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salantis [7]
3 years ago
10

True or False: The average atomic mass is always closer to the isotope with the largest mass

Physics
1 answer:
svlad2 [7]3 years ago
3 0
The awnser would be True. Hope I helped!
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A green light is submerged 2.70 m beneath the surface of a liquid with an index of refraction 1.31. What is the radius of the ci
Anettt [7]

Answer:

The radius is  r = 3.1905 \ m

Explanation:

From the question we are told that

        The  distance  beneath the liquid  is  d =  2.70 \ m

        The refractive index of the liquid is  n_i  =  1.31

Now the critical value is mathematically represented as

         \theta =  sin ^{-1} [\frac{1}{n_i} ]

substituting values

         \theta =  sin ^{-1} [\frac{1}{131} ]

         \theta  =  49.76^o

Using SOHCAHTOA rule we have that

         tan \theta =  \frac{ r}{d}

=>     r =  d * tan \theta

substituting values  

        r =  2.7 * tan (49.76)

        r = 3.1905 \ m

         

5 0
3 years ago
At take off, a plane flies 100 km north before turning to fly 200 km east. How far is its destination from where the plane took
Snowcat [4.5K]
The answer is 300 km

To the destination
4 0
3 years ago
Where should a force be applied on a lever arm to produce the most torque
nignag [31]

The answer is,

<u>Farthest from the axis of rotation</u>

Please rate <u>Brainliest</u> (:

3 0
3 years ago
The predictions of Rutherford's scattering formula failed to correspond with experimental data when the energy of the incoming a
Vera_Pavlovna [14]

Answer:

The radius of the gold nucleus is 7.1x10⁻¹⁵m

Explanation:

The nearest distance is:

r=\frac{kze^{2} }{mpV^{2} } (eq. 1)

Where

z = atomic number of gold = 79

e = electron charge = 1.6x10⁻¹⁹C

k = electrostatic constant = 9x10⁹Nm²C²

energy of the particle = 32 MeV = 5.12x10⁻¹²J

At the potential energy is zero, all the energy will be kinetic energy:

E_{k} =\frac{1}{2} m V^{2}

Where

m = 4 mp = mass of proton

5.12x10^{-12} =\frac{1}{2} *4*m_{p}* V^{2} \\m_{p}* V^{2} = 2.56x10^{-12}

Replacing in equation 1

r=\frac{9x10^{9}*79*(1.6x10^{-19})^{2}   }{2.56x10^{-12} } =7.1x10^{-15} m

8 0
2 years ago
The driver of a 1000 kg car traveling at a speed of 16.7 m/s applies the brakes. If the brakes provide a force of - 8000 N to st
RSB [31]

Answer:

Given:

m=1000kg

u= 16.7m/s

v=0m/s

F=8000N

Required:

s=?

Solution:

F=m × a

8000N=1000kg × a

a=8m/s^2

Since it decelerate a= -8m/s^2

v^2 = u^2 + 2as

s=v^2 - u^2 / 2a

s= 0 - (16.7m/s)^2 / 2 × -8m/s^2

s= -278.89/-16

s= 17.43m

The car travels approximately 17.43m before it stops

Please like and follow me

6 0
2 years ago
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