A child, who is 45 m from the bank of a river, is being carried helplessly downstream by the river's swift current of 1.0 m/s. A
s the child passes a lifeguard on the river's bank, the lifeguard starts swimming in a straight line (Fig. 3–46) until she reaches the child at a point downstream. If the lifeguard can swim at a speed of 2.0 m/s relative to the water, how long does it take her to reach the child? How far downstream does the lifeguard intercept the child?
1 answer:
Answer:
The lifeguard takes 25.9 seconds to reach the child, at 25.9 meters from the start point downstream.
Explanation:
As the image shows, the child trajectory, the lifeguard trajectory and the distance from the bank form a triangle. This triangle is formed by the distances, an we already know the distance from the bank and the speed of child, and the speed of the lifeguard. So we have unknom time in common. Lets see the equations:
Using phitagoras theorem
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Answer:
(a) The horizontal ground reaction force
(b) The vertical ground reaction force
(c) The resultant ground reaction force
Explanation:
Given
John mass , m = 65 kg
Horizontal acceleration ,
Vertical acceleration ,
(a) Using Newton's 2nd law in horizontal direction
=>
Thus the horizontal ground reaction force
(b) Using Newton's 2nd law in vertical direction
=>
=>
Thus the vertical ground reaction force
(c) Resultant ground reaction force is
=>
=>
Thus the resultant ground reaction force
Answer:
a) F₃₁ = 63.0 μN
b) F₃₂ = - 14.0 μN
c) q₂ = - 5.0 nC
Explanation:
a)
- Assuming that the three charges can be taken as point charges, the forces between them must obey Coulomb's Law, and can be found independent each other, applying the superposition principle.
- So, we can find the force that q₁ exerts along the x-axis on q₃, as follows:
b)
- Since total force exerted by q₁ and q₂ on q₃ is 49.0 μN, we can find the force exerted only by q₂ (which is along the x-axis only too) just by difference, as follows:
c)
- Finally, in order to find the value of q₂, as we know the value and sign of F₃₂, we can apply again the Coulomb's Law, solving for q₂, as follows:
Answer:
No the given statement is not necessarily true.
Explanation:
We know that the kinetic energy of a particle of mass 'm' moving with velocity 'v' is given by
Similarly the momentum is given by
For 2 particles with masses and moving with velocities
respectively the respective kinetic energies is given by
Similarly For 2 particles with masses and moving with velocities
respectively the respective momenta are given by
Now since it is given that the two kinetic energies are equal thus we have
Thus we infer that the moumenta are not equal since the ratio on right of 'i' is not 1 , and can be 1 only if the velocities of the 2 particles are equal which becomes a special case and not a general case.