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cluponka [151]
3 years ago
7

What is the final step in the fourth stage of technological design

Physics
1 answer:
Ierofanga [76]3 years ago
5 0

Answer:

after a product has been improved and approved? reporting the results finding ways to lower costs selling a prototype determining criteria.

Explanation:

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11. A cyclist accelerates from 0 m/s to 10 m/s in 3 seconds. What is his acceleration ? Is this acceleration higher than that of
Marat540 [252]

a =  \frac{v - u}{t}

v = final velocity

u = initial velocity

t = time taken

the acceleration of the cyclist is

\frac{10 - 0}{3}  = 3.333333....

approximately 3.33 m/s^2

the acceleration of the car is

\frac{40 - 0 }{8}  = 5.0

5.0 m/s^2

5.0 > 3.33 \\ so \:  the \: answer  \: is \: no

6 0
3 years ago
38. You are fishing and catch a fish with a mass of
adoni [48]

Answer:

1.7333333m/s²

Explanation:

Tension of the line = the weight + force from pulling up the fish

30N = mg + ma

30 = (6)(9.8) + (6)a

10.4 = 6a

∴ a = 1.7333333m/s²

6 0
3 years ago
Why might an electromagnet be used to pick up old cars in junkyards?
siniylev [52]
B) Hope it helps ,Have a nice day :)
7 0
3 years ago
Read 2 more answers
I need help with Science! ASAP!!! ASAP!!! ASAP!!!
Trava [24]
Possibilities . . .

-- nuclear reaction
-- nuclear fission
-- nuclear fusion
-- radioactive decay.

Any of these makes it a true statement.
5 0
3 years ago
A water main pipe of diameter 10 cm enters a house 2 m below ground. A smaller diameter pipe carries water to a faucet 5 m above
Lemur [1.5K]

Explanation:

Given that,

Diameter = 10 cm

Distance = 2 m

Speed v_{1}= 2\ m/s

Speed v_{2}=7\ m/s

Pressure in main pipe P_{1}=2\times10^{5}\ Pa

(I). We need to calculate the diameter

Using equation of continuity

Av_{1}=Av_{2}

\pi(\dfrac{d_{1}}{2})^2\times v_{1}=\pi(\dfrac{d_{2}}{2})^2\times v_{2}

(\dfrac{10}{2})^2\times2=(\dfrac{d_{2}}{2})^2\times7

d_{2}=\sqrt{\dfrac{25\times2\times4}{7}}

d_{2}=5.345\ cm

(II). We need to calculate the pressure the gauge pressure

Using Bernoulli equation

P_{1}+\dfrac{1}{2}\rho v_{1}^2+\rho gh_{1}=P_{2}+\dfrac{1}{2}\rho v_{2}^2+\rho g h_{2}

P_{2}=P_{1}+\dfrac{1}{2}\rho(v_{1}^2-v_{2}^2)-\rho g(h_{1}-h_{2})

P_{2}=2\times10^{5}+\dfrac{1}{2}\times1000(4-49)-1000\times 9.8\times(5)

P_{2}=1.28500\times10^{5}\ Pa

(III).  If it is possible to carry water to a faucet 17 m above ground,

Using Bernoulli equation

P_{1}+\dfrac{1}{2}\rho v_{1}^2+\rho gh=P_{3}+\dfrac{1}{2}\rho v_{3}^2+\rho g h_{3}

P_{3}=P_{1}+\dfrac{1}{2}\rho v_{1}^2-\rho g(h_{1}-h_{3})

Here, h_{3}=0

Put the value in the equation

P_{3}=2\times10^{5}+\dfrac{1}{2}\times1000\times4-1000\times 9.8\times17

P_{3}=3.5400\times10^{5}\ Pa

Hence, This is required solution.

7 0
3 years ago
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