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2 years ago

Which action is due to field forces?

Physics

1 answer:

Stella [2.4K]2 years ago

8 0

Answer:

an apple falling from a tree

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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

2 years ago

A 7.59-m-tall container is filled to the brim, partway with mercury and the rest of the way with water. The container is open to

hoa [83]

Answer:

0.22 m

Explanation:

$P_{o}$ = Atmospheric pressure = 101325 Pa

$P$ = Pressure at the bottom = tex]2 P_{o}[/tex] = 2 (101325) = 202650 Pa

$h$ = height of the container = 7.59 m

$x$ = depth of the mercury

Pressure at the bottom = Atmospheric pressure + Pressure due to mercury + Pressure due to water

$P = P_{o} + \rho _{w} g (h - x) + \rho _{hg} g x\\2 P_{o} = P_{o} + \rho _{w} g (h - x) + \rho _{hg} g x\\P_{o} = \rho _{w} g (h - x) + \rho _{hg} g x\\101325 = (1000) (9.8) (7.59 - x) + (13600) (9.8) x\\x = 0.22 m$

5 0

3 years ago

Read 2 more answers

Answer it and get 100 points.

Alex Ar [27]

Answer:

because metals a good conductor of electricity.

Explanation:

Metal particles are held together by strong metallic bonds, which is why they have high melting and boiling points. The free electrons in metals can move through the metal, allowing metals to conduct electricity.

4 0

2 years ago

Read 2 more answers

The part of a circult that increases the electric potential of the electrons is the

Tanya [424]

Answer:

The part of a circuit that increases the electric potential of the electrons is the battery.

6 0

2 years ago

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One-half second after starting from rest, a freely falling object will have a speed of about:_________

snow_tiger [21]

One-half second after starting from rest, a freely falling object will have a speed of about 5 m/s.

<h3>What do you mean by falling object?</h3>

Falling things include goods, tools, garbage, and rubbish that are blown, dropped, or displaced from heights.

Physical science research frequently involves the study of gravity. Of course, what makes things fall is the force of gravity. A force of attraction between two objects is constant. This pulling force is similar to the pull of gravity.

given free fall from rest i.e.

initial velocity u=0 m/s

acceleration a = 10m/s2

time t=0.5 seconds

we know final velocity v = u + at = 0 + (10 *0.5) = 5m/s

To learn more about the falling object, visit: brainly.com/question/13299152

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7 0

1 year ago