Print that out and do it lol but seriously

How much work, in Newtons, is required to lift a 20.4-kg (45lb) plate from the ground to a stand that is 1.50 meters up?

nataly862011 [7]

Answer:

Explanation:

Work, U, is equal to the force times the distance:

U = F · r

Force needed to lift the weight, is equal to the weight: F = W = m · g

so:

U = m · g · r

= 20.4kg · 9.81 $\frac{N}{kg}$ · 1.50m

= 35.316 $\frac{N}{m}$

= 35.316 W

4 0

2 years ago

A furnace wall composed of 200 mm, of fire brick. 120 mm common brick 50mm 80% magnesia and 3mm of steel plate on the outside. I

Liula [17]

Answer:

fire brick / common brick : 1218 °C
common brick / magnesia : 1019 °C
magnesia / steel : 90.06 °C
heat loss: 4644 kJ/m^2/h

Explanation:

The thermal resistance (R) of a layer of thickness d given in °C·m²·h/kJ is ...

R = d/k

so the thermal resistances of the layers of furnace wall are ...

R₁ = 0.200/4 = 0.05 °C·m²·h/kJ

R₂ = 0.120 2.8 = 3/70 °C·m²·h/kJ

R₃ = 0.05/0.25 = 0.2 °C·m²·h/kJ

R₄ = 0.003/240 = 1.25×10⁻⁵ °C·m²·h/kJ

So, the total thermal resistance is ...

R₁ +R₂ +R₃ +R₄ = R ≈ 0.29286 °C·m²·h/kJ

__

The rate of heat loss is ΔT/R = (1450 -90)/0.29286 = 4643.70 kJ/(m²·h)

__

The temperature drops across the various layers will be found by multiplying this heat rate by the thermal resistance for the layer:

fire brick: (4543.79 kJ/(m²·h))(0.05 °C·m²·h/kJ) = 232 °C

so, the fire brick interface temperature at the common brick is ...

1450 -232 = 1218 °C

For the next layers, the interface temperatures are ...

common brick to magnesia = 1218 °C - (3/70)(4643.7) = 1019 °C

magnesia to steel = 1019 °C -0.2(4643.7) = 90.06 °C

_____

<em>Comment on temperatures</em>

Most temperatures are rounded to the nearest degree. We wanted to show the small temperature drop across the steel plate, so we showed the inside boundary temperature to enough digits to give the idea of the magnitude of that.

5 0

3 years ago

Cite the phases that are present and the phase compositions for the following alloys: (a) 15 wt% Sn - 85 wt% Pb at 100 o C. (b)

Novosadov [1.4K]

Answer:

a) ∝ and β

The phase compositions are :

C $_{\alpha }$ = 5wt% Sn - 95 wt% Pb

C $_{\beta }$ = 98 wt% Sn - 2wt% Pb

b)

The phase is; ∝

The phase compositions is; 82 wt% Sn - 91.8 wt% Pb

Explanation:

a) 15 wt% Sn - 85 wt% Pb at 100⁰C.

The phases are ; ∝ and β

The phase compositions are :

C $_{\alpha }$ = 5wt% Sn - 95 wt% Pb

C $_{\beta }$ = 98 wt% Sn - 2wt% Pb

b) 1.25 kg of Sn and 14 kg Pb at 200⁰C

The phase is ; ∝

The phase compositions is; 82 wt% Sn - 91.8 wt% Pb

Csn = 1.25 * 100 / 1.25 + 14 = 8.2 wt%

Cpb = 14 * 100 / 1.25 + 14 = 91.8 wt%

6 0

3 years ago

Identify the right components for gsm architecture that consists of the hardware or physical equipment such as digital signal pr

sergiy2304 [10]

The right components for gsm architecture that consists of the hardware or physical equipment such as digital signal processors, radio transceiver, display, battery, case and sim card is the Mobile station.

<h3>What are the 4 main components?</h3>

In GSM, a cell station includes 4 fundamental additives: Mobile termination (MT) - gives not unusualplace features consisting of: radio transmission and handover, speech encoding and decoding, blunders detection and correction, signaling and get right of entry to to the SIM. The IMEI code is connected to the MT.

Under the GSM framework, a cell tele cell smartphone is called a Mobile Station and is partitioned into wonderful additives: the Subscriber Identity Module (SIM) and the Mobile Equipment (ME).