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3 years ago

Which of these has the most inertia?

Physics

1 answer:

VashaNatasha [74]3 years ago

3 0

Answer:

A. a parked bus

Explanation:

Because a parked bus probably has the most mass out of these 4 and as we know, mass is that quantity that is solely dependent upon the inertia of an object. The more inertia that an object has, the more mass that it has. A more massive object has a greater tendency to resist changes in its state of motion.

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Which of the following have the same density. number 26

marusya05 [52]

1 cubic cm is the same as 1 mL, so the answer would be C.

8 0

3 years ago

A cat dozes on a stationary merry-go-round, at a radius of 5.7 m from the center of the ride. Then the operator turns on the rid

tatyana61 [14]

Answer:

0.76

Explanation:

we are given:

radius (r) =5.7 m

speed (s) = 1 revolution in 5.5 seconds

acceleration due to gravity (g) = 9.8 m/s^{2}

coefficient of friction (Uk) = ?

we can get the minimum coefficient of friction from the equation below

centrifugal force = frictional force

m x r x ω^{2} = Uk x m x g

r x ω^{2} = Uk x g

Uk = $\frac{ r x ω^{2} }{g}$

where ω (angular velocity) = $\frac{2π}{time}$

= $\frac{2π }{5.5}$ = 1.14

Uk = $\frac{ 5.7 x 1.14^{2} }{9.8}$ = 0.76

6 0

3 years ago

Argon gas enters steadily an adiabatic turbine at 900 kPa and 450C with a velocity of 80 m/s and leaves at 150 kPa with a veloc

Crazy boy [7]

Answer:

Temperature at the exit = $267.3 C$

Explanation:

For the steady energy flow through a control volume, the power output is given as

$W_{out}= -m_{f}(h_{2}-h_{1} + \frac{v_{2}^{2}}{2} - \frac{v_{1}^{2}}{2})$

Inlet area of the turbine = $60cm^{2}= 0.006m^{2}$

To find the mass flow rate, we can apply the ideal gas laws to estimate the specific volume, from there we can get the mass flow rate.

Assuming Argon behaves as an Ideal gas, we have the specific volume $v_{1}$

$v_{1}=\frac{RT_{1}}{P_{1}}=\frac{0.2081\times723}{900}=0.1672m^{3}/kg$

$m_{f}=\frac{1}{v_{1}}\times A_{1}V_{1} = \frac{1}{0.1672}\times(0.006)(80)=2.871kg/sec$

for Ideal gasses, the enthalpy change can be calculated using the formula

$h_{2}-h_{1}=C_{p}(T_{2}-T_{1})$

hence we have

$W_{out}= -m_{f}((C_{p}(T_{2}-T_{1}) + \frac{v_{2}^{2}}{2} - \frac{v_{1}^{2}}{2})$

$250= -2.871((0.5203(T_{2}-450) + \frac{150^{2}}{2\times 1000} - \frac{80^{2}}{2\times 1000})$

<em>Note: to convert the Kinetic energy term to kilojoules, it was multiplied by 1000</em>

evaluating the above equation, we have $T_{2}=267.3C$

Hence, the temperature at the exit = $267.3 C$

5 0

3 years ago

What is an easy way to balance chemical equations?

Lorico [155]

write out what you have on both sides, then just use basic multiplication to try and even out both sides. I can help if you need me to balance some for you!!

8 0

3 years ago

What is described by the terms body-centered cubic and face-centered cubic?

Montano1993 [528]

In physical chemistry, the terms body-centered cubic (BCC) and face-centered cubic (FCC) refer to the cubic crystal system of a solid. Each solid is made up simple building blocks called lattice units. There are different layouts of a lattice unit.

It is better understood using 3-D models shown in the picture. A BCC unit cell has one lattice point in the center, together with eight corner atoms which represents 1/8 of an atom. Therefore, there are 1+ 8(1/8) = 2 atoms in a BCC unit cell. On the other hand, a FCC unit cell is composed of half of an atom in each of its faces and 1/8 of an atom in its corners. Therefore, there are (1/2)6 + (1/8)8 = 4 atoms in a FCC unit cell.

7 0

3 years ago