Answer: Δθ = 127.4 K
Explanation: by using the law of conservation of energy, the kinetic energy of the bullet equals the heat energy on the plate.
Kinetic energy of bullet = mv²/2
Heat energy = mcΔθ
Where m = mass of bullet = 0.09kg, v = velocity of bullet = 182 m/s, c = specific heat capacity of lead bullet = 130 j/kgk
Δθ = change in temperature
mv²/2 = mcΔθ
With 'm' on both sides of the equation, they cancel out each other, hence we have that
v²/2 = cΔθ
v² = 2cΔθ
Δθ= v²/2c
Δθ = (182)²/2×130
Δθ = 33124/260
Δθ = 127.4 K
Answer:
A) continue to move to the right, with its speed increasing with time.
Explanation:
As long as force is positive , even when it is decreasing , it will create positive increase in velocity . Hence the body will keep moving with increasing velocity towards the right . The moment the force becomes zero on continuously decreasing , the increase in velocity stops and the body will be moving with the last velocity uniformly towards right . When the force acting on it becomes negative , even then the body will keep on going to the right till negative force makes its velocity zero . D uring this period , the body will keep moving towards right with decreasing velocity .
Hence in the present case A , is the right choice.
The correct answer is C) type of medium. Electromagnetic waves travel faster in solids than in liquids, and faster in liquids than in gases.
Saludos!
Respuesta:28,64 m/s.
Explicación:Datos:
Altura o distancia recorrida: 40 m
Vo: 6 m/s
Aceleración de la gravedad: 9,81 m/s²
El ejercicio puede ser resuelto facilmente utilizando la siguiente formula, sin embargo es posible realizarlo utilizando formulas diferentes.
Entonces tenemos que:
![Vf ^{2} -Vo ^{2} =2 x g x h](https://tex.z-dn.net/?f=Vf%20%5E%7B2%7D%20-Vo%20%5E%7B2%7D%20%3D2%20x%20g%20x%20h)
Es importante saber que al estar lanzando el ladrillo hacia abajo, el sentido del movimiento sigue el sentido de la gravedad, es decir es necesario que tomes el valor de la gravedad como positivo (+) y no negativo (-) como normalmente se usa.
Sustituyendo tenemos que:
![Vf x^{2} =(6 m/s) ^{2} +(2)x(9,81 m/s ^{2})x(40m) \\ Vf ^{2} =820,8 m^{2} /s ^{2} \\ \sqrt{Vf ^{2} } = \sqrt{820,8m^{2}/s ^{2} } \\ Vf=28,64 m/s](https://tex.z-dn.net/?f=Vf%20x%5E%7B2%7D%20%3D%286%20m%2Fs%29%20%5E%7B2%7D%20%2B%282%29x%289%2C81%20m%2Fs%20%5E%7B2%7D%29x%2840m%29%20%5C%5C%20Vf%20%5E%7B2%7D%20%3D820%2C8%20m%5E%7B2%7D%20%2Fs%20%5E%7B2%7D%20%20%5C%5C%20%20%5Csqrt%7BVf%20%5E%7B2%7D%20%7D%20%3D%20%5Csqrt%7B820%2C8m%5E%7B2%7D%2Fs%20%5E%7B2%7D%20%20%7D%20%20%5C%5C%20Vf%3D28%2C64%20m%2Fs)
Que tengas un buen día!