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2 years ago

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Compare and contrast chemical reactions and nuclear reactions in terms of particles involved . Select all the statements that ap

ply Both chemical and nuclear equations are written with equal signs. Nuclear reactions create a difference in the number of protons, neutrons, and electrons after the reaction. A chemical reaction is the rearranging of atoms to form new substances. E=mc^ aleph 2 applies to both chemical and nuclear reactions. In a nuclear reactionparticles can transmute into others. Chemical reactions produce energy on levels comparable to nuclear reactions The total number of protons, neutrons, and electrons stay the same throughout the chemical reaction.

1 answer:

Anna11 [10]2 years ago

5 0

Answer:

. A chemical reaction is the rearranging of atoms to form new substances. E=mc^ aleph

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Newton's second law problem. The spring constant can be determined from the balance of forces in equilibrium. Suppose the spring

mario62 [17]

Answer:

K =6.697 Kg/s²

Explanation:

Given:

delta m =41 g = 0.041 kg

delta x = 6cm = 0.06m

g = 9.8 m/s²

according to the given formula

K = delta m g /delta x

K = (0.041 kg × 9.8 m/s²) / 0.06m

K =6.697 Kg/s²

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3 years ago

A package of mass m is released from rest at a warehouse loading dock and slides down a 3.0-m-high frictionless chute to a waiti

LuckyWell [14K]

Answer:

The speed of the package of mass m right before the collision $= 7.668\ ms^-1$

Their common speed after the collision $= 2.56\ ms^-1$

Height achieved by the package of mass m when it rebounds $= 0.33\ m$

Explanation:

Have a look to the diagrams attached below.

a.To find the speed of the package of mass m right before collision we have to use law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

where $K$ is Kinetic energy and $U$ is Potential energy.

$K= \frac{mv^2}{2}$ and $U= mgh$

Considering the fact $K_{initial} = 0\ and U_{final} =0$ we will plug out he values of the given terms.

So $V_{1}{(initial)} =\sqrt{2gh} = \sqrt{2\times9.8\times3} = 7.668\ ms^-1$

Keypoints:

Sum of energies and momentum are conserved in all collisions.
Sum of KE and PE is also known as Mechanical energy.
Only KE is conserved for elastic collision.
for elastic collison we have $e=1$ that is co-efficient of restitution.

<u>KE = Kinetic Energy and PE = Potential Energy</u>

b.Now when the package stick together there momentum is conserved.

Using law of conservation of momentum.

$m_1V_1(i) = (m_1+m_2)V_f$ where $V_1{i} =7.668\ ms^-1$ .

Plugging the values we have

$m\times 7.668 = (3m)\times V_{f}$

Cancelling m from both sides and dividing 3 on both sides.

$V_f = 2.56\ ms^-1$

Law of conservation of energy will be followed over here.

c.Now the collision is perfectly elastic $e=1$

We have to find the value of $V_{f}$ for m mass.

As here $V_{f}=-2.56\ ms^-1$ we can use that if both are moving in right ward with $2.56$ then there is a $-2.56$ velocity when they have to move leftward.

The best option is to use the formulas given in third slide to calculate final velocity of object $1$ .

So

$V_{1f} = \frac{m_1-m_2}{m_1+m_2} \times V_{1i}= \frac{m-2m}{3m} \times7.668=\frac{-7.668}{3} = -2.56\ ms^-1$

Now using law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

$\frac{m\times V(f1)^2}{2} + 0 = 0 +mgh$

$\frac{v(f1)^2}{2g} = h$

$h= \frac{(-2.56)^2}{9.8\times 3} =0.33\ m$

The linear momentum is conserved before and after this perfectly elastic collision.

So for part a we have the speed $=7.668\ ms^-1$ for part b we have their common speed $=2.56\ ms^-1$ and for part c we have the rebound height $=0.33\ m$ .

3 0

3 years ago

How does an increase in greenhouse gases in the atmosphere cause an increase in global temperature?

daser333 [38]

Answer:

<h2>Greenhouse gases absorb and emit radiation and less heat dissipates to space</h2>

Explanation:

when heat is trapped it is conducted to the trapped area and gradually heats up its surrounding

8 0

3 years ago

A uniform rod is set up so that it can rotate about a perpendicular axis at one of its ends. The length and mass of the rod are

igor_vitrenko [27]

Answer:

0.231 N

Explanation:

To get from rest to angular speed of 6.37 rad/s within 9.87s, the angular acceleration of the rod must be

$\alpha = \frac{\theta}{t} = \frac{6.37}{9.87} = 0.6454 rad/s^2$

If the rod is rotating about a perpendicular axis at one of its end, then it's momentum inertia must be:

$I = \frac{mL^2}{3} = \frac{1.27*0.847^2}{3} = 0.303kgm^2$

According to Newton 2nd law, the torque required to exert on this rod to achieve such angular acceleration is

$T = I\alpha = 0.303*0.6454 = 0.196 Nm$

So the force acting on the other end to generate this torque mush be:

$F = \frac{T}{L} = \frac{0.196}{0.847} = 0.231 N$

4 0

3 years ago

In a chemical equation, where do the reactants appear?

Bess [88]

Answer: A chemical equation describes a chemical reaction. Reactants are starting materials and are written on the left-hand side of the equation. Products are the end-result of the reaction and are written on the right-hand side of the equation.

Explanation:

8 0

3 years ago