Answer:-683 cal
Explanation:
Given
Heat released by system Q=-255 cal
as heat released is taken as negative and vice-versa
Work done by system W=428 cal
From First law of thermodynamics
=change in internal Energy

Distance, since distance represents how far something has travelled, which would be in our case 2.5m.
Mass = 0.201kg
Energy = 15J
temperature change = 10C
Energy(E) = mass(m) × specific heat capacity(c) × temperature change(θ)
we can rearrange this to make specific heat capacity the subject
c =

c =

c =7.46268657
Answer:
306 m/s
Explanation:
Law of conservation of momentum
m1v1 + m2v2 = (m1+m2)vf
m1 is the bullet's mass so it is 0.1 kg
v1 is what we're trying to solve
m2 is the target's mass so it is 5.0 kg
v2 is the targets velocity, and since it was stationary, its velocity is zero
vf is the velocity after the target is struck by the bullet, so it is 6.0 m/s
plugging in, we get
(0.1 kg)(v1) + (5.0 kg)(0 m/s) = (0.1 kg + 5.0 kg)(6.0 m/s)
(0.1)(v1) + 0 = 30.6
(0.1)(v1) = 30.6
v1 = 306 m/s