Answer:
Explanation:
Given
Initial velocity u = 200m/s
Final velocity = 4m/s
Distance S = 4000m
Required
Acceleration
Substitute the given parameters into the formula
v² = u²+2as
4² = 200²+2a(4000)
16 = 40000+8000a
8000a = 16-40000
8000a = -39,984
a = - 39,984/8000
a = -4.998m/s²
Hence the acceleration is -4.998m/s²
m = mass of the box
N = normal force on the box
f = kinetic frictional force on the box
a = acceleration of the box
μ = coefficient of kinetic friction
perpendicular to incline , force equation is given as
N = mg Cos30 eq-1
kinetic frictional force is given as
f = μ N
using eq-1
f = μ mg Cos30
parallel to incline , force equation is given as
mg Sin30 - f = ma
mg Sin30 - μ mg Cos30 = ma
"m" cancel out
a = g Sin30 - μ g Cos30
inserting the values
1.20 = (9.8) Sin30 - (9.8) Cos30 μ
μ = 0.44
Mass of gold m₁ = 47 g
Initial temperature of gold T₁ = 99 C
Specific heat of gold C₁ = 0.129 J/gC
final temperature T₂ = 38 C
Heat needed by the gold to cool down
Q =m₁ * C₁* ( T₁ - T₂)
Q = (47)(0.129)(99-38)
Q = 369.843 J
This heat will be given by the water
we need to find out mass of water m₂
and initial temperature of water is T₃ = 25 C
Specific heat of water C₂ = 4.184 J/gC
Q = m₂*C₂*(T₂ - T₃)
369.843 = m₂(4.184)(38-25)
m₂ = 6.8 g