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ser-zykov [4K]
3 years ago
8

A sphere of radius r = 5cm carries a uniform volume charge density rho = 400 nC/m^3. Q. What is the total charge Q of the sphere

?
Physics
1 answer:
Tanzania [10]3 years ago
6 0

Answer:

The total charge Q of the sphere is 2.094\times10^{-10}\ C.

Explanation:

Given that,

Radius = 5 cm

Charge density J= 400\ nC/m^3

We need to calculate the total charge Q of the sphere

Using formula of charge

q=\rho V

Where, \rho = charge density

V = volume

Put the value into the formula

q=\rho\times(\dfrac{4}{3}\pi r^3)

Put the value into the formula

q=\dfrac{4}{3}\times\pi\times400\times10^{-9}\times(5\times10^{-2})^3

q=2.094\times10^{-10}\ C

Hence, The total charge Q of the sphere is 2.094\times10^{-10}\ C.

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What is the minimal mass of helium (density 0.18 kg/m3) needed to lift a balloon carrying two people in a basket, if the total m
Sergio039 [100]

Answer:   
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 R[b] is the density of gas inside the baloon.   
 ==================================== 
 Let V be the volume of helium required. 
 Upthrust on helium = Weight of the volume of air displaced = Density of air * g * Volume of helium = 1.225 * g * V   
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 ---- 
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 So minimal F would be 2549.7 
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3 0
3 years ago
A 1 kg mass is attached to a spring with spring constant 7 Nt/m. What is the frequency of the simple harmonic motion? What is th
Scorpion4ik [409]

1. 0.42 Hz

The frequency of a simple harmonic motion for a spring is given by:

f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}

where

k = 7 N/m is the spring constant

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Substituting these numbers into the formula, we find

f=\frac{1}{2\pi}\sqrt{\frac{7 N/m}{1 kg}}=0.42 Hz

2. 2.38 s

The period of the harmonic motion is equal to the reciprocal of the frequency:

T=\frac{1}{f}

where f = 0.42 Hz is the frequency. Substituting into the formula, we find

T=\frac{1}{0.42 Hz}=2.38 s

3. 0.4 m

The amplitude in a simple harmonic motion corresponds to the maximum displacement of the mass-spring system. In this case, the mass is initially displaced by 0.4 m: this means that during its oscillation later, the displacement cannot be larger than this value (otherwise energy conservation would be violated). Therefore, this represents the maximum displacement of the mass-spring system, so it corresponds to the amplitude.

4. 0.19 m

We can solve this part of the problem by using the law of conservation of energy. In fact:

- When the mass is released from equilibrium position, the compression/stretching of the spring is zero: x=0, so the elastic potential energy is zero, and all the mechanical energy of the system is just equal to the kinetic energy of the mass:

E=K=\frac{1}{2}mv^2

where m = 1 kg and v = 0.5 m/s is the initial velocity of the mass

- When the spring reaches the maximum compression/stretching (x=A=amplitude), the velocity of the system is zero, so the kinetic energy is zero, and all the mechanical energy is just elastic potential energy:

E=U=\frac{1}{2}kA^2

Since the total energy must be conserved, we have:

\frac{1}{2}mv^2 = \frac{1}{2}kA^2\\A=\sqrt{\frac{m}{k}}v=\sqrt{\frac{1 kg}{7 N/m}}(0.5 m/s)=0.19 m

5. Amplitude of the motion: 0.44 m

We can use again the law of conservation of energy.

- E_i = \frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2 is the initial mechanical energy of the system, with x_0=0.4 m being the initial displacement of the mass and v_0=0.5 m/s being the initial velocity

- E_f = \frac{1}{2}kA^2 is the mechanical energy of the system when x=A (maximum displacement)

Equalizing the two expressions, we can solve to find A, the amplitude:

\frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2=\frac{1}{2}kA^2\\A=\sqrt{x_0^2+\frac{m}{k}v_0^2}=\sqrt{(0.4 m)^2+\frac{1 kg}{7 N/m}(0.5 m/s)^2}=0.44 m

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We can use again the law of conservation of energy.

- E_i = \frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2 is the initial mechanical energy of the system, with x_0=0.4 m being the initial displacement of the mass and v_0=0.5 m/s being the initial velocity

- E_f = \frac{1}{2}mv_{max}^2 is the mechanical energy of the system when x=0, which is when the system has maximum velocity, v_{max}

Equalizing the two expressions, we can solve to find v_{max}, the maximum velocity:

\frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2=\frac{1}{2}mv_{max}^2\\v_{max}=\sqrt{\frac{k}{m}x_0^2+v_0^2}=\sqrt{\frac{7 N/m}{1 kg}(0.4 m)^2+(0.5 m/s)^2}=1.17 m/s m

4 0
3 years ago
Read 2 more answers
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Explanation:

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Learn more:

Balanced equation brainly.com/question/2612756

#learnwithBrainly

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