How do motion and Newton's laws apply to your everyday life?

A hollow spherical shell has mass 8.20 kg and radius 0.220 m. It is initially at rest and then rotates about a stationary axis t

Likurg_2 [28]

Answer:

8.91 J

Explanation:

mass, m = 8.20 kg

radius, r = 0.22 m

Moment of inertia of the shell, I = 2/3 mr^2

= 2/3 x 8.2 x 0.22 x 0.22 = 0.265 kgm^2

n = 6 revolutions

Angular displacement, θ = 6 x 2 x π = 37.68 rad

angular acceleration, α = 0.890 rad/s^2

initial angular velocity, ωo = 0 rad/s

Let the final angular velocity is ω.

Use third equation of motion

ω² = ωo² + 2αθ

ω² = 0 + 2 x 0.890 x 37.68

ω = 8.2 rad/s

Kinetic energy,

$K = \frac{1}{2}I\omega ^{2}$

K = 0.5 x 0.265 x 8.2 x 8.2

K = 8.91 J

6 0

2 years ago

A 5.00 kilogram block slides along a horizontal,frictionless surface at 10.0 meters per second. for 4.00 seconds. The magnitude

finlep [7]

In this question a lot of information's are provided. Among the information's provided one information and that is the time of 4 seconds is not required for calculating the answer. Only the other information's are required.
Mass of the block that is sliding = 5.00 kg
Distance for which the block slides = 10 meters/second
Then we already know that
Momentum = Mass * Distance travelled
= (5 * 10) Kg m/s
= 50 kg m/s
So the magnitude of the blocks momentum is 50 kg m/s. The correct option among all the given options is option "b".

5 0

2 years ago

Se lanza verticalmente una esfera con una rapidez de 30m/se. Determinar la rapidez de la esfera a una altura de 40m (g=10m/s2)

sergeinik [125]

$v^2-{v_0}^2=2a(x-x_0)$

dónde $v$ es la velocidad de la esfera, $v_0$ es suya velocidad inicial, $a=-g$ la aceleración debida a la gravedad, $x$ la posición, y $x_0$ la posición inicial. Tomamos $x_0=0\,\mathrm m$ a referirse a la posición de la esfera en el momento que la esfera fue lanzada.

Entonces

$v^2-\left(30\,\dfrac{\mathrm m}{\mathrm s}\right)^2=2\left(-10\,\dfrac{\mathrm m}{\mathrm s^2}\right)(40\,\mathrm m)$

$\implies v^2=100\,\dfrac{\mathrm m^2}{\mathrm s^2}\implies v=\pm10\,\dfrac{\mathrm m}{\mathrm s}$

Esto nos dice que la esfera alcanza una altura de 40 m en dos momentos - una vez hacia arriba y una vez hacia abajo. Sin embargo, independientemente del signo de la velocidad, sabemos que suya magnitud es 10 m/s, y así tenemos una rapidez de 10 m/s también en ambos momentos.

4 0

2 years ago

The more light bulbs you add to a series circuit, the _______ the lights will become.

Virty [35]

Answer:

brighter

Explanation:

the more light bulbs you add to a series of circuits, the brighter the room will be.

7 0

1 year ago

Read 2 more answers

What is the wavelength of a sound wave<br> with a speed of 331 m/s and a frequency<br> of 500 Hz?

ludmilkaskok [199]

Answer:

0.777m

Explanation:

The sound wave has a wavelength of 0.773m.

Explanation:

To solve this problem we have to use the wave equation that is given below:

We know the frequency and the velocity, both of which have good units. All we have to do is rearrange the equation and solve for

λ :

λ = v f

Let's plug in our given values and see what we get!

λ = 340 m s

440 s − 1

λ = 0.773 m

Hope this helps, Mark as brainliest if u want

4 0

2 years ago