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zhuklara [117]
3 years ago
5

Does uranium use steam?

Physics
1 answer:
Neko [114]3 years ago
4 0

Answer:

Yes.

Explanation:

Reactors use uranium for nuclear fuel. The uranium is processed into small ceramic pellets and stacked together onto sealed metal tubes called fuel rods. The heat created by fission turns the water into steam.

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El dormitorio de Pablo es rectangular, y sus lados miden 3 y 4 metros. Ha decidido dividirlo en dos partes triangulares con una
Paraphin [41]

Answer:

 c = 5 m

Explanation:

this exercise you want to divide the rectangular room into two triangular rooms

                 

the area of ​​triangles is

           A = ½ base height

           A = ½ 4 3

          A = 6 m²

the length of the curtain can be found using the Pythagorean theorem

           c² = b² + a²

           c = √ (4² + 3²)

           c = 5 m

this is the length of the curtain

5 0
3 years ago
What is an easy way to balance chemical equations?
Lorico [155]

write out what you have on both sides, then just use basic multiplication to try and even out both sides. I can help if you need me to balance some for you!!

8 0
3 years ago
If the pressure in a gas is doubled while its volume is held constant, by what factor do vrms change
Nat2105 [25]

Answer is given below

Explanation:

given data

pressure = double

volume = constant

solution

As we know that an Average velocity and rms velocity is directly proportional to square root of PV ..................1

so if we take P is doubled while keeping V constant

than Velocity increases by a factor \sqrt{2}  

so that Factor = 1.414 for both the cases

8 0
3 years ago
-. A 2kg cart moving to the right at 5m/s collides with an 8kg cart at rest. As a
bulgar [2K]

Answer:

<em>The velocity of the carts after the event is 1 m/s</em>

Explanation:

<u>Law Of Conservation Of Linear Momentum </u>

The total momentum of a system of bodies is conserved unless an external force is applied to it. The formula for the momentum of a body with mass m and speed v is  

P=mv.  

If we have a system of bodies, then the total momentum is the sum of the individual momentums:

P=m_1v_1+m_2v_2+...+m_nv_n

If a collision occurs and the velocities change to v', the final momentum is:

P'=m_1v'_1+m_2v'_2+...+m_nv'_n

Since the total momentum is conserved, then:

P = P'

In a system of two masses, the equation simplifies to:

m_1v_1+m_2v_2=m_1v'_1+m_2v'_2

If both masses stick together after the collision at a common speed v', then:

m_1v_1+m_2v_2=(m_1+m_2)v'

The common velocity after this situation is:

\displaystyle v'=\frac{m_1v_1+m_2v_2}{m_1+m_2}

The m1=2 kg cart is moving to the right at v1=5 m/s. It collides with an m2= 8 kg cart at rest (v2=0). Knowing they stick together after the collision, the common speed is:

\displaystyle v'=\frac{2*5+8*0}{2+8}=\frac{10}{10}=1

The velocity of the carts after the event is 1 m/s

3 0
2 years ago
A point charge (–5.0 µC) is placed on the x axis at x = 4.0 cm, and a second charge (+5.0 µC) is placed on the x axis at x = –4.
AveGali [126]

Answer:

The magnitude of electric force is  7.2\times10^{-3} N

Explanation:

Coulomb's Law:

The force of attraction or repletion is

  • directly proportional to the products of charges i.e F\propto q_1q_2
  • inversely proportional to the square of distance i.e F\propto \frac{1}{r^2}

\therefore F\propto \frac{q_1q_2}{r^2}

\Rightarrow F=k \frac{q_1q_2}{r^2}    [ k is proportional constant=9×10⁹N m²/C²]

There are two types of force applied on Q=+2.5 μC=2.5×10⁻⁶ C

Let F₁ force be applied on Q =+2.5 μC by q₁= -5.0 μC = - 5.0×10⁻⁶ C

and F₂ force  be applied on Q=+2.5 μC by q₂= 5.0 μC= 5.0×10⁻⁶ C

Since the magnitude of F₁ and F₂ are same. Therefore their y component cancel.

If we draw a line from q₁ to Q .

The it forms a triangle whose base = 4.0 cm and altitude =3.0 cm.

Let hypotenuse = r

Therefore, r=\sqrt{altitude^2+base^2} =\sqrt{3^2+4^2} =5

we know,

cos \theta = \frac{base }{hypotenuse}

\Rightarrow cos \theta = \frac{4 }{r}

Total force F_Q = 2.F_1 cos\theta \hat{i}

                         =2k\frac{Qq_1}{r^2} cos\theta \hat i

                         =2\ \frac{9\times1 0^9\times2.5 \times 5\times 10^{-12}}{r^2} \frac{4}{r} \hat i

                         =8\ \frac{9\times10^9\times2.5 \times 5\times 10^{-12}}{5^3} \hat i     [ r=5]

                         =7.2\times10^{-3}\hat i   N

The magnitude of electric force is  7.2\times10^{-3} N

                         

3 0
3 years ago
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