All categories

2 years ago

A CAR ACCElerates FROM REST To

Physics

1 answer:

Ilia_Sergeevich [38]2 years ago

5 0

Answer:

666.6 kW

Explanation:

Power Formula

Power = Force × Velocity

Calculating Force

F = ma
F = m(v - u / t) [From the equation v = u + at]
F = 2,000 (50/7.5)
F = 2,000 (20/3)
F = 40,000/3
F = 13333.3 N

Solving for Power

P = 13333.3 × 50
P = 666666.6 W
P = 666.6 kW

You might be interested in

Identify the energy transformation that takes when place when you apply the brakes on a bicycle.

valina [46]

Um Kinetic, mechanical, potential idk

The rider is riding the nick so kinetic then mechanical making the bike move then potential stopping the bike

4 0

3 years ago

dangina [55]

Answer:

C 2000v its obviously ans because if o is 1000 2 vo is 2000v

7 0

3 years ago

The density of Mercury is 1.36 × 10 by 4 Kgm - 3 at 0 degrees. Calculate its value at 100 degrees and at 22 degrees. Take cubic

Drupady [299]

a) Density at 100 degrees: $1.34\cdot 10^4 kg/m^3$

Explanation:

The density of mercury at 0 degrees is $d=1.36\cdot 10^4 kg/m^3$

Let's take 1 kg of mercury. Its volume at 0 degrees is

$V=\frac{m}{d}=\frac{1 kg}{1.36\cdot 10^4 kg/m^3}=7.35\cdot 10^{-5} m^3$

The formula to calculate the volumetric expansion of the mercury is:

$\Delta V= \alpha V \Delta T$

where

$\alpha=180\cdot 10^{-6} K^{-1}$ is the cubic expansivity of mercury

V is the initial volume

$\Delta T$ is the increase in temperature

In this part of the problem, $\Delta T=100 C-0 C=100 C=100 K$

So, the expansion is

$\Delta V= \alpha V \Delta T=(180\cdot 10^{-6} K^{-1})(7.35\cdot 10^{-5} m^3)(100 K)=1.3\cdot 10^{-6} m^3$

So, the new density is

$d'=\frac{m}{V+\Delta V}=\frac{1 kg}{7.35\cdot 10^{-5} m^3+1.3\cdot 10^{-6} m^3}=1.34\cdot 10^4 kg/m^3$

b) Density at 22 degrees: $1.355\cdot 10^4 kg/m^3$

We can apply the same formula we used before, the only difference here is that the increase in temperature is

$\Delta T=22 C-0 C=22 C=22 K$

And the volumetric expansion is

$\Delta V= \alpha V \Delta T=(180\cdot 10^{-6} K^{-1})(7.35\cdot 10^{-5} m^3)(22 K)=2.9\cdot 10^{-7} m^3$

So, the new density is

$d'=\frac{m}{V+\Delta V}=\frac{1 kg}{7.35\cdot 10^{-5} m^3+2.9\cdot 10^{-7} m^3}=1.355\cdot 10^4 kg/m^3$

8 0

3 years ago

What is happening to the earth's field right now.

Pie

It is bracking down because of all the pollutoin

8 0

3 years ago

¿De dónde obtienen energías las plantas y los animales por debajo del mar?

soldier1979 [14.2K]

Explanation:

<h2>audio John of synergies list 1st year animal Court judgement and for judgement and</h2>

5 0

3 years ago