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A spinning wheel on a fireworks display is initially rotating in a counterclockwise direction. The wheel has an angular acceleration of -4.46 rad/s2. Because of this acceleration, the angular velocity of the wheel changes from its initial value to a final value of -31.4 rad/s. While this change occurs, the angular displacement of the wheel is zero. (Note the similarity to that of a ball being thrown vertically upward, coming to a momentary halt, and then falling downward to its initial position.) Find the time required for the change in the angular velocity to occur.

Answers

Answer 1

Answer:

Answer:

The time for the change in the angular velocity to occur is 14.08 secs

Explanation:

From the question,

the angular acceleration is - 4.46 rad/s²

Angular acceleration is given by the formula below

$\alpha =(\omega -\omega _(o) )/(t - t_(o) )$

Where $\alpha$ is the angular acceleration

$\omega$ is the final angular velocity

$\omega _(o)$ is the initial angular velocity

$t$ is the final time

$t_(o)$ is the initial time

From the question

$\alpha$ = - 4.46 rad/s²

$\omega _(o)$ = 0 rad/s (starting from rest)

$\omega$ = -31.4 rad/s

$t_(o)$ = 0 s

Now, we will determine t

From $\alpha =(\omega -\omega _(o) )/(t - t_(o) )$ , then

$-4.46 = (-31.4 - 0)/(t - 0)$

$-4.46 = (-31.4)/(t)$

$t = (-31.4)/(-4.46)$

t = 7.04 secs

This is the time spent in one direction,

Since the angular displacement of the wheel is zero ( it returned to its initial position), then the time required for the change in the angular velocity will be twice this time, that is 2t

Hence,

The time is 2×7.04 secs = 14.08 secs

This is the time for the change in the angular velocity to occur.

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Find a glass jar with a screw-top metal lid. Close the lid snugly and put the jar into the refrigerator. Leave it there for about 10 minutes and then take the jar out and try to open the lid. (a) Did the lid become tighter or looser? Explain your observation.

Answers

Answer:

The lid becomes tighter

It becomes tighter because metals have a lower heat capacity than glass meaning their temperature drops (or increases) much faster than glass for the same energy change. So in this example, the metal will contract faster than the glass causing it to be more tighter around the glass.

. Using your knowledge of circular (centripetal) motion, derive an equation for the radius r of the circular path that electrons follow in terms of the magnetic field B, the electrons' velocity v, charge e, and mass m. You may assume that the electrons move at right angles to the magnetic field.2. Recall from electrostatics, that an electron obtains kinetic energy when accelerated across a potential difference V. Since we can directly measure the accelerating voltage V in this expierment, but not the electrons' velocity v, replace velocity in your previous equation with an expression containing voltage. The electron starts at rest. Now solve this equation for e/m.

You should obtain e/m = 2V/(B^2)(r^2)

3. The magnetic field on the axis of a circular current loop a distance z away is given by

B = mu I R^2 / 2(R^2 + z^2)^ (3/2)

where R is the radius of the loops and I is the current. Using this result , calculate the magnetic field at the midpoint along the axis between the centers of the two current loops that make up the Helmholtz coils, in terms of their number of turns N, current I, and raidus R.Helmholtz coils are separated by a distance equal to their raidus R. You should obtain:

|B| = (4/5)^(3/2) *mu *NI/R = 9.0 x 10^-7 NI/R

where B is magnetic field in tesla, I is in current in amps, N is number of turns in each coil, and R is the radius of the coils in meters

Answers

Answer:

Explanation:

Magnetic field creates a force perpendicular to a moving charge in its field which is equal to Bev where B is magnetic field , e is amount of charge on the moving charge and v is the velocity of charge particle .

This force provides centripetal force for creation of circular motion. If r be the radius of the circular path

Bev = mv² / r

r = mv / Be

2 ) If an electron is accelerated by an electric field created by potential difference V then electric field

= V / d where d is distance between two points having potential difference v .

force on charged particle

electric field x charge

= V /d x e

work done by field

= force x distance

= V /d x e x d

V e

This is equal to kinetic energy created

V e = 1/2 mv²

= 1/2 m (r²B²e² / m² )

V = r²B²e/ 2 m

e / m = 2 V/ r²B²

3 )

B = $(\mu* I* R^2)/(2(R^2+Z^2)^(3)/(2) )$

In Helmholtz coils , distance between coil is equal to R so Z = R/2

B = $(\mu* I* R^2)/(2(R^2+(R^2)/(4) )^(3)/(2) )$

For N turns of coil and total field due to two coils

B = $(\mu* I* N)/(R*((5)/(4))^(3)/(2) )$

= $(\mu* I* N)/(R)* ((4)/(5))^(3)/(2)$

= 9.0 x 10^-7 NI/R

What problem did Katherine face in checking other people’s calculations in the movie ¨Hidden figures¨

Answers

Answer:

Information from the leading mathematicians was considered "classified".

