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3 years ago

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Object A has a mass m and a speed v, object B has a mass m/2 and a speed 4v, and object C has a mass 3m and a speed v/3. Rank th

e objects according to the magnitude of their momentum.

1 answer:

borishaifa [10]3 years ago

5 0

<h2>The rank is : $P_B>P_A=P_C$ .</h2>

We need to rank the objects according to the magnitude of their momentum.

We know momentum , $P = m\times v$ { here m is mass and v is velocity }

Momentum of object A , $P_A = m v.$

Momentum of object B , $P_B= \dfrac{m}{2}\times4v=2\ mv.$

Momentum of object C , $P_C= 3m\times\dfrac{v}{3}=mv.$

Now, we can rank them in order of their magnitude of momentum .

$P_B>P_A=P_C$ .

Hence, this is the required solution.

Learn More :

Momentum

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3 years ago

Nitrogen at 100 kPa and 25oC in a rigid vessel is heated until its pressure is 300 kPa. Calculate (a) the work done and (b) the

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Answer:

A. The work done during the process is W = 0

B. The value of heat transfer during the process Q = 442.83 $\frac{KJ}{kg}$

Explanation:

Given Data

Initial pressure $P_{1}$ = 100 k pa

Initial temperature $T_{1}$ = 25 degree Celsius = 298 Kelvin

Final pressure $P_{2}$ = 300 k pa

Vessel is rigid so change in volume of the gas is zero. so that initial volume is equal to final volume.

⇒ $V_{1}$ = $V_{2}$ ------------- (1)

Since volume of the gas is constant so pressure of the gas is directly proportional to the temperature of the gas.

⇒ P ∝ T

⇒ $\frac{P_{2} }{P_{1}}$ = $\frac{T_{2} }{T_{1}}$

⇒ Put all the values in the above formula we get the final temperature

⇒ $T_{2}$ = $\frac{300}{100}$ × 298

⇒ $T_{2}$ = 894 Kelvin

(A). Work done during the process is given by W = P × ( $V_{2} -V _{1}$ )

From equation (1), $V_{1}$ = $V_{2}$ so work done W = P × 0 = 0

⇒ W = 0

Therefore the work done during the process is zero.

Heat transfer during the process is given by the formula Q = m $C_{v}$ ( $T_{2}$ - $T_{1}$ )

Where m = mass of the gas = 1 kg

$C_{v}$ = specific heat at constant volume of nitrogen = 0.743 $\frac{KJ}{kg k}$

Thus the heat transfer Q = 1 × 0.743 × ( 894- 298 )

⇒ Q = 442.83 $\frac{KJ}{kg}$

Therefore the value of heat transfer during the process Q = 442.83 $\frac{KJ}{kg}$

6 0

3 years ago

A concave mirror has a focal length of 13.5 cm. This mirror forms an image located 37.5 cm in front of the mirror. Find the magn

77julia77 [94]

Explanation:

It is given that,

Focal length of the concave mirror, f = -13.5 cm

Image distance, v = -37.5 cm (in front of mirror)

Let u is the object distance. It can be calculated using the mirror's formula as :

$\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$

$\dfrac{1}{u}=\dfrac{1}{f}-\dfrac{1}{v}$

$\dfrac{1}{u}=\dfrac{1}{(-13.5)}-\dfrac{1}{(-37.5)}$

u = -21.09 cm

The magnification of the mirror is given by :

$m=\dfrac{-v}{u}$

$m=\dfrac{-(-37.5)}{(-21.09)}$

m = -1.77

So, the magnification produced by the mirror is (-1.77). Hence, this is the required solution.

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3 years ago