All categories

3 years ago

The billiard ball model was the working model of the atom until thomson discovered the

Physics

2 answers:

V125BC [204]3 years ago

7 0

The right answer is D, electron.

zysi [14]3 years ago

3 0

<span>d. electron

J J Thomson discovered the electron, and it was put in his model of the atom.</span>

You might be interested in

A vector always consits of

Kazeer [188]

Answer:

Size and Direction

Explanation:

8 0

3 years ago

The site<br> of proteins is called what

ValentinkaMS [17]

Answer: Ribosomes are the site of protein synthesis, and they convert the coded information in messenger RNA into an actual protein molecule.

I hope this helped! OvO

8 0

3 years ago

Read 2 more answers

A straight line means the object travels at ........... speed

Georgia [21]

If it's a distance graph, then it's a constant speed.

5 0

4 years ago

Read 2 more answers

Which of the following is an example of chemical change?

Maru [420]

Answer:

Explanation:

all the others r physical

8 0

3 years ago

Read 2 more answers

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