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Elena L [17]
4 years ago
6

(b) Suppose at a certain instant the kinetic energy is twice the elastic potential energy. Write an equation describing this sit

uation, using only the variables for the mass m, velocity v, spring constant k, and position x. (Submit a file with a maximum size of 1 MB.)
Physics
1 answer:
timurjin [86]4 years ago
8 0

Answer:

1/2mv² = ke²

Explanation:

Let's suppose the material in question is a spring with spring constant k, mass m and position k, the kinetic energy possessed by the string will be;

K.E = 1/2mass×velocity² i.e 1/2mv²

Its elastic potential energy will be the work done on the spring when stretched which is equal to 1/2kx²

E.P = 1/2kx²

The equation describing the case where the kinetic energy is twice the elastic potential energy will be;

K.E = 2EP... 1)

Substituting the KE and EP formula into (1), we have;

1/2mv² = 2(1/2ke²)

1/2mv² = ke² which gives the required equation

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Suppose the electrons and protons in 1g of hydrogen could be separated and placed on the earth and the moon, respectively. Compa
MAXImum [283]

Answer:

The gravitational force is 3.509*10^17 times larger than the electrostatic force.

Explanation:

The Newton's law of universal gravitation and Coulombs law are:

F_{N}=G m_{1}m_{2}/r^{2}\\F_{C}=k q_{1}q_{2}/r^{2}

Where:

G= 6.674×10^−11 N · (m/kg)2

k =  8.987×10^9 N·m2/C2

We can obtain the ratio of these forces dividing them:

\frac{F_{N}}{F_{C}}=\frac{Gm_{1}m_{2}}{kq_{1}q_{2}}=0.742\times10^{-20}\frac{C^{2}}{kg^{2}}\frac{m_{1}m_{2}}{q_{1}q_{2}}   --- (1)

The mass of the moon is 7.347 × 10^22 kilograms

The mass of the earth is  5.972 × 10^24 kg

And q1=q2=Na*e=(6.022*10^23)*(1.6*10^-19)C=9.635*10^4 C

Replacing these values in eq1:

\frac{F_{N}}{F_{C}}}}=0.742\times10^{-20}\frac{C^{2}}{kg^{2}}\frac{7.347\times5.972\times10^{46}kg^{2}}{(9.635\times10^{4})^{2}}

Therefore

\frac{F_{N}}{F_{C}}}}=3.509\times10^{17}

This means that the gravitational force is 3.509*10^17 times larger than the electrostatic force, when comparing the earth-moon gravitational field vs 1mol electrons - 1mol protons electrostatic field

7 0
3 years ago
What is the magnitude of the torque about his shoulder if he holds his arm straight out to his side, parallel to the floor
const2013 [10]

Complete Question

An athlete at the gym holds a 3.0 kg steel ball in his hand. His arm is 70 cm long and has a mass of 4.0 kg. Assume, a bit unrealistically, that the athlete's arm is uniform.

What is the magnitude of the torque about his shoulder if he holds his arm straight out to his side, parallel to the floor? Include the torque due to the steel ball, as well as the torque due to the arm's weight.

Answer:

The torque is  \tau = 34.3 \  N\cdot m

Explanation:

From the question we are told that

   The mass of the steel ball is  m  =  3.0 \  kg

    The length of arm is  l =  70 \ cm  = 0.7 \  m

    The mass of the arm is m_a  = 4.0 \  kg

Given that the arm of the athlete is uniform them the distance from the shoulder to the center of gravity of the arm is mathematically represented as

       r = \frac{l}{2}

=>    r = \frac{ 0.7}{2}  

=>    r = 0.35 \ m  

Generally the magnitude of torque about the athlete shoulder is mathematically represented as

      \tau =  m_a * g * r  + m * g *  L

=>    \tau =  4 * 9.8 * 0.35 + 3 * 9.8 *  0.70

=>    \tau = 34.3 \  N\cdot m

5 0
3 years ago
You throw a tennis ball straight up (neglect air resistance). It takes 7.0 seconds to go up and then return to your hand. How fa
sattari [20]

Answer:

Velocity of throwing = 34.335 m/s

Explanation:

Time taken by the tennis ball to reach maximum height, t = 0.5 x 7 = 3.5 seconds.

Let the initial velocity be u, we have acceleration due to gravity, a = -9.81 m/s² and final velocity = 0 m/s

Equation of motion result we have v = u + at

Substituting

             0 = u - 9.81 x 3.5

             u = 34.335 m/s

Velocity of throwing = 34.335 m/s

6 0
4 years ago
What heavy element was the chernobyl nuclear accelarators supposed to make?​
Gala2k [10]

Answer

plutonium

Explanation:

brainliest please

6 0
3 years ago
What is the angle of the 2nd order bright fringe produced by two slits that are 8.25x10-5m apart if the wavelength of the incide
dexar [7]

In order to calculate the angle, we can use the formula below for a constructive interference (the interference is constructive because the fringe is bright):

d\sin\theta=m\lambda

Where d is the distance between the slits, m is the order of the interference and lambda is the wavelength.

So, using d = 8.25 * 10^-5, m = 2 and lambda = 4.5 * 10^-7, we have:

\begin{gathered} 8.25\cdot10^{-5}\cdot\sin\theta=2\cdot4.5\cdot10^{-7}\\ \\ \sin\theta=\frac{9\cdot10^{-7}}{8.25\cdot10^{-5}}\\ \\ \sin\theta=1.091\cdot10^{-2}\\ \\ \theta=0.625° \end{gathered}

Therefore the correct option is the second one.

5 0
1 year ago
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