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3 years ago

Can someone pls help me with this? Its due in 25 minutes

Physics

1 answer:

QveST [7]3 years ago

5 0

The angle of reflection is 42 degrees because it says the angle of incidence is equal to the angle of reflection

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A man climbs a wall that has a height of 8.4 meters and gave the potential energy of 4620 joules. His mass is about_____ kilogra

Dmitrij [34]

Gravitational potential energy = mass x acceleration due to gravity x height
GPE=mgh
4620=mx9.81x8.4
4620/(9.81x8.4)=m=56.1 kg

7 0

3 years ago

a 60g tennis ball travelling at 30m/s hi a wall and bounce back at 20m/s.calculate (a)momentum of the ball before impact (b)mome

frez [133]

Explanation:

the formula for momentum is denoted by p=mv where p is momentum, m is mass and v is velocity. thus, the velocity before impact would be 0.060 x 30 = 1.8 kg/ms

the second one would just be 0.060 x 20 0.72kg/ms

I'm not 100 percent sure this is correct but yeah

6 0

3 years ago

What is the difference between stretching a t-shirt and stretching a rubber band

Rashid [163]

B.the rubber band will return to its original shape the t shirt wont.

7 0

3 years ago

Read 2 more answers

A mystery element has three isotopes of the following masses and percent abundances: 93.597 amu at 23.63 % abundance, 96.191 amu

iogann1982 [59]

Answer:

92.397amu

Explanation: The exact amu of the mystery element is obtained by multiplying the relative abundance of each individual isotope by its respective amu and then summing the results.

The sum of the total relative abundance for all the isotopes should be 100%.

However, the relative abundance of the isotope with 95.502amu is not given; therefore to obtain it we subtract the sum of the known relative abundances from 100% as follows:

Relative abundance of isotope with 95.502amu = 100-(23.63+30.53) = 42.84%

8 0

3 years ago

A cloud mass moving across the ocean at an altitude of 2000 m encounters a coastal mountain range. As it rises to a height of 35

GREYUIT [131]

Answer:

snow

Explanation:

Since the process undergoes adiabatic expansion, hence q = 0 and ΔU = w.

We can sole this problem using the following derivation:

$ln(\frac{T_2}{T_1} )=-(\gamma -1)ln(\frac{V_f}{V_i} )=-(\gamma -1)ln(\frac{T_2}{T_1}\frac{P_i}{P_f} )\\=-(\gamma -1)ln(\frac{T_2}{T_1})-(\gamma -1)ln(\frac{P_i}{P_f})\\=-(\frac{\gamma -1}{\gamma})ln(\frac{P_i}{P_f})\\=-(\frac{\frac{C_{p,m}}{C_{p,m}-R} -1}{\frac{C_{p,m}}{C_{p,m}-R}})ln(\frac{P_i}{P_f})\\\\ln(\frac{T_2}{T_1} )==-(\frac{\frac{C_{p,m}}{C_{p,m}-R} -1}{\frac{C_{p,m}}{C_{p,m}-R}})ln(\frac{P_i}{P_f})\\\\Substituting\ values:\\\\$

$ln(\frac{T_2}{T_1} )=-(\frac{\frac{28.86}{28.86-8.314} -1}{\frac{28.86}{28.86-8.314}})ln(\frac{0.802\ atm}{0.602\ atm})=-0.0826\\\\ln(\frac{T_2}{T_1} )=-0.0826\\\\Taking\ exponential\ of\ both \ sides:\\\\\frac{T_2}{T_1} =e^{-0.0826}\\\\T_2=0.9207T_1\\\\T_2=0.9207*288\\\\T_2=265\ K\\$

Since T2 = 265 K, we should expect a snow

4 0

3 years ago