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1 year ago

What did you notice about the angle of incidence and the angle of reflection?

Physics

1 answer:

Readme [11.4K]1 year ago

6 0

Answer:

They are equal

Explanation:

angle of incidence = angle of reflection

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Two friends, Al and Jo, have a combined mass of 195 kg. At the ice skating rink, they stand close together on skates, at rest an

ddd [48]

Answer:

Al's mass is 102.92 kg

Explanation:

As there are no external forces in the horizontal direction, the horizontal net force must be zero:

$F_{net} = 0$

As the force is the derivative in time of the momentum, this means that the horizontal momentum is constant:

$F_{net} = \frac{dp_{horizontal}}{dt} = 0$

$p_{horizontal_i }= p_{horizontal_f}$

where the suffix i and f means initial and final respectively.

The initial momentum will be:

$p_{horizontal_}i = m_{Al} \ v_{Al_i} + m_{Jo} \ v_{Jo_i}$

But, as they are at rest, initially

$p_{horizontal_i} = m_{Al} * 0 + m_{Jo} * 0$

$p_{horizontal_i} = 0$

So, this means:

$p_{horizontal_f} = m_{Al} \ v_{Al_f} + m_{Jo} \ v_{Jo_f} = 0$

We know that the have an combined mass of 195 kg:

$m_{total} = m_{Al} + m_{Jo} = 195 \ kg$ .

so:

$m_{Jo} = 195 \ kg - m_{Al}$ .

$m_{Al} \ v_{Al_f} + (195 \ kg - m_{Al}) \ v_{Jo_f} = 0$

$m_{Al} \ v_{Al_f} - m_{Al} \ v_{Jo_f}= - 195 \ kg \ v_{Jo_f}$

$m_{Al} \ (v_{Al_f} - v_{Jo_f})= - 195 \ kg \ v_{Jo_f}$

$m_{Al} = \frac{ - 195 \ kg \ v_{Jo_f} } { v_{Al_f} - v_{Jo_f} }$

$m_{Al} = \frac{195 \ kg \ v_{Jo_f} } { v_{Jo_f} - v_{Al_f} }$

Now, we can use the values:

$v_{Al_f}= 10.2 \frac{m}{s}$

$v_{Jo_f}= - 11.4 \frac{m}{s}$

where the minus sign appears as they are moving at opposite directions

$m_{Al} = \frac{195 \ kg ( - 11.4 \frac{m}{s} ) } { (- 11.4 \frac{m}{s}) - 10.2 \frac{m}{s} }$

$m_{Al} = 102.92 \ kg$

and this is the Al's mass.

5 0

2 years ago

A photon with an energy E = 2.12 GeV creates a proton-antiproton pair in which the proton has a kinetic energy of 96.0 MeV. What

Oksanka [162]

Answer:

The kinetic energy of the anti proton is 147.4 MeV.

Explanation:

Given that,

Energy = 2.12 GeV

Kinetic energy = 96.0 MeV

We need to calculate the kinetic energy of the anti proton

Using formula of energy

$E_{photon}=m_{p}c^2+m_{np}c^2+K.E_{p}+K.E_{np}$

We know that,

$m_{p}c^2=m_{np}c^2$

So, $E_{photon}=2mc^2+K.E_{p}+K.E_{np}$

$K.E_{np}=E_{photon}-(2mc^2+K.E_{p})$

Put the value into the formula

$K.E_{np}=2.12\times10^{9}-2\times938.3\times10^{6}-96\times10^{6}$

$K.E_{np}=147.4\ MeV$

Hence, The kinetic energy of the anti proton is 147.4 MeV.

6 0

2 years ago

How to do this question plz answer

Monica [59]

Answer:

I can't see the attachment it's too blury I can see bout two words and thats it

Explanation:

Think i got an eye infection or bad eye sight

3 0

2 years ago

What are the two varieties of friction?

murzikaleks [220]

There are two main types of friction, static friction and kinetic friction. Static friction operates between

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3 years ago

30 POINTS

Travka [436]

Answer:

to be honest I'm not sure

3 0

2 years ago

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