The alpha particle is emitted at 4235 m/s
Explanation:
We can use the law of conservation of momentum to solve the problem: the total momentum of the original nucleus must be equal to the total momentum after the alpha particle has been emitted. Therefore:
where:
is the mass of the original nucleus
is the initial velocity of the nucleus
is the mass of the alpha particle
is the final velocity of the alpha particle
is the mass of the daughter nucleus
is the final velocity of the nucleus
Solving for , we find the final velocity of the alpha particle:
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Answer:
Option B. O because the net force was 5 N in Alfredo's direction
Explanation:
To know the the correct answer to the question given above, we shall determine the net force acting on the bat. This can be obtained as follow:
Force of pull by Mason (Fₘ) = 15 N
Force of pull by Alfredo (Fₐ) = 20 N
Net force (Fₙ) =?
Fₙ = 20 – 15
Fₙ = 5 N in Alfredo's direction
From the calculation made above, we can see that the net force is 5N in Alfredo's direction. This is the reason why Alfredo end up having the bat.
Answer:
Spring constant in N / m = 6,000
Explanation:
Given:
Length of spring stretches = 5 cm = 0.05 m
Force = 300 N
Find:
Spring constant in N / m
Computation:
Spring constant in N / m = Force/Distance
Spring constant in N / m = 300 / 0.05
Spring constant in N / m = 6,000
Responder:
<h2>
0.7Hertz
</h2>
Explicación:
Usando la fórmula para calcular la velocidad de onda que se expresa como se muestra.
Velocidad de una onda = frecuencia * longitud de onda
v = fλ
Dada la velocidad de onda = 14 m / sy longitud de onda = 20 metros
frecuencia f = v / λ
f = 14/20
f = 0.7Hertz
La frecuencia de la onda es de 0.7 Hertz.
Answer:
The amount of kilograms of ice at -20.0°C that must be dropped into the water to make the final temperature of the system 40.0°C = 0.0674 kg
Explanation:
Heat gained by ice in taking the total temperature to 40°C = Heat lost by the water
Total Heat gained by ice = Heat used by ice to move from -20°C to 0°C + Heat used to melt at 0°C + Heat used to reach 40°C from 0°C
To do this, we require the specific heat capacity of ice, latent heat of ice and the specific heat capacity of water. All will be obtained from literature.
Specific heat capacity of ice = Cᵢ = 2108 J/kg.°C
Latent heat of ice = L = 334000 J/kg
Specific heat capacity of water = C = 4186 J/kg.°C
Heat gained by ice in taking the total temperature to 40°C = mCᵢ ΔT + mL + mC ΔT = m(2108)(0 - (-20)) + m(334000) + m(4186)(40 - 0) = 42160m + 334000m + 167440m = 543600 m
Heat lost by water = mC ΔT = 0.25 (4186)(75 - 40) = 36627.5 J
543600 m = 36627.5
m = 0.0674 kg = 67.4 g of ice.
