Plz help it’s passed due!

5)

Zina [86]

Answer:

The current through the wire is equal to 0.8 A.

Explanation:

Given that,

The length of a copper wire = 2 m

Potential difference = 24 mV

The current through the wire is 0.40 A.

The new potential difference is 48 mV.

We need to find the current through the wire.

As the potential difference is doubled for second wire. So the new current will be :

I' = 2I

= 2 × 0.4

= 0.80 A

So, the current through the wire is equal to 0.8 A.

5 0

3 years ago

023 (part 1 of 2) 10.0 points

Annette [7]

Answer:

Part 1

The angular speed is approximately 1.31947 rad/s

Part 2

The change in kinetic energy due to the movement is approximately 675.65 J

Explanation:

The given parameters are;

The rotation rate of the merry-go-round, n = 0.21 rev/s

The mass of the man on the merry-go-round = 99 kg

The distance of the point the man stands from the axis of rotation = 2.8 m

Part 1

The angular speed, ω = 2·π·n = 2·π × 0.21 rev/s ≈ 1.31947 rad/s

The angular speed is constant through out the axis of rotation

Therefore, when the man walks to a point 0 m from the center, the angular speed ≈ 1.31947 rad/s

Part 2

Given that the kinetic energy of the merry-go-round is constant, the change in kinetic energy, for a change from a radius of of the man from 2.8 m to 0 m, is given as follows;

$\Delta KE_{rotational} = \dfrac{1}{2} \cdot I \cdot \omega ^2 = \dfrac{1}{2} \cdot m \cdot v ^2$

I = m·r²

Where;

m = The mass of the man alone = 99 kg

r = The distance of the point the man stands from the axis, r = 2.8 m

v = The tangential velocity = ω/r

ω ≈ 1.31947 rad/s

Therefore, we have;

I = 99 × 2.8² = 776.16 kg·m²

$\Delta KE_{rotational}$ = 1/2 × 776.16 kg·m² × (1.31947)² ≈ 675.65 J

3 0

3 years ago

During the time a compact disc (CD) accelerates from rest to a constant rotational speed of 477 rev/min, it rotates through an a

atroni [7]

Answer:

126.01 rad/s^2

Explanation:

since it starts from rest, initial angular speed ω' = 0 rad/s

angular speed N = 477 rev/min

angular speed in rad/s ω = $\frac{2\pi N}{60}$ = $\frac{2*3.142* 477}{60}$ = 49.95 rad/s

angular displacement ∅ = 1.5758 rev

angular displacement in rad/s = $2\pi N$ = 2 x 3.142 x 1.5758 = 9.9 rad

angular acceleration $\alpha$ = ?

using the equation of angular motion

ω^2 = ω'^2 + 2 $\alpha$ ∅

imputing values, we have

$49.95^{2} = 0^{2} + (2 *\alpha*9.9 )$

2495 = 19.8 $\alpha$

$\alpha$ = 2495/19.8 = 126.01 rad/s^2

8 0

3 years ago

The cabinet is mounted on coasters and has a mass of 45 kg. The casters are locked to prevent the tires from rotating. The coeff

stira [4]

Answer:

the force P required for impending motion is 132.3 N

the largest value of "h" allowed if the cabinet is not to tip over is 0.8 m

Explanation:

Given that:

mass of the cabinet m = 45 kg

coefficient of static friction μ = 0.30

A free flow body diagram illustrating what the question represents is attached in the file below;

The given condition from the question let us realize that ; the casters are locked to prevent the tires from rotating.

Thus; considering the forces along the vertical axis ; we have :

$\sum f_y =0$

The upward force and the downward force is :

$N_A+N_B = mg$

where;

$\mathbf { N_A \ and \ N_B}$ are the normal contact force at center point A and B respectively .

$N_A+N_B = 45*9.8$

$N_A+N_B = 441$ ------- equation (1)

Considering the forces on the horizontal axis:

$\sum f_x = 0$

$F_A +F_B = P$

where ;

$\mathbf{ F_A \ and \ F_B }$ are the static friction at center point A and B respectively.

which can be written also as:

$\mu_s N_A + \mu_s N_B = P$

$\mu_s( N_A + N_B) = P$

replacing our value from equation (1)

P = 0.30 ( 441)

P = 132.3 N

Thus; the force P required for impending motion is 132.3 N

b) Since the horizontal distance between the casters A and B is 480 mm; Then half the distance = 480 mm/2 = 240 mm = 0.24 cm

the largest value of "h" allowed for the cabinet is not to tip over is calculated by determining the limiting condition of the unbalanced torque whose effect is canceled by the normal reaction at $N_A$ and it is shifted to $N_B$ :