שאלה 2 - כתיבת קוד (35 נק')

If a front gear had 24 teeth, and a rear gear has 12 teeth:

zubka84 [21]

Answer:

4 times around

Explanation:

The total number of teeth involved will be the same for each gear. If the front gear is connected to the pedal and it goes around twice, then 2·24 = 48 teeth will have passed the reference point.

If the rear gear is attached to the wheel, and 48 teeth pass the reference point, then it will have made ...

(48 teeth)/(12 teeth/turn) = 4 turns

4 0

2 years ago

Is santa real or nah is santa real or nah

Elena L [17]

Answer:

nah

Explanation:

3 0

3 years ago

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Water at a pressure of 3 bars enters a short horizontal convergent channel at 3.5 m/s. The upstream and downstream diameters of

earnstyle [38]

Answer:

The pressure reduces to 2.588 bars.

Explanation:

According to Bernoulli's theorem for ideal flow we have

$\frac{P}{\gamma _{w}}+\frac{V^{2}}{2g}+z=constant$

Since the losses are neglected thus applying this theorm between upper and lower porion we have

$\frac{P_{u}}{\gamma _{w}}+\frac{V-{u}^{2}}{2g}+z_{u}=\frac{P_{L}}{\gamma _{w}}+\frac{V{L}^{2}}{2g}+z_{L}$

Now by continuity equation we have

$A_{u}v_{u}=A_{L}v_{L}\\\\\therefore v_{L}=\frac{A_{u}}{A_{L}}\times v_{u}\\\\v_{L}=\frac{d^{2}_{u}}{d^{2}_{L}}\times v_{u}\\\\\therefore v_{L}=\frac{2500}{900}\times 3.5\\\\\therefore v_{L}=9.72m/s$

Applying the values in the Bernoulli's equation we get

$\frac{P_{L}}{\gamma _{w}}=\frac{300000}{\gamma _{w}}+\frac{3.5^{2}}{2g}-\frac{9.72^{2}}{2g}(\because z_{L}=z_{u})\\\\\frac{P_{L}}{\gamma _{w}}=26.38m\\\\\therefore P_{L}=258885.8Pa\\\\\therefore P_{L}=2.588bars$

6 0

3 years ago

What size resistor would produce a current flow of 5 Amps with a battery voltage of 12.6 volts

Debora [2.8K]

Answer:

resistance = 2.52 ohms

Explanation:

from the formula

V =IR

Voltage = (current)(resistance)

Resistance =

R=

R= 2.52 ohms

5 0

2 years ago

When subject to an unknown torque, the shear stress in a 2 mm thick rectangular tube of dimension 100 mm x 200 mm was found to b

laila [671]

Answer:

The shear stress will be 80 MPa

Explanation:

Here we have;

τ = (T·r)/J

For rectangular tube, we have;

Average shear stress given as follows;

Where;

$\tau_{ave} = \frac{T}{2tA_{m}}$

$A_m$ = 100 mm × 200 mm = 20000 mm² = 0.02 m²

t = Thickness of the shaft in question = 2 mm = 0.002 m

T = Applied torque

Therefore, 50 MPa = T/(2×0.002×0.02)

T = 50 MPa × 0.00008 m³ = 4000 N·m

Where the dimension is 50 mm × 250 mm, which is 0.05 m × 0.25 m

Therefore, $A_m$ = 0.05 m × 0.25 m = 0.0125 m².

Therefore, from the following average shear stress formula, we have;

$\tau_{ave} = \frac{T}{2tA_{m}}$

Plugging in then values, gives;

$\tau_{ave} = \frac{4000}{2\times 0.002 \times 0.0125} = 80,000,000 Pa$

The shear stress will be 80,000,000 Pa or 80 MPa.

7 0

3 years ago