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Marianna [84]
3 years ago
5

sara and tory are out fishing on the lake on a hot summer day when they both decide to go for a swim. sara dives off the front o

f the boat with a force of 45° N, while Tory dives off the back of the boat with a force of 60° N. Draw a vector diagram of the situation. Find the resultant force on the boat.

Physics
1 answer:
crimeas [40]3 years ago
3 0

Here it is an application of Newton's III law

as we know by Newton's III law that every action has equal and opposite reaction

So here as we know that two boys jumps off the boat with different forces

from front side of the boat the boy jumps off with force 45 N which means as per Newton's III law if boy has a force of 45 N in forward direction then he must apply a reaction force on the boat in reverse direction of same magnitude

So boat must have an opposite force on front end with magnitude 45 N

Now similar way we can say

from back side of the boat the boy jumps off with force 60 N which means as per Newton's III law if boy has a force of 60 N in backward direction then he must apply a reaction force on the boat in reverse direction of same magnitude

So boat must have an opposite force on front end with magnitude 60 N

So here net force due to both jump on the boat is given by

F_{net} = F_1 - F_2

F_{net} = 60 - 45

F_{net} = 15 N

so boat will have net force F = 15 N in forward direction due to both jumps

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Answer:

1.125m/s^2

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Since acceleration is defined as the rate of change in velocity with respect to time. Mathematically

v^2= u^2+2as

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a = ?

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s = 100m

Substituting the values into the formula above

v^2= u^2+2as

15^2=0^2+2×a×100

225= 0+200a

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Divide both sides by 200

225/200 = 200a/200

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Hence the acceleration of the car is 1.125m/s^2.

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(1 point) A frictionless spring with a 3-kg mass can be held stretched 1.6 meters beyond its natural length by a force of 90 new
Nonamiya [84]

Answer:

x(t)=0.337sin((5.929t)

Explanation:

A frictionless spring with a 3-kg mass can be held stretched 1.6 meters beyond its natural length by a force of 90 newtons. If the spring begins at its equilibrium position, but a push gives it an initial velocity of 2 m/sec, find the position of the mass after t seconds.

Solution. Let x(t) denote the position of the mass at time t. Then x satisfies the differential equation  

m \frac{d^{2}x}{dt^{2}} +kx=0

Definition of parameters  

m=mass 3kg

k=force constant

e=extension ,m

ω =angular frequency

k=90/1.6=56.25N/m

ω^2=k/m= 56.25/1.6

ω^2=35.15625

ω=5.929

General solution will be

x(t)=c1cos(ωt)+c2Sin(ωt)

x(t)=c1cos(5.929t)+c2Sin(5.929t)

differentiating x(t)

dx(t)=-5.929c1sin(5.929t)+5.929c2cos(5.929t)

when x(0)=0, gives c1=0

dx(t0)=2m/s gives c2=0.337

Therefore, the position of the mass after t seconds is  

x(t)=0.337sin((5.929t)

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3 years ago
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