All categories

3 years ago

Does a stove have mechanical energy

1 answer:

6 0

-- Any object has gravitational potential energy relative to any place
lower than where the object is. The stove in the kitchen has potential
energy relative to the basement floor.

-- If an object is not moving, then it has no kinetic energy. The stove has
no kinetic energy unless you throw it or drop it out of a window.

You might be interested in

Two bicyclists are accelerating forward in a straight line, and Biker 1 has less mass than Biker 2. If the net force on the bike

Solnce55 [7]

If Biker 1 has less mass than Biker 2 it the follows that biker 1 has a greater acceleration than Biker 2.

<h3>What is the Newton second law?</h3>

The Newton second law gives the relationship between the force and the acceleration. We know that; acceleration = Force/mass.

Now if Biker 1 has less mass than Biker 2 it the follows that biker 1 has a greater acceleration than Biker 2.

Learn more about Newton's law:brainly.com/question/27573481

#SPJ1

4 0

1 year ago

Which statement accurately describes the outer planets?

tiny-mole [99]

The outer planets have a high gravity due to their large size

6 0

3 years ago

Read 2 more answers

why is time an independent variable and position is a dependent variable in a position versus time graph

zhuklara [117]

Because the position depends on the amount of time that has passed.

8 0

3 years ago

What is wrong about the picture below

Oliga [24]

what even is that ??

4 0

2 years ago

Cart A of inertia m has attached to its front end a device that explodes when it hits anything, releasing a quantity of energy E

Leviafan [203]

We need to write down momentum and energy conservation laws, this will give us a system of equation that we can solve to get our final answer. On the right-hand side, I will write term after the collision and on the left-hand side, I will write terms before the collision.
Let's start with energy conservation law:
$\frac{mv^2}{2}+\frac{2mv^2}{2}+0.75E=\frac{mv_{A}^2}{2}+\frac{2mv_{B}^2}{2}$
$\frac{3mv^2}{2}+0.75E=\frac{mv_{A}^2}{2}+mv_{B}^2$
This equation tells us that kinetic energy of two carts before the collision and 3 quarters of explosion energy is beign transfered to kinetic energy of the cart after the collision.
Let's write down momentum conservation law:
$mv+2mv=mv_A+2mv_B\\ 3mv=mv_A+2mv_B\\$
Because both carts have the same mass we can cancel those out:
$3v=v_A+2v_B$
Now we have our system of equation that we have to solve:
$\frac{3mv^2}{2}+0.75E=\frac{mv_{A}^2}{2}+mv_{B}^2\\ 3v=v_A+2v_B$
Part A
We need to solve our system for $v_a$ . We will solve second equation for $v_b$ and then plug that in the first equation.
$3v=v_A+2v_B\\ 3v-v_A=2v_B\\ v_B=\frac{3v-v_A}{2}$
Now we have to plug this in the first equation:
$\frac{3mv^2}{2}+0.75E=\frac{mv_{A}^2}{2}+mv_{B}^2\\v_B=\frac{3v-v_A}{2}\\$
We will multiply the first equation with 2 and divide by m:
$3v^2+\frac{3E}{2m}=v_{A}^2}+2v_{B}^2\\v_B=\frac{3v-v_A}{2}\\$
Now we plug in the second equation into first one:
$3v^2+\frac{3E}{2m}=v_{A}^2}+2v_{B}^2\\ 3v^2+\frac{3E}{2m}=v_{A}^2}+2\frac{(3v-v_A)^2}{4}\\ 3v^2+\frac{3E}{2m}=v_{A}^2}+\frac{9v^2-6v\cdot v_A+v_{A}^2}{2} /\cdot 2\\ 6v^2+\frac{3E}{m}=2v_{A}^2+9v^2-6v\cdot v_A+v_{A}^2}\\ 3v_A^2-6v\cdot v_a+3(v^2-\frac{E}{m})=0/\cdot\frac{1}{3}\\ v_A^2-3v\cdot v_A+ (v^2-\frac{E}{m})=0$
We end up with quadratic equation that we have to solve, I won't solve it by hand.
Coefficients are:
$a=1\\
b=-6v\\
c=v^2-\frac{E}{m}$
Solutions are:
$v_A=\frac{3v+\sqrt{5v^2+\frac{4E}{m}}}{2},\:v_A=\frac{3v-\sqrt{5v^2+\frac{4E}{m}}}{2}$
Part B
We do the same thing here, but we must express $v_a$ from momentum equation:
$3v=v_A+2v_B\\
v_A=3v-2v_B$
Now we plug this into our energy conservation equation:
$3v^2+\frac{3E}{2m}=v_{A}^2}+2v_{B}^2\\v_A={3v-v_B}\\
3v^2+\frac{3E}{2m}=(3v-v_B)^2+2v_B^2\\
3v^2+\frac{3E}{2m}=9v^2-6v\cdot v_B+v_B^2+2v_B^2\\
3v^2+\frac{3E}{2m}=3v_B^2-6v\cdot v_B+9v^2\\
3v_B^2-6v\cdot v_B+9v^2-3v^2-\frac{3E}{2m}=0\\
3v_B^2-6v\cdot v_B+(6v^2-\frac{3E}{2m})=0
$
Again we end up with quadratic equation. Coefficients are:
$a=3\\
b=-6v\\
c=6v^2-\frac{3E}{2m}$
Solutions are:
$v_B=\frac{6v+\sqrt{-36v^2+\frac{18E}{m}}}{6},\:v_B=\frac{6v-\sqrt{-36v^2+\frac{18E}{m}}}{6}$

8 0

3 years ago