2 A. What is the resistance of the iron?

3. A current of 0.2 A flows through an electric bell having a resistance of 65 ohms. What must be

the applied voltage in the circuit?

Answer:

**Answer:**

(1) 0.04 ohms (2) 55 ohms (3) 13 volt

**Explanation:**

(1) The resistance of an electric device is 40,000 microhms.

We need to convert it into ohms.

To covert 40,000 microhms to ohms, multiply 40,000 and 10⁻⁶ as follows :

(2) Voltage used, V = 110 V

Current, I = 2 A

We need to find the resistance of the iron. Using Ohms law to find it as follows :

V = IR, where R is resistance

(3) Current, I = 0.2 A

Resistance, R = 65 ohms

We need to find the applied voltage in the circuit. Using Ohms law to find it as follows :

V=IR

V = 0.2 × 65

**V = 13 volt**

Answer:

**Answer:**

1. 0.04 Ohms

2. 55 Ohms

3. 13 Volts

**Explanation:**

Penn Foster

A car is initially moving at 35 km/h along a straight highway. To pass another car, it speeds up to 135 km/h in 10.5 seconds at a constant acceleration.(a) how large was the acceleration in m/s ^2(b)how large was the acceleration, in units go g= 9.80 m/s ^2

The thickness of a $1 bill is 0.11 mm. If you have a stack of $1 bills 450 m tall, how much money do you have?

Consider a bird that flies at an average speed of 10.7 m/s and releases energy from its body fat reserves at an average rate of 3.70 W (this rate represents the power consumption of the bird). Assume that the bird consumes 4.00g of fat to fly over a distance db without stopping for feeding. How far will the bird fly before feeding again?Fat is a good form of energy storage because it provides the most energy per unit mass: 1.00 grams of fat provides about 9.40 (food) Calories, compared to 4.20 (food) Calories per 1.00 grams of carbohydrate. Remember that Calories associated with food, which are always capitalized, are not exactly the same as calories used in physics or chemistry, even though they have the same name. More specifically, one food Calorie is equal to 1000 calories of mechanical work or 4186 joules. Therefore, in this problem use the conversion factor 1Cal=4186J.

A string that passes over a pulley has a 0.341 kg mass attached to one end and a 0.625 kg mass attached to the other end. The pulley, which is a disk of radius 9.00 cm , has friction in its axle.What is the magnitude of the frictional torque that must be exerted by the axle if the system is to be in static equilibrium? (Answer should be in N m)

A playground merry-go-round of radius R = 2.20 m has a moment of inertia I = 275 kg · m2 and is rotating at 9.0 rev/min about a frictionless vertical axle. Facing the axle, a 23.0-kg child hops onto the merry-go-round and manages to sit down on the edge. What is the new angular speed of the merry-go-round? rev/min

The thickness of a $1 bill is 0.11 mm. If you have a stack of $1 bills 450 m tall, how much money do you have?

Consider a bird that flies at an average speed of 10.7 m/s and releases energy from its body fat reserves at an average rate of 3.70 W (this rate represents the power consumption of the bird). Assume that the bird consumes 4.00g of fat to fly over a distance db without stopping for feeding. How far will the bird fly before feeding again?Fat is a good form of energy storage because it provides the most energy per unit mass: 1.00 grams of fat provides about 9.40 (food) Calories, compared to 4.20 (food) Calories per 1.00 grams of carbohydrate. Remember that Calories associated with food, which are always capitalized, are not exactly the same as calories used in physics or chemistry, even though they have the same name. More specifically, one food Calorie is equal to 1000 calories of mechanical work or 4186 joules. Therefore, in this problem use the conversion factor 1Cal=4186J.

