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3 years ago

An object wants to maintain its state of motion because it has mass

Physics

1 answer:

-Dominant- [34]3 years ago

8 0

I don't know what the question is but if its true or false then it is true.

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Suppose the coefficient of static friction between a quarter and the back wall of a rocket car is 0.330. At what minimum rate wo

Helga [31]

Answer: $3.23 m/s^2$

Explanation:

Given

$\mu_s =0.330$

Frictional Force is balanced by force due to car acceleration

Frictional force $F_s$

$F_s=ma_{min}$

$\mu_sN=ma_{min}$

$\mu_s\cdot mg=ma_{min}$

$a_{min}=\mu_s \cdot g=0.330\times 9.8=3.23 m/s^2$

6 0

3 years ago

Using equations, determine the temperature, pressure and density of the air for a aircraft flying at 19.5 km. Is this aircraft s

Viefleur [7K]

Answer:

a) - 72.5°c

b) pressure = 3625.13 Pa

c) density = 0.063 kg/m^3

d) it is a subsonic aircraft

Explanation:

a) Determine Temperature

Temperature at 19.5 km ( 19500 m )

T = -131 + ( 0.003 * altitude in meters )

= -131 + ( 0.003 * 19500 ) = - 72.5°c

b) Determine pressure and density at 19.5 km altitude

Given :

Po (atmospheric pressure at sea level ) = 101kpa

R ( gas constant of air ) = 0.287 KJ/Kgk

T = -72.5°c ≈ 200.5 k

pressure = 3625.13 Pa

hence density = 0.063 kg/m^3

attached below is the remaining part of the solution

C) determine if the aircraft is subsonic or super sonic

Velocity ( v ) = $\sqrt{CRT}$ = $\sqrt{1.4*287*200.5 }$ = 283.8 m/s

hence it is a subsonic aircraft

4 0

3 years ago

If two particles have equal kinetic energies, are their momenta necessarily equal? explain.

Mandarinka [93]

Answer:

No the given statement is not necessarily true.

Explanation:

We know that the kinetic energy of a particle of mass 'm' moving with velocity 'v' is given by

$K.E=\frac{1}{2}mv^{2}$

Similarly the momentum is given by $m\times v$

For 2 particles with masses $m_{1},m_{2}$ and moving with velocities $v_{1},v_{2}$ respectively the respective kinetic energies is given by

$K.E_{1}=\frac{1}{2}m_{1}v_{1}^{2}$

$K.E_{2}=\frac{1}{2}m_{2}v_{2}^{2}$

Similarly For 2 particles with masses $m_{1},m_{2}$ and moving with velocities $v_{1},v_{2}$ respectively the respective momenta are given by

$p_{1}=m_{1}\times v_{1}$

$p_{2}=m_{2}\times v_{2}$

Now since it is given that the two kinetic energies are equal thus we have

$\frac{1}{2}m_{1}v_{1}^{2}=\frac{1}{2}m_{2}v_{2}^{2}\\\\(m_{1}v_{1})\times v_{1}=(m_{2}v_{2})\times v_{2}\\\\p_{1}\times v_{1}=p_{2}\times v_{2}\\\\\therefore \frac{p_{1}}{p_{2}}=\frac{v_{2}}{v_{1}}............(i)$

Thus we infer that the moumenta are not equal since the ratio on right of 'i' is not 1 , and can be 1 only if the velocities of the 2 particles are equal which becomes a special case and not a general case.

5 0

3 years ago

How much force is required to accelerate a 5 kg mass at 20 m/s^2

Studentka2010 [4]

Hello!

$\large\boxed{F = 100N}$