Answer:
A. 2.57 K
Explanation:
Using the specific heat capacity,
Q = cmΔT........................ Equation 1
Where Q = Heat energy absorbed by silver, m = mass of silver, c = specific heat capacity for silver, ΔT = temperature change of the silver.
Rearranging gives ΔT as the subject of the equation
ΔT = Q/cm................... Equation 2
Provided: Q = 300 J, m = 500 g = 0.5 kg
Constant: c = 233 J/kg.K
Inserting values into equation 2 yields
ΔT = 300/(0.5×233)
ΔT = 300/116.5
ΔT = 2.57 K
Therefore, the correct answer is A. 2.57 K
Answer:
0.128 rad/s², 7.66 rad/s
Explanation:
length, l = 66.4 cm
initial angular velocity, ωo = 0 rad/s
Let ω represent the final angular velocity.
Let α denote the angular acceleration.
number of revolutions, n = 36.6
time taken, t = 1 min = 60 seconds
Angle rotated, θ = 2πn = 2 x 3.14 x 36.6 = 229.85 rad
Apply the second equation of motion for angular dynamics
229.85 = 0 + 0.5 x α x 60 x 60
α = 0.128 rad/s²
Utilize the first equation of motion
ω = ωo + αt
ω = 0 + 0.128 x 60
ω = 7.66 rad/s
The acceleration, a = 9.8 m/s². Explanation: The phrase 'A ball is dropped from the top of a building' implies that the ball's initial velocity is zero, u = 0 m/s. After 2 seconds, the ball's velocity measures 19.6 m/s, with t = 2s, v = 19.6 m/s. Employing the equation
v = u + at yields 19.6 = 0 + 2a, hence a = 9.8 m/s².
If the object remains stationary at the onset of 10 seconds, it will descend 490 meters straight downward over 10 seconds.
(Note: This holds true for all items on Earth... rubber balls, feathers, grains of sand, school buses, battleships... everything. As long as air resistance is not an issue. Any object dropped from rest falls 490 meters in the initial 10 seconds.)
