Explanation:
a) m1u1 + m2u2 = v(m1+m2)
1000×6 + 5000×0 = v(1000+5000)
6000 + 0 = 6000v
v = 6000/6000
v = 1 m/s
b) ½ ×(m1+m2)v²
= 0.5×6000×1²
=0.5×6000
=3000J
=3KJ
c) solve for b4 collision and compare
Goodluck
Answer:
Distance 20 km and Displacement 0 km
His displaceent is 0 km because he ends his walk where he started. The total distance of his walk is 20 km because he walks 10 km to the store + 10km back home.
Answer:
binding energy is 99771 J/mol
Exlanation:
given data
threshold frequency = 2.50 ×
Hz
solution
we get here binding energy using threshold frequency of the metal that is express as
..................1
here E is the energy of electron per atom
and h is plank constant i.e.
and x is binding energy
and here N is the Avogadro constant =
so E will
E =
so put value in equation 1 we get
= 2.50 ×
×
solve it we get
x = 99770.99
so binding energy is 99771 J/mol
Answer:
75.6J
Explanation:
Hi!
To solve this problem we must use the first law of thermodynamics that states that the heat required to heat the air is the difference between the energy levels of the air when it enters and when it leaves the body,
Given the above we have the following equation.
Q=(m)(h2)-(m)(h1)
where
m=mass=1.3×10−3kg.
h2= entalpy at 37C
h1= entalpy at -20C
Q=m(h2-h1)
remember that the enthalpy differences for the air can approximate the specific heat multiplied by the temperature difference
Q=mCp(T2-T1)
Cp= specific heat of air = 1020 J/kg⋅K
Q=(1.3×10−3)(1020)(37-(-20))=75.6J
Answer:
Work done by friction along the motion is given as

Explanation:
As per work energy theorem we can say
Work done by all forces = change in kinetic energy of the system
so here car is moving from bottom to top
so here the change in kinetic energy is total work done on the car
so here we will have


now plug in all data in it


