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Colt1911 [192]
3 years ago
11

Jacques and Georgette meet in the middle of a lake while paddling in their canoes. They come to a complete stop and talk for a w

hile. When they are ready to leave, Jacques pushes Georgette's canoe with a force F to separate the two canoes. What is correct to say about the final momentum and kinetic energy of the system if we can neglect any resistance due to the water
Physics
1 answer:
emmainna [20.7K]3 years ago
8 0

Answer:

The correct answer is the final momentum is zero but the final kinetic energy is positive.

Explanation:

Solution

Given that:

The momentum is zero because momentum is the sum of the mass * velocity for each component. momentum is conserved

The Kinetic energy (KE) is positive because kinetic energy is the total of the 1/2 *mass * velocity^2.

Whereas the velocity can be positive and negative since it is directed (this is what makes the momentum zero because on is positive and one is negative), the velocity squared will always be positive.

Thus adding together two positives will always be a positive number.

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If you wanted to know the location of a vehicle that ran out of gas after taking a zigzag route through the city, which quantity
Papessa [141]

Explanation:

the vehicles displacement, since displacement deals with position

8 0
3 years ago
You are standing in a building whose height (40m) you throw a ball downward at a angle of -30 at a speed of (10m/s) acceleration
jeyben [28]

Answer: 3.41 s

Explanation:

Assuming the question is to find the time t the ball is in air, we can use the following equation:

y=y_{o}+V_{o}sin \theta t-\frac{1}{2}gt^{2}

Where:

y=0m is the final height of the ball

y_{o}=40 m is the initial height of the ball

V_{o}=10 m/s is the initial velocity of the ball

t is the time the ball is in air

g=9.8 m/s^{2} is the acceleration due to gravity  

\theta=30\°

Then:

0 m=40 m+(10 m/s)(sin(30\°))t-\frac{1}{2}9.8 m/s^{2}t^{2}

0 m=40 m+5m/s t-4.9 m/s^{2}t^{2}

Multiplying both sides of the equation by -1 and rearranging:

4.9 m/s^{2}t^{2}-5m/s t-40 m=0

At this point we have a quadratic equation of the form at^{2}+bt+c=0, which can be solved with the following formula:

t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}

Where:

a=4.9

b=-5

c=-40

Substituting the known values:

t=\frac{-(-5) \pm \sqrt{(-5)^{2}-4(4.9)(-40)}}{2(4.9)}

Solving the equation and choosing the positive result we have:

t=3.41 s  This is the time the ball is in air

5 0
3 years ago
Can work done=mass*acceleration*displacement(work=m*a*s)
Airida [17]

no, work is = force * distance or displacement


5 0
3 years ago
A torque of 0.77 N⋅m is applied to a bicycle wheel of radius 30 cm and mass 0.70 kg
Naddik [55]

Answer:

α = τ/I = 0.77 / (0.70(0.30²)) = 12.22222... = 12 rad/s²

Explanation:

4 0
3 years ago
In a shuffleboard game, the puck slides a total of 12 m before coming to rest. If the coefficient ofkinetic friction between the
Artist 52 [7]

Answer:

puck decelerates due to the kinetic frictional force μk mg

Explanation:

given data

total distance = 12 m

coefficient of kinetic friction = 0.28

solution

we will apply equation of motion that is

v² - u² = 2 × a × s    ................1

we know acceleration will be

a = \frac{-u^2}{2\times S}  

Then we have

Force = mass × acceleration   .................2

m × \frac{-u^2}{2\times S} = -μk mg

The puck decelerates due to the kinetic frictional force μk mg  

and frictional force is negative as it opposes the motion.

so we get initial velocity of the puck which is strike.

4 0
3 years ago
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