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77julia77 [94]
3 years ago
10

8. The ratio of output power to input power, in percent?A. work. B. efficiency. C. horsepower. D. conductivity.

Physics
2 answers:
Tema [17]3 years ago
7 0
That ratio is the efficiency of whatever the power went into and came out of. There's no reason the efficiency has to be expressed as a percentage.
BARSIC [14]3 years ago
4 0
The ratio of output power to input power, expressed as a percentage is B. efficiency. 

Efficiency is expressed as a percentage less than 100%, as a machine could never be completely efficient.
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A 200-foot tall monument is located in the distance. From a window in a building, a person determines that the angle of elevatio
egoroff_w [7]

Answer:

665 ft

Explanation:

Let d be the distance from the person to the monument. Note that d is perpendicular to the monument and would make 2 triangles with the monuments, 1 up and 1 down.

The side length of the up right-triangle knowing the other side is d and the angle of elevation is 13 degrees is

dtan13^0 = 0.231d

Similarly, the side length of the down right-triangle knowing the other side is d and the angle of depression is 4 degrees

dtan4^0 = 0.07d

Since the 2 sides length above make up the 200 foot monument, their total length is

0.231d + 0.07d = 200

0.301 d = 200

d = 200 / 0.301 = 665 ft

7 0
3 years ago
A 23 kg body is moving through space in the positive direction of an x axis with a speed of 130 m/s when, due to an internal exp
babymother [125]

Answer:

a) Vx = 1088m/s

b) Vy = -162.93m/s

c) 5246745J

Explanation:

Mass of unbroken body = 23kg

Its velocity along +ve X-axis = 130m/s

Mass of first broken body, m1= 9.4kg

Its velocity along +ve X-axis = 130m/s

Nass of 2nd broken body, m2 = 6.1kg

Its velocity long-lived X - axis = -550m/s

Mass of 3rd broken body = ?

m3 = (23 - 9.4 - 6.1)kg

m3 = 7.5kg

Let velocity along the x-axis = Vx

Let the velocity along the x-axis = Vy

Applying law of conservation of momentum along x-axis

a) m1×0 + m2×(-550) + m3×(Vx) =M × 130

9.4 × 0 + 6.1× (-550) + 7.5(Vx) = 23 ×130

0 + (-5170) + 7.5Vx = 2990

2990 + 5170 = 7.5Vx

8160 = 7.5Vx

Vx = 8160/7.5

Vx = 1088m/s

b) Aplying conservation of momentum along the x-axis

(m1×130) + (m2 × 0) + (m3× Vy) = 0

(9.4 × 130) + (6.1 ×550) + 7.5Vy = 0

1222 + 0 + 7.5Vy = 0

1222 = -7.5Vy

Vy = 1222/(-7.5)

Vy = -262.93m/s

c) The energy released or change in KE is given by:

1/2[(m1v1^2) + (m2v2^2) +(m3Vx^2) ]= MV^2

Change in KE = 1/2[ 9.4× 130^2 + 6.1 × 550^2 + 7.5 × 1088^2 ] - 1/2(23 × 130^2)

Change in KE = 1/2[158860 + 1845250 + 8878080] - 1/2[388700]

Change in KE = 5441095 - 194350

Change in KE = 5246745J

4 0
3 years ago
Arrange an 8-, 12-, and 16-Ω resistor in a combination that has a total resistance of 8.89 Ω pls with de work
kolezko [41]


20 ohms in parallel with 16 ohm= 8.89

20x16/20+16. Product over sum


8 0
3 years ago
An elevator together with its passengers weights 5000 N. At a certain instant, the tension in its supporting cable is 6000 N. De
gayaneshka [121]
We have to forces acting on the system (elevator+passengers):
1) The weight (W=5000 N), acting downward
2) The cable's tension (T=6000 N), acting upward
So, the two forces have opposite direction. The resultant (in upward direction) will be
F=T-W
And for Newton's second law, the resultant of the forces acting on the system causes an acceleration on the system itself, given by
a= \frac{F}{m}
where m is the mass of the system.

So, we need to find F and m.
The resultant of the forces is
F=T-W=6000 N-5000 N=1000 N
To find m, we can use the weight of the system. In fact, the weight of an object is given by
W=mg
where g=9.81 m/s^2. Solving for m, and using W=5000 N, we find
m= \frac{W}{g}= \frac{5000 N}{9.81 m/s^2}=510 kg

and at this point, we can calculate the acceleration of the system (elevator+people):
a= \frac{F}{m}= \frac{1000 N}{510 kg}=1.96 m/s^2
and the acceleration has the same direction of the resultant force, so upward.
8 0
3 years ago
An electron is released from rest at the negative plate of a parallel plate capacitor. The charge per unit area on each plate is
dmitriy555 [2]

Answer:

v = 1.15*10^{7} m/s

Explanation:

given data:

charge/ unit area= \sigma = 1.99*10^{-7} C/m^2

plate seperation = 1.69*10^{-2} m

we know that

electric field btwn the plates isE = \frac{\sigma}{\epsilon}

force acting on charge is F = q E

Work done by charge q id\Delta X =\frac{ q\sigma \Delta x}{\epsilon}

this work done is converted into kinectic enerrgy

\frac{1}{2}mv^2 =\frac{ q\sigma \Delta x}{\epsilon}

solving for v

v = \sqrt{\frac{2q\Delta x}{\epsilon m}

\epsilon = 8.85*10^{-12} Nm2/C2

v = \sqrt{\frac{2 1.6*10^{-19}1.99*10^{-7}*1.69*10^{-2}}{8.85*10^{-12} *9.1*10^{-31}}

v = 1.15*10^{7} m/s

8 0
3 years ago
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