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Ksju [112]
3 years ago
5

1.

Physics
1 answer:
Gwar [14]3 years ago
4 0

<u>Answer:</u>

For 1: The correct option is Option C.

For 3: The final velocity of the opponent is 1m/s

<u>Explanation: </u>

During collision, the energy and momentum remains conserved. The equation for the conservation of momentum follows:

m_1u_1+m_2u_2=m_1v_1+m_2v_2      ...(1)

where,

m_1,u_1\text{ and }v_1 are the mass, initial velocity and final velocity of first object

m_2,u_2\text{ and }v_2 are the mass, initial velocity and final velocity of second object

<u>For 1:</u>

We are Given:

m_1=150g=0.15kg\\u_1=?m/s\\v_1=0.85m/s\\m_2=3500g=3.5kg\\u_2=0m/s\\v_2=0.85m/s

Putting values in equation 1, we get:

(0.15\times u_1)+(3.5\times 0)=(3.5+0.15)\times 0.85\\\\u_1=20.683\approx 21m/s

Hence, the correct answer is Option C.

  • <u>For 2: </u>

Impulse is defined as the product of force applied on an object and time taken by the object.

Mathematically,

J=F\times t

where,

F = force applied on the object

t = time taken

J = impulse on that object

Impulse depends only on the force and time taken by the object and not dependent on the surface which is stopping the object.

Hence, the impulse remains the same.

  • <u>For 3:</u>

Let the speed in right direction be positive and left direction be negative.

We are Given:

m_1=240kg\\u_1=0m/s\\v_1=-1m/s\\m_2=80kg\\u_2=-2m/s\\v_2=?m/s

Putting values in equation 1, we get:

(240\times 0)+(80\times (-2))=(240\times (-1))+(80\times v_2)\\\\v_2=1m/s

Hence, the final velocity of the opponent is 1m/s and has moved backwards to its direction of the initial velocity.

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Two disks are rotating about the same axis. Disk A has a moment of inertia of 3.3 kg · m2 and an angular velocity of +6.6 rad/s.
Licemer1 [7]

The angular momentum of a rotation object is the product of its moment of inertia and its angular velocity:

L = Iω

L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.

Apply the conservation of angular momentum. The total angular momentum before disks A and B are joined is:

L_{before} = (3.3)(6.6) + B(-9.3)

L_{before} = -9.3B+21.78

where B is the moment of inertia of disk B.

The total angular momentum after the disks are joined is:

L_{after} = (3.3+B)(-2.1)

L_{after} = -2.1B-6.93

L_{before} = L_{after}

-9.3B + 21.78 = -2.1B - 6.93

B = 4.0kg·m²

The moment of inertia of disk B is 4.0kg·m²

4 0
3 years ago
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bulgar [2K]
Hi,

I've found a link that should assist you or answer your question.
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4 years ago
tom does not reallt want to give away blue marbles and would like to change the probability that he chooses a blue marble to one
Llana [10]
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3 years ago
A grinding wheel 0.35 m in diameter rotates at 2600 rpm .
ddd [48]

It is calculated that a)The angular velocity of the wheel is 272.13 rad/s,

b)On the edge of the grinding wheel, the linear speed is 47.62 m/s,

and c) On the edge of the grinding wheel, the acceleration is 12958.08 m/s².

Calculation of angular velocity, linear speed & acceleration:

Provided that,

the diameter of the wheel = 0.35 m

So, the radius, r = 0.35/2 = 0.175 m

As 1 revolution = 2π rad

(a)the angular velocity, ω = 2600 rpm = \frac{2600 * 2\pi }{60} rad/s

⇒ω = 272.13 rad/s

So, the angular velocity is 272.13 rad/s.

(b)The linear speed, v = r * ω

⇒v = 0.175 * 272.13

⇒v= 47.62 m/s

(c)The angular acceleration, a=\frac{v^{2} }{r}

a = \frac{(47.62)^{2} }{0.175}

⇒a = 12958.08 m/s²

Learn more about angular velocity here:

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6 0
2 years ago
A high-temperature, gas-cooled nuclear reactor consists of a composite, cylindrical wall for which a thorium fuel element (kth =
WARRIOR [948]

Answer:

a) T_1 = 938 K , T_2 = 931 K

b) To prevent softening of the materials, which would occur below their  melting points, the reactor should not be operated much above:

                                      q = 3*10^8 W/m^3

Explanation:

Given:

- See the attachment for the figure for this question.

- Melting point of Thorium T_th = 2000 K

- Melting point of Thorium T_g = 2300 K

Find:

a) If the thermal energy is uniformly generated in the fuel element at a rate q = 10^8 W/m^3 then what are the temperatures T_1 and T_2 at the inner and outer surfaces, respectively, of the fuel element?

b) Compute and plot the temperature distribution in the composite wall for selected values of q.  What is the maximum allowable value of q.

Solution:

part a)

- The outer surface temperature of the fuel, T_2, may be determined from the rate equation:

                                 q*A_th = T_2 - T_inf / R'_total

Where,

           A_th: Area of the thorium section

           T_inf: The temperature of coolant = 600 K

           R'_total: The resistance per unit length.

- Calculate the resistance per unit length R' from thorium surface to coolant:

           R'_total = Ln(r_3/r_2) / 2*pi*k_g + 1 / 2*pi*r_3*h

Plug in values:

           R'_total = Ln(14/11) / 2*pi*3 + 1 / 2*pi*0.014*2000

           R'_total = 0.0185 mK / W

- And the heat rate per unit length may be determined by applying an energy balance to a control surface  about the fuel element. Since the interior surface of the element is essentially adiabatic, it follows that:

           q' = q*A_th = q*pi*(r_2^2 - r_1^2)

           q' = 10^8*pi*(0.011^2 - 0.008^2) = 17,907 W / m

Hence,

           T_2 = q' * R'_total + T_inf

           T_2 = 17,907*0.0185 + 600

          T_2 = 931 K

- With zero heat flux at the inner surface of the fuel element, We will apply the derived results for boundary conditions as follows:

 T_1 = T_2 + (q*r_2^2/4*k_th)*( 1 - (r_1/r_2)^2) - (q*r_1^2/2*k_th)*Ln(r_2/r_1)

Plug values in:

 T_1 = 931+(10^8*0.011^2/4*57)*( 1 - (.8/1.1)^2) - (10^8*0.008^2/2*57)*Ln(1.1/.8)

 T_1 = 931 + 25 - 18 = 938 K

part b)

The temperature distributions may be obtained by using the IHT model for one-dimensional, steady state conduction in a hollow tube. For the fuel element (q > 0),  an adiabatic surface condition is  prescribed at r_1 while heat transfer from the outer surface at r_2 to the coolant is governed by the thermal  resistance:

                              R"_total = 2*pi*r_2*R'_total

                              R"_total = 2*pi*0.011*0.0185 = 0.00128 m^2K/W

- For the graphite ( q = 0 ), the value of T_2 obtained from the foregoing solution is prescribed as an inner boundary condition at r_2, while a convection condition is prescribed at the outer surface (r_3).

- For 5*10^8 < q and q > 5*10^8, the distributions are given in attachment.

The graphs obtained:

- The comparatively large value of k_t yields small temperature variations across the fuel element,  while the small value of k_g results in large temperature variations across the graphite.

Operation  at q = 5*10^8 W/^3  is clearly unacceptable, since the melting points of thorium and graphite are exceeded  and approached, respectively. To prevent softening of the materials, which would occur below their  melting points, the reactor should not be operated much above:

                                      q = 3*10^8 W/m^3

6 0
3 years ago
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