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4 years ago

What is the correct current value if the selector is in the 0.5 dcma position?

1 answer:

8 0

1. The Red Meter lead/Probe is Positive 2. The Black Meter lead/Probe is Negative 3. There are three meter scales you are working with. Not only this but, the 10 range is used for 10V and the 1000 Volt selector switch positions the 50 range is used for 0.5V and the 50 Volt selector switch positions the 250 range is used for 2.5V and the 250 Volt selector switch positions this is the position you start with.

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A 32.0 kg wheel, essentially a thin hoop with radius 1.20 m, is rotating at 280 rev/min. It must be brought to a stop in 15.0 s.

elena-14-01-66 [18.8K]

Answer:

A) Must be done 19806.62 joules of work.

B) The average power is 1320.44 Watts.

Explanation:

A) First, we're going to use the work-energy theorem that states total work ( $W$ ) done on an object is equal to the change in its kinetic energy ( $\Delta K$ ):

$W=\Delta K = K_{f}-K_{i}$ (1)

So, all we must do is to find the change on kinetic energy. Because we're working with rotational body, we should use the equation $K=\frac{I\omega^{2}}{2}$ for the kinetic energy so:

$\Delta K=\frac{I(\omega_{f})^{2}}{2}-\frac{I(\omega_{i})^{2}}{2}$ (2)

with $\omega_{i}$ the initial angular velocity, $\omega_{f}$ the final angular velocity (is zero because the wheel stops) and I the moment of inertia that for a thin hoop is $I=MR^{2}$ , using those on (2)

$\Delta K=0-\frac{MR^{2}(\omega_{i})^{2}}{2}$ (3)

By (3) on (1):

$W= \frac{MR^{2}(\omega_{i})^{2}}{2} = \frac{(32.0)(1.2)^{2}(29.32)^{2}}{2}$

$W=19806.62\,J$

B) Average power is work done divided by the time interval:

$P=\frac{W}{\Delta t}=\frac{19806.62}{15.0}$

$P=1320.44\,W$

NOTE: We use the relation $1rpm*\frac{2\pi}{60s}=\frac{rad}{s}$ to convert 280 rev/min(rpm) to 29.32 rad/s

4 0

4 years ago

How do you add scientific notation if the problem is like this: 3.72 x 10^9 + 5.46 x 10^8?

Crazy boy [7]

You can only add or subtract numbers in scientific notation if they have
the same power of 10. If they're different, then you have to change one
to match the other one.

Here are both ways to do your example:

3.72x 10^9 = 37.2x10^8

(37.2 x 10^8)+(5.46 x 10^8) = (37.2+5.46) x 10^8 = 42.66x10^8 = 4.266x10^9

=======================

5.46 x 10^8 = 0.546 x 10^9

3.72 x 10^9 + 0.546 x 10^9 = (3.72 + 0.546) x 10^9 = 4.266 x 10^9

3 0

4 years ago

An ammeter is connected in series with a resistor of unknown resistance R and a DC power supply of known emf e. A student uses t

rosijanka [135]

Answer:

B:The actual power dissipated by the resistor is less than P because the ammeter had some resistance.

Explanation:

Here,power has been calculated using current I and total EMF \ε . So,P=EMF*current= ε I will represent total power dissipated in resistor and ammeter.

Now, this total power P has been dissipated in both resistor and ammeter. So, power dissipated in resistor must be less than P as some power is also dissipated in ammeter because it has non-zero resistance.

So, the answer is B:The actual power dissipated by the resistor is less than P because the ammeter had some resistance.

Note that option A,C and E are ruled out as they state power dissipated by resistor is greater than or equal to P which is false.

Also,option D is ruled out as ammeter is connected in series.

8 0

4 years ago

The inclined plane in the figure above has two sections of equal length and different roughness. The dashed line shows where sec

saveliy_v [14]

The static friction exerted on the block by the incline is $\mu _ s _1 Mgcos \ \theta$ .

The given parameters;

mass of the block, = M
coefficient of static friction in section 1, = $\mu_s_1$
angle of inclination of the plane, = θ

The normal force on the block is calculated as follows;

Fₙ = Mgcosθ

The static friction exerted on the block by the incline is calculated as follows;

$F_s = \mu_s F_n\\\\F_s = \mu _s_1(Mg cos\ \theta)\\\\F_s = \mu _s_1 Mgcos\ \theta$

Thus, the static friction exerted on the block by the incline is $\mu _ s _1 Mgcos \ \theta$

Learn more here:brainly.com/question/17237604

3 0

3 years ago

g Two long parallel wires are a center-to-center distance of 2.50 cm apart and carry equal anti-parallel currents of 2.70 A. Fin

schepotkina [342]

Answer:

864 mT

Explanation:

The magnetic field due to a long straight wire B = μ₀i/2πR where μ₀ = permeability of free space = 4π × 10⁻⁷ H/m, i = current in wire, and R = distance from center of wire to point of magnetic field.

The magnitude of magnetic field due to the first wire carrying current i = 2.70 A at distance R which is mid-point between the wires is B = μ₀i/2πR.

Since the other wire also carries the same current at distance R, the magnitude of the magnetic field is B = μ₀i/2πR.

The resultant magnetic field at B is B' = B + B = 2B = 2(μ₀i/2πR) = μ₀i/πR

Now R = 2.50 cm/2 = 1.25 cm = 1.25 × 10⁻² m and i = 2.70 A.

Substituting these into B' = μ₀i/πR, we have

B' = 4π × 10⁻⁷ H/m × 2.70 A/π(1.25 × 10⁻² m)

B = 10.8/1.25 × 10⁻⁵ T

B = 8.64 × 10⁻⁵ T

B = 864 × 10⁻³ T

B = 864 mT

4 0

3 years ago