can someone please take there time and answer this for me :)

Answer:

a) T = 1.69 s, b)  k = 0.825 N / m, c)  v = 1.46 feet/s, d) a = 5.41 ft / s²,

e)   v = - 1,319 ft / s,    a = - 2.70 ft / s², f) K = 4.8 10⁻³ J, U = 1.49 10⁻³ J

Explanation:

In a mass-spring system with simple harmonic motion, the angular velocity is

         w = [tex]\sqrt{\frac{k}{m} }[/tex]

a) find the period

angular velocity, frequency, and period are related

         w = 2π f = 2π / T

          f = 1 / T

          T = 1 / f

           T = 1 / 0.59

           T = 1.69 s

b) the spring constant

         w = 2π f

         w = 2π 0.59

         w = 3.70 rad / s

         w² = k / m

          k = w² m

          k = 3.70² 0.060

          k = 0.825 N / m

c) the maximum speed

simple harmonic movement is described by the expression

          x = A cos (wt + Ф)

speed is defined by

         v =[tex]\frac{dx}{dt}[/tex]

          v = -A w sin (wt + fi)

the speed is maximum when the cosine is ± 1

          v = A w

          v = 0.394 3.70

          v = 1.46 feet/s

d) maximum acceleration

            a = [tex]\frac{dv}{dt}[/tex]

            a = - A w² cos wt + fi

the acceleration is maximum when the cosine is ±1

            a = A w²

            a = 0.394 3.70²

            a = 5.41 ft / s²

e) velocity and acceleration for x = 6 cm

let's reduce the cm to feet

            x = 6 cm (1 foot / 30.48 cm) = 0.1969 foot

Before doing this part we must find the phase angle (Ф), the most common way to start the movement is to move the spring a small distance and release it, so its initial speed is zero for t = 0 s

let's use the expression for the velocity

           v = -A w sin (0 + Фi)

           0 = - A w sin Ф

so sin Ф = 0 which implies that Фi = 0

the equation of motion is

            x = A cos wt

            x = 0.394 cos 3.70t

we substitute

           0.1969 = 0.394 cos 370t

           3.70 t = cos⁻¹ (0.1969 / 0.394)

let's not forget that the angle is in radians

           3.70, t = 1.047

           t = 1.047 / 3.70

           t = 0.2826 s

we substitute this time in the equation for velocity and acceleration

           v = - Aw sin wt

           v = - 0.394 3.70 sin 3.70 0.2826

           v = - 1,319 ft / s

           a = - A w² cos wt

           a = - 0.394 3.70² cos 3.70 0.2826

           a = - 2.70 ft / s²

f) the kinetic and potential energy at this point

           K = ½ m v²

let's slow down to the SI system

           v = 1.319 ft / s (1 m / 3.28 ft) = 0.402 m / s

           K = ½ 0.060 0.402²

           K = 4.8 10⁻³ J

           U = ½ k x²

           U = ½ 0.825 0.06²

           U = 1.49 10⁻³ J

Answer:

T = 0.71 seconds

Explanation:

Given data:

mass m = 1Kg, spring constant K = 78 N/m, time period of oscillation T = 0.71 seconds.

We have to calculate time period when this same spring-mass system oscillates vertically.

As we know

[tex]T = 2\pi \sqrt{\frac{m}{K} }[/tex]

This relation of time period is true under every orientation of the spring-mass system, whether horizontal, vertical, angled or inclined. Therefore, time period of the same spring-mass system oscillating  vertically too remains the same.

Therefore, T = 0.71 seconds

Answer:

Earth has a much greater mass than objects on its surface

Answer:

24 J

Explanation:

[tex]K = \frac{1}{2} mv^{2} = \frac{1}{2} (3kg)(4m/s)^{2} = 24 J[/tex]

Answer:

i dont know but i should know try g o o g l e

Explanation:

Answer:

Truck [tex]\dfrac{g}{10}[/tex]

Road [tex]-\dfrac{g}{10}[/tex]

Explanation:

[tex]a_1[/tex] = Acceleration of truck = [tex]-\dfrac{g}{2}[/tex]

[tex]\mu[/tex] = Coefficient of friction = [tex]\dfrac{2}{5}[/tex]

Frictional force is given by

[tex]f=-\mu mg\\\Rightarrow f=-\dfrac{2}{5}mg\\\Rightarrow ma_2=-\dfrac{2}{5}mg\\\Rightarrow a_2=-\dfrac{2}{5}g[/tex]

Net acceleration is given by

[tex]a=a_2-a_1\\\Rightarrow a=-\dfrac{2}{5}g+\dfrac{g}{2}\\\Rightarrow a=\dfrac{g}{10}[/tex]

The acceleration of the box relative to the truck is [tex]\dfrac{g}{10}[/tex] and [tex]-\dfrac{g}{10}[/tex] relative to the road.

Answer:

Ability of the Keck telescope to capture more distant object despite been atop Mauna kea that Hale Telescope may not capture even if it is atop Palomar mountain in California

Explanation:

If increasing the Diameter of a Telescope beyond a given value will increase the ability of the telescope to capture more light and also capture astronomical objects located in a very distant position without improving resolution.

Hence the superiority of Keck telescope atop Mauna Kea over Hale Telescope atop Palomar mountain in California is the ability of the Keck telescope to capture more distant object despite been atop Mauna kea that Hale Telescope may not capture even if it is atop Palomar mountain in California

Answer:

trueeeeeeee..........mmmm...........

