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Inessa05 [86]
2 years ago
5

In a fusion reaction, the nuclei of two atoms join to form a single atom of a different element. In such a reaction, a fraction

of the rest energy of the original atoms is converted to kinetic energy of the reaction products. A fusion reaction that occurs in the Sun converts hydrogen to helium. Since electrons are not involved in the reaction, we focus on the nuclei. Hydrogen and deuterium (heavy hydrogen) can react to form helium plus a high-energy photon called a gamma ray:^1 H + ^2 H \longrightarrow ^3 He + \gammaObjects involved in the reaction:
Physics
1 answer:
aliya0001 [1]2 years ago
8 0

Answer:

looks like you answered your question

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Question #2
Contact [7]

Answer:

Distance 20 km and Displacement 0 km

His displaceent is 0 km because he ends his walk where he started. The total distance of his walk is 20 km because he walks 10 km to the store + 10km back home.

8 0
2 years ago
"In analyzing distances by apply ing the physics of gravitational forces, an astronomer has obtained the expression
zavuch27 [327]

Answer:

The value of R is 1.72\times10^{11}\ m.

(B) is correct option.

Explanation:

Given that,

In analyzing distances by apply ing the physics of gravitational forces, an astronomer has obtained the expression

R=\sqrt{\dfrac{1}{(\dfrac{1}{140\times10^{9}})^2-(\dfrac{1}{208\times10^{9}})^2}}

We need to calculate this for value of R

R=\sqrt{\dfrac{1}{(\dfrac{1}{140\times10^{9}})^2-(\dfrac{1}{208\times10^{9}})^2}}

R=1.89\times10^{11}\ m

So, The nearest option of the value of R is 1.72\times10^{11}\ m

Hence, The value of R is 1.72\times10^{11}\ m.

6 0
2 years ago
A pet-store supply truck moves at 25.0 m/s north along a highway. inside, a dog moves at 1.75 m/s at an angle of 35.0° east of
koban [17]

<u>Answer:</u>

 Velocity of the dog relative to the road = 26.04 m/s 3.15⁰ north of east.

<u>Explanation:</u>

  Let the east point towards positive X-axis and north point towards positive Y-axis.

  Speed of truck = 25 m/s north = 25 j m/s

  Speed of dog = 1.75 m/s at an angle of 35.0° east of north = (1.75 cos 35 i + 1.75 sin 35 j)m/s

                          = (1.43 i + 1.00 j) m/s

    Velocity of the dog relative to the road = 25 j + 1.43 i + 1.00 j = 1.43 i + 26.00 j

    Magnitude of velocity = 26.04 m/s

    Angle from positive horizontal axis = 86.85⁰

 So Velocity of the dog relative to the road = 26.04 m/s 86.85⁰ east of north = 26.04 m/s 3.15⁰ north of east.

4 0
2 years ago
A gas of helium atoms at 273 k is in a cubical container with 25.0 cm on a side. (a) what is the minimum uncertainty in momentum
qwelly [4]

wave function of a particle with mass m is given by ψ(x)={ Acosαx −

π

2α

≤x≤+

π

2α

0 otherwise , where α=1.00×1010/m.

(a) Find the normalization constant.

(b) Find the probability that the particle can be found on the interval 0≤x≤0.5×10−10m.

(c) Find the particle’s average position.

(d) Find its average momentum.

(e) Find its average kinetic energy −0.5×10−10m≤x≤+0.5×10−10m.

6 0
2 years ago
In a 5000 m race, the athletes run 12 1/2 laps; each lap is 400 m.Kara runs the race at a constant pace and finishes in 17.9 min
Ksju [112]

Answer:

No. of laps of Hannah are 7 (approx).

Solution:

According to the question:

The total distance to be covered, D = 5000 m

The distance for each lap, x = 400 m

Time taken by Kara, t_{K} = 17.9 min = 17.9\times 60 = 1074 s

Time taken by Hannah, t_{H} = 15.3 min = 15.3\times 60 = 918 s

Now, the speed of Kara and Hannah can be calculated respectively as:

v_{K} = \frac{D}{t_{K}} = \frac{5000}{1074} = 4.65 m/s

v_{H} = \frac{D}{t_{H}} = \frac{5000}{918} = 5.45 m/s

Time taken in each lap is given by:

(v_{H} - v_{K})t = x

(5.45 - 4.65)\times t = 400

t = \frac{400}{0.8}

t = 500 s

So, Distance covered by Hannah in 't' sec is given by:

d_{H} = v_{H}\times t

d_{H} = 5.45\times 500 = 2725 m

No. of laps taken by Hannah when she passes Kara:

n_{H} = \frac{d_{H}}{x}

n_{H} = \frac{2725}{400} = 6.8 ≈ 7 laps

3 0
1 year ago
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