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1 year ago

What are the products of a fusion reaction? Check all that apply. Lighter atoms energy heavier atoms a neutron a proton.

1 answer:

6 0

The process by which two or more tiny nuclei unite to generate a bigger nucleus is known as a nuclear fusion reaction. Heavier atoms are products of a fusion reaction.

<h3 /><h3>What is nuclear fusion?</h3>

The process by which two or more tiny nuclei unite to generate a bigger nucleus is known as a nuclear fusion reaction.

For example, the fusion of two hydrogen atoms produces more energy than the fusion of one helium atom, and surplus energy is expelled into space upon binding.

Hence heavier atoms are e products of a fusion reaction.

To learn more about nuclear fusion refer to the link;

brainly.com/question/14019172

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PLEASE HELP ASAP!!

kirza4 [7]

Answer:

A. Distance over which the force is applied

Explanation:

As we know that in pulley system the mass of the car is balanced by the tension in the string

so here we will have

$T = r \times F$

so here in order to decrease the force needed to lift the car we have to increase Distance over which the force is applied

So here if we increase the distance over which force is applied then it will reduce the effort applied by us in this pulley system as the torque will be more if the distance is more.

7 0

3 years ago

If a high jumper needs to make his center of gravity rise 1.50 m, how fast must he be able to sprint? Assume all of his kinetic

sergiy2304 [10]

Answer:

v = 5.42 m/s

Explanation:

given,

height of the jumper = 1.5 m

velocity of sprinter = ?

kinetic energy can be transformed into potential energy

$m g h = \dfrac{1}{2}mv^2$

$g h = \dfrac{1}{2}v^2$

$v =\sqrt{2gh}$

$v =\sqrt{2\times 9.8 \times 1.5}$

v = 5.42 m/s

Speed of the sprinter is equal to v = 5.42 m/s

7 0

3 years ago

A student throws a 130 g snowball at 6.5 m/s at the side of the schoolhouse, where it hits and sticks. What is the magnitude of

Alex73 [517]

Answer:

4.7 N

Explanation:

130 g = 0.13 kg

The momentum of the snowball when it's thrown at the wall is

$p = mv = 0.13*6.5 = 0.845 kgm/s$

Which is also the impulse. From here we can calculate the magnitude of the average force F knowing the duration of the collision is 0.18 s

$p = F\Delta t$

$F*0.18 = 0.845$

$F = 0.845 / 0.18 = 4.7 N$

8 0

3 years ago

What force keeps the outside of a bicycle wheel from flying off?

4vir4ik [10]

Answer:

Centripetal force

Explanation:

8 0

2 years ago

I need an answer for this plzz!!<br>number 2 <br>anybody can help ??

Anuta_ua [19.1K]

2.1) (i) W = mg downwards
(ii) N = R = Normal Reaction from the ground upwards
(iii) Fe = Force of engine towards the right
(iv) f = friction towards the left
(v) ma = Constant acceleration towards right.
2.2.1)
v = 25 m/s
u = 0 m/s
∆v = v - u = (25 - 0) m/s = 25 m/s
x = X
∆t = 50 s
$a \: = \: \frac{dv}{dt} \: = \: \frac{25 \: \frac{m}{s} }{50 \: s} \: = \: 0.5 \: \frac{m}{ {s}^{2} }$
a = 0.5 m/s².
2.2.2)
F = ma = 900 kg × 0.5 m/s² = 450 N.
2.2.3)
$2ax \: = \: {v}^{2} \: - \: {u}^{2}$
$x \: = \: \frac{ {v}^{2} \: - \: {u}^{2} }{2a} \: = \: \frac{{(25 \: \frac{m}{s})}^{2} \: - \: {(0 \: \frac{m}{s} )}^{2} }{2 \: \times \: 0.5 \: \frac{m}{ {s}^{2} } } \: = \: 625 \: m$
2.3)
Fe = f + ma
Fe - f = ma
For velocity to be constant,
a should be 0, or, a = 0,
Fe = f = 270 N
2.4.1)
v = 0
u = 25 m/s
a = -0.5 m/s²
v = u + at
t = -u/a = -(25)/(-0.5) = 50 s.
2.4.2)
x = -625/(2×(-0.5)) = 625 m.

8 0

3 years ago