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marishachu [46]
2 years ago
9

A rock is dropped from the top of a vertical cliff and takes 3.00 s to reach the ground below the cliff, A second rock is thrown

vertically from the cliff, and it takes this rock 2.00 s to reach the ground below the cliff from the time it is released. With what velocity was the second rock thrown, assuming no air resistance?
Physics
1 answer:
Tcecarenko [31]2 years ago
3 0

Answer:

12.25m/s

Explanation:

d=v_ot+\dfrac{1}{2}at^2

Since the initial velocity of the dropped rock is 0, you can write this as:

d=\dfrac{1}{2}(9.8)(3)^2=44.1m

Now, you can set up the equation for the thrown rock:

44.1=v_o(2)+\dfrac{1}{2}(9.8)(2)^2 \\\\\\44.1=2v_o+19.6 \\\\\\2v_o=24.5 \\\\\\v_o=12.25m/s

Hope this helps!

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Calculate curls of the following vector functions (a) AG) 4x3 - 2x2-yy + xz2 2
aleksandr82 [10.1K]

Answer:

The curl is 0 \hat x -z^2 \hat y -4xy \hat z

Explanation:

Given the vector function

\vec A (\vec r) =4x^3 \hat{x}-2x^2y \hat y+xz^2 \hat z

We can calculate the curl using the definition

\nabla \times \vec A (\vec r ) = \left|\begin{array}{ccc}\hat x&\hat y&\hat z\\\partial/\partial x&\partial/\partial y&\partial/\partial z\\A_x&X_y&A_z\end{array}\right|

Thus for the exercise we will have

\nabla \times \vec A (\vec r ) = \left|\begin{array}{ccc}\hat x&\hat y&\hat z\\\partial/\partial x&\partial/\partial y&\partial/\partial z\\4x^3&-2x^2y&xz^2\end{array}\right|

So we will get

\nabla  \times \vec A (\vec r )= \left( \cfrac{\partial}{\partial y}(xz^2)-\cfrac{\partial}{\partial z}(-2x^2y)\right) \hat x - \left(\cfrac{\partial}{\partial x}(xz^2)-\cfrac{\partial}{\partial z}(4x^3) \right) \hat y + \left(\cfrac{\partial}{\partial x}(-2x^2y)-\cfrac{\partial}{\partial y}(4x^3) \right) \hat z

Working with the partial derivatives we get the curl

\nabla  \times \vec A (\vec r )=0 \hat x -z^2 \hat y -4xy \hat z

6 0
3 years ago
A policeman is chasing a criminal across a rooftop at 10 m/s. He decides to jump to the next building which is 2 meters across f
valina [46]

Answer:

They will meet at a distance of 7.57 m

Given:

Initial velocity of policeman in the x- direction, u_{x} = 10 m/s

The distance between the buildings, d_{x} = 2.0 m

The building is lower by a height, h = 2.5 m

Solution:

Now,

When the policeman jumps from a height of 2.5 m, then his initial velocity, u was 0.

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From the second eqn of motion, we can write:

h = ut + \frac{1}{2}gt^{2}

h = \frac{1}{2}gt^{2}

2.5 = \frac{1}{2}\times 10\times t^{2}

t = 0.707 s

Now,

When the policeman was chasing across:

d_{x} = u_{x}t + \frac{1}{2}gt^{2}

d_{x} = 10\times 0.707 + \frac{1}{2}\times 10\times 0.5 = 9.57 m

The distance they will meet at:

9.57 - 2.0 = 7.57 m

   

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Answer:

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SVETLANKA909090 [29]

Answer:

The correct answer is B.

Explanation:

Step 1:

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Here, the predict height is dependent variable and the  age is in-dependent variable.

Intercept = 0.29

Slope      = 0.48

The given regression equation indicates the y on x model and the intercept coefficients of the regression equation is 0.29 and the slope is 0.48.

Step 2:

The height increases, an average, by 0.48 m per year.

Because co-efficient of slope variable indicate the positive sign and we increase 1 year in age then automatically height increased is 0.48 m.

<h3></h3><h3>The height increases, on average, by 0.48 meter each year.</h3>
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