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zvonat [6]
3 years ago
13

This image shows a stream of positively charged particles being directed at gold foil. The positively charged particles are call

ed "alpha particles” and each one is like a nucleus without any electrons.
What is the best explanation for why a particle is striking point X?

Alpha particles are mostly empty space, so they move in random directions, and one of these alpha particles can strike point X.
The gold atoms contain negative electrons, so when alpha particles strike the gold they keep moving but in various directions.
If an alpha particle hits the gold foil, a gold nucleus splits and a particle from it flies out at point X.
When the dense, positive alpha particle passes close to a positive nucleus of gold, the alpha particle repels and hits the screen at point X.

Physics
2 answers:
coldgirl [10]3 years ago
8 0

Answer:

D. When the dense, positive alpha particle passes close to a positive nucleus of gold, the alpha particle repels and hits the screen at point X.

Explanation:

nikklg [1K]3 years ago
3 0

Answer:

It is D, youre right. thanks for being awesome

Explanation:

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In a fit of rage about distance learning, you throw your computer out the window with an initial velocity of 3 m/s and it breaks
lesya692 [45]

Answer:

9 m

Explanation:

i did the test and got 100%

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PLEASE HELP! Compare and contrast between reflection and refraction. Be sure to be specific in your explanation and state in you
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4 years ago
A car accelerates from 300 km/h to 140 km/h in 2.53 seconds. what is the distance covered?
Snowcat [4.5K]

Answer:

Acceleration = Change in Velocity/Time

Change in Velocity = 36-18 = 18 km/h=5 m/s

Time= 5 Seconds

Acceleration = 5/5= 1 m/s2

Equation of motion,s=ut+(1/2)at2

u=18 km/h=5 m/s

t=5 s

a=1 m/s2

s= (5*5)+(1/2*1*5*5)

s=25+12.5 i.e., s=37.5 m

Hope you are clear with my explanations

7 0
3 years ago
What will be the value of both charges if they are 5 cm apart and suffer a
Mrac [35]

Answer:

\boxed{q = 1.2 \times  {10}^{ - 6} C}

Explanation:

f_e =   \frac{{q}^{2}k }{ {r}^{2} }  \\ q =  \sqrt{ \frac{f_e( {r}^{2} )}{k} }  =  \sqrt{ \frac{5.2(5 \times  {10}^{ - 2} )^{2} }{9 \times  {10}^{9} } }  \\ q =\sqrt{ \frac{5.2(5 \times  {10}^{ - 2} )^{2} }{9 \times  {10}^{9} } }  =  \sqrt{ \frac{0.013}{9 \times  {10}^{9} } }  \\ q = 1.2 \times  {10}^{ - 6}

3 0
3 years ago
A rock is dropped from the edge of a cliff into a pool of water. Assume free-fall acceleration is 10 m/s per second, and air res
il63 [147K]

Answer:

20 m

Explanation:

From the equation of motion,

S = ut+1/2gt²................................. Equation 1

Where S = Height, u = initial velocity, t = time, g = acceleration due to gravity.

Note: Because the rocked is being dropped from a height, acceleration due to gravity is positive (g), and initial velocity (u) is negative

Given: t = 2.0 s, g = 10 m/s², u = 0 m/s (dropped from height)

Substituting into equation 1

S = 0(2) + 1/2(10)(2)²

S = 5(4)

S = 20 m

Hence the height of the the cliff above the pool is 20 m

7 0
3 years ago
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