Diamonds have a density of 3.5 g/cm3. How big is a diamond that has a mass of 0.10

The rotor in a certain electric motor is a flat, rectangular coil with 84 turns of wire and dimensions 2.61 cm by 3.64 cm. The r

ira [324]

Given Information:

Number of turns = N = 84

Area of Rectangular coil = 2.61x3.64 cm = 0.0261x0.0364 m

Magnetic field = B = 0.80 T

Current = I = 10.5 mA = 0.0105 A

Angular speed = ω = 3.54x10³ rev/min

Required Information:

(a) Maximum torque = τmax = ?

(b) Peak output power = Ppeak = ?

(c) Work done = W = ?

(d) Average power = Pavg?

Answer:

(a) Maximum torque = 0.00067 N.m

(b) Peak output power = 0.248 W

(c) Work done = 0.00189 J

(d) Average power = 0.1115 W

Explanation:

(a) The toque τ acting on the rotor is given by,

τ = NIABsin(θ)

Where N is the number of turns, I is the current, A is the area of rectangular coil and B is the magnetic field

A = 0.0261x0.0364

A = 0.00095 m²

The maximum toque τ is achieved when θ = 90°

τmax = NIABsin(90°)

τmax = 84*0.0105*0.00095*0.80*1

τmax = 0.00067 N.m

(b) The peak output power of the motor is given by,

Pmax = τmax*ω

ω = 3.54x10³ x 2π/60

ω = 370.7 rad/sec

Pmax = 0.00067*370.7

Pmax = 0.248 W

(c) The amount of work done by the magnetic field on the rotor in every full revolution is given by

W = 2∫NIABωsin(ωt) dt

W = -2NIABcos(ωt)

Evaluating limits,

W = -2NIABcos(π) - (-2NIABcos(0))

W = 2NIAB + 2NIAB

W = 4NIAB

W = 4*84*0.0105*0.00067*0.80

W = 0.00189 J

(d) Average power of the motor is given by

Pavg = W/t

t = 2π/ω

t = 2π/370.7

t = 0.01694 sec

Pavg = W/t

Pavg = 0.00189/0.01694

Pavg = 0.1115 W

4 0

3 years ago

A 500 kg dragster accelerates from rest to a final speed of 100 m/s in 400 m (about a quarter of a mile) and encounters an avera

Fantom [35]

In order to accelerate the dragster at a speed $v_f = 100 m/s$ , its engine must do a work equal to the increase in kinetic energy of the dragster. Since it starts from rest, the initial kinetic energy is zero, so the work done by the engine to accelerate the dragster to 100 m/s is
$W= K_f - K_i = K_f = \frac{1}{2}mv_f^2=2.5 \cdot 10^6 J$

however, we must take into account also the fact that there is a frictional force doing work against the dragster, and the work done by the frictional force is:
$W_f = F_f d = -(1200 N)(400 m)= -4.8 \cdot 10^5 J$
and the sign is negative because the frictional force acts against the direction of motion of the dragster.

This means that the total work done by the dragster engine is equal to the work done to accelerate the dragster plus the energy lost because of the frictional force, which is $-W_f$ :
$W_t = W + (-W_f)=2.5 \cdot 10^6 J+4.8 \cdot 10^5 J=2.98 \cdot 10^6 J$

So, the power delivered by the engine is the total work divided by the time, t=7.30 s:
$P= \frac{W}{t}= \frac{2.98 \cdot 10^6 J}{7.30 s}=4.08 \cdot 10^6 W$

And since 1 horsepower is equal to 746 W, we can rewrite the power as
$P=4.08 \cdot 10^6 W \cdot \frac{1 hp}{746 W} =547 hp$

3 0

4 years ago

Read 2 more answers

A girl on a spinning amusements park is 12m from the center of the ride and has a centripetal acceleration of 17 m/s^2. What is

Tanya [424]

Answer: 14.28 m/s

Explanation:

Assuming the girl is spinning with uniform circular motion, her centripetal acceleration $a_{c}$ is given by the following equation:

$a_{c}=\frac{V^{2}}{r}$ (1)

Where:

$a_{c}=17 m/s^{2}$ is the centripetal acceleration

$V$ is the tangential speed

$r=12 m$ is the radius of the circle

Isolating $V$ from (1):

$V=\sqrt{a_{c}r}$ (2)

$V=\sqrt{(17 m/s^{2})(12 m)}$

Finally:

$V=14.28 m/s$ This is the girl's tangential speed

3 0

3 years ago

How does the surface tension of water compared with the surface tension of most other liquids?

Anuta_ua [19.1K]

Surface tension is the direct measure of the cohesive forces between liquid molecules that allows them to form "film", making it more difficult to move an object through the liquid surface. Compared to other known liqiuds, the surface tension of water is much greater.

5 0

3 years ago

This chart lists four examples of two objects that are in contact.

Oxana [17]

Before coming into conclusion first we have to understand the direction of heat flow.

Heat is the transferred thermal energy from one body to another body due to the temperature difference just like water flows from higher level to lower level.

Whenever two bodies having different temperature come closer to each other heat will flow from hotter body to cooler one if no external work is done. The heat flow may be through any of the ways i.e conduction,radiation or convection. Hence temperature difference is the parameter which gives the direction of heat flow.

The temperature is also considered as a measure of average kinetic energy of the substance.The thermal energy does not give the direction heat flow. Heat may flow from the body having low thermal energy but at higher temperature to the body having higher thermal energy but at low temperature. The reverse does not happen naturally .

In example 1 there is fire and air. Obviously fire is at high temperature and air at low temperature.So heat will flow from object 1 to object 2.

In example 2 there is a metal at 80 degree Celsius and another metal at 12 degree Celsius .So heat will flow from object 1 to object 2

In example 3 we have cooler ocean and warm air. So the heat will flow from object 2 to object 1.

In example 4 we have a tool with high thermal energy and a material with little thermal energy.We already know that thermal energy can not determine the direction of heat flow. Here the temperature of each substance is not given.The kinetic energy is part of thermal energy.So there is the chance of higher kinetic energy of the tool for having higher thermal energy .At that time the heat will flow object 1 to object 2.Otherwise the reverse will occur. So it is a special case.

As per the question only option 4 is correct which tells that heat will flow from object 1 to object 2 in examples 1,2,4, and heat will flow from object 2 to 1 in example 3. Other options violate the fundamental law of thermodynamics.

7 0

3 years ago