Answer:
t = o.6 s
Explanation:
Let ball thrown from below be A and ball dropped from above be B.
A and B meet when they both are same level above the ground. Then let A moved up a distance d and B dropped a distance h. Then you know
d + h = 15 m ---------------(1)
Now apply s = ut + 
at²
To A upwards,
d = 25t - gt² -----------------(2)
To B downwards,
h = 0 + gt² ----------------(3)
(1) = (2) + (3) ⇒ 15 = 25t
t = 0.6 s
Explanation:
Solution:
Let the time be
t1=35min = 0.58min
t2=10min=0.166min
t3=45min= 0.75min
t4=35min= 0.58min
let the velocities be
v1=100km/h
v2=55km/h
v3=35km/h
a. Determine the average speed for the trip. km/h
first we have to solve for the distance
S=s1+s2+s3
S= v1t1+v2t2+v3t3
S= 100*0.58+55*0.166+35*0.75
S=58+9.13+26.25
S=93.38km
V=S/t1+t2+t3+t4
V=93.38/0.58+0.166+0.75+0.58
V=93.38/2.076
V=44.98km/h
b. the distance is 93.38km
Answer: 225N to the right
Explanation:
The pedaling force is to the right and of 325N.
The air resistance is to the opposite side (in this case to the left) of 100N.
To find Net force of an object with two opposite forces acting on them then we have to subtract the smaller force from the bigger force and but the direction of the bigger force, so
325N-100N=225N
The greater force is towards the right proving the direction of the Net force is to the right.
Therefore the Net force is 225N to the right.
Please mark me brainliest.
Answer: 45.3°
Explanation:
Given,
Length of ladder = l
Weight of ladder = w
Coefficient of friction = μs = 0.495
Smallest angle the ladder makes = θ
If we assume the forces in the vertical direction to be N1, and the forces in the horizontal direction to be N2, then,
N1 = mg and
N2 = μmg
Moment at a point A in the clockwise direction is
N2 Lsinθ - mg.(L/2).cosθ = 0
μmgLsinθ - mg.(L/2).cosθ = 0
μmgLsinθ = mg.(L/2).cosθ
μsinθ = cosθ/2
sin θ / cos θ = 1 / 2μ
Tan θ = 1 / 2μ
Substituting the value of μ = 0.495, we have
Tan θ = 1 / 2 * 0.495
Tan θ = 1 / 0.99
Tan θ = 1.01
θ = tan^-1(1.01)
θ = 45.3°