A string that passes over a pulley has a 0.341 kg mass attached to one end and a 0.625 kg mass attached to the other end. The pulley, which is a disk of radius 9.00 cm , has friction in its axle.What is the magnitude of the frictional torque that must be exerted by the axle if the system is to be in static equilibrium? (Answer should be in N m)

A playground merry-go-round of radius R = 2.20 m has a moment of inertia I = 275 kg · m2 and is rotating at 9.0 rev/min about a frictionless vertical axle. Facing the axle, a 23.0-kg child hops onto the merry-go-round and manages to sit down on the edge. What is the new angular speed of the merry-go-round? rev/min

The individual calcium atom has a positive and not negative, 2 charge

**Answer:**

The individual calcium atom has a positive, not negative, 2 charge.

**Explanation:**

Did the quiz also had it on the unit test on edgunity.

Hope this helps guys!

**Explanation:**

The given data is as follows.

Mass, m = 62 kg, Initial speed, = 6.90 m/s

Length of rough patch, L = 4.50 m, coefficient of friction, = 0.3

Height of inclined plane, h = 2.50 m

According to energy conservation equation,

(Final kinetic energy) + (Final potential energy) = Initial kinetic energy + Initial potential energy - work done by the friction

Since, final potential energy is equal to zero. Therefore, the equation will be as follows.

Cancelling the common terms in the above equation, we get

=

= 36.055 - 13.23

= 22.825

= 6.75 m/s

Thus, we can conclude that** the skier is moving at a speed of 6.75 m/s when she gets to the bottom of the hill.**

**Answer:**

**Explanation:**

mass, m = 62 kg

initial velocity, u = 6.9 m/s

length, l = 4.5 m

height, h = 2.5 m

coefficient of friction, μ = 0.3

Final kinetic energy + final potential energy = initial kinetic energy + initial potential energy + wok done by friction

Let the final velocity is v.

0.5 mv² + 0 = 0.5 mu² + μmgl + mgh

0.5 v² = 0.5 x 6.9 x 6.9 + 0.3 x 9.8 x 4.5 + 9.8 x 2.5

0.5 v² = 23.805 + 13.23 + 24.5

v² = 123.07

v = 11.1 m/s

Answer:

a) W = 25.872 J

b) - 35.28 J

c) - 9.408

Explanation:

a) The amount of work done by the force of gravity on the ball = Change in potential energy between the two vertical points = - mg (H₂ - H₁)

F = - mg (gravity is acting downwards)

F = - 0.6 × 9.8 = - 5.88 N

(H₂ - H₁) = (1.6 - 6) = - 4.4 m

W = (-5.88)(-4.4) = 25.872 J

b) Gravitational-potential energy of the ball when it was released relative to the ground = (- mg) H₁ = (- 0.6 × 9.8) × 6 = - 35.28 J

c) Gravitational-potential energy of the ball when it is caught relative to the ground = (-mg)(H₂) = -0.6 × 9.8 × 1.6 = - 9.408 J

**Answer:**

k = 15.62 MN/m

**Explanation:**

**Given:-**

**- **The viscous damping constant, **c = 1.8 KNs/m**

- The floor oscillation magnitude, **Yo = 3 mm**

- The frequency of floor oscillation, **f = 18 Hz.**

- The combined weight of the grinding machine and the wheel, **W = 4200 N**

- Two springs of identical stiffness k are attached in parallel arrangement.

**Constraints:-**

- The stiffness k > 3.25 MN/m

- The grinding machine’s steady-state amplitude of oscillation to at most 10 mm. ( Xo ≤ 10 mm )

**Find:-**

What is the minimum required stiffness of each of the two springs as per the constraints given.

