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1 year ago

What are the two forces that keep a pendulum swinging?

1 answer:

5 0

The correct answer would be the third option. The two forces that keep a pendulum swinging would be tension and gravity. The force of gravity causes the pendulum to keep it swinging and tension blocks all resistances in order for the motion to continue.

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A uniform disk has a mass of 3.7 kg and a radius of 0.40 m. The disk is mounted on frictionless bearings and is used as a turnta

yuradex [85]

Answer:

1.25 kgm²/sec

Explanation:

Disk inertia, Jd =

Jd = 1/2 * 3.7 * 0.40² = 0.2960 kgm²

Disk angular speed =

ωd = 0.1047 * 30 = 3.1416 rad/sec

Hollow cylinder inertia =

Jc = 3.7 * 0.40² = 0.592 kgm²

Initial Kinetic Energy of the disk

Ekd = 1/2 * Jd * ωd²

Ekd = 0.148 * 9.87

Ekd = 1.4607 joule

Ekd = (Jc + 1/2*Jd) * ω²

Final angular speed =

ω² = Ekd/(Jc+1/2*Jd)

ω² = 1.4607/(0.592+0.148)

ω² = 1.4607/0.74

ω² = 1.974

ω = √1.974

ω = 1.405 rad/sec

Final angular momentum =

L = (Jd+Jc) * ω

L = 0.888 * 1.405

L = 1.25 kgm²/sec

5 0

1 year ago

A toy rocket engine is securely fastened to a large puck that can glide with negligible friction over a horizontal surface, take

Ad libitum [116K]

Answer:

$F=(3i+3.6j)\ N$

Explanation:

It is given that,

Mass of the puck, m = 4.8 kg

Initial velocity of the puck, $u=(1i+0j)\ m/s$

After 8 seconds, final velocity of the puck, $v=(6i+6j)\ m/s$

Let the x and y component of force is given by $F_x\ and\ F_y$ .

x component of force is given by :

$F_x=m\times \dfrac{v-u}{t}$

$F_x=4.8\times \dfrac{6-1}{8}$

$F_x=3\ N$

y component of force is given by :

$F_y=m\times \dfrac{v-u}{t}$

$F_y=4.8\times \dfrac{6-0}{8}$

$F_y=3.6\ N$

So, the component of the force is $F=(3i+3.6j)\ N$ . Hence, this is the required solution.

7 0

2 years ago

A copper wire has radius 0.800 mm and carries current I at 20.0°C. A silver wire with radius 0.500 mm carries the same current a

IgorC [24]

Answer:

Ecu/Eag = 0.46

Explanation:

E = PI/A

Ecu = Pcu × I/A

Pcu = 1.72×10^-8 ohm-meter

r = 0.8 mm = 0.8/1000 = 8×10^-4 m

A = πr^2 = π×(8×10^-4)^2 = 6.4×10^-7π

Ecu = 1.72×10^-8I/6.4×10^-7π = 0.026875I/1

Eag = Pag × I/A

Pag = 1.47×10^-8 ohm-meter

r = 0.5 mm = 0.5/1000 = 5×10^-4 m

A = πr^2 = π × (5×10^-4)^2 = 2.5×10^-7π

Eag = 1.47×10^-8I/2.5×10^-7π = 0.0588I/π

Ecu/Eag = 0.026875I/π × π/0.0588I = 0.46

7 0

2 years ago

Reset the PhET simulation (using the button in the lower right) and set it up in the following manner: select Oscillate, select

lubasha [3.4K]

Answer:

Check Explanation

Explanation:

This is a question that is as a result of an experimental procedure.

The Phet simulation is put on the settings given in the question;

- select Oscillate

- Select No End

- Use the parameters in parentheses by sliding the bars for Amplitude (1.00 cm), Frequency (1.40 Hz)

- Damping (none)

- Tension (highest)

I'll attach an interface of the Phet simulation to this solution.

Once all of these settings have been fixed, the simulation gives a wave pattern whose wavelength can be read from the ruler attached to the background of the simulation.

The wavelength is the distance from crest to crest or from trough to trough.

From the simulation example I have attached, the distance from crest to crest is from the green indicator on one crest to the green indicator on the next crest, that is about 5 to 5.1 cm

The velocity of a wave, v, is related to the frequency, f, and wavelength, λ of the wave through

v = fλ

For the photon, the velocity of the wave is the speed of light,

v = c = (3.00 × 10⁸) m/s

The wavelength computed from the simulation = λ = 5.1 cm = 0.051 m

c = fλ

frequency = (c/λ) = (3.00 × 10⁸) ÷ 0.051 = (5.88 × 10⁹) Hz

So, this step can be used to obtaim rhe required frequency of the photon, just follow these steps and use the calculation method too. You should be able to obtain the frequency of the photon in your experiment.

Hope this Helps!!!