The transfer of energy between two objects by electromagnetic waves is called

Explain what will happen if an object has high momentum or low momentum and a force is applied?

Korvikt [17]

-- If the force is applied in the same direction as the object is moving, then the object's momentum in that direction will increase.

-- If the force is applied in the direction OPPOSITE to the way the object is moving, then the object's momentum will decrease.

-- In either case, the CHANGE in the object's momentum will be

(strength of the force) x (length of time the force is applied) .

This quantity is also called "impulse".

5 0

4 years ago

In a perfectly elastic collision between two perfectly rigid objects

ipn [44]

Both the total momentum and the total kinetic energy are conserved

Explanation:

- In a collision between two or more objects, if there are no external forces acting on the system (isolated system), the total momentum of the objects is always conserved. This is called principle of conservation of momentum, and can be written as follows:

$mu+MU = mv+MV$

where

m, M are the masses of the two objects

u, U are the initial velocities of the two objects

v, V are the final velocities of the two objects

- The total kinetic energy, however, is not always conserved. In fact, we have two types of collision:

1) In a perfectly elastic collision, the total kinetic energy of the objects is conserved. This means that we can write the following equation:

$\frac{1}{2}mu^2 + \frac{1}{2}MU^2 = \frac{1}{2}mv^2+\frac{1}{2}MV^2$

2) In an inelastic collision, the total kinetic energy of the object is NOT conserved. This means that part of the total kinetic energy is "lost", converted into other forms of energy (mainly thermal energy, due to the presence of frictional forces within the system). The most extreme case is called perfectly inelastic collision, in which the two objects stick together after the collision, and there is the maximum loss of kinetic energy.

Learn more about collisions:

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7 0

4 years ago

A ball is thrown vertically into the air and when it returns after an interval of 2 seconds, it is caught. Which one of the foll

baherus [9]

Answer:

Last statement option: "The acceleration after it leaves the hand is 10 m/s/s downwards."

Explanation:

At every instant of its motion, the ball is under the effects of the acceleration due to gravity (assumed to be 10 m/s^2). This is true at whatever altitude the ball is. The acceleration due to gravity is always pointing down (not up).

In the absence of air resistance, the motion is described kinematically by a parabola with the branches pointing down as a function of time (motion under constant acceleration), with the vertex indicating the maximum altitude the ball reaches. Both branches (representing motion upwards and downwards) are equidistant from the vertex, so the time going up equals the time coming down.

Therefore, the only statement option that is correct is the last one: "The acceleration after it leaves the hand is 10 m/s/s downwards."

8 0

4 years ago

Calculate the separation between the two lowest levels for an O2 molecule in a one-dimensional container of length 5.0 cm. At wh

MariettaO [177]

Answer:

The separation between the two lowest levels = $1.24 * 10^{-39}J$

The values of n where the energy of molecule reaches 1/2 kT at 300K = $2.2 * 10^{9}$

The separation at this level = 1.8 * $10^{-30}$ J

Explanation:

Knowing the formula

En = $\frac{n^{2} h^{2} }{8 mL^{2} }$

Mass of oxygen molecule

m (O2) = 32 amu * $\frac{1.6605 * 10^{-27 kg} }{1 amu}$

So the energy diference between the two lowest levels:

E2 - E1 = $\frac{3h^{2} }{8mL^{2} }$

E2 - E1 = $\frac{3 * (6.626 * 10^{-34} Js)^{2} } {8 * 32 amu * (\frac{1.6605 * 10^{-27 kg} }{1 amu})* (5*10^{-2})^{2} } = 1.24 * 10^{-39}J$

Now we should find n where the energy of molecule reaches 1/2 kT

En = $\frac{n^{2} h^{2} }{8 mL^{2} }$ = $\frac{1}{2}kT$

$\frac{h^{2} }{8 mL^{2} }$ = $4.13 * 10^{-14}J$

$n^{2} * (4.13 * 10^{-14}J) = \frac{1}{2} (1.38 * 10^{-23}JK^{-1}) * 300K$

$n = 2.2 * 10^{9}$

by the end is necessary to calculate the separation of the level

En - En-1 = $(n^{2} - (n - 1)^{2}) * \frac{h^{2} }{8 mL^{2} }$

= 1.8 * $10^{-30}$ J

4 0

3 years ago

What would happen if the moon were twice as big

IgorLugansk [536]

The universal gravity formula is F = G(Me)(Mm)/r^2. Lets use this formula to help us calculate what would happen if the Moon was twice as big.
Mn = new mass = 2*Mm
Fn = G(Me)(Mn)/r^2
Fn = G(Me)(2Mm)/r^2
Fn = 2*G(Me)(Mm)/r^2
Fn = 2*F
So 2 times the force it was before. The force should be the same, but you never know the moon and space as well know it can work in very mysterious ways.

7 0

4 years ago

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