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DaniilM [7]
3 years ago
13

WHAT IS THE MOST POWERFUL PART IN A CAR

Engineering
2 answers:
STatiana [176]3 years ago
7 0
I think it would be the cage
DiKsa [7]3 years ago
6 0
The cage because it protects us
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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 single fixed pulley is used to lift a load of 400N by the application of an effort of 480N in 10s through a vertical height of
Allushta [10]

Answer:

(a) the velocity ratio of the machine (V.R) = 1

(b) The mechanical advantage of the machine (M.A) = 0.833

(c) The efficiency of the machine (E) = 83.3 %

Explanation:

Given;

load lifted by the pulley, L = 400 N

effort applied in lifting the, E = 480 N

distance moved by the effort, d = 5 m

(a) the velocity ratio of the machine (V.R);

since the effort applied moved downwards through a distance of d, the load will also move upwards through an equal distance 'd'.

V.R = distance moved by effort / distance moved by the load

V.R = 5/5 = 1

(b) The mechanical advantage of the machine (M.A);

M.A = L/E

M.A = 400 / 480

M.A = 0.833

(c) The efficiency of the machine (E);

E = \frac{M.A}{V.R} \times 100\%\\\\E = 0.833 \ \times \ 100\%\\\\ E = 83.3 \ \%

4 0
3 years ago
A glass plate is subjected to a tensile stress of 40 MPa. If the specific surface energy is 0.3 J/m2 and the modulus of elastici
Pavlova-9 [17]

Answer:

8.24μm

Explanation:

The theory of brittle fracture was used to solve this problem.

And if you follow through with the attachment made a the subject of the formula

Such that,

a = 2x(69x10⁹)x0.3/pi(40x10⁶)²

= 4.14x10¹⁰/5.024x10¹⁵

= 8.24x10^-06

= 8.24μm

This is the the maximum length of the surface flaw

4 0
3 years ago
To produce cooling the refrigerant much
lawyer [7]
Yes that is correct good job ❤️
7 0
4 years ago
a ten station assembly machine has ideal cycle time of 6 sec. the fraction defect rate at each station 0.005 and defect always j
MAXImum [283]

Answer:

T_{P}=(2.6667)(10^{-3})h

Explanation:

Let's write the equation of the production rate for the assembly machine :

T_{P}=T_{C}+(n).(m).(p).(T_{D})

Where T_{P} is the production rate for the assembly machine.

Where T_{C} is the ideal cycle time

Where n is the number of stations.

Where m is the number stations that get jam when the defect occurs.

Where p is the defect rate at each station.

And where T_{D} is the average downtime per breakdown

We are looking for the hourly production rate ⇒

1h=60min\\1min=60s ⇒

1h=3600s ⇒

6s=\frac{(6s)(1h)}{(3600s)}= \frac{1}{600}h

60min=1h ⇒

1.2min=\frac{(1.2min)(1h)}{(60min)}=0.02h

T_{P}=\frac{1}{600}h+(10)(1.0)(0.005)(0.02h)=\frac{1}{375}h=(2.6667)(10^{-3})h

m = 1.0 in the equation.

3 0
3 years ago
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