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enyata [817]
2 years ago
10

A runner begins a race from the starting line and accelerates to a speed of 8.9 m/s. If it takes the runner 3 seconds to reach h

is maximum speed, what was his acceleration?
Physics
2 answers:
stellarik [79]2 years ago
4 0

Answer:

i believe its 26.7

Explanation:

if the runner goes 8.9 m/s each second while accelerating for 3 seconds to reach top speed, the top speed would be 26.7 m/s

ale4655 [162]2 years ago
3 0

Answer:

it will be 8.9 over 3 m/s/s or m/s square

Explanation:

acceleration equal to speed over time

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Janet was pulling a box that weighed 20N across 5m. How much work did she exert?
Galina-37 [17]

Answer:

100nm

Explanation:

Work=Force×distance

=20×5

=100nm

6 0
2 years ago
The ability of your joints and muscles to move in their full range of motion is called
Studentka2010 [4]

Answer:

Dynamic flexibility

Explanation:

Dynamic flexibility can be generally defined as the ability of the body muscles and joints to move in full range of motion. High flexibility in these joints and muscles leads to the decreasing pain and injury in different parts of the body.

Proper warm up exercises are needed to be carried out that involves both the combination of controlling movements and stretching of the body, and this directly enhances the dynamic flexibility of the body.

The athletes and sports persons possesses a good dynamic flexibility of their body as they carry our different types of body exercises.

6 0
3 years ago
What a neurology professor does = _______ brains
Natasha_Volkova [10]
That’s really easy ask your teacher and also peace happy
3 0
3 years ago
How much force is required to accelerate a 2 kg mass at 3 m/s2
swat32

Force = (mass) x (acceleration)                  Newton's second law of motion.

Force = (2 kg) x (3 m/s²)  =  6 newtons.

3 0
4 years ago
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
4 years ago
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