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Nutka1998 [239]
2 years ago
11

A spring stretches by 15cm when a mass of 300g hangs down from it,if the spring is then stretched an additional 10cm and release

d, calculate;the spring constant,the angular velocity, amplitude of oscillation, maximum velocity, maximum acceleration of the mass,period, frequency​
Physics
1 answer:
bearhunter [10]2 years ago
4 0

Answer:

0.1 m

Explanation:

It is given that,

Mass of the object, m = 350 g = 0.35 kg

Spring constant of the spring, k = 5.2 N/m

Amplitude of the oscillation, A = 10 cm = 0.1 m

Frequency of a spring mass system is given by :

Time period:

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A 1-m-long monopole car radio antenna operates in the AM frequency of 1.5 MHz. How muchcurrent is required to transmit 4 W of po
Zanzabum

Answer:

The current needed to transmit Power of 4 W is 28.47 A

Solution:

As per the question:

Length of the antenna, L_{a} = 1 m

Frequency, \vartheta = 1.5 MHz = 1.5\times 10^{6} Hz

Power transmitted, P_{t} = 4 W

Now,

For a monopole antenna:

\lambda_{a} = \frac{c}{\vartheta}

where

\lambda_{a} = wavelength transmitted by the antenna

c = speed of light in vacuum

\lambda_{a} = \frac{3\times 10^{8}}{1.5\times 10^{6}} = 200 m

Now,

Since, the value of \lambda_{a} >> L_{a} thus the monopole is a Hertian monopole.

The resistance is calculated as:

R = \frac{1}{2}(\frac{dL_{a}}{\lambda_{a}})^{2}\times 80\pi^{2}

R = \frac{1}{2}(\frac{1}{200)^{2}\times 80\pi^{2} = 9.869\times 10^{- 3} = 9.869 m\Omega

P_{radiated} = P_{t}

P_{radiated} = \frac{R}{I^{2}}

Now, the current I is given by:

I = \sqrt{\frac{2P_{t}}{R}} = \sqrt{\frac{2\times 4}{9.869\times 10^{- 3}}} = 28.47 A

5 0
2 years ago
If the light strikes the first mirror at an angle θ1, what is the reflected angle θ2? express your answer in terms of θ1.
Alexus [3.1K]

Answer:

θ₂ = 90° - θ₁

Explanation:

When the light falls on a mirror it bounces back. This is know as reflection. The incident angle is equal to the angle of reflection.

Here, the light strikes the mirror at an angle = θ₁

To find the angle of reflection we first need to understand angle of incidence. The angle of incidence is the angle made between the incident ray and normal. Normal is an imaginary line drawn perpendicular line on the boundary of the mirror.

Since the light strikes the mirror at angle of θ₁, which is the angle between light ray and the mirror.

Angle of incidence = 90° - θ₁.

Thus, angle of reflection, θ₂ = 90° - θ₁

3 0
2 years ago
A cyclist traveling at 5m/s uniformly accelerates up to 10 m/s in 2 seconds. Each tire of the bike has a 35 cm radius, and a sma
konstantin123 [22]

Answer:

a=2.5\ m/s^2

Explanation:

Given that,

Initial speed, u = 5 m/s

Final speed, v = 10 m/s

Time, t = 2 s

The radius of the tire of the bike, r = 35 cm

We need to find the angular acceleration of the pebble during those two seconds. It can be calculated as follows.

a=\dfrac{v-u}t{}\\\\a=\dfrac{10-5}{2}\\\\a=2.5\ m/s^2

So, the required angular acceleration of the pebble is equal to 2.5\ m/s^2.

8 0
1 year ago
A bug starts at point A, crawls 8.0 cm east, then 5.0 cm south, 3.0 west, and 4.0 cm north to point B.
Sholpan [36]

Answer:

5cm east& 1cm west from A

Explanation:

https://brainly.ph/question/2753392

7 0
1 year ago
Two long conducting cylindrical shells are coaxial and have radii of 20 mm and 80 mm. The electric potential of the inner conduc
xxMikexx [17]

Answer: 14.52*10^6 m/s

Explanation: In order to explain this problem we have to consider the energy conservation for the electron within the coaxial cylidrical wire.

the change in potential energy for the electron; e*ΔV is  equal to energy kinetic gained for the electron so:

e*ΔV=1/2*m*v^2  v^=(2*e*ΔV/m)^1/2= (2*1.6*10^-19*600/9.1*10^-31)^1/2=14.52 *10^6 m/s

3 0
2 years ago
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