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Naddik [55]
3 years ago
6

A 3-kg object is attached to a spring and moving in simple harmonic motion. Its angular frequency is 20 rads/sec. When the mass-

spring system is 0.10 m from its equilibrium position it has a velocity of v=3 m/s. What is the potential energy when the mass reaches amplitude? A. 0 J
B. 6 J
C. 13.5 J
D. 8 J
E. 19.5 J
Physics
1 answer:
Trava [24]3 years ago
7 0

Answer:19.5 J

Explanation:

Given

mass of block=3 kg

angular frequency=20 rad/sec

spring constant k=\omega _n^2m=1200 N/m

we know total energy remain conserved

E_T at x=0.1 m

E_T=E_P+E_K

Where E_K=kinetic energy

E_P=potential Energy

E_P=\frac{1}{2}kx^2

E_P=600\times 0.01=6 J

E_K=\frac{1}{2}mv^2

E_K=\frac{1}{2}\times 3\times 3^2=13.5 J

E_T=13.5+6=19.5 J

When mass reaches amplitude its velocity becomes zero

there is only potential energy which is equal to Total energy

E_T=E_P=19.5 J

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An fm radio station broadcasts at 98. 6 mhz. What is the wavelength of the radiowaves?.
victus00 [196]

The wavelength of the radio waves is 3.04 cm.

<h3>Calculation:</h3>

λf = c

λ = c/f

where,

λ = wavelength

c = speed of light

f = frequency

Given,

f = 98.6 MHz = 98.6 × 10⁶

c = 3 × 10⁸

To find,

λ =?

Put the values in the formula,

λ = c/f

λ = 3 × 10⁸/98.6 × 10⁶

  = 0.0304 × 10²  m

  = 3.04 cm

Therefore, the wavelength of the radio waves is 3.04 cm.

Learn more about the calculation of wavelength here:

brainly.com/question/8422432

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3 0
2 years ago
An incompressible fluid (water) is flowing through a pipe of diameter 20 cm with
sergey [27]

Answer:

115 kPa

Explanation:

Use Bernoulli equation:

P₁ + ½ ρ v₁² + ρgh₁ = P₂ + ½ ρ v₂² + ρgh₂

Assuming no elevation change, h₁ = h₂.

P₁ + ½ ρ v₁² = P₂ + ½ ρ v₂²

Plugging in values:

(582,000 Pa) + ½ (1000 kg/m³) (1.28 m/s)² = P + ½ (1000 kg/m³) (30.6 m/s)²

P = 115,000 Pa

P = 115 kPa

3 0
3 years ago
An object has a height of 0.066 m and is held 0.210 m in front of a converging lens with a focal length of 0.140 m. (Include the
Anvisha [2.4K]

Explanation:

Given that,

Size of object, h = 0.066 m

Object distance from the lens, u = 0.210 m (negative)

Focal length of the converging lens, f = 0.14 m

If v is the image distance from the lens, we can find it using lens formula as follows :

\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}\\\\\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{0.14 }+\dfrac{1}{(-0.21)}\\\\v=0.42\ m

(a) Magnification,

m=\dfrac{v}{u}\\\\m=\dfrac{0.42}{(-0.21)}\\\\m=-2

(b) Magnification, m=\dfrac{h'}{h}

h' is image height

-2=\dfrac{h'}{0.066}\\\\h'=-2\times 0.066\\\\h'=-0.132\ m

Hence, this is the required solution.

4 0
3 years ago
7. A block of copper of unknown mass has an initial temperature of 65.4oC. The copper is immersed in a beaker containing 95.7g o
dolphi86 [110]

Answer:

37.34372 kg

Explanation:

m = Mass

\Delta T = Change in temperature

1 denotes water

2 denotes copper

c = Heat capacity

Heat is given by

Q=mc\Delta T

In this case the heat transfer will be equal

m_1c_1\Delta T_1=m_2c_2\Delta T_2\\\Rightarrow m_2=\frac{m_1c_1\Delta T_1}{c_2\Delta T_2}\\\Rightarrow m_2=\frac{95.7\times 4.18(24.2-22.7)}{0.39(65.4-24.2)}\\\Rightarrow m_2=37.34372\ kg

Mass of copper block is 37.34372 kg

5 0
3 years ago
A dentist’s drill starts from rest. After 4.28 s of constant angular acceleration it turns at a rate of 28940 rev/min. Find the
mylen [45]

Answer:

Angular acceleration, is 708.07\ rad/s^2

Explanation:

Given that,

Initial speed of the drill, \omega_i=0

After 4.28 s of constant angular acceleration it turns at a rate of 28940 rev/min, final angular speed, \omega_f=28940\ rev/min=3030.58\ rad/s

We need to find the drill’s angular acceleration. It is given by the rate of change of angular velocity.

\alpha =\dfrac{\omega_f}{t}\\\\\alpha =\dfrac{3030.58\ rad/s}{4.28\ s}\\\\\alpha =708.07\ rad/s^2

So, the drill's angular acceleration is 708.07\ rad/s^2.

4 0
3 years ago
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