Answer
given,
mass = 100 kg
acceleration = 10 m/s²
A mass 20 kg slides over 100 kg block
acceleration = 3 m/s²
horizontal friction exerted by the 100 kg block on 20 kg
using newton's second law
F - f = 0
F = f
f = ma
f = 20 × 3
f = 60 N
now net force acting on the 100 kg block
F_net = m a
F_net = 100 x 10
F_net = 1000 N
after 20 kg block falls the acceleration of the bock
F = 1000 +60
F = 1060 N
acceleartion on the block


a = 10.60 m/s²
Answer:
Volume of balloon = 1000 cm^3
Explanation:
The head of a normal person can be assumed as a sphere with radius 10 cm.
Volume of sphere
, where r is the radius.
We have approximate radius = 10 cm.
Approximate volume of head 
In the given options the closest value to the approximate volume is 1000 cm^3.
So, volume of head = Volume of balloon = 1000 cm^3
The magnitude of the sum of the frictional forces acting on the bike and its rider is 400N.
<h3>What is friction force?</h3>
The friction force is the opposing force which acts on the object which is in relative motion.
The driving force is equal and opposite to the friction force acting between road and bicycle.
Friction force = 400N
The friction force between rider and bike is zero.
So the magnitude of sum of friction force = 400N +0 = 400N
Thus, the magnitude of the sum of the frictional forces acting on the bike and its rider.
Learn more about friction force.
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Answer:
Explanation:
We shall represent displacement in vector form .Consider east as x axes and north as Y axes west as - ve x axes and south as - ve Y axes . 255 km can be represented by the following vector
D₁ = - 255 cos 49 i + 255 sin49 j
= - 167.29 i + 192.45 j
Let D₂ be the further displacement which lands him 125 km east . So the resultant displacement is
D = 125 i
So
D₁ + D₂ = D
- 167.29 i + 192.45 j + D₂ = 125 i
D₂ = 125 i + 167.29 i - 192.45 j
= 292.29 i - 192.45 j
Angle of D₂ with x axes θ
tan θ = -192.45 / 292.29
= - 0.658
θ = 33.33 south of east
Magnitude of D₂
D₂² = ( 192.45)² + ( 292.29)²
D₂ = 350 km approx
Tan