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3 years ago

Nuclear fusion differs from nuclear fission because nuclear fusion reactions

Physics

2 answers:

Ksivusya [100]3 years ago

6 0

Answer:

they are different because nuclear fission is one heavy and unstable nuclear breaks into two parts, or half, and produce very high energy. nuclear fusion two small fuse and produce very high amount of energy

Explanation:

kirill [66]3 years ago

4 0

Answer:

Nuclear fusion reactions produce more energy.( basic difference)

Explanation:

Nuclear fusion reactions are different from nuclear fusion reactions :- In nuclear fission one heavy and unstable nuclear breaks into two parts(smaller) and produce energy( very high energy than normal reaction). whereas in nuclear fusion two small and unstable nuclei fuse and produce very high amount of energy( even higher than nuclear fusion).

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In an experiment, a variable, position-dependent force F(x)F(x) is exerted on a block of mass 1.0kg1.0kg that is moving on a hor

leonid [27]

Answer:

The function F(x) for 0 < x < 5, the block's initial velocity, and the value of F(f).

Explanation:

Given that,

Mass of block = 1.0 kg

Dependent force = F(x)

Frictional force = F(f)

Suppose, the following information would students need to test the hypothesis,

(A) The function F(x) for 0 < x < 5 and the value of F(f).

(B) The function a(t) for the time interval of travel and the value of F(f).

(D) The function a(t) for the time interval of travel, the time it takes the block to move 5 m, and the value of F(f).

(E) The block's initial velocity, the time it takes the block to move 5 m, and the value of F(f).

We know that,

The work done by a force is given by,

$W=\int_{x_{0}}^{x_{f}}{F(x)\ dx}$ .....(I)

Where, $F(x)$ = net force

We know, the net force is the sum of forces.

So, $\sum{F}=ma$

According to question,

We have two forces F(x) and F(f)

So, the sum of these forces are

$F(x)+(-F(f))=ma$

Here, frictional force is negative because F(f) acts against the F(x)

Now put the value in equation (I)

$W=\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}$

We need to find the value of $\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}$

Using newton's second law

$\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\int_{x_{0}}^{x_{f}}{ma\ dx}$ ...(II)

We know that,

Acceleration is rate of change of velocity.

$a=\dfrac{dv}{dt}$

Put the value of a in equation (II)

$\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\int_{x_{0}}^{x_{f}}{m\dfrac{dv}{dt}dx}$

$\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\int_{v_{0}}^{v_{f}}{mv\ dv}$

$\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\dfrac{mv_{f}^2}{2}+\dfrac{mv_{0}^2}{2}$

Now, the work done by the net force on the block is,

$W=\dfrac{mv_{f}^2}{2}+\dfrac{mv_{0}^2}{2}$

The work done by the net force on the block is equal to the change in kinetic energy of the block.

Hence, The function F(x) for 0 < x < 5, the block's initial velocity, and the value of F(f).

(C) is correct option.

7 0

3 years ago

Convert <img src="https://tex.z-dn.net/?f=%5Cfrac%7B%280.779mg%29%28min%29%7D%7BL%7D" id="TexFormula1" title="\frac{(0.779mg)(mi

Orlov [11]

The number converted is $0.0467 \frac{(kg)(s)}{m^3}$

Explanation:

In order to convert from the original units to the final units, we have to keep in mind the following conversion factors:

$1 kg = 1000 g = 10^6 mg$

$1 min = 60 s$

$1 m^3 = 1000 L$

The original unit that we have is

$\frac{mg\cdot min}{L}$

Therefore, it can be rewritten as:

$=\frac{mg \frac{1}{10^6 mg/kg}\cdot min\cdot 60 s/min}{L\frac{1}{1000L/m^3}}=0.06 \frac{(kg)(s)}{m^3}$

Therefore, since the initial number was 0.779, the final value is

$0.779\cdot 0.06 \frac{(kg)(s)}{m^3}=0.0467 \frac{(kg)(s)}{m^3}$

#LearnwithBrainly

5 0

4 years ago

A body of mass 40kg is accelerating at a

jok3333 [9.3K]

Momentum = (mass) x (speed)

Mass is constant, so the rate of change of momentum is
(mass) x (rate of change of speed) .

But (rate of change of speed ) is just acceleration.

So the rate of change of momentum is (mass) x (acceleration).

But (mass) x (acceleration) is Force.

So Force is the rate of change of momentum. Verrrrrrrry interesting !

In this problem, Force = (40 kg) x (9 m/s²) = 360 newtons.

One 'Newton' is one kilogram-meter per second² .
Unit of momentum is (kilogram)-(meter per second), so 'newton'
is also a unit of time rate of change of momentum.

Rate of change of momentum is 360 momentum units per second.

8 0

4 years ago

Read 2 more answers

The acceleration of a baseball pitcher's hand as he delivers a pitch is extreme. For a professional player, this acceleration ph

SashulF [63]

Answer:

116 N.

Explanation:

Given,

mass of the ball, m = 0.145 Kg

initial speed = 0 m/s

Final speed = 40 m/s

time = 50 ms = 0.05 s

Force is equal to change in momentum per unit time

$F = \dfrac{m\Delta v}{\Delta t}$

$F = \dfrac{0.145(40-0)}{0.05}$

F = 116 N

Force of the pitcher hand on the ball is equal to 116 N.

4 0

3 years ago

the net force on a vehicle is accelerating at a rate of 1.5 milliseconds is 1800 Newtons. What is the mass of the vehicles neare

Goryan [66]

Answer:

The mass is 1200 kilograms

Explanation:

Because Force is equal to mass times acceleration (F=m×a)

F=m×a

1800N=?×1.5

1800÷1.5=1200

1800N=1200Kg×1.5

7 0

4 years ago