A pitcher is in 85° of abduction, holding a 1.4 N baseball at point C, 65 cm from the joint axis at point O • The center of gravity of his arm is 25 cm from the joint axis of shoulder abduction at point O • The weight of the arm W is 0.06 of the pitcher’s weight of 100 N • Deltoids muscles are at an angle θ of 15° with respect to the humerus and insert 15 cm from the joint axis at point A • Determine the force applied by the Deltoid muscles and the joint reaction force at the shoulder joint and its orientation β

Answers

I attached a Diagram for this problem.

We star considering the system is in equlibrium, so

Fm makes $90-(\theta+5)$ with vertical

Fm makes 70 with vertical

Applying summatory in X we have,

$\sum F_x = 0$

$W+1.4-Fm cos(70)$

We know that W is equal to

$W= 0.06*100N = 6N$

Substituting,

$Fm cos (70) = W+1.4N$

$Fm cos (70) = 6N + 1.4N$

$Fm = (7.4)/(cos(70))$

$Fm = 21.636N$

For the second part we know that the reaction force Fj on deltoid Muscle is equal to Fm, We can assume also that $\beta = \theta$

Suppose that the speedometer of a truck is set to read the linear speed of the truck, but uses a device that actually measures the angular speed of the tires. If larger-diameter tires are mounted on the truck, will the reading on the speedometer be correct? If not, will the reading be greater than or less than the true linear speed of the truck? Why?

Answers

The measurement will be significantly affected.

Recall that the relationship between linear velocity and angular velocity is subject to the formula

$v = \omega r$ ,

Where r indicates the radius and $\omega$ the angular velocity.

As the radius increases, it is possible that the calibration is delayed and a higher linear velocity is indicated, that to the extent that the velocity is directly proportional to the radius of the tires.

Two resistors are to be combined in parallelto form an equivalent resistance of 400Ω. The resistors are takenfrom available stock on hand as acquired over the years. Readily available are two common resistorsrated at 500±50 Ωand two common resistors rated at 2000 Ω±5%. What isthe uncertainty in an equivalent 400 Ωresistance?(Hint: the equivalent resistance connected in parallel can be obtained by 1212TRRRRR=+)

Answers

Answer:

Δ $R_(e)$ = 84 Ω, $R_(e)$ = (40 ± 8) 10¹ Ω

Explanation:

The formula for parallel equivalent resistance is

1 / $R_(e)$ = ∑ 1 / Ri

In our case we use a resistance of each

R₁ = 500 ± 50 Ω

R₂ = 2000 ± 5%

This percentage equals

0.05 = ΔR₂ / R₂

ΔR₂ = 0.05 R₂

ΔR₂ = 0.05 2000 = 100 Ω

We write the resistance

R₂ = 2000 ± 100 Ω

We apply the initial formula

1 / $R_(e)$ = 1 / R₁ + 1 / R₂

1 / $R_(e)$ = 1/500 + 1/2000 = 0.0025

$R_(e)$ = 400 Ω

Let's look for the error (uncertainly) of Re

$R_(e)$ = R₁R₂ / (R₁ + R₂)

R’= R₁ + R₂

$R_(e)$ = R₁R₂ / R’

Let's look for the uncertainty of this equation

Δ $R_(e)$ / $R_(e)$ = ΔR₁ / R₁ + ΔR₂ / R₂ + ΔR’/ R’

The uncertainty of a sum is

ΔR’= ΔR₁ + ΔR₂

We substitute the values

Δ $R_(e)$ / 400 = 50/500 + 100/2000 + (50 +100) / (500 + 2000)

Δ $R_(e)$ / 400 = 0.1 + 0.05 + 0.06

Δ $R_(e)$ = 0.21 400

Δ $R_(e)$ = 84 Ω

Let's write the resistance value with the correct significant figures

$R_(e)$ = (40 ± 8) 10¹ Ω

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