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3 years ago

How to get the answer

Physics

1 answer:

Jobisdone [24]3 years ago

4 0

I think you need to include the Ebola question

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what happens if you partially block the stream of water coming out a hose? use your answer to describe in detail why a nozzle ma

stealth61 [152]

Answer:the hose would explode and you would mess the water system out

Explanation:

7 0

3 years ago

Read 2 more answers

A bird flies overhead from where you stand at an altitude of 270.0ĵ m and at a velocity horizontal to the ground of 14.0î m/is.

Varvara68 [4.7K]

Answer:

L = 8694 Kg.m²/s

Explanation:

r = 270 ĵ m

v = 14 î m/s

m = 2.3 kg

θ = 90º

L = ?

We can apply the equation

L = m*v*r*Sin θ

L = (2.3 kg)*(14 m/s)*(270 m)*Sin 90º = 8694 Kg.m²/s

8 0

3 years ago

7. Water flows trough a horizontal tube of diameter 2.5 cm that is joined to a second horizontal tube of diameter 1.2 cm. The pr

Usimov [2.4K]

To solve this problem it is necessary to apply the concepts related to the continuity of fluids in a pipeline and apply Bernoulli's balance on the given speeds.

Our values are given as

$d_1 = 2.5cm \rightarrow r_1=1.25cm=1.25*10^{-2}m$

$d_2 = 1.2cm \rightarrow r_2 = 0.6cm = 0.6*10^{-2}m$

From the continuity equations in pipes we have to

$A_1V_1 = A_2 V_2$

Where,

$A_{1,2}$ = Cross sectional Area at each section

$V_{1,2}$ = Flow Velocity at each section

Then replacing we have,

$(\pi r_1^2) v_1 = (\pi r_2^2) v_2$

$(1.25*10^{-2})^2 v_1 = ( 0.06*10^{-2})^2 v_2$

$v_2 = \frac{(1.25*10^{-2})^2 }{0.6*10^{-2})^2} v_1$

From Bernoulli equation we have that the change in the pressure is

$\Delta P = \frac{1}{2} \rho (v_2^2-v_1^2)$

$7.3*10^3 = \frac{1}{2} (1000)([ \frac{(1.25*10^{-2})^2 }{0.6*10^{-2})^2} v_1 ]^2-v_1^2)$

$7300= 8919.01 v_1^2$

$v_1 = 0.9m/s$

Therefore the speed of flow in the first tube is 0.9m/s

6 0

4 years ago

A person is standing on a level floor. His head, upper

BabaBlast [244]

Answer:

$y_{cg} = 1.03 m$

Explanation:

Given data:

weigh (head+arms + head) w_1 = 438 N

centre of gravity y_1= 1.28 m

weigh (upper leg) w_2 = 144 N

Center of gravity y_2 = 0.760 m

weigh ( lower leg + feet) = 87 N

centre of gravity = y_3 = 0.250 m

location of center of gravity $= \frac{w_1 y_1 + w_2 y_2 + w_3 y_3}{w_1 +W_2 +w_3}$

$y_{cg} = \frac{438 \times 1.28 + 144\times 0.760 + 87 \times 0.250}{438+144+87}$

$y_{cg} = 1.03 m$

8 0

3 years ago

Which of the following are correct statements about the way an atom is put

Ann [662]

Answer:

valenc e shell

Explanation:

4 0

3 years ago