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loris [4]
2 years ago
9

A funnel-shaped low pressure closure

Physics
1 answer:
cupoosta [38]2 years ago
6 0
What is the rest ?? What about the funnel shaped low pressure closure??

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In a closed energy system, energy is?​
Alisiya [41]

Answer:

Explanation:

A closed system can exchange energy but not matter, with its surroundings. An isolated system cannot exchange any heat, work, or matter with the surroundings, while an open system can exchange energy and matter.

Hope this helped you!

3 0
2 years ago
You throw a baseball straight up into the air with a speed of 24.5 m/s. How long does it take the baseball to reach its highest
galben [10]
Time=speed/acceleration
Gravitaional Acceleration=9.8 m/s^2
Speed=24.5 m/s
Time=24.5/9.8=2.5 s
3 0
3 years ago
Read 2 more answers
A girl is sitting in a sled sliding horizontally along some snow (there is friction present). The mass of the girl is 29.8 kg an
jonny [76]

Answer:

28.81 m

Explanation:

Ff = -123

m * a  = -123

(29.8+10.3) * a = -123

a = -123/40.1 = -3.07

We know,

v^2 = u^2 + 2as

0^2 = 13.3^2 + 2*(-3.07)*s

s = 176.89/6.14 = 28.81

[ If there's a problem with the solution, pleaase let me know ]

6 0
3 years ago
An airplane flies with a constant speed of 840km/h. how far can it travel in 1 hour?
arlik [135]
Distance = speed X time

In this example, the speed of the airplane = 840km. The time (that the question is asking)is how far can it travel in 1 hour.

So just plug in your numbers.

Distance = 840km X 1 hour = 840km/hour or 840km for short.
3 0
3 years ago
Four copper wires of equal length are connected in series. Their cross-sectional areas are 0.7 cm2 , 2.5 cm2 , 2.2 cm2 , and 3 c
Travka [436]

Answer:

22.1 V

Explanation:

We are given that

A_1=0.7 cm^2=0.7\times 10^{-4} m^2

A_2=2.5 cm^2=2.5\times 10^{-4} m^2

A_3=2.2 cm^2=2.2\times 10^{-4} m^2

A_4=3 cm^2=3\times 10^{-4} m^2

Using 1cm^2=10^{-4} m^2

We know that

R=\frac{\rho l}{A}

In series

R=R_1+R_2+R_3+R_4

R=\frac{\rho l}{A_1}+\frac{\rho l}{A_2}+\frac{\rho l}{A_3}+\frac{\rho l}{A_4}

R=\frac{\rho l}{\frac{1}{A_1}+\frac{1}{A_2}+\frac{1}{A_3}+\frac{1}{A_4}}

Substitute the values

R=\rho A(\frac{1}{0.7\times 10^{-4}}+\frac{1}{2.5\times 10^{-4}}+\frac{1}{2.2\times 10^{-4}}+\frac{1}{3\times 10^{-4}})

R=\rho l(2.62\times 10^4)

V=145 V

I=\frac{V}{R}=\frac{145}{\rho l(2.62\times 10^4)}

Voltage across the 2.5 square cm wire=IR=I\times \frac{\rho l}{A_2}

Voltage across the 2.5 square cm wire=\frac{145}{\rho l(2.62\times 10^4)}\times \frac{\rho l}{2.5\times 10^{-4}}=22.1 V

Voltage across the 2.5 square cm wire=22.1 V

6 0
3 years ago
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