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3 years ago

The angle between an incidentray and the mirror is 40°.1) What is the angle of reflection?

Physics

1 answer:

Vlad1618 [11]3 years ago

5 0

Answer:

1) 50°

Explanation:

We need to find the angle of incidence first before finding the angle of reflection.

Angle of incidence = 90° - 40°

= 50°

Since the angle of incidence is the same as the angle of reflection, the angle of reflection here would be 50°.

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A 1kg cart slams into a stationary 1kg cart at 2 m/s. The carts stick together and move forward at a speed of 1 m/sl. Determine

finlep [7]

Answer:

No, it is not conserved

Explanation:

Let's calculate the total kinetic energy before the collision and compare it with the total kinetic energy after the collision.

The total kinetic energy before the collision is:

$K_i = K_1 + K_2 = \frac{1}{2}mv_1^2 + \frac{1}{2}mv_2^2=\frac{1}{2}(1 kg)(2 m/s)^2+\frac{1}{2}(1 kg)(0)^2=2 J$

where m1 = m2 = 1 kg are the masses of the two carts, v1=2 m/s is the speed of the first cart, and where v2=0 is the speed of the second cart, which is zero because it is stationary.

After the collision, the two carts stick together with same speed v=1 m/s; their total kinetic energy is

$K_f = \frac{1}{2}(m_1+m_2)v^2=\frac{1}{2}(1 kg+1kg)(1 m/s)^2=1 J$

So, we see that the kinetic energy was not conserved, because the initial kinetic energy was 2 J while the final kinetic energy is 1 J. This means that this is an inelastic collision, in which only the total momentum is conserved. This loss of kinetic energy does not violate the law of conservation of energy: in fact, the energy lost has simply been converted into another form of energy, such as heat, during the collision.

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3 years ago

A proton, an alpha particle (a bare helium nucleus), and a singly ionized helium atom are accelerated through a potential differ

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Answer:

The correct question is:

"Find the energy each gains"

The energy gained by a charged particle accelerated through a potential difference is given by

$\Delta U = q\Delta V$

where

q is the charge of the particle

$\Delta V$ is the potential difference

For a proton,

$q=+e=1.6\cdot 10^{-19}C$

And since $\Delta V=100 V$

The energy gained by the proton is

$\Delta U=(1.6\cdot 10^{-19})(100)=1.6\cdot 10^{-17}J$

For an alpha particle,

$q=+2e=3.2\cdot 10^{-19}C$

Therefore, the energy gained is

$\Delta U=(3.2\cdot 10^{-19})(100)=3.2\cdot 10^{-17}J$

Finally, for a singly ionized helium nucleus (a helium nucleus that has lost one electron)

$q=+e=1.6\cdot 10^{-19}C$

So the energy gained is the same as the proton:

$\Delta U=(1.6\cdot 10^{-19})(100)=1.6\cdot 10^{-17}J$

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3 years ago

A hot air balloon is hovering in the air when it drops a 40 Kg food package to some lost golfers. If the package is dropped from

UNO [17]

We can calculate this with the law of conservation of energy. Here we have a food package with a mass m=40 kg, that is in the height h=500 m and all of it's energy is potential. When it is dropped, it's potential energy gets converted into kinetic energy. So we can say that its kinetic and potential energy are equal, because we are neglecting air resistance:

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(1/2)*v²=g*h, and we multiply by 2 both sides of the equation

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Let's say I perform an experiment and get very exciting results. I'm a good scientist, so the next thing I want to do is to publish a complete description of how I did my experiment, and include all of my results. That way, scientists around the world can read about what I did, they can find any mistakes that I made, and they can even repeat my experiment for themselves and see if they get the same results.

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The angle between an incidentray and the mirror is 40°.1) What is the angle of reflection?​

The angle between an incidentray and the mirror is 40°.1) What is the angle of reflection?