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3 years ago

A β− and a β+ particle meet and interact within a tissue. Discuss what would be the likely outcome of this interaction.

Physics

1 answer:

Arturiano [62]3 years ago

3 0

When a β− and a β+ particle meet and interact they are annihilated.

Every matter has an antimatter. An antimatter of a particular particle is another particle that has exactly the same properties as the initial particle but carries an opposite charge.

If we look at β− and β+ particles, we will notice that they are matter and antimatter. The interaction of matter and antimatter leads to annihilation. The energy produced by annihilation of matter and antimatter is enough to destroy the tissue.

So, when β− and a β+ particle meet and interact they are annihilated.

Learn more: brainly.com/question/23266051

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A helicopter’s velocity increases from 35 m/s to 70 m/s in 10 seconds. What is the acceleration of this helicopter?

Alenkinab [10]

3.5 meters per second second is the acceleration because we know that acceleration is change in velocity over time and the change is velocity here is 35 and the time is 10 so we can simply divide 35 by 10 which is 3.5 m/s squared

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3 years ago

an airplane releases a ball as it flies parallel to the ground at a height of 235m. if the ball lands on the ground exactly at 2

Oksanka [162]

When the question says the ball lands a distance of 235 meters from the release point, we can assume this means the horizontal distance is 235 meters. Let's calculate the time for the ball to fall 235 meters to the ground. y = (1/2)gt^2 t^2 = 2y / g t = sqrt{ 2y / g } t = sqrt{ (2) (235 m) / (9.81 m/s^2) } t = 6.9217 s We can use the time t to find the horizontal speed. v = d / t v = 235 m / 6.9217 s v = 33.95 m/s Since the horizontal speed is the speed of the plane, the speed of the plane is 33.95 m/s

7 0

3 years ago

A hot-air balloon of diameter 12 mm rises vertically at a constant speed of 14 m/s. A passenger accidentally drops his camera fr

fgiga [73]

Answer:

The balloon is 66.62 m high

Explanation:

Combined Motion

The problem has a combination of constant-speed motion and vertical launch. The hot-air balloon is rising at a constant speed of 14 m/s. When the camera is dropped, it initially has the same speed as the balloon (vo=14 m/s). The camera has an upward movement for some time until it runs out of speed. Then, it falls to the ground. The height of an object that was launched from an initial height yo and speed vo is

$\displaystyle y=y_o+v_o\ t-\frac{g\ t^2}{2}$

The values are

$\displaystyle y_o=15\ m$

$\displaystyle v_o=14\ m/s$

We must find the values of t such that the height of the camera is 0 (when it hits the ground)

$\displaystyle y=0$

$\displaystyle y_o+v_o\ t-\frac{g\ t^2}{2}=0$

Multiplying by 2

$\displaystyle 2y_o+2v_ot-gt^2=0$

Clearing the coefficient of $t^2$

$\displaystyle t^2-\frac{2\ V_o}{g}\ t-\frac{2\ y_o}{g}=0$

Plugging in the given values, we reach to a second-degree equation

$\displaystyle t^2-2.857t-3.061=0$

The equation has two roots, but we only keep the positive root

$\displaystyle \boxed {t=3.69\ s}$

Once we know the time of flight of the camera, we use it to know the height of the balloon. The balloon has a constant speed vr and it already was 15 m high, thus the new height is

$\displaystyle Y_r=15+V_r.t$

$\displaystyle Y_r=15+14\times3.69$

$\displaystyle \boxed{Y_r=66.62\ m}$

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3 years ago

The products of one turn of dark reaction cycle are called

harina [27]

Light Independent Reactions

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3 years ago

An automobile travels on a straight road for 40 km at 30 km/h. It then continues in the same direction for another 40 km at 60 k

Butoxors [25]

Answer:

The average velocity is 40km/h.

Explanation:

The average velocity is $\bar{v}=\frac{\Delta x }{\Delta t}$ , where $\Delta x$ is the distance traveled and $\Delta t$ the time elapsed.

The distance traveled is clearly 80km since it's all done in the same direction, we only need to know the time elapsed. For this we calculate the time elapsed on the first part, and add it to the time elapsed on the second part using always the formula $\Delta t=\frac{\Delta x }{v}$ , where v is the velocity on each part, which is constant.

The time elapsed for the first part is $\Delta t_1=\frac{40 km}{30km/h}=\frac{4}{3}h$ , and the time elapsed for the second part is $\Delta t_2=\frac{40 km}{60km/h}=\frac{2}{3}h$ , giving us a total time of $\Delta t_1+\Delta t_2=\frac{4}{3}h+\frac{2}{3}h$ =2h.

Finally, we can calculate the average velocity: $\bar{v}=\frac{80km}{2h}=40km/h$ .

6 0

3 years ago