Answer:

You get Day and Night

It takes 24 hour

Answer:

Explanation:

The Earth's orbit makes a circle around the sun. At the same time the Earth orbits around the sun, it also spins.Since the Earth orbits the sun and rotates on its axis at the same time we experience seasons, day and night, and changing shadows throughout the day.It only takes 23 hours, 56 minutes and 4.0916 seconds for the Earth to turn once on its axis.

Answer:

[tex]523269.9\ \text{N/m}[/tex]

Explanation:

q = Charge

r = Distance

[tex]q_1=25\ \text{C}[/tex]

[tex]r_1=3000\ \text{m}[/tex]

[tex]q_2=40\ \text{C}[/tex]

[tex]r_2=850\ \text{m}[/tex]

The electric field is given by

[tex]E=E_1+E_1\\\Rightarrow E=k(\dfrac{q_1}{r_1^2}+\dfrac{q_2}{r_2^2})\\\Rightarrow E=9\times 10^9\times (\dfrac{25}{3000^2}+\dfrac{40}{850^2})\\\Rightarrow E=523269.9\ \text{N/m}[/tex]

The electric field at the aircraft is [tex]523269.9\ \text{N/m}[/tex]

Answer:

  K_{total} = 8.91 J

Explanation:

In this exercise you are asked to find the kinetic energy of the can of coca-cola

         K_total = K_ {Translation} + K_ {rotation}

the translational kinetic energy is

         K_ {translation} = ½ m v²

the kinetic energy of rotation is

         K_ {rotation} = ½ I w²

The moment of inertia of a cylinder is

           I = ½ m r²

we substitute

          K_ {total} = ½ m v² + ½ (½ m r²) w²

angular and linear velocity are related

          v = w r

we substitute

          K_ {total} = ½ m v² + ¼ m r² v² / r²

          K_ {total} = m v² (½ + ¼)

          K_ {total} = ¾ m v²

let's calculate

          K_ {total} = ¾ 0.33 6.00²

           K_{total} = 8.91 J

Answer:

50j

Explanation:

Watts are units used to measure power. power can be defined as rate of energy transfer

500 watts means - 500 J of energy per second

in 1 second - 500 J of work is done

therefore within 10 seconds - 500 J/s x 10 s = 5000 J

work of 5000 J is carried out in 10 seconds

Answer:

Watts are units used to measure power. power can be defined as rate of energy transfer

500 watts means - 500 J of energy per second

in 1 second - 500 J of work is done

therefore within 10 seconds - 500 J/s x 10 s = 5000 J

work of 5000 J is carried out in 10 seconds

Explanation:

Answer:

Nuclear fusion

Explanation:

Answer:

[tex]10.54\ \text{rad/s}[/tex]

Explanation:

[tex]\omega_i[/tex] = Initial angular velocity = 500 rpm = [tex]500\times \dfrac{2\pi}{60}\ \text{rad/s}[/tex]

[tex]\omega_f[/tex] = Final angular velocity

t = Time = 34 s

[tex]\theta[/tex] = Angular displacement = 170 revs = [tex]170\times 2\pi\ \text{rad}[/tex]

[tex]\alpha[/tex] = Angulr acceleration

From the kinematic equations of angular motion we have

[tex]\theta=\omega_it+\dfrac{1}{2}\alpha t^2\\\Rightarrow \alpha=\dfrac{\theta-\omega_it}{\dfrac{1}{2}t^2}\\\Rightarrow \alpha=\dfrac{170\times 2\pi-500\times \dfrac{2\pi}{60}\times 34}{\dfrac{1}{2}\times 34^2}\\\Rightarrow \alpha=-1.23\ \text{rad/s}^2[/tex]

[tex]\omega_f=\omega_i+\alpha t\\\Rightarrow \omega_f=500\times \dfrac{2\pi}{60}+(-1.23)\times 34\\\Rightarrow \omega_f=10.54\ \text{rad/s}[/tex]

The rate at which the wheel is spinning when the power comes back on is [tex]10.54\ \text{rad/s}[/tex].

A new planet that supports life would have all the following characteristics except an orbiting moon. Hence, option B is correct.

An enormous, spherical celestial object known as a planet is neither a star nor its remains. The nebular hypothesis, which states how an interstellar cloud falls out of a nebula to produce a young protostar encircled by a protoplanetary disk, is now the best explanation for planet formation.

By gradually accumulating material under the influence of gravity, or accretion, planets develop in this disk.

The rocky planets Mercury, Venus, Earth, and Mars, as well as the giant planets Jupiter, Saturn, Uranus, and Neptune, make up the Solar System's minimum number of eight planets. These planets all revolve around axes that are inclined relative to their respective polar axes.

To know more about Planet:

https://brainly.com/question/14581221

#SPJ2

Answer: I think B.)

Explanation:

Answer:

[tex]4.12\times 10^{-5}\ J[/tex].

Explanation:

Given that,

Capacitance, [tex]C=1.4\times 10^{-7}\ F[/tex]

Charge stored in the capacitor, [tex]Q=3.4\times 10^{-6}\ C[/tex]

We need to find the electric potential energy stored in the capacitor. The formula for the electric potential energy stored in the capacitor is given by :

[tex]E=\dfrac{Q^2}{2C}[/tex]

Put all the values,

[tex]E=\dfrac{(3.4\times 10^{-6})^2}{2\times 1.4\times 10^{-7}}\\\\=4.12\times 10^{-5}\ J[/tex]

So, the required electric potential eenergy is equal to [tex]4.12\times 10^{-5}\ J[/tex].