(a) The system of interest if the acceleration of the child in the wagon is to be calculated are the wagon and the children outside the wagon.
(b) The acceleration of the child-wagon system is 0.33 m/s².
(c) Acceleration of the child-wagon system is zero when the frictional force is 21 N.
<h3>
Net force on the third child</h3>
Apply Newton's second law of motion;
∑F = ma
where;
- ∑F is net force
- m is mass of the third child
- a is acceleration of the third child
∑F = 96 N - 75 N - 12 N = 9 N
Thus, the system of interest if the acceleration of the child in the wagon is to be calculated are;
- the wagon
- the children outside the wagon
<h3>Free body diagram</h3>
→ → Ф ←
1st child friction wagon 2nd child
<h3>Acceleration of the child and wagon system</h3>
a = ∑F/m
a = 9 N / 27 kg
a = 0.33 m/s²
<h3>When the frictional force is 21 N</h3>
∑F = 96 N - 75 N - 21 N = 0 N
a = ∑F/m
a = 0/27 kg
a = 0 m/s²
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Answer:
D = 25 miles
Explanation:
To solve this problem, we just need to know how much time it took both bicyclists to collide and that will be the same amount of time that the bee flew at 25miles per hour. With those values we could calculate the distance it traveled.
Since both bicyclists collide, we know that Xa=Xb, so:
Xa = V*t = 10*t and Xb = 20 - V*t = 20 - 10*t
10*t = 20 - 10*t Solving for t:
t = 1 hour Now we can calculate the distance for the bee:
D = Vbee * t = 25 * 1 = 25 miles
Answer:
I think is D I'm sorry if I'm wrong
Answer:
Number of electrons are flowing per second is 2.42 x 10¹⁹
Explanation:
The electric current flows through a wire is given by the relation :
....(1)
Here I is current, e is electronic charge, v is drift velocity of electrons and A is the Area of the wire.
But electric current is also define as rate of electrons passing through junction times their charge, i.e. ,
....(2)
Here N is the rate of electrons passing through junction.
From equation (1) and (2).
But area of wire, 
Here d is diameter of wire.
So, 
Substitute 2.91 x 10⁻³ m for d, 0.000191 m/s for v and 6 x 10²⁸ m⁻³ for n in the above equation.
N = 2.42 x 10¹⁹ s⁻¹
Answer:
a. 2143 turns/m
b. 111.5 m
Explanation:
a. The minimum number of turns per unit length (N/L) can be found using the following equation:
Hence, the minimum number of turns per unit length is 2143 turns/m.
b. The total length of wire is the following:
Since each turn has length 2πr of wire, the total length is:
Therefore, the total length of wire required is 111.5 m.
I hope it helps you!
