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Talja [164]
3 years ago
11

Holding health care personal to lesser standards of care in emergencies is call the what statue?

Physics
1 answer:
olga55 [171]3 years ago
3 0

Hello there,

Holding health care personal to lesser standards of care in emergencies is call the what statue?

Answer: A. Good Samaritan


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Potential energy can be converted into kinetic energy, a good example of this is when a pole-vaulter bends the pole during a lea
lesantik [10]

Answer:

The pole stores elastic potential energy when the pole is bent because its shape is change from its natural shape and it will want to go back to its original form, just like a spring or stretched elastic material.

Question:

Why did you lie about being in college?

6 0
3 years ago
Read 2 more answers
The driver of a car slams on the brakes, causing the car to slow down at a rate of
sdas [7]

Answer:

A. The time taken for the car to stop is 3.14 secs

B. The initial velocity is 81.64 ft/s

Explanation:

Data obtained from the question include:

Acceleration (a) = 26ft/s2

Distance (s) = 256ft

Final velocity (V) = 0

Time (t) =?

Initial velocity (U) =?

A. Determination of the time taken for the car to stop.

Let us obtain an express for time (t)

Acceleration (a) = Velocity (V)/time(t)

a = V/t

Velocity (V) = distance (s) /time (t)

V = s/t

a = s/t^2

Cross multiply

a x t^2 = s

Divide both side by a

t^2 = s/a

Take the square root of both side

t = √(s/a)

Now we can obtain the time as follow

Acceleration (a) = 26ft/s2

Distance (s) = 256ft

Time (t) =..?

t = √(s/a)

t = √(256/26)

t = 3.14 secs

Therefore, the time taken for the car to stop is 3.14 secs

B. Determination of the initial speed of the car.

V = U + at

Final velocity (V) = 0

Deceleration (a) = –26ft/s2

Time (t) = 3.14 sec

Initial velocity (U) =.?

0 = U – 26x3.14

0 = U – 81.64

Collect like terms

U = 81.64 ft/s

Therefore, the initial velocity is 81.64 ft/s

7 0
3 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
3 years ago
A speedboat moves on a lake with initial velocity vector v1,x = 8.57 m/s and v1,y = -2.61 m/s, then accelerates for 6.67 seconds
KengaRu [80]
First, you find the velocity at each component. The general equation is:

a = (v2 - v1)/t

a,x = (v2,x - v1,x)/t
-0.105 = (v2,x - 8.57)/6.67
v2,x = 7.87 m/s

a,y = (v2,y - v1,y)/t
0.101 = (v2,y - -2.61)/6.67
v2,y = -1.94 m/s

To find the final speed, find the resultant velocity by taking the hypotenuse.

v^2 = (v2,x)^2 + (v2,y)^2
v^2 = (7.87)^2 + (-1.94)^2
v = 8.1 m/s
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What is the frequency of a wave if the speed is 4300 m/s and the wavelength is 32 m?
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