What is the maximum distance allowed between the center of hole #2 and datum B as seen in the front view?

Answers

Answer 1

Answer:

Answer:

4.003" (inches )

Explanation:

The maximum distance allowed between the center of hole #2 and datum B can be calculated by adding 4.000" + 0.003" ( perpendicularity of the of hole #2) as seen from the front view of the diagram .

Note :The hole 2 is sited below the workpiece when viewed from the front view while the Datum B is positioned on the left end of the workpiece also note that the diameter is

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A golf pro swings a golf club, striking a golf ball that has a mass of 55.0 g. The club is in contact with the ball for only 0.00340 s. After the collision, the ball leaves the club at a speed of 46.0 m/s. What is the magnitude of the average force (in N) exerted on the ball by the club?

An airplane is in level flight over Antarctica, where the magnetic field of the earth is mostly directed upward away from the ground. As viewed by a passenger facing toward the front of the plane, is the left or the right wingtip at higher potential? Does your answer depend on the direction the plane is flying?

Answers

Answer:

It does not depend on direction of plane and the left windtips more potential

Explanation:

Because if the Fleming right hand rule is applied the the right han is pointed in the direction of flight, and the fingers are curled in the direction of the magnetic lines. Thus , the lines are vertical and so are pointing down, thus by the right hand rule, the electrons move to the left hand side of the plane, although the potentials are equal and opposite in direction

A train moving west with an initial velocity of 20 m/s accelerates at 4 m/s2 for 10 seconds. During this time, the train moves a distance of meters.

Answers

So lets fill out what we have first:

Vi or initial velocity = 20 m/s

Acceleration or a = 4 m/s^2

Time for the motion = 10s

Now, using the four main kinematic equations we can deduce that the best kinematic equation to use in these terms is:

Δx = Vi(t) + 0.5at²

Plug all of our information in:

Δx = (20)(10) + (0.5)(4)(100)

Δx = 400 m

Answer:

400 m

Explanation:

answer on ed

Neon signs require about 12,000 V for their operation. Consider a neon-sign transformer that operates off 120- V lines. How many more turns should be on the secondary compared with the primary?

Answers

I can't give you the actual number of turns, because it's the RATIO
that counts.

However many turns the primary has, the secondary should have
about TEN TIMES that number. Then the transformer will multiply
the primary voltage by 10 ... 120 volts of AC at the primary will
become 1,200 volts of AC at the secondary.

Note that it HAS TObe AC. If the transformer is supplied with DC,
then 120 volts at the primary becomes zero volts at the secondary
and a big cloud of stinky smoke in the room.

An insulated rigid tank contains 3 kg of H2O in the form of a saturated mixture of liquid and vapor at a pressure of 150 kPa and a quality of 0.25. An electric heater inside the tank is turned on to heat this H2O until the pressure increases to 200 kPa. Please determine the change in total entropy of water during this process. Hint: See if you can find the electrical work consumed during this process.

Answers

Answer:

change in entropy is 1.44 kJ/ K

Explanation:

from steam tables

At 150 kPa

specific volume

Vf = 0.001053 m^3/kg

vg = 1.1594 m^3/kg

specific entropy values are

Sf = 1.4337 kJ/kg K

Sfg = 5.789 kJ/kg

initial specific volume is calculated as

$v_1 = vf + x(vg - vf)$

$= 0.001053 + 0.25(1.1594 - 0.001053)$

$v_1 = 0.20964 m^3/kg$

$s_1 = Sf + x(Sfg)$

$= 1.4337 + 0.25 * 5.7894 = 2.88 kJ/kg K$

FROM STEAM Table

at 200 kPa

specific volume

Vf = 0.001061 m^3/kg

vg = 0.88578 m^3/kg

specific entropy values are

Sf = 1.5302 kJ/kg K

Sfg = 5.5698 kJ/kg

constant volume so $v_1 - v_2 = 0.29064 m^3/kg$

$v_2 = v_1 = vf + x(vg - vf)$

$=0.29064 = x_2(0.88578 - 0.001061)$

$x_2 = 0.327$

$s_2 = 1.5302 + 0.32 * 5.5968 = 3.36035 kJ/kg K$

Change in entropy $\Delta s = m(s_2 - s_1)$

=3( 3.36035 - 2.88) = 1.44 kJ/kg

A 3.00N rock is thrown vertically into the air from the ground. At h=15.0m, v=25m/s upward. Use the work-energy theorem to find the initial speed of the rock.a. 3m/s
b. 30.3 m/s
c. None of the above

Answers

Answer:

so initial speed of the rock is 30.32 m/s

correct answer is b. 30.3 m/s

Explanation:

given data

h = 15.0m

v = 25m/s

weight of the rock m = 3.00N

solution

we use here work-energy theorem that is express as here

work = change in the kinetic energy ..............................1

so it can be written as

work = force × distance ...................2

and

KE is express as

K.E = 0.5 × m × v²

and it can be written as

F × d = 0.5 × m × (vf)² - (vi)² ......................3

here

m is mass and vi and vf is initial and final velocity

F = mg = m (-9.8) , d = 15 m and v{f} = 25 m/s

so put value in equation 3 we get

m (-9.8) × 15 = 0.5 × m × (25)² - (vi)²

solve it we get

(vi)² = 919

vi = 30.32 m/s

so initial speed of the rock is 30.32 m/s

A rock is pulled back in a slingshot as shown in the diagram below. The elastic on the slingshot is displaced 0.2 meters from its initial position. The rock is pulled back with a force of 10 newtons.When the rock is released, what is its kinetic energy?

Answers

The rock takes 8.16s to return to its release point. Given that the elastic band provides a speed of 40m/s to the rock in 10 cm stretch.

What will be the speed of the rock?

Initial speed of the rock, u = 40m/s

Final position of the rock s = 0m taking the release point as reference. The rock takes 8.16s to return to its release point. Given that the elastic band provides a speed of 40m/s to the rock in 10 cm stretch.

Nuclear energy is a useful source of power but has disadvantages. The disadvantage of nuclear energy is it produces dangerous waste.

Initial speed of the rock, u = 40m/s

Final position of the rock s = 0m taking the release point as reference

From the second equation of motion:

solving above we get:

t = 0s or t = 8.16s, t =0 seconds is neglected since it represents the initial position which is the same as the final position at t = 8.16s

So, the rock takes 8.16 seconds to return to the release point.

Therefore, The rock takes 8.16s to return to its release point. Given that the elastic band provides a speed of 40m/s to the rock in 10 cm stretch.

Learn more about speed of rock on:

brainly.com/question/11049671

#SPJ2

Answer:

Explanation:

i don't know

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As the captain of the scientific team sent to Planet Physics, one of your tasks is to measure g. You have a long, thin wire labeled 1.73 g/m and a 1.30 kg weight. You have your accurate space cadet chronometer but, unfortunately, you seem to have forgotten a meter stick. Undeterred, you first find the midpoint of the wire by folding it in half. You then attach one end of the wire to the wall of your laboratory, stretch it horizontally to pass over a pulley at the midpoint of the wire, then tie the 1.30 kg weight to the end hanging over the pulley. By vibrating the wire, and measuring time with your chronometer, you find that the wire's second harmonic frequency is 200 Hz . Next, with the 1.30 kg weight still tied to one end of the wire, you attach the other end to the ceiling to make a pendulum. You find that the pendulum requires 313 s to complete 200 oscillations. Pulling out your trusty calculator, you get to work.What value of g will you report back to headquarters?