Answer:
4 s
Explanation:
Given:
Δx = 12 m
v₀ = 6 m/s
v = 0 m/s
Find: t
Δx = ½ (v + v₀) t
12 m = ½ (0 m/s + 6 m/s) t
t = 4 s
Answer:
1. True WA > WB > WC
Explanation:
In this exercise they give work for several different configurations and ask that we show the relationship between them, the best way to do this is to calculate each work separately.
A) Work is the product of force by distance and the cosine of the angle between them
WA = W h cos 0
WA = mg h
B) On a ramp without rubbing
Sin30 = h / L
L = h / sin 30
WB = F d cos θ
WB = F L cos 30
WB = mf (h / sin30) cos 30
WB = mg h ctan 30
C) Ramp with rubbing
W sin 30 - fr = ma
N- Wcos30 = 0
W sin 30 - μ W cos 30 = ma
F = W (sin30 - μ cos30)
WC = mg (sin30 - μ cos30) h / sin30
Wc = mg (1 - μ ctan30) h
When we review the affirmation it is the work where there is rubbing is the smallest and the work where it comes in free fall at the maximum
Let's review the claims
1. True The work of gravity is the greatest and the work where there is friction is the least
2 False. The job where there is friction is the least
3 False work with rubbing is the least
4 False work with rubbing is the least
M = mass of the first sphere = 10 kg
m = mass of the second sphere = 8 kg
V = initial velocity of the first sphere before collision = 10 m/s
v = initial velocity of the second sphere before collision = 0 m/s
V' = final velocity of the first sphere after collision = ?
v' = final velocity of the second sphere after collision = 4 m/s
using conservation of momentum
M V + m v = M V' + m v'
(10) (10) + (8) (0) = (10) V' + (8) (4)
100 = (10) V' + 32
(10) V' = 68
V' = 6.8 m/s
Answer:
the rate of change in volume with time is 280πr² cm³/min
Explanation:
Data provided in the question:
Radius of the sphere as 'r'
= 70 cm/min
Volume of the sphere, V =
Surface area of the sphere as 4πr²
Now,
Rate of change in volume with time,
=
=
Substituting the value of 
=
= 280πr² cm³/min
Hence, the rate of change in volume with time is 280πr² cm³/min