Calculate the inhomogeneity of a 1.5 t magnet

A uniform disk, a thin hoop, and a uniform sphere, all with the same mass and same outer radius, are each free to rotate about a

olga nikolaevna [1]

Answer:

4 hoop, disk, sphere

Explanation:

Because

We are given data that

Hoop, disk, sphere have Same mass and radius

So let

And Initial angular velocity, = 0

The Force on each be F

And Time = t

Also let

Radius of each = r

So let's find the inertia shall we!!

I1 = m r² /2

= 0.5 mr² the his is for dis

I2 = m r² for hoop

And

Moment of inertia of sphere wiil be

I3 = (2/5) mr²

= 0.4 mr²

So

ωf = ωi + α t

= 0 + ( τ / I ) t

= ( F r / I ) t

So we can see that

ωf is inversely proportional to moment of inertia.

And so we take the

Order of I ( least to greatest ) :

I3 (sphere) , I1 (disk) , I2 (hoop) , ,

Order of ωf: ( least to greatest)

That of omega xf is the reverse of inertial so

hoop, disk, sphere

Option - 4

5 0

3 years ago

Given a force of 100 N and a acceleration of 5 m/s, what is the mass

SIZIF [17.4K]

Answer:

$force = mass \times acceleration \\ 100 = m \times 5 \\ m = \frac{100}{5} \\ m = 20 \: kg$

8 0

2 years ago

Read 2 more answers

Optical tweezers use light from a laser to move single atoms and molecules around. Suppose the intensity of light from the tweez

katrin [286]

Answer:

a= 4.4×10 m/s^2

Explanation:

pressure P = E/c

Where, E = 100 W/m^2 intensity of light

c= speed of light = 3×10^8 m/s

P = 1000/ 3×10^8

P = 3.33×10^(-6) Pa

Force F = P×A

P is the pressure and c= speed of light

F = 3.33×10^{-6}×6.65×10(-29)

= 2.22×10^{-6}

acceleration a = F/m = 2.22×10^{-6}/ 5.10×10^{-27}

a= 4.4×10 m/s^2

4 0

3 years ago

If the primary coil of a transformer has 200 turns and is supplied with 120 v ac power, how many turns must the secondary coil h

Oduvanchick [21]

The ratio of the turns to the voltage should be equal
i.e: 200/120 = t/12
so the secondary coil should have 20 turns

5 0

3 years ago

A car's bumper is designed to withstand a 5.04 km/h (1.4-m/s) collision with an immovable object without damage to the body of t

emmasim [6.3K]

Answer:

the magnitude of the average force on the bumper is 3189.8 N

Explanation:

Given the data in the question;

In terms of force and displacement, work done is;

W = $F^>$ × $x^>$

W = $Fxcos\theta$ ------- let this be equation 1

where F is force applied, x is displacement and θ is angle between force and displacement.

Now, since the displacement of the bumper and force acting on it is in the same direction,

hence, θ = 0°

we substitute into equation 1

W = $Fxcos($ 0° $)$

W = $Fx$ ------- let this be equation 2

Now, using work energy theorem,

total work done on the system is equal to the change in kinetic energy of the system.

$W_{net$ = ΔKE

= $\frac{1}{2}$ mv² - $\frac{1}{2}$ mu² --------- let this be equation 3

where m is mass of object, v is final velocity, u is initial velocity.

from equation 2 and 3

$Fx$ = $\frac{1}{2}$ mv² - $\frac{1}{2}$ mu²

we make F, the subject of formula

F = $\frac{m}{2x}$ ( v² - u² )

given that mass of car m = 830 kg, x = 0.255 m, v = 0 m/s, and u = 1.4 m/s

so we substitute

F = $\frac{830}{(2)(0.255)}$ ( (0)² - (1.4)² )

F = 1627.45098 ( 0 - 1.96 )

F = 1627.45098 ( - 1.96 )

F = -3189.8 N

The negative sign indicates that the direction of the force was in opposite compare to the direction of the velocity of the car.

Therefore, the magnitude of the average force on the bumper is 3189.8 N

6 0

2 years ago