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3 years ago

What do we measure energy in

Physics

2 answers:

olga2289 [7]3 years ago

5 0

We measure energy in joules

Goshia [24]3 years ago

3 0

Energy is measured by joule(j)

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A 3250-kg aircraft takes 12.5 min to achieve its cruising

scoundrel [369]

Answer:

effeciency n = = 49%

Explanation:

given data:

mass of aircraft 3250 kg

power P = 1500 hp = 1118549.81 watt

time = 12.5 min

h = 10 km = 10,000 m

v =85 km/h = 236.11 m/s

$n = \frac{P_0}{P}$

$P_o = \frac{total\ energy}{t} = \frac{ kinetic \energy + gravitational\ energy}{t}$

kinetic energy $= \frac{1}{2} mv^2 =\frac{1}{2} 3250* 236 = 90590389.66 kg m^2/s^2$

kinetic energy $= 90590389.66 kg m^2/s^2$

gravitational energy $= mgh = 3250*9.8*10000 = 315500000.00 kg m^2/s^2$

total energy $= 90590389.66 +315500000.00 = 409091242.28 kg m^2/s^2$

$P_o =\frac{409091242.28}{750} = 545454.99 j/s$

$effeciency\ n = \frac{P_o}{P} = \frac{545454.99}{1118549.81} = 0.49$

effeciency n = = 49%

8 0

2 years ago

The main difference between turbojets and rocket engines is the fact that

mafiozo [28]

Answer;

D. rocket engines are not dependent on oxygen from the air.

Explanation;

-Jet engines and rockets work on the same principle. They produce thrust through an internal pressure difference and, as explained by Newton’s Third Law of Motion, eject exhaust gases in an equal and opposite direction.

-The main difference between them is that jets get the oxygen to burn fuel from the air and rockets carry their own oxygen, which allows them to operate in space.

Additionally, Jet engines have two openings (an intake and an exhaust nozzle). Rocket engines only have one opening (an exhaust nozzle).

5 0

3 years ago

Read 2 more answers

Mass can be considered concentrated at the center of mass for rotational motion.Select one:TrueFalse

frez [133]

iIn this case the mass of a body cannot be considered to be concentrated at the centre of mass of the body for the purpose of computing the rotational motion

Therefore the answer is False

6 0

8 months ago

Two forces F1 and F2 of equal magnitude are applied to a brick of mass 20kg lying on the floor as shown in the figure above. If

jeka57 [31]

Let $F_{1}=F_{2}=F$ .

Normal force equals (using Newton's third law) $N=mg+F\sin30^{o}-F\sin37^{o}$ .

$F_{f}\leq \mu N = \mu(mg+F\sin30^{o}-F\sin37^{o})$ , but $F_{f}\leq F(\cos30^o+\cos37^o)$ for all $F_{f}$ (in order to start moving the break). Therefore $F(\cos 30^o+\cos37^o)\geq \mu(mg+F\sin30^{o}-F\sin37^{o})$ , solving for $F$ : $F\geq \frac{\mu mg}{\cos30^o+\cos37^o-\mu\sin30^o+\mu\sin37^o}\approx 46,91\; \textbf{N}$