Answer:
The higher the altitude the colder
Explanation:
Mountains are very high up obviously higher then the ground. It is colder so it makes it easier for mountain climbers to get frost bite because of the high altitude.
Hope this helps:)
Answer:
41.27m/s
Explanation:
According to law of conservation of momentum
m1u1+m2u2 = (m1+m2)v
m1 and m2 are the masses
u1 and u2 are the initial velocities
v is the velocity after impact
Given
m1 = 0.2kg
u1 = 43.7m/s
m2 = 45.9g = 0.0459kg
u2 = 30.7m/s
Required
Velocity after impact v
Substitute the given parameters into the formula
0.2(43.7)+0.0459(30.7) = (0.2+0.0459)v
8.74+1.409 = 0.2459v
10.149 = 0.2459v
v = 10.149/0.2459
v = 41.27m/s
Hence the speed of the golf ball immediately after impact is 41.27m/s
Answer:
a) Fc = 4.15 N, Fi = 435.65 N, (F1)a = 640 N, and F2 = 239.6 N,
b) Ha = 1863.75 N, nfs = 1 , length = 11.8 mm
Explanation:
Given that:
γ= 9.5 kN/m³ = 9500N/m3
b = 6 inches = 0.1524 m
t = 0.0013 mm
d = 2 inches = 0.0508 m
n = 1750 rpm

L = 9 ft = 2.7432 m
Ks = 1.25
g = 9.81 m/s²
a)







b)


dip = 
Explanation:
It is given that,
Mass of bumper car, m₁ = 202 kg
Initial speed of the bumper car, u₁ = 8.5 m/s
Mass of the other car, m₂ = 355 kg
Initial velocity of the other car is 0 as it at rest, u₂ = 0
Final velocity of the other car after collision, v₂ = 5.8 m/s
Let p₁ is momentum of of 202 kg car, p₁ = m₁v₁
Using the conservation of linear momentum as :


p₁ = m₁v₁ = -342 kg-m/s
So, the momentum of the 202 kg car afterwards is 342 kg-m/s. Hence, this is the required solution.
Answer:
(A) The correct answer is option (B) three halves that of the old unit.
(B) The answer is three fourth that of old unit
Explanation:
from the relation;
(A) Fromthe expression;
K = Qd/AΔT
Anew = 3/2 A(old)
(B)
K¹ = 2K(old), so we get
A(old) = A(old)/2
Combining with part A, we have;
Anew = 3/2 *A(old)/2
= 3/4A(old)
The answer is three fourth that of old unit