Three blocks are shown: Three blocks are shown. Block A has mass 3 kilograms, length 6 centimeters, height 4 centimeters, and wi
1 answer:
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Answer:
c = 0.4356 J/gK
Explanation:
Given the following data;
Mass = 450 grams
Initial temperature, T1 = 150°C
Final temperature, T2 = 53°C
Quantity of heat = 34500 Joules
To find the specific heat capacity of the metal;
Heat capacity is given by the formula;
Where;
Q represents the heat capacity or quantity of heat.
m represents the mass of an object.
c represents the specific heat capacity of water.
dt represents the change in temperature.
dt = T2 - T1
dt = 53 - 150
dt = -97°C
Converting the temperature in Celsius to Kelvin, we have;
dt = 273 + (-97) = 176 Kelvin
Making c the subject of formula, we have;
Substituting into the equation, we have;
c = 0.4356 J/gK
Answer:
electricfield at (0,0,0) is Et = 2 k q / a²
Explanation:
For the first part see the diagram , the field lines start from the positive charge and reach the negative charge, notice that no line should cross, some lines go to infinity
For the second part we use that the electric field is a vector quantity and therefore we add the field of each charge, using the equation
E = k q / r²
Point (0,0,0)
We calculate for the charge -q which is at a distance R = a
E1 = k (-q) / a²
E1 = - kq / a²
As the test charge is positive in the field it goes to the left, attractive force
We calculate for the charge that is also at R = a
E2 = k q / a²
This field goes to the left, repulsive force
We find the total electric field
Et = E1 + E2
Et = kq / a² + kq / a²
Et = 2 k q / a²
Point (0,0, R)
We use the same equations, but with another distance, for the charge -q the distance is R = R+a and for the charge + q the distance is R = R-a
E1 = k q / (R + a)²
E2 = kq / (R-a)²
Et = kq [1 / (R + a)² + 1 / (R-a)²]
Et= kq {[(R-a)² + (R + a)²] / [(R + a)² (R-a)²]}
Et= kq {2 (R² + a²) / [(R + a)² (R-a)²]}
If we use the condition that R> a we can despise in the patents "a"
(R² + a²) = R² (1+ a² / R²) ≈ R²
(R + a)² = R² (1 + a / R)² ≈ R²
(R- a)² = R² (1-a / R)² ≈ R²
Substituting in the total electric field
Et = kq {2 R²) / [R²R²]}
Et =kq 2 / R²
Answer:
given,
speed of the car = 20 m/s
final speed of car = 0 m/s
distance between car and the deer = 38 m
reaction time, t = 0.5 s
deceleration of the car = 10 m/s².
a) distance between deer and car
distance travel in the reaction time
d₁ = v x t
d₁ = 20 x 0.5 = 10 m
distance travel after you apply brake
using equation of motion
v² = u² + 2 a s
0 = 20² - 2 x 10 x s
s = 20 m
total distance traveled by the car
D = d₁ + d₂
D = 20 + 10 = 30 m
distance between car and the deer = 38 m - 30 m
= 8 m
b) now, maximum speed car.
distance travel in reaction time
d₁ = s x t
d₁ = 0.5 V
distance left between them
d₂ = 38 - d₁
d₂ = 38 - 0.5 V
distance travel after you apply brake
using equation of motion
v² = u² + 2 a d₂
0 = (V)² - 2 x 10 x (38 - 0.5 V)
V² + 10 V - 760 = 0
now, solving the quadratic equation
V = 23.01 , -33.01
rejecting the negative term.
hence, maximum speed of the car could be V = 23.01 m/s