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olga_2 [115]
3 years ago
5

PLEASE HURRY NEED ASAP

Physics
1 answer:
daser333 [38]3 years ago
3 0
I THINK THIS IS THOUGHTS AND FEELINGS
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In the equation for centripetal force, which expression represents the centripetal acceleration of the object? mv2 StartFraction
Sphinxa [80]

Answer: \frac{V^{2}}{r}

Explanation:

According to Newton's 2nd Law of motion the force F is proportional to the mass Fm and acceleration a:

F=m.a (1)

On the other hand, the equation for the Centripetal force is:

F=\frac{mV^{2}}{r} (2)

Where:

V is the velocity

r is the radius of the circular motion

Making (1) and (2) equal:

m.a=\frac{mV^{2}}{r} (3)

Hence:

a=\frac{V^{2}}{r} This is the expression for the centripetal acceleration

It should be noted, this acceleration is directed toward the center of the circumference of the circular motion (that's why it's called centripetal acceleration).

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3 years ago
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Một dây nhôm dài 10 m khi ở 25 độ C. Biết khi nhiệt độ tăng thêm 1 độ C thì chiều dài 1m dây nhôm sẽ tăng thêm 0,024mm.
kondor19780726 [428]

Answer:

??????? what language is this. um A then

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3 years ago
The _______ is a factor that's affected by changes in the independent variable. A. dependent variable B. hypothesis C. experimen
Irina18 [472]
A. dependent variable

As the the independent variable (i.e. number of cats being sold) increases, the dependent variable (i.e. money made) will also increase. You can have 1,000,000 cats for sale and make no money, but you cannot make money without having sold some cats first.
Hope my weird example helped!
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Beet juice is added to a unknown solution and changes color from red to purple. Curry powder, added to the same solution, turns
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4 years ago
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Water is circulating through a closed system of pipes in a two floor apartment. On the first floor, the water has a gauge pressu
anastassius [24]

Answer:

The value of gauge pressure at outlet = -38557.224 pascal

Explanation:

Apply Bernoulli' s Equation

\frac{P_{1}}{9810} + \frac{V_{1} ^{2}}{19.62} + h_{1} = \frac{P_{2}}{9810} + \frac{V_{2} ^{2}}{19.62} + h_{2} --------------(1)

Where

P_{1} =  Gauge pressure at inlet = 3.70105 pascal

V_{1} = velocity at inlet =  2.4 \frac{m}{sec}

P_{2} = Gauge pressure at outlet = we have to calculate

V_{2} = velocity at outlet = 3.5 \frac{m}{sec}

h_{2} - h_{1} = 3.6 m

Put all the values in equation (1) we get,

⇒ \frac{3.70105}{9810} + \frac{2.4 ^{2}}{19.62} = \frac{P_{2}}{9810} + \frac{3.5 ^{2}}{19.62} + 3.6

⇒ 0.294 = \frac{P_{2}}{9810} + 0.6244 + 3.6

⇒ \frac{P_{2}}{9810} = 0.294 - 0.6244 - 3.6

⇒ \frac{P_{2}}{9810} = - 3.9304

⇒ P_{2} = - 38557.224 pascal

This is the value of gauge pressure at outlet.

3 0
3 years ago
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