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2 years ago

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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: $^1 H + ^2 H \longrightarrow ^3 He + \gamma$ Objects involved in the reaction:

1 answer:

aliya0001 [1]2 years ago

8 0

Answer:

looks like you answered your question

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Consider a spring that does not obey Hooke’s law very faithfully. One end of the spring is fixed. To keep the spring stretched o

IRINA_888 [86]

Answer:

a) $W=-0.0103125\ J$

b) $W=0.0059375\ J$

c) Compressing is easier

Explanation:

Given:

Expression of force:

$F=kx-bx^2+cx^3$

where:

$k=100\ N.m^{-1}$

$b=700\ N.m^{-2}$

$c=12000\ N.m^{-3}$

$x$ when the spring is stretched

$x$ when the spring is compressed

hence,

$F=100x-700x^2+12000x^3$

a)

From the work energy equivalence the work done is equal to the spring potential energy:

here the spring is stretched so, $x=-0.05\ m$

Now,

The spring constant at this instant:

$j=\frac{F}{x}$

$j=\frac{100\times (-0.05)-700\times (-0.05)^2+12000\times (-0.05)^3}{-0.05}$

$j=-8.25\ N.m^{-1}$

Now work done:

$W=\frac{1}{2} j.x^2$

$W=0.5\times -8.25\times (-0.05)^2$

$W=-0.0103125\ J$

b)

When compressing the spring by 0.05 m

we have, $x=0.05\ m$

<u>The spring constant at this instant:</u>

$j=\frac{F}{x}$

$j=\frac{100\times (0.05)-700\times (0.05)^2+12000\times (0.05)^3}{0.05}$

$j=4.75\ N.m^{-1}$

Now work done:

$W=\frac{1}{2} j.x^2$

$W=0.5\times 4.75\times (0.05)^2$

$W=0.0059375\ J$

c)

Since the work done in case of stretching the spring is greater in magnitude than the work done in compressing the spring through the same deflection. So, the compression of the spring is easier than its stretching.

8 0

2 years ago

A stone is held at a height h above the ground. A second stone with four times the mass of the first one is held at the same hei

QveST [7]

gravitational potential energy is given by formula

$U = mgh$

here we need to compare the gravitational potential energy of stone 2 with respect to stone 1

so we will say

$\frac{U_2}{U_1} = \frac{m_2gh}{m_1gh}$

$\frac{U_2}{U_1} =\frac{m_2}{m_1}$

given that

$m_2 = 4 m_1$

now we have

$\frac{U_2}{U_1} = 4$

5 0

2 years ago

When the mass of the bottle is 0.125 kg, the KE is______ kg m2/s2.

tensa zangetsu [6.8K]

Answers are:
(1) KE = 1 kg m^2/s^2
(2) KE = 2 kg m^2/s^2
(3) KE = 3 kg m^2/s^2
(4) KE = 4 kg m^2/s^2

Explanation:

(1) Given mass = 0.125 kg
speed = 4 m/s

Since Kinetic energy = (1/2)*m*(v^2)

Plug in the values:
Hence:
KE = (1/2) * 0.125 * (16)
KE = 1 kg m^2/s^2

(2) Given mass = 0.250 kg
speed = 4 m/s

Since Kinetic energy = (1/2)*m*(v^2)

Plug in the values:
Hence:
KE = (1/2) * 0.250 * (16)
KE = 2 kg m^2/s^2

(3) Given mass = 0.375 kg
speed = 4 m/s

Since Kinetic energy = (1/2)*m*(v^2)

Plug in the values:
Hence:
KE = (1/2) * 0.375 * (16)
KE = 3 kg m^2/s^2

(4) Given mass = 0.500 kg
speed = 4 m/s

Since Kinetic energy = (1/2)*m*(v^2)

Plug in the values:
Hence:
KE = (1/2) * 0.5 * (16)
KE = 4 kg m^2/s^2

5 0

2 years ago

A sprinter accelerates from rest to a velocity of 12m/s in the first 6 seconds of the 100 meter dash .

GREYUIT [131]

Answer:

a) 36 m

b) 64 m

Explanation:

Given:

v₀ = 0 m/2

v = 12 m/s

t = 6 s

Find: Δx

Δx = ½ (v + v₀) t

Δx = ½ (12 m/s + 0 m/s) (6 s)

Δx = 36 m

The track is 100 m, so the sprinter still has to run another 64 m.

5 0

2 years ago

Sully is riding a snowmobile on a flat, snow-covered surface with a constant velocity of 10 meters/second. The total mass of the

Gelneren [198K]

Answer: 168N

16 - 10 = 6
6 / 10 = .6
F = 280 x .6 = 168

7 0

2 years ago

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