A clump of soft clay is thrown horizontally from 9.80 m above the ground with a speed of 20.0 m/s. Assume it sticks in place whe
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(a)The acceleration of the 1,400-kg boat will be 0.425 m/sec²
(b) If it starts from rest, the distance through which the boat moves in 20.0 s will be 85 m.
(c)Velocity at the end of that time will be 8.5 m/sec.
<h3>What is velocity?</h3>The change of distance with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.
Given data;
Forwad force acting on the boat,v₁ = 1,800-N
Resistive force acting on the boat,v₂ = 1,200-N
The acceleration of the boat,a
Mass of boat,m = 1,400-kg
Initial velocity of boat,u= 0 m/sec
Distance travelled by boat,S = ?
Time for the boat travels,t = 20.0 s
Final velocity,V = ? m/sec
The net force on the boat;
F = F₁ - F₂
F = 1800 N - 1200 N
F = 600 N
From the defination of force;
F= ma
a = F / m
a = 600 N / 1400 kg
a = 0.425 m/sec²
b)
The distance through which the boat moves is 20.0 s;
c)
The velocity at the end of that time is found as;
m/sec
Hence the acceleration of the boat, the distance through which the boat moves in 20.0 s, and velocity at the end of that time will be 0.425 m/sec²,85 m, and 8.5 m/sec.
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Answer:
Explanation:
The force on the point charge q exerted by the rod can be found by Coulomb's Law.
Unfortunately, Coulomb's Law is valid for points charges only, and the rod is not a point charge.
In this case, we have to choose an infinitesimal portion on the rod, which is basically a point, and calculate the force exerted by this point, then integrate this small force (dF) over the entire rod.
We will choose an infinitesimal portion from a distance 'x' from the origin, and the length of this portion will be denoted as 'dx'. The charge of this small portion will be 'dq'.
Applying Coulomb's Law:
The direction of the force on 'q' is to the right, since both charges are positive, and they repel each other.
Now, we have to write 'dq' in term of the known quantities.
Now, substitute this into 'dF':
Now we can integrate dF over the rod.