PLEASE HELP!! THANKS!! WILL MARK BRAINLIEST!!

A large grinding wheel in the shape of a solid cylinder of radius 0.330 m is free to rotate on a frictionless, vertical axle. A

Flura [38]

Answer:

The mass of the wheel is 2159.045 kg

Explanation:

Given:

Radius $r = 0.330$

m

Force $F = 290$ N

Angular acceleration $\alpha = 0.814 \frac{rad}{s^{2} }$

From the formula of torque,

Γ $= I\alpha$ (1)

Γ $= rF$ (2)

$rF = I \alpha$

Find momentum of inertia $I$ from above equation,

$I = \frac{rF}{\alpha }$

$I = \frac{0.330 \times 290}{0.814}$

$I = 117.56$ $Kg. m^{2}$

Find the momentum inertia of disk,

$I = \frac{1}{2} Mr^{2}$

$M = \frac{2I}{r^{2} }$

$M = \frac{2 \times 117.56}{(0.330)^{2} }$

$M = 2159.045$ Kg

Therefore, the mass of the wheel is 2159.045 kg

8 0

3 years ago

1. A Faraday ice pail similar to the one you will use in the experiment is used in this question. The principle of its operation

algol [13]

Answer:

your mommy

Explanation:

YOUR MAMMA

I LOVE MY DADDY

HE DOES ME REAL HARD

8 0

3 years ago

A chef places an open sack of flour on a kitchen scale. The scale reading of

Novay_Z [31]

Answer:

Explanation:

Given

Initial reading on scale =40 N

So, we can conclude that weight of the sack is 40 N

After this a 10 N force is applied upward on the sack such that the net force becomes (40-10) N downward (because downward force is more)

This net downward force is the resultant of earth graviational pull and the applied upward force.

So, this downward force acts on the machine which inturn applies an upaward force of same magnitude called Normal reaction.

This situation can be diagramatically represented by figure given below

4 0

3 years ago

A fisherman notices that his boat is moving up and down periodically, owing to waves on the surface of the water. It takes 2.5 s

gulaghasi [49]

(a) 0.96 m/s

The period of the wave corresponds to the time taken for one complete oscillation of the boat, from the highest point to the highest point again. Since the time between the highest point and the lowest point is 2.5 s, the period is twice this time:

$T=2\cdot 2.5 s=5.0 s$

The frequency of the waves is the reciprocal of the period:

$f=\frac{1}{T}=\frac{1}{5.0 s}=0.20 Hz$

The wavelength instead is just the distance between two consecutive crests, so

$\lambda=4.8 m$

And the wave speed is given by:

$v=\lambda f=(4.8 m)(0.20 Hz)=0.96 m/s$

(b) 0.265 m

The total distance between the highest point of the wave and its lowest point is

d = 0.53 m

The amplitude is just the maximum displacement of the wave from the equilibrium position, so it is equal to half of this distance. So, the amplitude is

$A=\frac{d}{2}=\frac{0.53 m}{2}=0.265 m$

(c) Amplitude: 0.15 m, wave speed: same as before

In this case, the amplitude of the wave would be lower. In fact,

d = 0.30 m

So the amplitude would be

$A=\frac{d}{2}=\frac{0.30 m}{2}=0.15 m$

Instead, the wave speed would not change, since neither the frequency nor the wavelength of the wave have changed.

8 0

3 years ago

An interference pattern is produced by light with a wavelength 520 nm from a distant source incident on two identical parallel s

Goshia [24]

Answer:

1) θ = 0.00118 rad, 2) θ = 0.00236 rad , 3) I / I₀ = 0.1738, 4) I / Io = 0.216

Explanation:

In the double-slit interference phenomenon it is explained for constructive interference by the equation

d sin θ = m λ

1) the first order maximum occurs for m = 1

sin θ = λ / d

θ = sin⁻¹ λ / d

let's reduce the magnitudes to the SI system

λ = 520 nm = 520 10⁻⁹ θ = 0.00118 radm

d = 0.440 mm = 0.440 10⁻³ m ³

let's calculate

θ = sin⁻¹ (520 10⁻⁹ / 0.44 10⁻³)

θ = sin⁻¹ (1.18 10⁻³)

θ = 0.00118 rad

2) the second order maximum occurs for m = 2

θ = sin⁻¹ (m λ / d)

θ = sin⁻¹ (2 5¹20 10⁻⁹ / 0.44 10⁻³)

θ = 0.00236 rad

3) To calculate the intensity of the interference spectrum, the diffraction phenomenon must be included, so the equation remains

I = I₀ cos² (π d sin θ /λ ) sinc² (pi b sin θ /λ )

where the function sinc = sin x / x

and b is the width of the slits

we caption the values

x = π 0.310 10⁻³ sin 0.00118 / 520 10⁻⁹)

x = 2.21

I / I₀ = cos² (π 0.44 10⁻³ sin 0.00118 / 520 10⁻⁹) (sin (2.21) /2.21)²

remember angles are in radians

I / I₀ = cos² (3.0945) [0.363] 2

I / I₀ = 0.9978 0.1318

I / I₀ = 0.1738

4) the maximum second intensity is

I / I₀ = cos² (π d sinθ / λ) sinc² (πb sin θ /λ)

x =π 0.310 10⁻³ sin 0.00236 / 520 10⁻⁹)

x = 4.41

I / Io = cos² (π 0.44 10⁻³ sin 0.00236 / 520 10⁻⁹) (sin 4.41 / 4.41)²

I / Io = cos² 6.273 0.216

I / Io = 0.216

.

7 0

2 years ago