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2 years ago

How much time is required for a car engine to do 278 kJ of work, if its maximum power is 95 kW?

Physics

1 answer:

zalisa [80]2 years ago

3 0

W= work = 278 kj

P = Power = 95 Kw

t= time

P = W / t

Isolate t

t = W /P

Replacing:

t= 278kj / 95Kw = 2.93 seconds

It is required 2.93 seconds

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Force X has a magnitude of 1260 pounds, and Force Y has a magnitude of 1530 pounds. They act on a single point at an angle of 4

weeeeeb [17]

Answer:

Fe= 2579.68 P

α= 24.8°

Explanation:

Look at the attached graphic

we take the forces acting on the x-y plane and applied at the origin of coordinates

FX = 1260 P , horizontal (-x)

FY = 1530 P , forming 45° with positive x axis

x-y components FY

FYx= - 1530*cos(45)° = - 1081.87 P

FYy= - 1530*sin(45)° = - 1081.87 P

Calculation of the components of net force (Fn)

Fnx= FX + FYx

Fnx= -1260 P -1081.87 P

Fnx= -2341.87 P

Fny=FYy

Fny= -1081.87 P

Calculation of the components of equilibrant force (Fe)

the x-y components of the equilibrant force are equal in magnitude but in the opposite direction to the net force components:

Fnx= -2341.87 P, then, Fex= +2341.87 P

Fny= -1081.87 P P, then, Fex= +1081.87 P

Magnitude of the equilibrant (Fe)

$F_{n} = \sqrt{(F_{nx})^{2} +(F_{ny})^{2} }$

$F_{e} =\sqrt{(2341.87)^{2}+(1081.87)^{2} }$

Fe= 2579.68 P

Calculation of the direction of equilibrant force (α)

$\alpha =tan^{-1} (\frac{F_{ny} }{F_{nx} } )$

$\alpha =tan^{-1} (\frac{1081.87 }{2341.87} )$

α= 24.8°

Look at the attached graphic

6 0

3 years ago

Hallar la distancia que recorre una movil al cabo

Kryger [21]

Answer:

Distancia, S = 136 metros

Explanation:

Dados los siguientes datos;

Aceleración, a = 3 m/s²

Velocidad inicial, u = 5 m/s

Tiempo, t = 8 segundos

Para encontrar la distancia recorrida, usaríamos la segunda ecuación de movimiento;

S = ut + ½at² Sustituyendo en la fórmula, tenemos;

S = 5 × 8 + ½ × 3 × 8²

S = 40 + 1,5 × 64

S = 40 + 96

Distancia, S = 136 metros

3 0

3 years ago

The precision of a laboratory instrument is ± 0.05 g. The accepted value for your measurement is 7.92 g. Which measurements are

skelet666 [1.2K]

Answer:

7.89 7.91

Explanation:

The ranges of measurement lie between 7.92-0.05 and 7.92+0.05

7.87g and 7.97g

3 0

3 years ago

Read 2 more answers

A 1.31 kg object is attached to a horizontal spring of force constant 2.70 N/cm and is started oscillating by pulling it 6.20 cm

goldfiish [28.3K]

Answer:

The answer is below

Explanation:

a) The change in energy is the difference between the final energy and the initial energy.

ΔE (energy change) = Ef (final energy) - Ei (initial energy)

$\Delta E=\frac{1}{2}kA_f^2 -\frac{1}{2}kA_i^2\\\\k=force\ constant=2.7\ N/cm=270\ N/m, A_f=final\ dispalacment= 3.7\ cm=0.037\ m,\\ A_i=initial \ displacement = 6.2\ cm=0.062\ m\\\\Hence:\\\\\Delta E=\frac{1}{2}(270)(0.037)^2 -\frac{1}{2}(270)(0.062)^2\\\\\Delta E=-0.334 \ J$

The negative sign shows that energy is lost to the environment. Hence 0.334 J is lost to the environment.

b) According to the law of conservation of energy, energy cannot be created or destroyed but transformed from one form to another.

The oscillating object loses energy due to wind resistance, friction between the spring and the object. Given that the air is frictionless, hence the energy loss is due to friction which is converted to heat.

6 0

3 years ago

The turbines can be seen inside this hydroelectric dam. Why are they located at that particular height?

Yakvenalex [24]

Answer:

Explanation:

the answer is number three

5 0

4 years ago

Read 2 more answers