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3 years ago

14

Please help me with this assignment!

2 answers:

makkiz [27]3 years ago

5 0

Answer:

lol you got the answer just need points sorry

Explanation:

Licemer1 [7]3 years ago

3 0

Answer:

Ice cube, molecules, melting, zero degrees, liquid, faster, temperature, 100 degrees, steam

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The moon Phobos orbits Mars

dlinn [17]

Answer:

The moon Phobos orbits Mars

(mass = 6.42 x 1023 kg) at a distance

of 9.38 x 106 m. What is its period of

orbit?

Explanation:

Answer: 27.9816 x 10^3 is the period of orbit

8 0

3 years ago

A large grinding wheel in the shape of a solid cylinder of radius 0.330 m is free to rotate on a frictionless, vertical axle. A

Flura [38]

Answer:

The mass of the wheel is 2159.045 kg

Explanation:

Given:

Radius $r = 0.330$

m

Force $F = 290$ N

Angular acceleration $\alpha = 0.814 \frac{rad}{s^{2} }$

From the formula of torque,

Γ $= I\alpha$ (1)

Γ $= rF$ (2)

$rF = I \alpha$

Find momentum of inertia $I$ from above equation,

$I = \frac{rF}{\alpha }$

$I = \frac{0.330 \times 290}{0.814}$

$I = 117.56$ $Kg. m^{2}$

Find the momentum inertia of disk,

$I = \frac{1}{2} Mr^{2}$

$M = \frac{2I}{r^{2} }$

$M = \frac{2 \times 117.56}{(0.330)^{2} }$

$M = 2159.045$ Kg

Therefore, the mass of the wheel is 2159.045 kg

8 0

4 years ago

Steam enters a well-insulated nozzle at 200 lbf/in.2 , 500F, with a velocity of 200 ft/s and exits at 60 lbf/in.2 with a velocit

Ede4ka [16]

Answer:

$386.2^{\circ}F$

Explanation:

We are given that

$P_1=200lbf/in^2$

$P_2=60lbf/in^2$

$v_1=200ft/s$

$v_2=1700ft/s$

$T_1=500^{\circ}F$

$Q=0$

$C_p=1BTU/lb^{\circ}F$

We have to find the exit temperature.

By steady energy flow equation

$h_1+v^2_1+Q=h_2+v^2_2$

$C_pT_1+\frac{P^2_1}{25037}+Q=C_pT_2+\frac{P^2_2}{25037}$

$1BTU/lb=25037ft^2/s^2$

Substitute the values

$1\times 500+\frac{(200)^2}{25037}+0=1\times T_2+\frac{(1700)^2}{25037}$

$500+1.598=T_2+115.4$

$T_2=500+1.598-115.4$

$T_2=386.2^{\circ}F$

7 0

4 years ago

does anyone know how to do this? the scenario is ""dominique reads that race cars have wide tires because the increased area of

Usimov [2.4K]

Answer:

Stupid

Explanation:

Because there is never a answer when we are trying to find one

7 0

3 years ago

A 1000-kg car is traveling east at 20\:m.s^{-1}20 m . s − 1 and a 1200-kg car is traveling west at 22\:m.s^{-1}\:22 m . s − 1. W

nydimaria [60]

Answer:

490,400 J

Explanation:

Mass of first car, m = 1000 kg

Mass of second car, M = 1200 kg

velocity of first car, u = 20 m/s east

velocity of second car, U = 22 m/s west

The formula for the kinetic energy is

$k = \frac{1}{2}mv^{2}$

where, m is the body and v be the velocity of the body.

Total kinetic energy is given by

$k = \frac{1}{2}mu^{2}+\frac{1}{2}MU^{2}$

$k = \frac{1}{2}\times1000\times20^{2}+\frac{1}{2}\times1200\times22^{2}$

$k = 200000 + 290400$

k = 490,400 J

Thus, the total kinetic energy of the system is 490,400 J.

3 0

3 years ago