Answer:
Q = 177J
Explanation:
Specific heat capacity of lead=0.13J/gc
Q=MCΔT
ΔT=T2-T1,where T1=22degrees Celsius and T2=37degree Celsius.
ΔT=37 - 22 = 15
Q = Change in energy
M = mass of substance
C= Specific heat capacity
Q = (91g) * (0.13J/gc) * (15c)= 177.45J
Approximately, Q = 177J
The answer is 6 and a half days
Answer:
a) 42.9 m/s
b) 41.6m
Explanation:
a)
solving the eq we get
v=42.9 m/s
b) <em>here,</em>
when put in y(t) it gives y=44.6 m so result is
h=44.6-3= 41.6m
Answer:
x = 0.67 m
Explanation:
For this problem, let's use the projectile launch equations, as the jug goes through the bar, it comes out with horizontal speed vx = 1.3 m / s, which does not decrease as there is no friction.
Let's find the time or it takes to get to the floor
y = y₀ + v_{oy} - ½ g t²
in this case I go = 0 and when I get to the floor y = 0
0 = y₀ + 0 - ½ g t²
t² = 2y₀ / g
t² = 2 1.3 / 9.8 = 0.2653
t = 0.515 s
now let's find the distance traveled in this time
x = vx t
x = 1.3 0.515
x = 0.6696 m
x = 0.67 m