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IrinaVladis [17]
3 years ago
10

Which image below best represents the process of nuclear fusion?

Physics
1 answer:
Hunter-Best [27]3 years ago
7 0
"Fusion" means separate objects hooking up to become ONE object.

The first image kind of hints at fusion.
The others don't suggest it at all.
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To view an interactive solution to a problem that is similar to this one, select Interactive Solution 7.24. A 0.0129-kg bullet i
ipn [44]

Answer:

t=0.42s

Explanation:

Here you have an inelastic collision. By the conservation of the momentum you have:

m_1v_1+m_2v_2=(m_1+m_2)v

m1: mass of the bullet

m2: wooden block mass

v1: velocity of the bullet

v2: velocity of the wooden block

v: velocity of bullet and wooden block after the collision.

By noticing that after the collision, both objects reach the same height from where the wooden block was dropped, you can assume that v is equal to the negative of v2. In other words:

m_1(-v_1)+m_2v_2=(m1+m2)(-v2)

Where you assumed that the negative direction is upward. By replacing and doing v2 the subject of the formula you get:

-(0.0129kg)(767m/s)+(1.17kg)v_2=(1.1829kg)(-v_2)\\\\v_2=4.20m/s

Now, with this information you can use the equation for the final speed of an accelerated motion and doing t the subject of the formula. IN other words:

v_2=v_o+gt\\\\t=\frac{v_2-v_o}{g}=\frac{4.2m/s-0m/s}{9.8m/s^2}=0.42s

hence, the time is t=0.42 s

4 0
3 years ago
If javier has a car thats 250,000 kg and a forse of 300 newtons.what would be the accleration.
const2013 [10]

Answer:

The acceleration would be 3.455.

3 0
3 years ago
"A steel rotating-beam test specimen has an ultimate strength of 120 kpsi. Estimate the life of the specimen if it is tested at
MrRissso [65]

Answer:

life (N) of the specimen is 117000  cycles

Explanation:

given data

ultimate strength Su = 120 kpsi

stress amplitude σa = 70 kpsi

solution

we first calculate the endurance limit of specimen Se i.e

Se = 0.5× Su   .............1

Se = 0.5 × 120

Se = 60 kpsi

and we know strength of friction f  = 0.82

and we take endurance limit Se is = 60 kpsi

so here coefficient value (a) will be

a = \frac{(f\times Su)^2}{Se}     ......................1  

put here value and we get

a = \frac{(0.82\times 120)^2}{60}  

a = 161.4  kpsi

so coefficient value (b) will be

b = -\frac{1}{3}log\frac{(f\times Su)}{Se}  

b =  -\frac{1}{3}log\frac{(0.82\times 120)}{60}  

b = −0.0716

so here number of cycle N will be  

N =  (\frac{ \sigma a}{a})^{1/b}

put here value  and we get

N =  (\frac{ 70}{161.4})^{1/-0.0716}

N = 117000

so life (N) of the specimen is 117000  cycles

7 0
3 years ago
Which statement(s) is/are indicative of a covalent bond?
Over [174]
Do you have any answer options?

The most I can help is to tell you that a covalent bond is a type of bond that shares electrons between the bonded atoms.

4 0
4 years ago
The curved section of a horizontal highway is a circular unbanked arc of radius 740 m. If the coefficient of static friction bet
kirill [66]
The answer would be 54 m/s as the maximum speed
8 0
4 years ago
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