Answer:
so initial speed of the rock is 30.32 m/s
correct answer is b. 30.3 m/s
Explanation:
given data
h = 15.0m
v = 25m/s
weight of the rock m = 3.00N
solution
we use here work-energy theorem that is express as here
work = change in the kinetic energy ..............................1
so it can be written as
work = force × distance ...................2
and
KE is express as
K.E = 0.5 × m × v²
and it can be written as
F × d = 0.5 × m × (vf)² - (vi)² ......................3
here
m is mass and vi and vf is initial and final velocity
F = mg = m (-9.8) , d = 15 m and v{f} = 25 m/s
so put value in equation 3 we get
m (-9.8) × 15 = 0.5 × m × (25)² - (vi)²
solve it we get
(vi)² = 919
vi = 30.32 m/s
so initial speed of the rock is 30.32 m/s
Answer:
577g
Explanation:
Given parameters:
Temperature change = 5.9°C
Amount of heat lost = 427J
Unknown:
Mass of the block = ?
Solution:
The heat capacity of a body is the amount of heat required to change the temperature of that body by 1°C.
H = m c Ф
H is the heat capacity
m is the mass of the block
c is the specific heat capacity
Ф is the temperature change
Specific heat capacity of lead is 0.126J/g°C
m = H / m Ф
m =
= 577g
Mass of the lead block is 577g