Answer:
Displacement and Velocity
when drawing electric field lines ___________ charges have vectors point away/out and______ charges have vectors point toward/in.
Answer:
Positive, Negative
Explanation:
The image I've attached shows that vectors point into the negative source and vectors point away from the positive source.
Water has a heat capacity of 4.184 J/g °C. If 50 g of water has a temperature of 30ºC and a piece of hot copper is added to the water causing the temperature to increase to 70ºC. What is the amount of heat absorbed by the water?
The amount of heat absorbed by the water will be 8368 J.
What are heat gain and heat loss?Heat gain is defined as the amount of heat required to increase the temperature of a substance by some degree of Celsius. While heat loss is inverse to heat gain.
It is given by the formula as ;
[tex]\rm Q= mcdt[/tex]
The given data in the problem is;
Equilibrium temperature = 30°C.
mass of water = 50 g ,
Temperature change = 70ºC
The specific heat of water =4.184 J//g °C
The amount of heat absorbed by the water is;
[tex]\rm Q= mcdt \\\\ Q=50 \times 4.184 \times (70^0 -30^0)C\\\\ Q= 8368 J[/tex]
Hence, the amount of heat absorbed by the water will be 8368 J.
To learn more about the heat gain refer to the link;
https://brainly.com/question/26268921
#SPJ2
As a glacier melts, the volume V of the ice, measured in cubic kilometers, decreases at a rate modeled by the differential equation dVdt=kV , where t is measured in years. The volume of the glacier is 400km3 at time t=0 . At the moment when the volume of the glacier is 300km3 , the volume is decreasing at the rate of 15km3 per year. What is the volume V in terms of time
Solve the differential equation:
dV/dt = k V → 1/V dV/dt = k
→ d/dt [ln(V)] = k
→ ln(V) = k t + C
→ V (t )= exp(k t + C ) = C exp(k t ) = C e ᵏᵗ
At t = 0, the glacier has volume 400 km³ of ice, so
V (0) = 400 → C e⁰ = C = 400
Find when the glacier's volume is 300 km³:
V (t ) = 400 e ᵏᵗ = 300 → e ᵏᵗ = 3/4
→ k t = ln(3/4)
→ t = 1/k ln(3/4)
At this time, the volume is decreasing at a rate of 15 km³/yr, so
V ' (t ) = C k e ᵏᵗ → V ' (1/k ln(3/4)) = 400 k exp(k × 1/k ln(3/4)) = -15
→ 3/4 k = -3/80
→ k = -1/20
Then the volume V (t ) of the glacier at time t is
V (t ) = 400 exp(-1/20 t )
The volume in terms of time will be "V(t) = 400 exp (-[tex]\frac{1}{20}[/tex] t)".
Differential equation and VolumeAccording to the question,
Glacier's volume, V(0) = 400 km³
→ [tex]\frac{dV}{dt}[/tex] = kV
[tex]\frac{1}{V}[/tex] [tex]\frac{dV}{dt}[/tex] = k
ln(V) = kt + C
Now,
V (t) = exp (kt + C)
= C exp (kt)
= C[tex]e^{kt}[/tex]
When volume of glacier be "300 km³",
V(t) = 400 [tex]e^{kt}[/tex] = 300
[tex]e^{kt}[/tex] = [tex]\frac{3}{4}[/tex]
By taking log,
kt = ln([tex]\frac{3}{4}[/tex])
t = [tex]\frac{1}{k}[/tex] ln([tex]\frac{3}{4}[/tex])
When volume decrease at 15 km³/yr, then
→ V'(t) = Ck[tex]e^{kt}[/tex]
= 400 k exp (k × [tex]\frac{1}{k}[/tex] kn ([tex]\frac{3}{4}[/tex]))
= -15
Now,
[tex]\frac{3}{4}[/tex] k = - [tex]\frac{3}{80}[/tex]
By applying cross-multiplication,
k = - [tex]\frac{1}{20}[/tex]
Thus the response above is correct.
Find out more information about volume here:
https://brainly.com/question/5398201
