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IgorC [24]
3 years ago
6

When do you produce more pressure on the ground...standing or laying down?

Physics
1 answer:
ycow [4]3 years ago
7 0
When you are Standing you produce more pressure on the ground
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Write one to two sentences that summarize Bernoulli's principle.
Ierofanga [76]

Answer and Explanation:

Bernoulli's Principle deals with Fluid dynamics and fluid includes both liquid and gas.

The Principle gives an inverse relation between the speed or velocity of the fluid and the pressure of the fluid. It gives the speed or velocity of the liquid over varying pressure.

Bernoulli's Principle states that for a given fluid,the low pressure region will have high speed or velocity and high pressure region will have low speed or velocity

4 0
3 years ago
Obligation of reasearchers to review true or fasle
MrMuchimi
I think if I was a researcher I would review my work or the sources I use, True
5 0
3 years ago
Please help me, this school work is not the easiest for me.
lidiya [134]

Answer:

A = mm2

M = km2

F = mg

Explanation:

7 0
2 years ago
Read 2 more answers
Ax = 44.4 m and Ay = 25.1 m<br> Find the magnitude of the<br> vector.
max2010maxim [7]

Answer:

The magnitude of the vector A is <u>51 m.</u>

Explanation:

Given:

The horizontal component of a vector A is given as:

A_x=44.4\ m

The vertical component of a vector A is given as:

A_y=25.1\ m

Now, we know that, a vector A can be resolved into two mutually perpendicular components; one along the x axis and the other along the y axis. The magnitude of the vector A can be written as the square root of the sum of the squares of each component.

Therefore, the magnitude of vector A is given as:

|\overrightarrow A|=\sqrt{A_{x}^2+A_{y}^2}

Now, plug in 44.4 for A_x, 25.1 for A_y and solve for the magnitude of A. This gives,

|\overrightarrow A|=\sqrt{(44.4)^2+(25.1)^2}\\|\overrightarrow A|=\sqrt{1971.36+630.01}\\|\overrightarrow A|=\sqrt{2601.37}\\|\overrightarrow A|=51\ m

Therefore, the magnitude of the vector A is 51 m.

6 0
3 years ago
Suppose quantity s is a length and quantity t is a time. Suppose the quantities vand aare defined by v = ds/dt and a = dv/dt. (a
saul85 [17]

Explanation:

(a) Velocity is given by :

v=\dfrac{ds}{dt}

s is the length of the distance

t is the time

The dimension of v will be, [v]=[LT^-1]      

(b) The acceleration is given by :

a=\dfrac{dv}{dt}

v is the velocity

t is the time

The dimension of a will be, [a]=[LT^{-2}]

(c) Since, d=\int\limits{v{\cdot}dt} =[LT^{-1}][T]=[L]

(d) Since, v=\int\limits{a{\cdot}dt} =[LT^{-2}][T]=[LT^{-1}]

(e)

\dfrac{da}{dt}=\dfrac{[LT^{-2}]}{[T]}

\dfrac{da}{dt}=[LT^{-3}]}

Hence, this is the required solution.

7 0
3 years ago
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