If gear a rotates with a constant angular acceleration of aa = 90 rad>s2, starting from rest, determine the time required for gear d to attain an angular velocity of 600 rpm. Also, find the number of revolutions of gear d to attain this angular velocity. Gears a, b, c, and d have radii of 15 mm, 50 mm, 25 mm, and 75 mm, respectively.

Answers

Answer 1

Answer:

solution:

$Given: (let a = alpha)aA = 90 rad/s^2(wd) = 600 rpm = 62.831 rad/s(w0) = 0 rpmra = .015 mrb = .05 mrc = .025 mrd = .075aB = aA*(ra/rb) = 90 * (.015/.05)aB = 27 rad/s^2aC = aB, Therefore aC = 27 rad/s^2aD = aC*(rc/rb) = 27 * (.025/.075)aD = 9 rad/s^2$

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The rms speed of the molecules in 1.3 g of hydrogen gas is 1600 m/s.Part A. What is the total translational kinetic energy of the gas molecules? Part B. What is the thermal energy of the gas? Part C. 500J of work are done to compress the gas while, in the same process, 2000J of heat energy are transferred from the gas to the environment. Afterward, what is the rms speed of the molecules?

A fully loaded, slow-moving freight elevator has a cab with a total mass of 1200 kg, which is required to travel upward 35 m in 3.5 min, starting and ending at rest. The elevator's counterweight has a mass of only 940 kg, so the elevator motor must help pull the cab upward. What average power is required of the force the motor exerts on the cab via the cable

Answers

Answer:

425.1 W

Explanation:

We are given;

Counter mass of elevator; m_c = 940 kg

Cab mass of elevator; m_d = 1200 kg

Distance from rest upwards; d = 35 m

Time to cover distance; t = 3.5 min

Now, this elevator will have 3 forces acting on it namely;

Force due to the counter weight of the elevator; F_c

Force due to the cab weight on the elevator; F_d

Force exerted by the motor; F_m

Now, from Newton's 2nd law of motion,

The force exerted by the motor on the elevator can be given by the relationship;

F_m = F_d - F_c

Now,

F_d = m_d × g

F_d = 1200 × 9.81

F_d = 11772 N

F_c = m_c × g

F_c = 940 × 9.81

F_c = 9221.4 N

Thus;

F_m = 11772 - 9221.4

F_m = 2550.6 N

Now, the average power required of the force the motor exerts on the cab via the cable is given by;

P_m = F_m × v

Where v is the velocity of the elevator.

The velocity is calculated from;

v = distance/time

v = 35/3.5

v = 10 m/min

Converting to m/s gives;

v = 10/60 m/s = 1/6 m/s

Thus;

P_m = 2550.6 × 1/6

P_m = 425.1 W

Please helpa car is driven 200 km west and then 80 km southwest. what is the displacement of the car from the point of orgin (magnitude and direction)? draw a diagram.

Answers

Let's take east and west to be positive and negative, respectively, and north and south to be positive and negative, respectively. Then in terms of vectors (using ijk notation), the car first moves 200 km west,

r = (-200 km) j

then 80 km southwest,

s = (-80/√2 km) i + (-80/√2 km) j

so that its total displacement is

r + s = (-80/√2 km) i + ((-200 - 80/√2) km) j

r + s ≈ (-56.6 km) i + (-256.6 km) j

This vector has magnitude

√((-56.6 km)² + (-256.6 km)²) ≈ 262.7 km

and direction θ such that

tan(θ) = (-256.6 km) / (-56.6 km) ==> θ ≈ -102.4º

relative to east, or about 12.4º west of south.

Using energy considerations, calculate the average force (in N) a 62.0 kg sprinter exerts backward on the track to accelerate from 2.00 to 6.00 m/s in a distance of 25.0 m, if he encounters a headwind that exerts an average force of 30.0 N against him.

