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3 years ago

12. Dies are turned using a special tool called a/an

Engineering

1 answer:

PIT_PIT [208]3 years ago

3 0

Answer:

the answer is die stock

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Two grenades, A and B, are thrown horizontally with different speeds from the top of a cliff 70 m high. The speed of A is 2.50 m

Aloiza [94]

Explanation:

uuui ielts k oshru with the best of my life u

3 0

3 years ago

A copper block receives heat from two different sources: 5 kW from a source at 1500 K and 3 kW from a source at 1000 K. It loses

LUCKY_DIMON [66]

Answer:

a) Zero

b) the rate of entropy generation in the system's universe = ds/dt = 0.2603 KW/K

Explanation:

a) In steady state

Net rate of Heat transfer = net rate of heat gain - net rate of heat lost

Hence, the rate of heat transfer = 0

b) In steady state, entropy generated

ds/dt = - [ Qgain/Th1 + Qgain/Th2 - Qlost/300 K]

Substituting the given values, we get –

ds/dt = -[5/1500 + 3/1000 – (5+3)/300]

ds/dt = - [0.0033 + 0.003 -0.2666]

ds/dt = 0.2603 KW/K

6 0

4 years ago

Describe how to mix and apply body filler?

malfutka [58]

This is all you need to mix body filler.
Start with a golf-ball size glob of filler.
Squeeze a ribbon of hardner across the filler.
Stir the mixture quickly, but do not "whip" it. ...
Mix until overall color is even.
Pick up an appropriate amount of filler for the task at hand.

8 0

3 years ago

A 12-ft circular steel rod with a diameter of 1.5-in is in tension due to a pulling force of 70-lb. Calculate the stress in the

padilas [110]

Answer:

The stress in the rod is 39.11 psi.

Explanation:

The stress due to a pulling force is obtained dividing the pulling force by the the area of the cross section of the rod. The respective area for a cylinder is:

$A=\pi*D^2/4$

Replacing the diameter the area results:

$A= 17.76 in^2$

Therefore the the stress results:

$σ = 70/17.76 lb/in^2 = 39.11 lb/in^2= 39.11 psi$

5 0

3 years ago

You plan to install an active, liquid-based solar heating system for hot water. There are four candidate collector systems. Your

olchik [2.2K]

Solution:

The given formula,

$x=F_{R} U_{L} \times \frac{P l}{F R_{1}} \times\left(T_{r e f}-\bar{T}_{a}\right) \Delta t \times \frac{A_{c}}{L}$

$y=F_{R}(\tau \alpha)_{n} x \frac{F_{R}^{\prime}}{F_{R}} \times \frac{(\bar{\tau} d)}{(T d)_{n}} \times \bar{H}_{T} N \times \frac{A C}{L}$

$\frac{x}{y}=\frac{ u_{L} \times\left(T_{x t}-\bar{T}_{a}\right) \times \Delta t}{\left(\tau_{x}\right)_{h} \times\left(\frac{\bar{\tau}_{d}}{\left.| \tau_{d}\right)_{n}}\right) \times \bar{H}+N}$

From the table,

$1) $\quad x=2 \cdot 87, \quad y=0.96$\\$\frac{x}{y}=\frac{2187}{0.96}$22895\\\\2) $x=3 \cdot 466 \cdot y=6 \cdot 998$\\$\frac{x}{y}=\frac{3 \cdot 466}{0.898}$$=3 \cdot 4729$$

$3$x=3 \cdot 229, y=1 \cdot 08$\\$\frac{x}{x}=\frac{3 \cdot 229}{1 \cdot 08}$\\=2.9898\)\\\\4) $x=6.525, y=1.094$\\$\frac{x}{y}=\frac{5.625}{1.094}$\\=5.0502$

8 0

4 years ago