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3 years ago

7

Real-world efficiencies are generally very high, in the 90 percent range. Please select the best answer from the choices provide

d. T F

1 answer:

Vesna [10]3 years ago

4 0

Answer:

i think it truth

Explanation:

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Define density<br><br> Please help!!

Romashka-Z-Leto [24]

, density is how compact an object is. Put another way, density is the mass of an object divided by its volume.

5 0

3 years ago

The mean diameters of Mars and Earth are 6.9 ✕ 103 km and 1.3 ✕ 104 km, respectively. The mass of Mars is 0.11 times Earth's mas

Roman55 [17]

Answer:

(a) Ratio of mean density is 0.735

(b) Value of g on mars 0.920 $m,/sec^2$

(c) Escape velocity on earth is $3.563\times 10^4m/sec$

Explanation:

We have given radius of mars $R_{mars}=6.9\times 10^3km=6.9\times 10^6m$ and radius of earth $R_{E}=1.3\times 10^4km=1.3\times 10^7m$

Mass of earth $M_E=5.972\times 10^{24}kg$

So mass of mars $M_m=5.972\times\times 0.11 \times 10^{24}=0.657\times 10^{24}kg$

Volume of mars $V=\frac{4}{3}\pi R^3=\frac{4}{3}\times 3.14\times (6.9\times 10^6)^3=1375.357\times 10^{18}m^3$

So density of mars $d_{mars}=\frac{mass}{volume}=\frac{0.657\times 10^{24}}{1375.357\times 10^{18}}=477.69kg/m^3$

Volume of earth $V=\frac{4}{3}\pi R^3=\frac{4}{3}\times 3.14\times (1.3\times 10^7)^3=9.198\times 10^{21}m^3$

So density of earth $d_{E}=\frac{mass}{volume}=\frac{5.972\times 10^{24}}{9.198\times 10^{21}}=649.271kg/m^3$

(A) So the ratio of mean density $\frac{d_{mars}}{d_E}=\frac{477.69}{649.27}=0.735$

(B) Value of g on mars

g is given by $g=\frac{GM}{R^2}=\frac{6.67\times 10^{-11}\times0.657\times 10^{24}}{(6.9\times 10^6)^2}=0.920m/sec^2$

(c) Escape velocity is given by

$v=\sqrt{\frac{2GM}{R}}=\sqrt{\frac{2\times 6.67\times 10^{-11}\times 0.657\times 10^{24}}{6.9\times 10^6}}=3.563\times 10^4m/sec$

5 0

3 years ago

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At a certain instant, coil A is in a 10-T external magnetic field and coil B is in a 1-T external magnetic field. Both coils hav

Neporo4naja [7]

Answer:

A) coil A

Explanation:

According to Faraday, Induced emf is given as;

E.M.F = ΔФ/t

ΔФ = BACosθ

where;

ΔФ is change in magnetic flux

θ is the angle between the magnetic field, B, and the normal to the loop of area A

A is the area of the loop

B is the magnetic field

From the equation above, induced emf depends on the strength of the magnetic field.

Both coils have the same area and are oriented at right angles to the field.

Coil A has a magnetic field strength of 10-T which is greater than 1 T of coil B, thus, coil A will have a greater emf induced in it.

7 0

3 years ago

An object 16.8 cm tall is placed in front of a converging lens. A real image, 46 cm tall, is formed on the other side of the len

worty [1.4K]

Answer:

2.74

Explanation:

Magnification = image distance/object distance

Mag = v/u

Given

v = 46cm

u = 16.8

Magnification = 46/16.8

Magnification = 2.74

Hence the magnification is 2.74

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3 years ago

What's the value of 57,281 joules in Btu? A. 54.3 Btu B. 14.2 Btu C. 28.9 Btu D. 37.7 Btu

Alex Ar [27]

I'm quite sure it is A

8 0

3 years ago

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