Answers

Answer:

69.68 N

Explanation:

Work done is equal to change in kinetic energy

W = ΔK = Kf - Ki = $(1)/(2) mv^(2) _(f) - (1)/(2) mv^(2) _(i)$

W = $F_(total) .d$

where m = mass of the sprinter

vf = final velocity

vi = initial velocity

W = workdone

kf = final kinetic energy

ki = initial kinetic energy

d = distance traveled

Ftotal = total force

vf = 8m/s

vi= 2m/s

d = 25m

m = 60kg

inserting parameters to get:

W = ΔK = Kf - Ki = $(1)/(2) mv^(2) _(f) - (1)/(2) mv^(2) _(i)$

$F_(total) .d =(1)/(2) mv^(2) _(f) - (1)/(2) mv^(2) _(i)$

$F_(total) = ((1)/(2) mv^(2) _(f) - (1)/(2) mv^(2) _(i))/(d)$

$F_(total=) ((1)/(2) X 62 X6^(2) -(1)/(2) X 62 X2^(2) )/(25)$

= 39.7

we know that the force the sprinter exerted F sprinter, the force of the headwind Fwind = 30N

$F_(sprinter) = F_(total) + F_(wind) = 39.7 + 30 = 69.68 N$

Answer:

Force exerted by sprinter = 69.68 N

Explanation:

From work energy theorem, we know that, work done is equal to change in kinetic energy.

Thus,

W = ΔK = Kf - Ki = (1/2)m•(v_f)² - (1/2)m•(v_i)² - - - - eq(1)

Now,

Work done is also;

W = Force x Distance = F•d - - - (2)

From the question, we are given ;

v_f = 6 m/s

v_i = 2 m/s

d = 25m

m = 62 kg

Equating equation 1 and 2,we get;

(1/2)m•(v_f)² - (1/2)m•(v_i)² = F•d

Plugging in the relevant values to obtain ;

(1/2)(62)[(6)² - (2)²] = F x 25

31(36 - 4) = 25F

992 = 25F

F = 39.68 N

The force the sprinter exerts backward on the track will be the sum of this force and the headwind force.

Thus,

Force of sprinter = 39.68 + 30 = 69.68N

A thin flashlight beam traveling in air strikes a glass plate at an angle of 52° with the plane of the surface of the plate. If the index of refraction of the glass is 1.4, what angle will the beam make with the normal in the glass?

Answers

To solve this problem it is necessary to apply Snell's law and thus be able to calculate the angle of refraction.

From Snell's law we know that

$n_1sin\theta_1 = n_2 sin\theta_2$

Where,

n_i = Refractive indices of each material

$\theta_1$ = Angle of incidence

$\theta_2$ = Refraction angle

Our values are given as,

$\theta_1 = 38\°$

$n_1 = 1$

$n_2 = 1.4$

Replacing

$1*sin38 = 1.4*sin\theta_2$

Re-arrange to find $\theta_2$

$\theta_2 = sin^(-1) (sin38)/(1.4)$

$\theta_2 = 26.088°$

Therefore the angle will the beam make with the normal in the glass is 26°

Water, initially saturated vapor at 4 bar, fills a closed, rigid container. The water is heated until its temperature is 360°C. For the water, determine the heat transfer, in kJ per kg of water. Kinetic and potential energy effects can be ignored.

Answers

Explanation:

Using table A-3, we will obtain the properties of saturated water as follows.

Hence, pressure is given as p = 4 bar.

$u_(1) = u_(g)$ = 2553.6 kJ/kg

$v_(1) = v_(g) = 0.4625 m^(3)/kg$

At state 2, we will obtain the properties. In a closed rigid container, the specific volume will remain constant.

Also, the specific volume saturated vapor at state 1 and 2 becomes equal. So, $v_(2) = v_(g) = 0.4625 m^(3)/kg$

According to the table A-4, properties of superheated water vapor will obtain the internal energy for state 2 at $v_(2) = v_(g) = 0.4625 m^(3)/kg$ and temperature $T_(2) = 360^(o)C$ so that it will fall in between range of pressure p = 5.0 bar and p = 7.0 bar.

Now, using interpolation we will find the internal energy as follows.

$u_(2) = u_{\text{at 5 bar, 400^(o)C}} + (\frac{v_(2) - v_{\text{at 5 bar, 400^(o)C}}}{v_{\text{at 7 bar, 400^(o)C - v_(at 5 bar, 400^(o)C)}}})(u_{at 7 bar, 400^(o)C - u_(at 5 bar, 400^(o)C)})$

$u_(2) = 2963.2 + ((0.4625 - 0.6173)/(0.4397 - 0.6173))(2960.9 - 2963.2)$

= 2963.2 - 2.005

= 2961.195 kJ/kg

Now, we will calculate the heat transfer in the system by applying the equation of energy balance as follows.

