Answer:
<h2>
3,343.68kJ </h2>
Explanation:
Heat energy used up can be calculated using the formula:
H = mcΔt
m = mass oof the object (in kg) = 20kg
c = specific heat capacity of water = 4179.6J/kg°C
Δt change in temperature = 80-40 = 40°C
H= 20* 4179.6 * 40
H = 3,343,680Joules
H = 3,343.68kJ
i do not have an answer because it depends on the size and the distance lol
The equation that relates distance, velocities, acceleration, and time is,
d = V₀t + 0.5gt²
where d is distance,
V₀ is the initial velocity,
t is time, and
g is the acceleration due to gravity (equal to 9.8 m/s²)
(1) Dropped rock,
(3 x 10² m ) = 0(t) + 0.5(9.8 m/s²)(t²)
The value of t from this equation is 24.73 s
(2) Thrown rock with V₀ = 26 m/s
(3 x 10² m) = (26)(t) + 0.5(9.8 m/s²)(t²)
The value of t from the equation is 5.61 s
The difference between the tim,
difference = 24.73 s - 5.61 s
difference = 19.12 s
<em>ANSWER: 19.12 s</em>
Answer:
The average acceleration of the ball during the collision with the wall is 
Explanation:
<u>Known Data</u>
We will asume initial speed has a negative direction, 
, final speed has a positive direction, 
, 
and mass 
.
<u>Initial momentum</u>
<u>final momentum</u>
<u>Impulse</u>
<u>Average Force</u>
<u>Average acceleration</u>
, so 
.
Therefore, 
k = 
k = (6.626×10-¹⁹/590 × 10-⁹ )^{2} /2 × 1.673 × 10-²⁷
k = (1.12 × 10-³⁰)^2/3.346×10-²⁷
k = 1.25 × 10-⁶⁰ /3.346×10-²⁷
k = 0
ldk why, my answer is coming this :(
