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andreev551 [17]
3 years ago
15

The engine of a 1520-kg automobile has a power rating of 75 kW. Determine the time required to accelerate this car from rest to

a speed of 100 km/h at full power on a level road. Is your answer realistic?
Physics
1 answer:
LUCKY_DIMON [66]3 years ago
3 0

Answer:

t=15.68 s

Explanation:

Given that

m = 1520 kg

P =75 KW

We know that

Power  ,P = F .v

F=force

v=velocity

v= 100 km/h

v=\dfrac{1000}{3600}\times 100\ m/s

v=27.77 m/s

75 x 1000  = F x 27.77

F= \dfrac{75000}{27.77}\ N

F= 2700.75 N

F= m a

m=mass

a=acceleration

2700.75 = 1520 x a

 a=1.77 m/s²

time t given as

v= u + a t

27.77 = 0 + 1.77 x t

t=15.68 s

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Answer:

False

Explanation:

Density is a physical property of matter.

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kolezko [41]

Potential energy is the energy that is possible to be realized, based on position, while  the kinetic energy is the energy that is present and due to motion

The location where the potential energy is converted into kinetic energy is the point (2)

Reason:

The mechanical energy, M.E., of the roller coaster, is given as follows;

M.E. = P.E. + K.E.

Where;

P.E. = The potential energy = m·g·h

K.E. = The kinetic energy = \dfrac{1}{2} \cdot m \cdot v^2

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v = The velocity

h = The height

m = The mass

g = Acceleration due to gravity

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The mechanical energy, M.E., is constant, therefore;

When the height, h = 0, we have, the kinetic energy is maximum, given that the mechanical energy is entirely kinetic energy

Therefore;

The locations where potential energy is converted to kinetic energy are lowest locations, where the height, <em>h</em>, is minimum, such as the point (2)

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brainly.com/question/18963960

7 0
2 years ago
Three forces act on a flange as shown below. Determine the magnitude of the unknown force F (in lb) such that the net force acti
Tems11 [23]

Answer:

The unknown force will be 18.116 lb.

Explanation:

Given that,

Three forces act on a flange as shown in figure.

The net force acting on the flange is a minimum.

\dfrac{dF_{net}}{df}=0

We need to calculate the unknown force

Using formula of net force

\vec{F_{net}}=\vec{F_{x}}+\vec{F_{y}}

Put the value into the formula

\vec{F_{net}}=(F\cos45+70\cos30-40)\hat{i}+(70\sin30-F\sin45)\hat{j}

\vec{F_{net}}=(F\cos45+70\times\dfrac{\sqrt{3}}{2})\hat{i}+(70\times\dfrac{1}{2}-F\sin45)\hat{j}

The magnitude of net force,

F_{net}=\sqrt{F_{x}^2+F_{y}^2}

F_{net}=\sqrt{(F\times\dfrac{1}{\sqrt{2}}+60.62)^2+(35-F\times\dfrac{1}{\sqrt{2}})^2}

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F_{net}=\sqrt{F^2+4899.78+36.232F}

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6 0
2 years ago
You drop an ice cube into an insulated flask full of water and wait for the ice cube to completely melt. The ice cube initially
3241004551 [841]

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