A mass is hung from a vertical spring and allowed to come to rest or its equilibrium position. The mass is then pulled down an additional 0.25 meters and released. As the mass oscillates it completes one full cycle in 3.0 seconds. Place the numbers below to correctly identify the mass's amplitude, full range of vertical motion, frequency, and period. The full range of vertical motion is the distance between the maximum and minimum heights of the mass.
e amplitude of the spring is______ m.
The full range of vertical motion is _____m.
The frequency of the spring is______ Hz.
The period of the spring is_______ s.

Answer:

The speed (magnitude of the velocity) is 14.4 m/s

Explanation:

Vertical Launch Upwards

It occurs when an object is launched vertically up without taking into consideration any kind of friction with the air.

If vo is the initial speed and g is the acceleration of gravity, the speed vf at any time is calculated by:

[tex]v_f=v_o-g.t[/tex]

A tennis ball is launched vertically up with an initial speed of vo=34 m/s. At time t=2 s, its speed is:

[tex]v_f=34-9.8*2[/tex]

[tex]v_f=34-19.6[/tex]

[tex]v_f=14.4\ m/s[/tex]

The speed (magnitude of the velocity) is 14.4 m/s

Answer:

The direction could change

Answer:

So increasing the voltage increases the charge in direct proportion to the voltage. If the voltage exceeds the capacitors rated voltage, the capacitor may fail due to breakdown of the dielectric between the two plates that make up the capacitor.

Explanation:

A option.

The gravitational force exerted by the moon on the satellite is such that

F = G M m / R ² = m a   →   a = G M / R ²

where a is the satellite's centripetal acceleration, given by

a = v ² / R

The satellite travels a distance of 2πR about the moon in complete revolution in time T, so that its tangential speed is such that

v = 2πR / T   →   a = 4π ² R / T ²

Substitute this into the first equation and solve for T :

4π ² R / T ² = G M / R ²

4π ² R ³ = G M T ²

T ² = 4π ² R ³ / (G M )

T = √(4π ² R ³ / (G M ))

T = 2πR √(R / (G M ))

The correct expression for the time T it takes the satellite to make one complete revolution around the moon is [tex] T = 2\pi R\sqrt{\frac{R}{GM}} [/tex].

We can find the period T (the time it takes the satellite to make one complete revolution around the moon) from the gravitational force:

[tex] F = \frac{GmM}{R^{2}} [/tex]    (1)

Where:

G: is the gravitational constant = 6.67x10⁻¹¹ Nm²/kg²

R: is the distance between the satellite and the center of the moon

m: is the satellite's mass

M: is the moon's mass

The gravitational force is also equal to the centripetal force:

[tex] F = ma_{c} [/tex]   (2)

The centripetal acceleration ([tex]a_{c}[/tex]) is equal to the tangential velocity (v):

[tex] a_{c} = \frac{v^{2}}{R} [/tex]   (3)

And from the tangential velocity we can find the period:

[tex] v = \omega R = \frac{2\pi R}{T} [/tex]   (4)

Where:

ω: is the angular speed = 2π/T

By entering equations (4) and (3) into (2), we have:

[tex] F = m\frac{v^{2}}{R} = m\frac{(\frac{2\pi R}{T})^{2}}{R} = \frac{mR(2\pi)^{2}}{T^{2}} [/tex]   (5)

By equating (5) and (1), we get:

[tex] \frac{mR(2\pi)^{2}}{T^{2}} = \frac{GmM}{R^{2}} [/tex]

[tex] T^{2} = \frac{R^{3}(2\pi)^{2})}{GM} [/tex]  

[tex] T = \sqrt{\frac{R^{3}(2\pi)^{2})}{GM}} [/tex]

[tex] T = 2\pi R\sqrt{\frac{R}{GM}} [/tex]

Therefore, the expression for the time T is [tex] T = 2\pi R\sqrt{\frac{R}{GM}} [/tex].

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I hope it helps you!

Answer:

4m/s^2

Explanation:

Answer:

33.5J

Explanation:

Given:

Mass of the canister= 3.1 kg

Initial velocity of canister= v(i)= 4.4i m/s

Final velocity of canister= v(f)= 6.4j m/s

Force magnitude( xy plane)= 5 N

The magnitude of vector V'= Vxi + Vyj + Vzk

|V|= √( Vx^2 + Vy^2 + Vz^2

From Kinectic energy and work theorem.

Net work = Kinectic energy of the canister

ΔK= W

(Kf - Ki)= W

Where Kf= final Kinectic energy

= 1/2 mv^2

If we input the given values we have,

= 1/2 × 3.1 ×√(4.4^2 + 0^2 + 0^2)^2 = 30J

Ki= initial Kinectic energy

= 1/2 mv^2

If we substitute the given values we have

=1/2 × 3.1 ×√(0^2 + 6.4^2 + 0^2)^2 = 63.5 J

Work done by canister = (final Kinectic energy - initial energy)

= 63.5- 30

=33.5J

Hence, work done on the canister 33.5J

Answer:

weight = 900 N

acceleration  = 11.25 m/s²

Explanation:

given data

mass m1 = 75 kg

mass m2 = 80 kg

gravitational acceleration = 12 m/s²

solution

As we know weight of a mass that is

weight of mass = Mass × Acceleration due to gravity .................1

so Astro 1 weight is

weight = 75kg  × 12 m/s²

weight = 900 N

and

so, when Astro 2 needs this much weight the planet on which he is will have the acceleration

acceleration = Weight ÷ Mass of Astro 2        .....................2

acceleration  = 900 ÷ 80 m/s²

acceleration  = 11.25 m/s²

Answer:

a. P.E = 3430Joules.

b. Workdone = 3430Nm

Explanation:

Given the following data;

Mass = 70kg

Distance = 5m

We know that acceleration due to gravity is equal to 9.8m/s²

To find the potential energy;

Potential energy = mgh

P.E = 70*9.8*5

P.E = 3430J

b. To find the workdone;

Workdone = force * distance

But force = mass * acceleration

Force = 70*9.8

Force = 686 Newton.

Workdone = 686 * 5

Workdone = 3430Nm

Answer:

x = 1382.9 km

Explanation:

The speed of the wave is constant, so we can use the uniform motion relationships

p wave

          [tex]v_p[/tex] = x / t₁

          t₁ = x /v_p

S wave

          v_s = x / t₂

           t₂ = x / v_s

indicate that the time difference between the two waves is

          t₂ - t₁ = 3.25 min (60 s / 1 min)

          t₂ -t₁ = 195 s

let's substitute

         [tex]\frac{x}{v_s} - \frac{x}{v_p}[/tex] = 195

         x ([tex]\frac{1}{v_s} - \frac{1}{v_p}[/tex] = 195

let's calculate

         x [tex]( \frac{1}{4.2} - \frac{1}{10.3} )[/tex] = 195

         x (0.1410) = 195

         x = 195 /0.141

         x = 1382.9 km