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3 years ago

Which energy transformation occurs in the core of a nuclear reactor?

1 answer:

5 0

The correct answer among the choices given is option B. The energy transformation that occurs in the core of a nuclear reactor is from nuclear energy to thermal energy. In a power plant nuclear fission which involves nuclear energy to heat up water around it. This part is the core of the process.

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Research on the Big Five has indicated that approximately _____ percent of traits appear to be inherited.

Evgen [1.6K]

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3 years ago

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A 10.0-kg mass is placed on a 25.0o incline and friction keeps it from sliding. The coefficient of static friction in this case

vovangra [49]

The frictional force while the mass is sliding will be 46.2 N.

<h3>What is friction force?</h3>

Opposition forces on the surface cause heat loss during the motion of an object known as the friction force.

Given data:

m(mass)= 10.0-kg

Θ (Inclination angle)=25.0o

Coefficient of sliding friction, $\rm \mu_k$ =0.520

Coefficient of static friction, $\rm \mu_s=0.520$

The friction force, F=?

Resolve the force in the inclined plane;

$\rm F=\mu_s mg cos25^0 \\\\ F=0.520 \times 10 \times 9.81 \times cos 25 ^0 \\\\ F= 46.2 \ N$

Hence, the frictional force while the mass is sliding will be 46.2 N.

To know more about friction force refer to the link;

brainly.com/question/1714663

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5 0

2 years ago

PLEASE HELP ME ASAP PLEASEEEEE

AleksAgata [21]

I believe its the law of inertia

6 0

4 years ago

Q) A farmer moves along the boundary of a

Alja [10]

The answer will be 8 because kedks

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3 years ago

An ideal spring hangs from the ceiling. A 1.95 kg mass is hung from the spring, stretching the spring a distance d=0.0865 m from

uranmaximum [27]

Answer:

kinetic energy = 0.1168 J

Explanation:

From Hooke's law, we know that ;

F = kx

k = F/x

We are given ;

Mass; m = 1.95 kg

Spring stretch; d = x = 0.0865

So, Force = mg = 1.95 × 9.81

k = 1.95 × 9.81/0.0865 = 221.15 N/m

Now, initial energy is;

E1 = mgL + ½k(x - L)²

Also, final energy; E2 = ½kx² + ½mv²

From conservation of energy, E1 = E2

Thus;

mgL + ½k(x - L)² = ½kx² + ½mv²

Making the kinetic energy ½mv² the subject, we have;

½mv² = mgL + ½k(x - L)² - ½kx²

We are given L=0.0325 m

Plugging other relevant values, we have ;

½mv² = (1.95 × 9.81 × 0.0325) + (½ × 221.15(0.0865 - 0.0325)² - ½(221.15 × 0.0865²)

½mv² = 0.62170875 + 0.3224367 - 0.82734979375

½mv² = 0.1168 J

7 0

3 years ago