What Grade will you give my One Pager so Far?

(b) Figure 4 shows a car travelling on a motorway.

Alik [6]

Answer:

To calculate anything - speed, acceleration, all that - we need data. The more data we have, and the more accurate that data is, the more accurate our calculations will be. To collect that data, we need to measure it somehow. To measure anything, we need tools and a method. Speed is a measure of distance over time, so we'll need tools for measuring time and distance, and a method for measuring each.

Conveniently, the lamp posts in this problem are equally spaced, and we can treat that spacing as our measuring stick. To measure speed, we'll need to bring time in somehow too, and that's where the stopwatch comes in. A good method might go like this:

Press start on the stopwatch right as you pass a lamp post
Each time you pass another lamp post, press the lap button on the stopwatch
Press stop after however many lamp posts you'd like, making sure to hit stop right as you pass the last lamp post
Record your data
Calculate the time intervals for passing each lamp post using the lap data
Calculate the average of all those invervals and divide by 40 m - this will give you an approximate average speed

Of course, you'll never find an *exact* amount, but the more data points you have, the better your approximation will become.

5 0

3 years ago

Consider the vector field. f(x, y, z) = xy2z2i x2yz2j x2y2zk (a) find the curl of the vector field?

Marat540 [252]

Observe that the given vector field is a gradient field:

Let $f(x,y,z)=\nabla g(x,y,z)$ , so that

$\dfrac{\partial g}{\partial x} = x y^2 z^2$

$\dfrac{\partial g}{\partial y} = x^2 y z^2$

$\dfrac{\partial g}{\partial z} = x^2 y^2 z$

Integrating the first equation with respect to $x$ , we get

$g(x,y,z) = \dfrac12 x^2 y^2 z^2 + h(y,z)$

Differentiating this with respect to $y$ gives

$\dfrac{\partial g}{\partial y} = x^2 y z^2 + \dfrac{\partial h}{\partial y} = x^2 y z^2 \\\\ \implies \dfrac{\partial h}{\partial y} = 0 \implies h(y,z) = i(z)$

Now differentiating $g$ with respect to $z$ gives

$\dfrac{\partial g}{\partial z} = x^2 y^2 z + \dfrac{di}{dz} = x^2 y^2 z \\\\ \implies \dfrac{di}{dz} = 0 \implies i(z) = C$

Putting everything together, we find a scalar potential function whose gradient is $f$ ,

$f(x,y,z) = \nabla \left(\dfrac12 x^2 y^2 z^2 + C\right)$

It follows that the curl of $f$ is 0 (i.e. the zero vector).

5 0

2 years ago

A ball, with a mass of 5.9kg, is thrown directly upwards. It reaches a maximum height of 10m from the point at which it was rele

katrin2010 [14]

Answer:

14 m/s

Explanation:

We can solve the problem by using the law of conservation of energy.

At the beginning, when the ball is thrown from the ground, it has only kinetic energy, which is given by

$K=\frac{1}{2}mv^2$

where m = 5.9 kg is the mass of the ball and v is its initial speed.

As the ball goes up, its speed decreases, so its kinetic energy decreases and converts into gravitational potential energy. When the ball reaches its maximum height, the speed has become zero, and all the kinetic energy has been converted into gravitational potential energy, given by:

$U=mgh$

where g = 9.8 m/s^2 is the gravitational acceleration and h = 10 m is the maximum height reached by the ball.

Since we can ignore air resistance, energy must be conserved, so the initial kinetic energy must be equal to the final potential energy of the ball, so we can write:

$K=U\\\frac{1}{2}mv^2=mgh$