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saw5 [17]
3 years ago
9

Before taking off a plane travels at a speed of 1/4 km per second. The runaway is 5 km. How many seconds does it take the plane

to get to the end of the runaway?
Engineering
1 answer:
Vikentia [17]3 years ago
8 0

Answer:

1 5segundos

Explanation:

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What are three types of land reform ​
s344n2d4d5 [400]

Answer:

Abolition of intermediaries (rent collectors under the pre-Independence land revenue system); Tenancy regulation (to improve the contractual terms including the security of tenure); A ceiling on landholdings (to redistributing surplus land to the landless);

5 0
2 years ago
Read 2 more answers
A bicycle has tires that are 26 inches in diameter. The while in motion, the tires rotate through 125 complete revolutions. How
Stolb23 [73]

Answer:

850.8480103 feet

Explanation:

First you take the diameter and find the circumference, which is (2)(pi)(r) plug in your r which is 26/2= 13 so 2(13)(pi) and multiply taht by 125 after that take your answer and divide by 12which is 850.8480103

3 0
2 years ago
What is the answer ?
tigry1 [53]
Is the control group
4 0
3 years ago
21.Why are throttling devices commonly used in refrigeration and air-conditioning<br> applications?
Sloan [31]

Answer is given below

Explanation:

we know that some common types of throttling devices are

  • Hard -throttling devices
  • Capillary valve
  • Constant pressure throttling devices
  • Thermostatic expansion valve
  • Float expansion valve

so here throttling devices commonly used in refrigeration and air-conditioning because

  • To reduce the coolant pressure, the high pressure of the refrigerant from the condenser is necessary to reduce the evaporation to obtain evaporation at the right temperature  
  • To meet the refrigerated load, the throttling valve flows through the coolant to cool the load at high temperatures.
5 0
3 years ago
Given the vector current density J = 10rho2zarho − 4rho cos2 φ aφ mA/m2:
Xelga [282]

Answer:

(a) Current density at P is J(P)=180.\textbf{a}_{\rho}-9.\textbf{a}_{\phi} \ (mA/m^2)\\.

(b) Total current I is 3.257 A

Explanation:

Because question includes symbols and formulas it can be misunderstood. In the question current density is given as below;

J=10\rho^2z.\textbf{a}_{\rho}-4\rho(\cos\phi)^2\textbf{a}_{\phi}\\

where \textbf{a}_{\rho} and \textbf{a}_{\phi} unit vectors.

(a) In order to find the current density at a specific point <em>(P)</em>, we can simply replace the coordinates in the current density equation.  Therefore

J(P(\rho=3, \phi=30^o,z=2))=10.3^2.2.\textbf{a}_{\rho}-4.3.(\cos(30^o)^2).\textbf{a}_{\phi}\\\\J(P)=180.\textbf{a}_{\rho}-9.\textbf{a}_{\phi} \ (mA/m^2)\\

(b) Total current flowing outward can be calculated by using the relation,

I=\int {\textbf{J} \, \textbf{ds}

where integral is calculated through the circular band given in the question. We can write the integral as below,

I=\int\{(10\rho^2z.\textbf{a}_{\rho}-4\rho(\cos\phi)^2\textbf{a}_{\phi}).(\rho.d\phi.dz.\textbf{a}_{\rho}})\}\\\\I=\int\{(10\rho^2z).(\rho.d\phi.dz)\}\\\\\\

due to unit vector multiplication. Then,

I=10\int\(\rho^3z.dz.d\phi

where \rho=3,\ 0. Therefore

I=10.3^3\int_2^{2.8}\(zdz.\int_0^{2\pi}d\phi\\I=270(\frac{2.8^2}{2}-\frac{2^2}{2} )(2\pi-0)=3257.2\ mA\\I=3.257\ A

4 0
3 years ago
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