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Dimas [21]
3 years ago
11

10 POINT PLS HELP

Physics
1 answer:
Hitman42 [59]3 years ago
5 0

Answer: A heat engine cannot be 100% efficient because, according to the law, energy added as heat is always greater than energy removed as heat.

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If there is a 25 [lbs] of force acting on a 5 [lbs] of mass, what is the acceleration of that mass?
mihalych1998 [28]

Answer:

f = 25 lbs

m = 5 lbs

a =?

f = ma

25 = 5 a( divide both sides by 5)

a = 5(lbs)

5 0
3 years ago
A 2.00 kg block hangs from a spring balance calibrated in Newtons that is attached to the ceiling of an elevator.(a) What does t
tresset_1 [31]

Answer:

Part a)

Reading = 2.00 kg

Part b)

Reading = 2.00 kg

Part c)

Reading = 4.04 kg

Part d)

from t = 0 to t = 4.9 s

so the reading of the scale will be same as that of weight of the block

Then its speed will reduce to zero in next 3.2 s

from t = 4.9 to t = 8.1 s

The reading of the scale will be less than the actual mass

Explanation:

Part a)

When elevator is ascending with constant speed then we will have

F_{net} = 0

T - mg = 0

T = mg

So it will read same as that of the mass

Reading = 2.00 kg

Part b)

When elevator is decending with constant speed then we will have

F_{net} = 0

T - mg = 0

T = mg

So it will read same as that of the mass

Reading = 2.00 kg

Part c)

When elevator is ascending with constant speed 39 m/s and acceleration 10 m/s/s then we will have

F_{net} = ma

T - mg = ma

T = mg + ma

Reading is given as

Reading = \frac{mg + ma}{g}

Reading = 2.00\frac{9.81 + 10}{9.81}

Reading = 4.04 kg

Part d)

Here the speed of the elevator is constant initially

from t = 0 to t = 4.9 s

so the reading of the scale will be same as that of weight of the block

Then its speed will reduce to zero in next 3.2 s

from t = 4.9 to t = 8.1 s

The reading of the scale will be less than the actual mass

3 0
3 years ago
A high speed train is traveling at a speed of 44.7 m/s when the engineer sounds the 415 Hz warning horn. The speed of sound is 3
il63 [147K]

Answer:

Explanation:

Speed of the source of sound = v = 44.7 m/s

Speed of sound = V = 343 m/s

a) Apparent  frequency as the train approaches = f =  [V /(V -v) ] × f

= [343 / (343 - 44.7) ] × 415  = 477.18 Hz

Wave length =  λ = v / f = 343 / 477.18 = 0.719 m

b) Frequency heard as the train leaves = f ' =  [V / ( V + v) ] f

                                                                     = [343 / { 343 + 44.7 ) ] x 415

                                                                      = 367.2 Hz

Wavelength when leaving = v / f = 343 / 367.2 = 0.934 m

8 0
3 years ago
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