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Mamont248 [21]
3 years ago
14

O que é necessário se conhecer para a determinação de uma força concentrada em um ponto?

Engineering
1 answer:
skelet666 [1.2K]3 years ago
7 0

Answer: its c

Explanation:

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(d) Suppose two students are memorizing a list according to the same model dL dt = 0.5(1 − L) where L represents the fraction of
Neporo4naja [7]

Answer:

Rate of learning =0

Explanation:

Please see attachment

8 0
3 years ago
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
Discuss the difference between the observed and calculated values. Is this error? If yes, what is the source?
Scrat [10]

Answer and Explanation:

In any experiment, the observed values are the actual values obtained in any experiment.

The calculated values are the values that are measured by using the observed values in a formula.

The observed values are primary values whereas the calculated values are the secondary values as calaculations are made using observed values.

Yes, if the observed values are of low accuracy.

The values should be recorded with proper care and attention in order to avoid any error.

8 0
3 years ago
A flat plate 1.5 m long and 1.0 m wide is towed in water at 20 o C in the direction of its length at a speed of 15 cm/s. Determi
beks73 [17]

Answer:

15.8

0.0944

Explanation:

L = 1.5

B = 1.0

Speed of water = 15cm

Temperature = 20⁰C

At 20⁰C

Specific weight = 9790

Kinematic viscosity v = 1.00x10^-4m²/s

Dynamic viscosity u = 1.00x10^-3

Density p = 998kg/m²

Reynolds number

= 0.15x1.5/1.00x10^-4

= 225000

S = 5

5x1.5/225000^1/2

= 0.0158

= 15.8mm

Resistance on one side of plate

F = 0.664x1x1.0x10^-3x0.15x225000^1/2

= 0.04724N

Total resistance

= 2N

= 2x0.04724

= 0.0944N

3 0
3 years ago
A piece of aluminum wire is 500 ft long and has a diameter of 0.03 inches. What is the resistance of the piece of wire?​
dexar [7]

Answer:

8.85 Ω

Explanation:

Resistance of a wire is:

R = ρL/A

where ρ is resistivity of the material,

L is the length of the wire,

and A is the cross sectional area.

For a round wire, A = πr² = ¼πd².

For aluminum, ρ is 2.65×10⁻⁸ Ωm, or 8.69×10⁻⁸ Ωft.

Given L = 500 ft and d = 0.03 in = 0.0025 ft:

R = (8.69×10⁻⁸ Ωft) (500 ft) / (¼π (0.0025 ft)²)

R = 8.85 Ω

5 0
3 years ago
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