All categories

3 years ago

A 1200-kg car moving at 20 km/h is accelerated

Engineering

1 answer:

Grace [21]3 years ago

4 0

Answer:

Force on the car will be 4800 N and time required to cover this distance 13.75 sec

Explanation:

We have given mass of the car = 1200 kg

Initial velocity u = 20 km/h

Final velocity v = 75 km/h

Acceleration $a=4m/sec^2$

From the first equation of motion we know that

v = u+at, here v is final velocity, u is initial velocity, a is acceleration and t is time

So $75=20+4\times t$

t = 13.75 sec

From second law of motion we know that $F=ma$

So force $F=1200\times 4=4800N$

You might be interested in

Water is stored in a tank which has vent open to the atmosphere. The water level is 1.0 m below the top the tank and the water i

sp2606 [1]

Answer:

6.99 x 10⁻³ m³ / s

Explanation:

Th e pressure difference at the two ends of the delivery pipe due to atmospheric pressure and water column will cause flow of water.

h = difference in the height of water column at two ends of delivery pipe

6 - 1 = 5 m

Velocity of flow of water

v = √2gh

= √ (2 x 9.8 x 5)

= 9.9 m /s

Volume of water flowing per unit time

velocity x cross sectional area

= 9.9 x 3.14 x .015²

= 6.99 x 10⁻³ m³ / s

7 0

3 years ago

Please help this is due today!!!!!

White raven [17]

Answer:

1:c 2:False

Explanation:

7 0

3 years ago

2. What is the Function of the Camshaft in an Internal Combustion Engine?

mamaluj [8]

Answer:

camshaft, in internal-combustion engines, rotating shaft with attached disks of irregular shape (the cams), which actuate the intake and exhaust valves of the cylinders.

Explanation:

I'm taking an engineering/tech class. I hope this helps! :)

8 0

3 years ago

A sheet of steel 3-mm thick has nitrogen atomospheres on both sides at 900 C and is permitted to achieve a steady-state di usion

kati45 [8]

Answer:

$X_B = 1.8 \times 10^{-3} m = 1.8 mm$

Explanation:

Given data:

Diffusion constant for nitrogen is $= 1.85\times 10^{-10} m^2/s$

Diffusion flux $= 1.0\times 10^{-7} kg/m^2-s$

concentration of nitrogen at high presuure = 2 kg/m^3

location on which nitrogen concentration is 0.5 kg/m^3 ......?

from fick's first law

$J = D \frac{C_A C_B}{X_A X_B}$

Take C_A as point on which nitrogen concentration is 2 kg/m^3

$x_B = X_A + D\frac{C_A -C_B}{J}$

Assume X_A is zero at the surface

$X_B = 0 + ( 12\times 10^{-11} ) \frac{2-0.5}{1\times 10^{-7}}$

$X_B = 1.8 \times 10^{-3} m = 1.8 mm$

4 0

3 years ago

Pipe Diameter and Reynolds Number. An oil is being pumped inside a 10.0-mm-diameter pipe at a Reynolds number of 2100. The oil d

alexdok [17]

Answer:

The velocity in the pipe is 5.16m/s. The pipe diameter for the second fluid should be 6.6 mm.

Explanation:

Here the first think you have to consider is the definition of the Reynolds number ( $Re$ ) for flows in pipes. Rugly speaking, the Reynolds number is an adimensonal parameter to know if the fliud flow is in laminar or turbulent regime. The equation to calculate this number is:

$Re=\frac{\rho v D}{\mu}$

where $\rho$ is the density of the fluid, $\mu$ is the viscosity, D is the pipe diameter and v is the velocity of the fluid.

Now, we know that Re=2100. So the velocity is:

$v=\frac{Re*\mu}{\rho*D} =\frac{2100*2.1x10^{-2}Pa*s }{855kg/m^3*0.01m} =5.16m/s$

For the second fluid, we want to keep the Re=2100 and v=5.16m/s. Therefore, using the equation of Reynolds number the diameter is:

$D=\frac{Re*\mu}{\rho*v} =\frac{2100*1.5x10^{-2}Pa*s}{925kg/m^3*5.16m/s}=6.6 mm$

8 0

3 years ago