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3 years ago

True or False: Chemical energy stored in food cannot be transformed into mechanical energy

Physics

2 answers:

Novosadov [1.4K]3 years ago

7 0

The answer to this question is false.

nordsb [41]3 years ago

3 0

It's false. it can be transform

help me to promote as brainliest

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The masses of the two moons are determined to be 2M2M for Moon AA and MM for Moon BB . It is observed that the distance between

seraphim [82]

Answer:

F_A = 8 F_B

Explanation:

The force exerted by the planet on each moon is given by the law of universal gravitation

F = $G \frac{m M}{r^{2} }$

where M is the mass of the planet, m the mass of the moon and r the distance between its centers

let's apply this equation to our case

Moon A

the distance between the planet and the moon A is r and the mass of the moon is 2m

F_A = G \frac{2m M}{r^{2} }

Moon B

F_B = G \frac{m M}{(2r)^{2} }

F_B = G \frac{m M}{4 r^{2} }

the relationship between these forces is

F_B / F_A = $\frac{1}{2 \ 4 }$ = 1/8

F_A = 8 F_B

7 0

3 years ago

Read 2 more answers

Oh lord help me please

Zina [86]

Answer: Friction

Explanation:

The friction force is the force exerted by a surface as an object moves across it or makes an effort to move across it. There are at least two types of friction force - sliding and static friction. Though it is not always the case, the friction force often opposes the motion of an object.

7 0

3 years ago

What is a pioneer species?

Eddi Din [679]

B.
the first species that colonize new or undisturbed land

4 0

3 years ago

Ablok slides on ahorizonted sur fors which has been lubricated with heavy oil such that the blok soffers a viscous resistance th

svetoff [14.1K]

Answer:

$v = (v_0^{\frac{3}{2}}-\frac{3cx}{2m})^\frac{2}{3}$

$x = \frac{2m}{c}*v_0^{\frac{3}{2}}$

Explanation:

Given

$f(v) =- cv^\frac{1}{2}$

To start with, we begin with

$F = ma$

Substitute the expression for F

$-cv^\frac{1}{2} = ma$

$-ma = cv^\frac{1}{2}$

Acceleration (a) is:

$a = \frac{dv}{dt}$

So, the expression becomes:

$m\frac{dv}{dt} = -cv^\frac{1}{2}$

-----------------------------------------------------------------------------------------

Velocity (v) is:

$v = \frac{dx}{dt}$ --- distance/time

$v * \frac{dv}{dx}= \frac{dx}{dt}* \frac{dv}{dx}$

$v * \frac{dv}{dx}= \frac{dv}{dt}$

$\frac{dv}{dt} = v * \frac{dv}{dx}$

--------------------------------------------------------------------------------------------

So, we have:

$m\frac{dv}{dt} = -cv^\frac{1}{2}$

$mv * \frac{dv}{dx} = -cv^\frac{1}{2}$

Divide both sides by $v^\frac{1}{2}$

$mv^{1-\frac{1}{2}} * \frac{dv}{dx} = -c$

$mv^{\frac{1}{2}} * \frac{dv}{dx} = -c$

Divide both sides by m

$v^{\frac{1}{2}} * \frac{dv}{dx} = -\frac{c}{m}$

$v^{\frac{1}{2}} * dv = -\frac{c}{m} * dx$

Integrate:

$\int\limits^v_{v_0} {v^{\frac{1}{2}}} \, dv = -\frac{c}{m}\int\limits^x_0 {}} \, dx$

$\frac{2}{3}v^{\frac{3}{2}}|\limits^v_{v_0} = -\frac{c}{m}x|\limits^x_0$

$\frac{2}{3}(v^{\frac{3}{2}} - v_0^{\frac{3}{2}} ) = -\frac{cx}{m}$

$v^{\frac{3}{2}} - v_0^{\frac{3}{2}} = -\frac{3cx}{2m}$

$v^{\frac{3}{2}} = v_0^{\frac{3}{2}}-\frac{3cx}{2m}$

$v = (v_0^{\frac{3}{2}}-\frac{3cx}{2m})^\frac{2}{3}$

Next, is to get the maximum velocity by distance x.

To do this, we find the derivation by x

$\frac{dv}{dx} = 0$

$\frac{2}{3}(v_0^{\frac{3}{2}}-\frac{3cx}{2m})^{\frac{2}{3} - 1} * -\frac{3c}{2m} = 0$

$\frac{2}{3}(v_0^{\frac{3}{2}}-\frac{3cx}{2m})^{\frac{2-3}{3}} * -\frac{3c}{2m} = 0$

$\frac{2}{3}(v_0^{\frac{3}{2}}-\frac{3cx}{2m})^{\frac{-1}{3}} * -\frac{3c}{2m} = 0$

$(v_0^{\frac{3}{2}}-\frac{3cx}{2m})^{\frac{-1}{3}} * -\frac{c}{m} = 0$

Divide both sides by $-\frac{c}{m}$

$(v_0^{\frac{3}{2}}-\frac{3cx}{2m})^{\frac{-1}{3}} = 0$

Take cube roots of both sides

$(v_0^{\frac{3}{2}}-\frac{3cx}{2m})^{-1} = 0$

$v_0^{\frac{3}{2}}-\frac{3cx}{2m} = 0$

$\frac{3cx}{2m} = v_0^{\frac{3}{2}}$

$x = \frac{2m}{c}*v_0^{\frac{3}{2}}$

7 0

3 years ago

A net force of magnitude 36N gives an object

slega [8]

In working toward the complete solution of this problem,
I'm going to use addition, subtraction, multiplication, and
division. Also, I think I'll be using Newtons 2nd law of motion
like at least fifteen times, so I'll write it here for our review:

  <span> Force = (mass) x (acceleration).</span>

-- When 36 N of force acts on m₁ it accelerates at 6 m/s² .

   36 N = (m₁) x (6 m/s²)

Divide each side by 6 m/s² :   m₁ = (36 N) / (6 m/s²) = 6 kg .

-- The same force acting on (m₁+m₂) accelerates them at 2 m/s² .

   36 N = (6kg + m₂) x (2 m/s²)

Divide each side by 2 m/s² : 6kg + m₂ = (36N) / (2m/s²) = 18 kg

Subtract 6kg from each side:   m₂ = 12 kg .

-- The same net force acts on m₂ alone:

   36 N = (12 kg) x (acceleration)

Divide each side by 12 kg :   Acceleration = (36 N) / (12 kg) = <span>3 m/s² </span>.

Hope this helps alot

5 0

3 years ago