Refer to the diagram shown below.
Still-water speed = 9.5 m/s
River speed = 3.75 m/s down stream.
The velocity of the swimmer relative to the bank is the vector sum of his still-water speed and the speed of the river.
The velocity relative to the bank is
V = √(9.5² + 3.75²) = 10.21 m/s
The downstream angle is
θ = tan⁻¹ 3.75/9.5 = 21.5°
Answer: 10.2 m/s at 21.5° downstream.
According to the net force, the acceleration of the book is 16.47 m/s².
We need to know about force to solve this problem. According to second Newton's Law, the force applied to an object will be proportional to mass and acceleration. Hence, it can be written as
∑F = m . a
where F is force, m is mass and a is acceleration
From the question above, we know that
m = 3 kg
g = 9.8 m/s²
F1 = 20 N
Find the net force
∑F = F1 + W
∑F = 20 + m . g
∑F = 20 + 3 . 9.8
∑F = 20 + 29.4
∑F = 49.4 N
Find the acceleration
∑F = m . a
49.4 = 3 . a
a = 16.47 m/s²
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Answer:
Explanation:
A ) When gymnast is motionless , he is in equilibrium
T = mg
= 63 x 9.81
= 618.03 N
B )
When gymnast climbs up at a constant rate , he is still in equilibrium ie net force acting on it is zero as acceleration is zero.
T = mg
= 618.03 N
C ) If the gymnast climbs up the rope with an upward acceleration of magnitude 0.600 m/s2
Net force on it = T - mg , acting in upward direction
T - mg = m a
T = mg + m a
= m ( g + a )
= 63 ( 9.81 + .6)
= 655.83 N
D ) If the gymnast slides down the rope with a downward acceleration of magnitude 0.600 m/s2
Net force acting in downward direction
mg - T = ma
T = m ( g - a )
= 63 x ( 9.81 - .6 )
= 580.23 N
Answer:
Part a)
Part B)
Part C)
Explanation:
Part A)
As we know that ball is hanging from the top and its angle with the vertical is 20 degree
so we will have
Part B)
Here we can use energy theorem to find the distance that it will move
Part C)
At terminal speed condition we know that
