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kirza4 [7]
3 years ago
15

How did rutherford's experiment show that thomson's plum pudding model was incorrect?

Physics
1 answer:
Aleksandr [31]3 years ago
5 0
He said that the atom is mostly empty space with a positive nucleus in the center with electrons around it.
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How does the force of gravity affects an objects acceleration
MrRissso [65]

Answer:

When objects fall to the ground, gravity causes them to accelerate.

Explanation:

Gravity causes an object to fall toward the ground at a faster and faster velocity the longer the object falls.

3 0
3 years ago
Statement I: At the same temperature lighter gas molecules have a higher average velocity than heavier gas molecules.
PilotLPTM [1.2K]

Answer:

Statement 1 and statement 2 are correct and statement 2 is the correct explanation of statement 1

Explanation:

Both the velocity and kinetic energy of a gas molecule depends on its relative molecular mass according to Graham's law of diffusion in gases. Hence, the greater the relative molecular mass of the gas, the lesser its average velocity and kinetic energy.

Hence we can see that statement 2 vividly explains the postulation of statement 1 and makes the points more easily comprehensible.

6 0
4 years ago
A narrow beam of light containing red (660 nm) and blue (470 nm) wavelengths travels from air through a 2.60 cm thick flat piece
Brut [27]

Answer:

A: 28°

B. 1x10^-3M

Explanation:

See attached file

5 0
4 years ago
Types of mechanical waves include
USPshnik [31]
I think surface waves
4 0
4 years ago
A 62 kg skier is moving at 6.5 m/s on frictionless horizontal snow-covered plateau when she encounters a rough patch 3.50 m long
Degger [83]

Answer:

(A). The work done by friction in crossing the patch is -637.98 J.

(B). The speed of skier is 10.57 m/s.

Explanation:

Given that,

Mass of skier = 62 kg

Speed = 6.5 m/s

Length = 3.50 m

Coefficient kinetic friction = 0.30

Height = 2.5 m

(A) we need to calculate the work done by friction in crossing the patch

Using formula of work done

W=-\mu mg\times l

Put the value into the formula

W=-0.30\times62\times9.8\times3.50

W=-637.98\ J

The work done by friction in crossing the patch is -637.98 J.

(B) we need to calculate the speed of skier

Using conservation of energy

K.E_{i}+U_{i}-W_{friction}=K.E_{f}+U_{f}

\dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2+U_{f}

Final potential energy is zero

So, \dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2

\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}v_{1}^2+gh-\mu gl

Put the value into the formula

\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}\times6.5^2+9.8\times2.5+0.30\times9.8\times3.50

v_{2}=\sqrt{2\times55.915}

v_{2}=10.57\ m/s

The speed of skier is 10.57 m/s.

Hence,  (A).The work done by friction in crossing the patch is -637.98 J.

(B).The speed of skier is 10.57 m/s.

6 0
3 years ago
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