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2 years ago

A boy spins a top on a table. How can he INCREASE the total energy of this system?

Physics

1 answer:

babunello [35]2 years ago

8 0

He can increase the total energy of the system by spinning the top faster.

<h3>What is energy?</h3>

The term energy refers to the ability to do work. The total energy of the system refers to the gravitational potential energy and the rotational energy of the top.

Hence, he can increase the total energy of the system by spinning the top faster.

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A laser emits two wavelengths (λ1 = 420 nm; λ2 = 630 nm). When these two wavelengths strike a grating with 450 lines/mm, they pr

Westkost [7]

A) Order of the first laser: 3, order of the second laser: 2

B) The overlap occurs at an angle of $34.9^{\circ}$

Explanation:

The formula that gives the position of the maxima (bright fringes) for a diffraction grating is

$d sin \theta = m \lambda$

where

d is spacing between the lines in the grating

$\theta$ is the angle of the maximum

m is the order of diffraction

$\lambda$ is the wavelength of the light

For laser 1,

$d sin \theta = m_1 \lambda_1$

For laser 2,

$d sin \theta = m_2 \lambda_2$

where

$\lambda_1 = 420 nm\\\lambda_2 = 630 nm$

Since the position of the maxima in the two cases overlaps, then the term $d sin \theta$ on the left is the same for the two cases, therefore we can write:

$m_1 \lambda_1 = m_2 \lambda_2\\\frac{m_1}{m_2}=\frac{\lambda_2}{\lambda_1}=\frac{630}{420}=\frac{3}{2}$

Therefore:

$m_1 = 3$

$m_2 = 2$

In order to find the angle at which the overlap occurs, we use the 1st laser situation:

$d sin \theta = m_1 \lambda_1$

where:

N = 450 lines/mm = 450,000 lines/m is the number of lines per unit length, so the spacing between the lines is

$d=\frac{1}{N}=\frac{1}{450,000}=2.2\cdot 10^{-6} m$

$m_1 = 3$ is the order of the maximum

$\lambda_1 = 420 nm = 420\cdot 10^{-9} m$ is the wavelength of the laser light

Solving for $\theta$ , we find the angle of the maximum:

$sin \theta = \frac{m_1 \lambda_1}{d}=\frac{(3)(420\cdot 10^{-9})}{2.2\cdot 10^{-6}}=0.572$

So the angle is

$\theta=sin^{-1}(0.572)=34.9^{\circ}$

Learn more about diffraction:

brainly.com/question/3183125

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5 0

3 years ago

Republicans are called the Liberal Party while the Democrats are called Conservative Party.

Fudgin [204]

Answer:

Explanation:

False

5 0

3 years ago

Read 2 more answers

What is noble gases<br>

Marysya12 [62]

Answer:

helium, neon, argon,krypton, xenon, and radonoccupying Group 0 (18) of the periodic table. They were long believed to be totally unreactive but compounds of xenon, krypton, and radon are now known.

6 0

3 years ago

A particle with a mass of 0.500 kg is attached to a horizontal spring with a force constant of 50.0 N/m. At the moment t = 0, th

svp [43]

a) $x(t)=2.0 sin (10 t) [m]$

The equation which gives the position of a simple harmonic oscillator is:

$x(t)= A sin (\omega t)$

where

A is the amplitude

$\omega=\sqrt{\frac{k}{m}}$ is the angular frequency, with k being the spring constant and m the mass

t is the time

Let's start by calculating the angular frequency:

$\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{50.0 N/m}{0.500 kg}}=10 rad/s$

The amplitude, A, can be found from the maximum velocity of the spring:

$v_{max}=\omega A\\A=\frac{v_{max}}{\omega}=\frac{20.0 m/s}{10 rad/s}=2 m$

So, the equation of motion is

$x(t)= 2.0 sin (10 t) [m]$

b) t=0.10 s, t=0.52 s

The potential energy is given by:

$U(x)=\frac{1}{2}kx^2$

While the kinetic energy is given by:

$K=\frac{1}{2}mv^2$

The velocity as a function of time t is:

$v(t)=v_{max} cos(\omega t)$

The problem asks as the time t at which U=3K, so we have:

$\frac{1}{2}kx^2 = \frac{3}{2}mv^2\\kx^2 = 3mv^2\\k (A sin (\omega t))^2 = 3m (\omega A cos(\omega t))^2\\(tan(\omega t))^2=\frac{3m\omega^2}{k}$

However, $\frac{m}{k}=\frac{1}{\omega^2}$ , so we have

$(tan(\omega t))^2=\frac{3\omega^2}{\omega^2}=3\\tan(\omega t)=\pm \sqrt{3}\\$

with two solutions:

$\omega t= \frac{\pi}{3}\\t=\frac{\pi}{3\omega}=\frac{\pi}{3(10 rad/s)}=0.10 s$

$\omega t= \frac{5\pi}{3}\\t=\frac{5\pi}{3\omega}=\frac{5\pi}{3(10 rad/s)}=0.52 s$

c) 3 seconds.

When x=0, the equation of motion is:

$0=A sin (\omega t)$

so, t=0.

When x=1.00 m, the equation of motion is:

$1=A sin(\omega t)\\sin(\omega t)=\frac{1}{A}=\frac{1}{2}\\\omega t= 30\\t=\frac{30}{\omega}=\frac{30}{10 rad/s}=3 s$

So, the time needed is 3 seconds.

d) 0.097 m

The period of the oscillator in this problem is:

$T=\frac{2\pi}{\omega}=\frac{2\pi}{10 rad/s}=0.628 s$

The period of a pendulum is:

$T=2 \pi \sqrt{\frac{L}{g}}$

where L is the length of the pendulum. By using T=0.628 s, we find

$L=\frac{T^2g}{(2\pi)^2}=\frac{(0.628 s)^2(9.8 m/s^2)}{(2\pi)^2}=0.097 m$

5 0

3 years ago

Ryan is driving his car to band practice. His speed is 55 mph. Splat! A bug smashes against the windshield, and bug "guts" are e

MrRa [10]

Answer:

The answer is C.

Explanation:

Every point mass attracts every single other point mass by a force acting along the line intersecting both points. The force is proportional to the product of the two masses and inversely proportional to the square of the distance between them. In magnitude, the force they apply each other is the same. Therefore, the force that the windshield exerts on the bug and the force that the bug exerts on the windshield are the same magnitude.

3 0

3 years ago

Read 2 more answers