Answer:
The magnitude of the acceleration is equal to 19.6m/s² and the acceleration is directed upwards though the magnitude of the charge has doubled. This is because the electric force is directed upwards and from newton's second law of motion the charge will have acceleration in the same direction as the electric force on the charge.
Explanation:
The detailed solution can be found in the attachment below.
Thank you for reading and I hope this is helpful to you.
q = 1156363.6W/m².
To calculate the heat flux per unit area (W/m²) of a sheet made of metal:
q = -k(ΔT/Δx)
q = -k[(T₂ - T₁)/Δx]
Where, k is the thermal conductivity of the metal, ΔT is the temperature difference and Δx is the thick.
With Δx = 11 mm = 11x10⁻³m, T₂ = 350°C and T₁ = 110°C, and k = 53.0 W/m-K:
q = -53.0W/m-K[(110°C - 350°C)/11x10⁻³m
q = 1156363.6W/m²
:
Please give me brainly,
The thermal energy of an object depends on three things: 4 the number of molecules in the object 4 the temperature of the object (average molecular motion) 4 the arrangement of the object's molecules (states of matter). The more molecules an object has at a given temperature, the more thermal energy it has.
Hello There!!
<h3><u>What is kinematics?</u></h3>
<h3>Answer</h3>
=> Let The Answer be "x"
________________________________
<h3><u>Name it's </u><u>father</u></h3>
<h3>Answer</h3>
=> He don't have a child so he is not a father
________________________________
<h3><u>Explain its </u><u>History</u></h3>
<h3>Answer</h3>
=> Kinematics is a part of science not history.
__________________________________
Hope this helps :D
a fórmula do calor sensível é Q = m.c.ΔT
m = massa
c = calor específico do corpo
ΔT = variação de temperatura (Tf - Ti)
quando ocorre o equilíbrio térmico, a soma dos calores trocados entre os corpos é 0:
QA + QB + QC = 0
sabendo que a capacidade térmica é o produto m.c e substituindo os demais valores fornecidos no enunciado:
QA = calor sensível do calorímetro = (6 cal/°C).(42-18)
QB = calor sensível do líquido = (85g).c.(42-18)
QC = calor sensível do bloco = (120g).(0,094 cal/g°C).(42-100)
substituindo na expressão para calcular c, o calor específico do líquido :
(6 cal/°C).(42°C-18°C) + (85g).c.(42°C-18°C) + (120g).(0,094 cal/g°C).(42°C-100°C) = 0
144cal + 2040c - 654,2cal = 0
2040c = 510cal
c = 510cal/2040g.°C
c = 0,25cal/g.°C