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Novay_Z [31]
3 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]3 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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La mascota de felipe esta jugando en el parque despues de un tiempo esta agotado de tanto correr. Felipe conoce que el perro con
MatroZZZ [7]

Responder:

6.704 m / s

Explicación:

Se dice que el trabajo se realiza cuando la fuerza aplicada a un objeto hace que el objeto se mueva. Primero necesitamos calcular la distancia recorrida por el perro usando la fórmula del trabajo realizado.

Trabajo realizado = Fuerza × distancia

Distancia = Trabajo realizado / Fuerza

Distancia = W / mg

S = 176/8 × 9,81

S = 176 / 78,48

S = 2,24 m

Dada la velocidad inicial u = 3.6km / h

Convertir a m / s

= 3.6km × 1000m / 1h × 3600

= 3600/3600

= 1 m / s

u = 1 m / s

Usando la ecuación de movimiento

v² = u² + 2gS para obtener la velocidad final v:

v² = 1² + 2 (9,81) (2,24)

v² = 1 + 43,9488

v² = 44,9488

v = √44,9488

v = 6,704 m / s

Por tanto, la rapidez final del perro es de 6,704 m / s

6 0
3 years ago
A table-tennis ball is thrown at a stationary bowling ball. The table-tennis ball makes a one-dimensional elastic collision and
Alecsey [184]

<u>Option b. </u>A smaller magnitude of momentum and more kinetic energy.

<h3>What is a momentum?</h3>
  • In Newtonian physics, an object's linear momentum, translational momentum, or simply momentum is defined as the product of its mass and velocity.
  • It has both a magnitude and a direction, making it a vector quantity. The object's momentum, p, is defined as: p=mv if m is the object's mass and v is its velocity (also a vector quantity).
  • The kilogram metre per second (kg m/s), or newton-second in the International System of Units (SI), is the unit used to measure momentum.
  • The rate of change of a body's momentum is equal to the net force exerted on it, according to Newton's second law of motion.

To know more about momentum, refer:

brainly.com/question/1042017

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7 0
2 years ago
A real object is 10.0 cm to the left of a thin, diverging lens having a focal length of magnitude 16.0 cm. What is the location
amm1812

Answer:

A)6.15 cm to the left of the lens

Explanation:

We can solve the problem by using the lens equation:

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}

where

q is the distance of the image from the lens

f is the focal length

p is the distance of the object from the lens

In this problem, we have

f=-16.0 cm (the focal length is negative for a diverging lens)

p=10.0 cm is the distance of the object from the lens

Solvign the equation for q, we find

\frac{1}{q}=\frac{1}{-16.0 cm}-\frac{1}{10.0 cm}=-0.163 cm^{-1}

q=\frac{1}{-0.163 cm^{-1}}=-6.15 cm

And the sign (negative) means the image is on the left of the lens, because it is a virtual image, so the correct answer is

A)6.15 cm to the left of the lens

6 0
3 years ago
The velocity of a 1.3 kg remote-controlled car is plotted on the graph. The work of segment A is J.
tamaranim1 [39]

Answer: 585 J

Explanation:

We can calculate the work done during segment A by using the work-energy theorem, which states that the work done is equal to the gain in kinetic energy of the object:

W=K_f -K_i

where Kf is the final kinetic energy and Ki the initial kinetic energy. The initial kinetic energy is zero (because the initial velocity is 0), while the final kinetic energy is

K_f =\frac{1}{2}mv^2

The mass is m=1.3 kg, while the final velocity is v=30 m/s, so the work done is:

W=K_f = \frac{1}{2}(1.3 kg)(30 m/s)^2=585 J

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