What two things influence the strength of gravitation?

You have a two-wheel trailer that you pull behind your ATV. Two children with a combined mass of 77.2 kg hop on board for a ride

lions [1.4K]

To solve this problem it is necessary to apply the kinematic equations of motion and Hook's law.

By Hook's law we know that force is defined as,

$F= kx$

Where,

k = spring constant

x = Displacement change

PART A) For the case of the spring constant we can use the above equation and clear k so that

$k= \frac{F}{x}$

$k = \frac{mg}{x}$

$k= \frac{77.2*9.8}{0.0637}$

$k = 11876.92N/m$

Therefore the spring constant for each one is 11876.92/2 = 5933.46N/m

PART B) In the case of speed we can obtain it through the period, which is given by

$T = \frac{2\pi}{\omega}$

Re-arrange to find \omega,

$\omega = \frac{2\pi}{T}$

$\omega = \frac{2\pi}{2.14}$

$\omega = 2.93rad/s$

Then through angular kinematic equations where angular velocity is given as a function of mass and spring constant we have to

$\omega^2 = \frac{k}{m}$

$m = \frac{k}{\omega^2}$

$m = \frac{ 11876.92}{2.93}$

$m = 4093.55Kg$

Therefore the mass of the trailer is 4093.55Kg

PART C) The frequency by definition is inversely to the period therefore

$f = \frac{1}{T}$

$f = \frac{1}{2.14}$

$f = 0.4672 Hz$

Therefore the frequency of the oscillation is 0.4672 Hz

PART D) The time it takes to make the route 10 times would be 10 times the period, that is

$t_T = 10*T$

$t_T = 10 *2.14s$

$t_T = 21.4s$

Therefore the total time it takes for the trailer to bounce up and down 10 times is 21.4s

5 0

3 years ago

Consider a model of a hydrogen atom in which an electron is in a circular orbit of radius r = 5.92×10−11 m around a stationary p

DaniilM [7]

Answer:

2.068 x 10^6 m / s

Explanation:

radius, r = 5.92 x 10^-11 m

mass of electron, m = 9.1 x 10^-31 kg

charge of electron, q = 1.6 x 10^-19 C

As the electron is revolving in a circular path, it experiences a centripetal force which is balanced by the electrostatic force between the electron and the nucleus.

centripetal force = $\frac{mv^{2}}{r}$

Electrostatic force = $\frac{kq^{2}}{r^{2}}$

where, k be the Coulombic constant, k = 9 x 10^9 Nm^2 / C^2

So, balancing both the forces we get

$\frac{kq^{2}}{r^{2}}=\frac{mv^{2}}{r}$

$v=\sqrt{\frac{kq^{2}}{mr}}$

$v=\sqrt{\frac{9\times 10^{9}\times1.6\times 10^{-19}\times 1.6\times 10^{-19}}{9.1\times 10^{-31}\times 5.92\times10^{-11}}}$

v = 2.068 x 10^6 m / s

Thus, the speed of the electron is give by 2.068 x 10^6 m / s.

6 0

3 years ago

A copper telephone wire has essentially

Lunna [17]

Answer:

128.21 m

Explanation:

The following data were obtained from the question:

Initial temperature (θ₁) = 4 °C

Final temperature (θ₂) = 43 °C

Change in length (ΔL) = 8.5 cm

Coefficient of linear expansion (α) = 17×10¯⁶ K¯¹)

Original length (L₁) =.?

The original length can be obtained as follow:

α = ΔL / L₁(θ₂ – θ₁)

17×10¯⁶ = 8.5 / L₁(43 – 4)

17×10¯⁶ = 8.5 / L₁(39)

17×10¯⁶ = 8.5 / 39L₁

Cross multiply

17×10¯⁶ × 39L₁ = 8.5

6.63×10¯⁴ L₁ = 8.5

Divide both side by 6.63×10¯⁴

L₁ = 8.5 / 6.63×10¯⁴

L₁ = 12820.51 cm

Finally, we shall convert 12820.51 cm to metre (m). This can be obtained as follow:

100 cm = 1 m

Therefore,

12820.51 cm = 12820.51 cm × 1 m / 100 cm

12820.51 cm = 128.21 m

Thus, the original length of the wire is 128.21 m

5 0

3 years ago

What parts are found in an electric generator? Check all that apply.

Blizzard [7]

Answer:

an armature a permanent magnet brushes slip rings

Explanation:

6 0

3 years ago

Read 2 more answers

g determine what frequency is required of a source powering a 100 uf capacitor a 500 ohm resistor and a s50 mH inductor in serie

polet [3.4K]

Answer: $71.16\ Hz$

Explanation:

Given

Capacitance $C=100\ \mu F$

Resistance $R=500\ \Omega$

Inductance $L=50\ mH$

In LCR circuit, current is maximum at resonance frequency i.e.

$X_L=X_C\ \text{and}\ \omega_o=\dfrac{1}{\sqrt{LC}}$

Insert the values

$\Rightarrow \omega_o=\dfrac{1}{\sqrt{50\times 10^{-3}\times 100\times 10^{-6}}}\\\\\Rightarrow \omega_o=\dfrac{1}{\sqrt{5}\times 10^{-3}}\\\\\Rightarrow \omega_o=0.447\times 10^{3}$

Also, frequency is given by

$\Rightarrow 2\pi f=\omega_o\\\\\Rightarrow f=\frac{\omega_o}{2\pi}$

$\Rightarrow f=\dfrac{1}{2\pi}\times 0.447\times 10^3\\\\\Rightarrow f=71.16\ Hz$

8 0

3 years ago