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lara [203]
2 years ago
8

In Paul Hewitt's book, he poses this question: "If the forces that act on a bullet and the recoiling gun from which it is fired

are equal in magnitude, why do the bullet and gun have very different accelerations?" Think about Newton's Second Law (F = ma) and answer his question.
Physics
1 answer:
Sauron [17]2 years ago
4 0
They have different accelerations because of their masses. According to Newton's Second Law, an objects acceleration is inversely proportional to its mass. Therefore the object with the larger mass, in this case the gun, will have a smaller acceleration. In the same way, the less massive object, being the bullet, will have a higher acceleration.

Hope this helps :)
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A car with a mass of 1,324 kilograms, traveling at a speed of 20 meters/second, crashes into a wall and stops. What is the kinet
Valentin [98]
I think the correct answer from the choices listed above is option A. The kinetic energy after the perfectly inelastic collision would be zero Joules. <span>A </span>perfectly inelastic collision<span> occurs when the maximum amount of kinetic energy of a system is lost. Hope this answers the question.</span>
7 0
3 years ago
Read 2 more answers
How much heat is released when 432 g of water cools down from 71'c to 18'c?
maria [59]
The heat released by the water when it cools down by a temperature difference \Delta T is
Q=mC_s \Delta T
where
m=432 g is the mass of the water
C_s = 4.18 J/g^{\circ}C is the specific heat capacity of water
\Delta T =71^{\circ}C-18^{\circ}C=53^{\circ} is the decrease of temperature of the water

Plugging the numbers into the equation, we find
Q=(432 g)(4.18 J/g^{\circ}C)(53^{\circ}C)=9.57 \cdot 10^4 J
and this is the amount of heat released by the water.
7 0
2 years ago
1. Si tengo medio kilo de fruta y te doy un cuarto y tú me das tres cuartos de kilo, ¿cuánto tengo? 2. Si en una carrera te qued
Kipish [7]

Answer:

1. Tienes 1 kg de fruta.

2. Queda por recorrer 1/4 km.

3. Ambos pesan lo mismo.

Explanation:

1. Tienes 1/2 kg y cuando te doy 1/4 te queda:

m = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}

Ahora cuando te doy 3/4 kg te queda en total:

m_{T} = \frac{1}{4} + \frac{3}{4} = 1 kg

Por lo tanto, tienes 1 kg de fruta al final.

2. Si falta por recorrer la mitad de la mitad, tenemos:

d = \frac{1/2}{2} = \frac{1}{4}

Entonces, queda por recorrer 1/4 km.

3. El peso (P) del hierro es:

P = m*g    

P = (1 + 1/2)kg*9.81 m/s^{2} = 14.72 N

Y el peso de la paja es:

P = 3/2 kg*9.81 m/s^{2} = 14.72 N

Por lo tanto, ambos pesan lo mismo.

Espero que te sea de utilidad!

6 0
2 years ago
1.5 kg of air within a piston-cylinder assembly executes a Carnot power cycle with maximum and minimum temperatures of 800 K and
liraira [26]

Answer:

Explanation:

Carton cycle consists of four thermodynamic processes . The first is isothermal expansion at higher temperature , then adiabatic expansion which lowers the temperature of gas . The third process is isothermal compression at lower temperature and the last process is adiabatic compression which increases the temperature of the gas to its original temperature .

So the given process of isothermal compression must have been done at the temperature of 300K  , keeping the temperature constant .

Work done on gas at isothermal compression is equal to heat transfer .

work done on gas = 80 x 10³ J

work done on gas = n RT ln v₁ / v₂

n is number of moles v₁ and v₂ are initial and final volume

molecular weight of gas = 28.97 g

1.5 kg = 1500 / 28.97 moles

= 51.77 moles

work done on gas = n RT ln v₁ / v₂

Putting the values in the equation above

80 x 10³ = 51.78 x 8.31 x 300 x ln v₁ / .2

ln v₁ / .2 = .62

v₁ / .2 = 1.8589

v₁ = 0.37 m³

3 0
2 years ago
A car with speed v and an identical car with speed 2v both travel the same circular section of an unbanked road. If the friction
yawa3891 [41]

Answer:

F'=\dfrac{F}{4}

Explanation:

Let m is the mass of both cars. The first car is moving with speed v and the other car is moving with speed 2v. The only force acting on both cars is the centripetal force.

For faster car on the road,

F=\dfrac{mv^2}{r}

v = 2v

F=\dfrac{m(2v)^2}{r}

F=4\dfrac{m(v)^2}{r}..........(1)

For the slower car on the road,

F'=\dfrac{mv^2}{r}............(2)

Equation (1) becomes,

F=4F'

F'=\dfrac{F}{4}

So, the frictional force required to keep the slower car on the road without skidding is one fourth of the faster car.

8 0
2 years ago
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