**Solution:-**

- The floor experiences some harmonic excitation due to the unbalanced engine running in the vicinity of the grinding wheel. The amplitude "Yo" and the frequency "f" of the floor excitation is given

- The floor is excited with a harmonic displacement of the form:

Where,

Yo : The amplitude of excitation = **3 mm**

w : The excited frequency = 2*π*f = 2*π*18 = **36π**

- The harmonic excitation of the floor takes the form:

- The equation of motion for the floor excitation of mass-spring-damper system is given as follows:

Where,

m: The combined mass of the rigid body ( wheel + grinding wheel body) c : The viscous damping coefficient

k_eq: The equivalent spring stiffness of the system ( parallel )

x : The absolute motion of mass ( free vibration + excitation )

- We will use the following substitutions to determine the general form of the equation of motion:

Where,

w_n: The natural frequency

p = ζ = damping ratio = c / cc , damping constant/critical constant

- The Equation of motion becomes:

- The steady solution of a damped mass-spring system is assumed to be take the form of harmonic excitation of floor i.e:

Where,

X_o : The amplitude of the steady-state vibration.

α: The phase angle ( α )

- The steady state solution is independent from system's initial conditions and only depends on the system parameters and the base excitation conditions.

- The general amplitude ( X_o ) for a damped system is given by the relation:

Where,

r = Frequency ratio =

- We will use the one of the constraints given to limit the amplitude of steady state oscillation** ( Xo ≤ 10 mm )**:

- We will use the expression for steady state amplitude of oscillation ( Xo ) and determine a function of frequency ratio ( r ) and damping ratio ( ζ ):

- Solve the inequality ( quadratic ):

- The equivalent stiffness of the system is due to the parallel arrangement of the identical springs:

- Therefore,

- The minimum stiffness of spring is minimum of the two values:

**k = 15.62 MN/m**

**Answer:**

E₁ / E₂ = M / m

**Explanation:**

Let the electric field be E₁ and E₂ for ions and electrons respectively .

Force on ions = E₁ e where e is charge on ions .

Acceleration on ions a = E₁ e / M . Let initial velocity of both be u . Final velocity v = 0

v² = u² - 2as

0 = u² - 2 x E₁ e d / M

u² = 2 x E₁ e d / M

Similarly for electrons

u² = 2 x E₂ e d / m

Hence

2 x E₁ e d / M = 2 x E₂ e d / m

E₁ / E₂ = M / m

The ratio of the magnitude of the electric field the ions travel through to the magnitude of the electric field found in part (a) is M/m.

The ratio of the magnitude of the electric field the ions travel through to the magnitude of the electric field found in part (a) can be determined using the concept of mechanical energy conservation. Since the ions come to a stop, their initial kinetic energy must be equal to the work done by the electric field on them. The work done is given by the equation:

**Work = Change in kinetic energy**

The change in kinetic energy can be calculated using the formula:

**Change in kinetic energy = (1/2)Mv****2**** - (1/2)mv****2**

where M and m are the masses of the ions and electrons respectively, and v is their initial speed. Solving for the ratio, we get:

**Ratio = (1/2)M/(1/2)m = M/m**

So, the ratio of the magnitude of electric field the ions travel through to the magnitude of the electric field found in part (a) is **M/m**.

#SPJ3

**Answer:**

Speed of the wind is 48.989 mph

**Explanation:**

We have given each trip is of 200 miles

So total distance = 200 +200 = 400 miles

Speed of the airplane = 120 mph

Let the speed of the wind = x mph

So the speed of the airplane with wind = 120+x

So time taken by airplane with wind =

Speed of the airplane against the wind = 120 - x

So time taken by the airplane against the wind

Total time is given as t= 4 hour

So

x = 48.989 mph

**Answer:**

**Explanation:**

Type Distance Rate Time

Headwind 200 120-r 200/120-r

Tailwind 200 120 - r 200/120 - r

We know the times add to 4, so we write the equation:

200/120−r + 200/120 + r = 4

We multiply both sides by the LCD and simplify to get:

(120−r)(120+r) ((200/120 -r ) + 200/120+r) = 4(120 -r) (120 +r)

200(120−r)+200(120+r)=4(120−r)(120+r)

Factor the 200 and simplify inside the parentheses to find:

200(120−r+120+r)=4(1202−r2)

200(240)=4(1202−r2)

200(60)=120^2−r^2

12,000=14,400−r^2

−2,400= −r^2

49 ≈ r

The speed of the wind is 49mph.