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2 years ago

A kayak is traveling across a pond at a speed

Physics

1 answer:

cricket20 [7]2 years ago

3 0

Answer:

Vy = -3.13j m/s

Vx = -7.36i m/s

Explanation:

Check the attachment for the diagram.

Velocity of Kaya = 8m/s

Direction of travel = S67° W

In component form, the velocity of Kaya will be resolved along both x components and y component

If she moves along the south, she will be moving in the -ve y-direction.

Vy = Vcostheta

Vy = -8 cos67°j

Vy = -3.13j m/s

If she moves along the west, she will be moving in the -ve x-direction

Vx = Vsintheta

Vx = -8sin 67°i

Vx = -7.36i m/s

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2 years ago

Before leaving the house in the morning, you plop some stew in your slow cooker and turn it on Low. The slow cooker has a 160 Oh

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Answer:

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2 years ago

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7. A mother pushes her 9.5 kg baby in her 5kg baby carriage over the grass with a force of 110N @ an angle

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Weight of the carriage $=(m+M)g =142.1\ N$

Normal force $=Fsin(\theta) + W = 197.1\ N$

Frictional force $=\mu N=27.59\ N$

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Explanation:

We have to look into the FBD of the carriage.

Horizontal forces and Vertical forces separately.

To calculate Weight we know that both the mass of the baby and the carriage will be added.

So Weight(W) $=(m+M)\times g =(9.5+5)\ kg \times 9.8 =142.1\ Newton\ (N)$

To calculate normal force we have to look upon the vertical component of forces, as Normal force is acting vertically.We have weight which is a downward force along with $F_x$ , force of $110\ N$ acting vertically downward.Both are downward and Normal is upward so Normal force $=Summation\ of\ both\ forces$

Normal force (N) $= Fsin(\theta)+W=110sin(30) + 142.1 =197.1\ N$

Frictional force (f) $=\mu N=0.14\times 197.1 =27.59\ N$

To calculate acceleration we will use Newtons second law.

That is Force is product of mass and acceleration.

We can see in the diagram that $F_y=Horizontal$ and $F_x=Vertical$ component of forces.

So Fnet = Fy(Horizontal) - f(friction) $= m\times a$

Acceleration (a) = $\frac{Fcos(\theta)-\mu N}{mass(m)} =\frac{(95.26-27.59)}{14.5}= 4.66\ m\ s{^2 }$

So we have the weight of the carriage, normal force,frictional force and acceleration.

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