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Korvikt [17]
3 years ago
11

An undamped 1.48 kg horizontal spring oscillator has a spring constant of 35.4 N/m. While oscillating, it is found to have a spe

ed of 3.90 m/s as it passes through its equilibrium position. What is its amplitude A of oscillation
Physics
1 answer:
givi [52]3 years ago
5 0

Answer: 0.798 m

Explanation:

Given

Mass of the spring oscillator, m = 1.48 kg

Force constant of the spring, k = 35.4 N/m

Speed of oscillation, v = 3.9 m/s

Kinetic Energy = 1/2 mv²

Kinetic Energy = 1/2 * 1.48 * 3.9²

KE = 0.5 * 22.5108

KE = 11.26 J

Using the law of conservation of Energy. The Potential Energy of the system is equal to Kinetic Energy of the system

KE = PE

PE = 1/2kx²

11.26 = 1/2 * 35.4 * x²

11.26 = 17.7x²

x² = 11.26 / 17.7

x² = 0.6362

x = √0.6362

x = 0.798 m

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A physics student throws a ball thrown horizontally from the top of a building with a speed of 8 m/s. The student measures the h
Alekssandra [29.7K]

Answer:

Following are the solution to the given question:

Explanation:

Please find the complete question in the attached file.

a_x = 0\\\\a_y = -g = -9.81\ \ \frac{m}{s^2} \\\\v_{iy} = 0 \\\\y = v_{iy} \times t+ \frac{1}{2ay} \times t^2 \\\\ 15 = \frac{1}{2} \times 9.8 \times t^2 \\\\t = \sqrt{(2 \times \frac{15}{9.8})} \\\\

  = 1.75 \ sec \\\\

Distance:

x = v_{ix} \times t \\\\

  = 8 \times 1.75\\\\= 14 m

4 0
3 years ago
3. Land iguanas in the Galápagos are examples of ____________
jek_recluse [69]

Answer:

There are three species of land iguana found in the Galapagos Islands. The well-known yellowish land iguanas include Conolophus subcristatus, native to six islands, and Conolophus pallidus, found only on the island of Santa Fe.

8 0
3 years ago
Three adjacent keys on a piano (F, F-sharp, and G) are struck simultaneously, producing frequencies of 349, 370, and 392 Hz. Wha
PSYCHO15rus [73]

Answer:

21 Hz, 43 Hz and 22 Hz

Explanation:

The computation of the beat frequencies that are generated by this discordant combination is as follows:

As we know that

beat frequencies = |f_1  - f_2|

So

For the first one

= |349 Hz - 370 Hz|

= 21 Hz

For the second one

= |349 Hz - 392 Hz|

= 43 Hz

And, for the third one

= |370 Hz - 392 Hz|

= 22 Hz

5 0
3 years ago
Se dispara un proyectil con una velocidad inicial de 50 m/s y un ángulo de 30°, por encima de la horizontal. Calcular: a) Posici
dmitriy555 [2]

Answer:

a) Posición y velocidad después de los 6s

i) Posición = -26.58m

ii) velocidad = -33.86m/s

b) Tiempo para alcanzar la altura máxima

= 2.55s

c) Alcance horizontal

= 220.7m

Explanation:

a) Posición y velocidad después de los 6s

i) Posición

y = (usinθ)t – 1/2 gt²

u = 50m/s

θ = 30°

g = 9.81m/s²

t = 6s

y = (50 × sin 30)6 - 1/2 × 9.81 × 6²

y = 150 - 176.58

y = -26.58m

ii) velocidad

v = u sinθ–gt

u = 50m/s

θ = 30°

g = 9.81m/s²

t = 6s

v = 50 × sin 30 - 9.81 × 6

v = 25 - 58.86

v = -33.86m/s

b) Tiempo para alcanzar la altura máxima

usinθ /g

u = 50m/s

θ = 30°

g = 9.81m/s²

= 50 × sin 30/ 9.81

= 25/9.81

= 2.5484199796s

≈ 2.55s

c) Alcance horizontal

R = u²sin2θ/g

u = 50m/s

θ = 30°

g = 9.81m/s²

R = 50² ×( sin 2 ×30°) /9.81

R = 220.69964419m

≈ 220.7m

7 0
3 years ago
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