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4 years ago

Why did some alpha particles bounce off the gold foil?

Physics

1 answer:

Viktor [21]4 years ago

3 0

Because alpha particles are positive charged. And they bounce back because of the electromagnetic force between them and the gold foils protons.

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What is the Net Force?

marshall27 [118]

It is 800 N FN = 600N + 200 N = 800 N Answer to your question: The net force is all Newton's second law. It is the force that acts on a body or a particle. for example: It is the force we make when we push a car or something heavy that is in a straight line. .

3 0

4 years ago

You are standing on a sheet of ice that covers the football stadium parking lot in Buffalo; there is negligible friction between

Anni [7]

Answer:

0.074m/s

Explanation:

We need the formula for conservation of momentum in a collision, this equation is given by,

$m_1u_1+m_2u_2 = m_1v_1+m_2v_2$

Where,

$m_1$ = mass of ball

$m_2$ = mass of the person

$u_1$ = Velocity of ball before collision

$u_2$ = Velocity of the person before collision

$v_1$ = velocity of ball afer collision

$v_2$ = velocity of the person after collision

We know that after the collision, as the person as the ball have both the same velocity, then,

$v_1 = v_2$

$m_1u_1 + m_2u_2 = (m_1+m_2)v_2$

Re-arrenge to find $v_2$ ,

$v_2 = \frac{m_1u_1+m_2u_2}{m_1+m_2}$

Our values are,

$m_1$ = 0.425kg

$u_1$ = 12m/s

$m_2$ = 68.5kg

$u_2$ = 0m/s

Substituting,

$v_2 = \frac{(0.425)(12)+(68.5)(0)}{0.425+68.5}$

$v_2 = 0.074m/s$

<em />

<em>The speed of the person would be 0.074m/s after the collision between him/her and the ball</em>

7 0

3 years ago

A football is kicked from the ground with a velocity of 38m/s at an angle of 40 degrees and eventually lands at the same height.

Anastasy [175]

Initially, the velocity vector is $\langle 38cos(40^{\circ}),38sin(40^{\circ}) \rangle=\langle 29.110, 24.426 \rangle$ . At the same height, the x-value of the vector will be the same, and the y-value will be opposite (assuming no air resistance). Assuming perfect reflection off the ground, the velocity vector is the same. After 0.2 seconds at 9.8 seconds, the y-value has decreased by $4.9(0.2)^2$ , so the velocity is $\langle 29.110, 24.426-0.196 \rangle = \langle 29.110, 24.23 \rangle$ .

Converting back to direction and magnitude, we get $\langle r,\theta \rangle=\langle \sqrt{29.11^2+24.23^2},tan^{-1}(\frac{29.11}{24.23}) \rangle = \langle 37.87,50.2^{\circ}\rangle$

4 0

3 years ago

If you're driving towards the sun late in the afternoon, you can reduce the glare from the road by wearing sunglasses that permi

AveGali [126]

Correct answer choice is:

C. Polarized in a vertical plane.

Explanation:

Polarized sunglasses give excellent glare shield, particularly on the water. Polarized lenses include a specific filter that prevents this type of strong reflected light, diminishing glare.

This is because when you angle one polarized lens to different perpendicularly, they prevent glare both horizontally and vertically. The polarized lenses are enduringly tinted sunglasses that exceedingly decrease glare.

7 0

4 years ago

A particle with charge − 2.74 × 10 − 6 C −2.74×10−6 C is released at rest in a region of constant, uniform electric field. Assum

s2008m [1.1K]

Answer:

241.7 s

Explanation:

We are given that

Charge of particle= $q=-2.74\times 10^{-6} C$