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Novay_Z [31]
2 years ago
9

In his famous experiment, Rutherford fired alpha particles at a thin gold film. Most of the alpha particles went through the fir

m and a very few bounced back. Suppose instead that about one-half the alpha particles bounced back and one-half went through. How would this have changed his conclusion about the structure of the atom?
1. The nucleus is about 1/2 the size of the atom.
2. The atom has no nucleus.
3. None of these
4. The nucleus covers the entire atom.
5. The atom is mostly empty space.
Physics
1 answer:
ale4655 [162]2 years ago
5 0

Answer:

1. The nucleus is about 1/2 the size of the atom

Explanation:

Alpha particles are positive charge particles and they are bounced back by the nucleus because nucleus is also same size

Now in present experiment Rutherford found that very few alpha particles are bounced back along same path which shows that very small region inside the nucleus is having positive charge and rest part of the atom is empty.

Now if we found that half of the alpha particles are bounced back then it shows that size of the nucleus is very large now as compare to previous one because only nucleus can bounce back the alpha particles

so correct answer will be

1. The nucleus is about 1/2 the size of the atom

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The least count of stopwatch is 0.2s.The time of 20 oscillations of a pendulum was measured to be 25s.Find the percentage error
Svet_ta [14]

Answer:

0.8%

Explanation:

We are given;

Number of oscillations; n = 20

Time taken; t = 25 s

Formula for period of oscillation;

T = t/n = 25/20 = 1.25 s

We are told that the least count is 0.2 s. Thus, error is; ΔT = 0.2 s

percentage error in the measurement of time is given by;

(0.2/(20 × 1.25)) × 100% = 0.8%

4 0
2 years ago
1. The choice of materials for an exciting playground slide should __________.
gavmur [86]
  1. have a large coefficient of kinetic friction between cloth and the slide material
  2. 40(0.3+0.5)=40(0.8)=32N
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  4. When the tire is locked, the frictional force is greater but the car cannot be steered
4 0
2 years ago
Read 2 more answers
How long must a flute be in order to have a fundamental frequency of 262 Hz (this frequency corresponds to middle C on the evenl
spin [16.1K]

Answer:

L=0.654 m

Explanation:

<u>Concepts and Principles  </u>

1- The speed of sound in air is expressed as a function of the temperature of air as follows:  

 v=(331 m/s)√(1+T_C/273°C)                        (1)

where 331 m/s is the speed of sound in air at temperature 0°C and Tc is the temperature of air in Celsius.  

<u>Standing Wave Patterns in Pipes:  </u>

A pipe open at both ends can have standing wave patterns with resonant frequencies:  

f=v/λ=nv/2L                     n=1,2,3.........

where v is the speed of sound in air.  

<u>Given Data </u>

f_1 (fundamental frequency of the flute) = 262 Hz

T (temperature of the air) = 20°C  

The flute is open at both ends.  

<u>Required Data </u>

We are asked to determine the length of the tube.  

<u>Solution</u><u>  </u>

The speed of sound in air at temperature T = 20°C is found from Equation (1):

 v=(331 m/s)√(1+T_C/273°C)  

 =342.91 m/s

The fundamental frequency of the flute is found by substituting n = 1 into Equation (2):  

f=v/2L

Solve for L:  

L=v/2f_1

L=0.654 m

7 0
2 years ago
(3.16_Q2) Which weights would you use on a single thread to create a 6.86 N force? Question 2 options: Weight IDs A, B, C, D Wei
Tomtit [17]

1. E,F

2. D,E,F

3. B,C,E,G

4. A,B,C,D

I did the test! :)

5 0
2 years ago
2. An athlete of average size is hanging from the end of a 20 m long rope, which has a mass of 4 kg and is attached to a hook in
a_sh-v [17]

Answer:

  t = 0.319 s

Explanation:

With the sudden movement of the athlete a pulse is formed that takes time to move along the rope, the speed of the rope is given by

             v = √T/λ

Linear density is

           λ = m / L

           λ = 4/20

           λ = 0.2 kg / m

The tension in the rope is equal to the athlete's weight, suppose it has a mass of m = 80 kg

           T = W = mg

           T = 80 9.8

           T = 784 N

The pulse rate is

          v = √(784 / 0.2)

          v = 62.6 m / s

The time it takes to reach the hook can be searched with kinematics

          v = x / t

          t = x / v

          t = 20 / 62.6

          t = 0.319 s

7 0
3 years ago
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