \ J$

Explanation:

From the question we are told that

The third charge is $q_3 = 55 nC = 55 *10^{-9} C$

The position of the third charge is $x = -1.220 \ m$

The first charge is $q_1 = -16 nC = -16 *10^{-9} \ C$

The position of the first charge is $x_1 = -1.650m$

The second charge is $q_2 = 32 nC = 32 *10^{-9} C$

The position of the second charge is $x_2 = 0 \ m$

The distance between the first and the third charge is

$d_{1-3} = -1.650 -(-1.220)$

$d_{1-3} = -0.43 \ m$

The force exerted on the third charge by the first is

$F_{1-3} = \frac{k q_1 q_3}{d_{1-3}^2}$

Where k is the coulomb's constant with a value $9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.$

substituting values

$F_{1-3} = \frac{9*10^{9}* 16 *10^{-9} * (55*10^{-9})}{(-0.43)^2}$

$F_{1-3} = 4.28 *10^{-5} \ N$

The distance between the second and the third charge is

$d_{2-3} = 0- (-1.22)$

$d_{2-3} =1.220 \ m$

The force exerted on the third charge by the first is mathematically evaluated as

$F_{2-3} = \frac{k q_2 q_3}{d_{2-3}^2}$

substituting values

$F_{2-3} = \frac{9*10^{9} * (32*10^{-9}) *(55*10^{-9})}{(1.220)^2}$

$F_{2-3} = 1.06*10^{-5} N$

The net force is

$F_{net} = F_{1-3} -F_{2-3}$

substituting values

$F_{net} = 4.28 *10^{-5} - 1.06*10^{-5}$

$F_{net}= 3.22*10^{-5} \ J$

6 0

2 years ago

Be sure to answer all parts. Compare the wavelengths of an electron (mass = 9.11 × 10−31 kg) and a proton (mass = 1.67 × 10−27 k

Harrizon [31]

Explanation:

Given that,

(a) Speed, $v=6.66\times 10^6\ m/s$

Mass of the electron, $m_e=9.11\times 10^{-31}\ kg$

Mass of the proton, $m_p=1.67\times 10^{-27}\ kg$

The wavelength of the electron is given by :

$\lambda_e=\dfrac{h}{m_ev}$

$\lambda_e=\dfrac{6.63\times 10^{-34}}{9.11\times 10^{-31}\times 6.66\times 10^6}$

$\lambda_e=1.09\times 10^{-10}\ m$

The wavelength of the proton is given by :

$\lambda_p=\dfrac{h}{m_p v}$

$\lambda_p=\dfrac{6.63\times 10^{-34}}{1.67\times 10^{-27}\times 6.66\times 10^6}$

$\lambda_p=5.96\times 10^{-14}\ m$

(b) Kinetic energy, $K=1.71\times 10^{-15}\ J$

The relation between the kinetic energy and the wavelength is given by :

$\lambda_e=\dfrac{h}{\sqrt{2m_eK}}$

$\lambda_e=\dfrac{6.63\times 10^{-34}}{\sqrt{2\times 9.11\times 10^{-31}\times 1.71\times 10^{-15}}}$

$\lambda_e=1.18\times 10^{-11}\ m$

$\lambda_p=\dfrac{h}{\sqrt{2m_pK}}$

$\lambda_p=\dfrac{6.63\times 10^{-34}}{\sqrt{2\times 1.67\times 10^{-27}\times 1.71\times 10^{-15}}}$

$\lambda_p=2.77\times 10^{-13}\ m$

Hence, this is the required solution.

6 0

3 years ago

40 POINTS!!! + BRAINLIEST!!!

Sergeu [11.5K]

Answer:

I think it is better if you read and shortly write my explanation

Explanation:

simple pendulum with no friction, mechanical energy is conserved. Total mechanical energy is a combination of kinetic energy and gravitational potential energy. As the pendulum swings back and forth, there is a constant exchange between kinetic energy and gravitational potential energy.

8 0

3 years ago

The potential at location A is 382 V. A positively charged particle is released there from rest and arrives at location B with a

jarptica [38.1K]

Answer: 247.67 V

Explanation:

Given

Potential At A $V_a=382\ V$

Potential at $V_c=785\ V$

when particle starts from A it reaches with velocity $v_b$ at Point while when it starts from C it reaches at point B with velocity $2v_b$

Suppose m is the mass of Particle

Change in Kinetic Energy of particle moving under the Potential From A to B

$q\cdot \left ( V_a-V_b\right )=0.5m\cdot (v_b)^2----1$

Change in Kinetic Energy of particle moving under the Potential From C to B

$q\cdot \left ( V_c-V_b\right )=0.5m\cdot (2v_b)^2-----2$

Divide 1 and 2 we get

$\frac{V_a-V_b}{V_c-V_b}=\frac{v_b^2}{4v_b^2}$

on solving we get

$V_b=\frac{4}{3}\cdot V_a-\frac{1}{3}\cdot V_c$

$V_b=\frac{743}{3}=247.67\ V$

4 0

3 years ago