Q - W = $\Delta U + \Delta K.E + \Delta P.E$ ......... (1)

Since, the container is rigid so work will be equal to zero and the effects of both kinetic energy and potential energy can be ignored.

$\Delta K.E = \Delta P.E$ = 0

Now, equation will be as follows.

Q - W = $\Delta U + \Delta K.E + \Delta P.E$

Q - 0 = $\Delta U + 0 + 0$

Q = $\Delta U$

Now, we will obtain the heat transfer per unit mass as follows.

$(Q)/(m) = \Delta u$

$(Q)/(m) = u_(2) - u_(1)$

= (2961.195 - 2553.6)

= 407.595 kJ/kg

Thus, we can conclude that the heat transfer is 407.595 kJ/kg.

Final answer:

The heat transfer is 227.4 kJ per kg of water.

Explanation:

Water, initially saturated vapor at 4 bar, fills a closed, rigid container. The water is heated until its temperature is 360°C. To determine the heat transfer in kJ per kg of water, we need to calculate the heat absorbed by the water as it reaches 360°C.

Using the specific heat capacity of water (4,186 J/kg°C) and the change in temperature (360°C - 100°C), we can calculate the heat transfer:

Qw = mw * cw * AT = (1 kg) * (4186 J/kg°C) * (360°C - 100°C) = 227,440 J = 227.4 kJ

Therefore, the heat transfer is 227.4 kJ per kg of water.

Heat transfer is the process by which thermal energy moves from one object or substance to another due to a difference in temperature. This fundamental phenomenon occurs through three main mechanisms: conduction, convection, and radiation. Conduction involves the direct transfer of heat through a material, such as metal. Convection is the transfer of heat through the movement of fluids (liquids or gases). Radiation is the emission of electromagnetic waves carrying heat energy. Understanding heat transfer is essential in various fields, including physics, engineering, and environmental science, as it governs temperature regulation, climate dynamics, and the functioning of countless technological devices.

Learn more about Heat transfer here:

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An unladen swallow that weighs 0.03 kg flies straight northeast a distance of 125 km in 4.0 hours. With the x x direction due east and the y y direction due north, what is the average momentum of the bird (in unit vector notation)?

Answers

The average momentum of the bird (in unit vector notation) is (0.1842i + 0.1842j) kgm/s.

Total displacement

Since the unladen swallow that weighs 0.03 kg flies straight northeast (that is at a bearing of 45°) a distance of 125 km in 4.0 hours.

Its position vector after 4.0 hours is d = (125kmcos45)i + (125kmsin45)j = (125000 × 1/√2)i + (125000 × 1/√2)j

= (62500√2)i + (62500√2)j.

If the initial position of the swallow is d' = 0i + 0j, then its total displacement after 4 hours is, D = d - d'

= (62500√2)i + (625000√2)j - (0i + 0j)

= (62500√2)i + (62500√2)j m

Average velocity

The unladen swallow's average velocity, v = D/t where

D = total displacement = (62500√2)i + (62500√2)j m and
t = time = 4.0 hours = 4 × 60 min/hr × 60 s/min = 14400 s

So, v = [(62500√2)i + (62500√2)j m]/14400 s = (88388.35)i/14400 + (88388.35)j /1440

= 6.14i + 6.14j m/s

Average momentum

The average momentum of the unladen swallow is p = mv where

m = mass of unladen swallow = 0.03 kg and
v = average velocity = 6.14i + 6.14j m/s

So, p = mv

p = 0.03 kg × (6.14i + 6.14j m/s)

p = (0.1842i + 0.1842j) kgm/s

So, the average momentum of the bird (in unit vector notation) is (0.1842i + 0.1842j) kgm/s.

Learn more about average momentum here:

brainly.com/question/25941500

Answer:

The average momentum of the bird is 0.26 kgm/s

Explanation:

The formula to be used here is that of momentum which is

momentum (in kgm/s) = mass (in kg) × velocity (in m/s)

The velocity of the bird is

velocity (in m/s) = distance (in meter) ÷ time (in seconds)

distance in meters = 125km × 1000 = 125,000 m

time in seconds = 4 hrs × 60 × 60 = 14,400 secs

velocity = 125000/14400

velocity = 8.68 m/s

momentum (p) = 0.03 × 8.68

p = 0.26 kgm/s

The average momentum of the bird is 0.26 kgm/s

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