When6-2 He He-6 undergoes beta decay, the daughter is?​

Answer:

materials that allow electricity to pass through them are called conductors, some examples of conductors are many metals, such as copper, iron and steel.

Answer:

In metals there are free electrons at normal temperature so when we increase temperature it resistivity gets increases,so conductivity decreases,while in semiconductor the electrons are not free so when we increase the temperature the covalent bonds begin to break and the electron becomes free so conductivity get.

Explanation:

Answer:

864 mT

Explanation:

The magnetic field due to a long straight wire B = μ₀i/2πR where μ₀ = permeability of free space = 4π × 10⁻⁷ H/m, i = current in wire, and R = distance from center of wire to point of magnetic field.

The magnitude of magnetic field due to the first wire carrying current i = 2.70 A at distance R which is mid-point between the wires is B = μ₀i/2πR.

Since the other wire also carries the same current at distance R, the magnitude of the magnetic field is B = μ₀i/2πR.

The resultant magnetic field at B is B' = B + B = 2B = 2(μ₀i/2πR) = μ₀i/πR

Now R = 2.50 cm/2 = 1.25 cm = 1.25 × 10⁻² m and i = 2.70 A.

Substituting these into B' = μ₀i/πR, we have

B' = 4π × 10⁻⁷ H/m × 2.70 A/π(1.25 × 10⁻² m)

B = 10.8/1.25 × 10⁻⁵ T

B = 8.64 × 10⁻⁵ T

B = 864 × 10⁻³ T

B = 864 mT

This question involves the concept of the magnetic field due to two current-carrying wires in the same direction, parallel to each other.

The magnitude of the magnetic field at the point P, which is equidistant from the wires is "8.64 x 10⁻⁵ T".

The following formula is used to find the magnetic field at the center distance between two parallel current-carrying wires in the same direction:

[tex]B = \frac{\mu_oI_1}{2\pi r}+\frac{\mu_oI_2}{2\pi r}\\\\But,\ I_1=I_2=I\\\\B = \frac{\mu_oI}{\pi r}[/tex]

where,

B = magnetic field at required point = ?

μ₀ = permeability of free space = 4π x 10⁻⁷ H/m

I = current = 2.7 A

r = distance from wires to the point = 2.5 cm/2 = 1.25 cm = 0.0125 m

Therefore,

[tex]B=\frac{(4\pi\ x\ 10^{-7}\ H/m)(2.7\ A)}{\pi (0.0125\ m)}[/tex]

B = 8.64 x 10⁻⁵ T

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Answer:

0.002 m/s

Explanation:

27.13(60) = 1,627.8 seconds

3.12/1,627.8 = 0.00191 ≈ 0.002 = s

Are you sure you're not looking for cm/s?

Answer:

1) 1000Nm

2)  95,625Nm

3) 1.05%

Explanation:

Mechanical Advantage is the ratio of the load  to the effort applied to an object.

MA = Load/Effort

1) Workdone on the load = Force(Load) * distance covered by the load

Workdone on the load = 500N * 2m

Workdone on the load = 1000Nm

2)  work done by the effort = Effort * distance moves d by effort

work done by the effort = 2125 * 45

work done by the effort = 95,625Nm

3) Efficiency = Workdone on the load/ work done by the effort * 100

Efficiency = 1000/95625 * 100

Efficiency = 1.05%

Hence the efficiency of the system is 1.05%

Answer:

1115560000 J

Explanation:

1/2 * 80,000 * 167^2 m/s = 1115560000 J

Answer:

The right solution is "126 Psi".

Explanation:

The given values are:

P₁ = 130 psig

i.e.,

   = [tex]130\times 6.894[/tex]

   = [tex]896.22 \ Kpa[/tex]

or,

   = [tex]896.22\times 10^3 \ Pa[/tex]

Z₂ = 10ft

    = 3.05 m

[tex]\delta[/tex] = 1000 kg/m³

According to the question,

Z₁ = 0

V₁ = V₂

As we know,

⇒  [tex]\frac{P_1}{\delta_g} +\frac{V_1^2}{2g} +Z_1=\frac{P_2}{\delta_g} +\frac{V_2^2}{2g} +Z_2[/tex]

On substituting the values, we get

⇒  [tex]\frac{P_1}{\delta_g} +0+0=\frac{P_2}{\delta_g} +0+Z_2[/tex]

⇒  [tex]\frac{896.22\times 10^3}{1000\times 9.8} =\frac{P_2}{1000\times 9.8} +3.05[/tex]

⇒  [tex]P_2=866330 \ P_a[/tex]

i.e.,

⇒       [tex]=866330\times 0.000145[/tex]

⇒       [tex]=126 \ Psi[/tex]

Answer:

5.0/s

Explanation:

Answer:

b and a it is this that abewsr

Answer:

1.61 N/kg

Explanation:

Take the universal gravitational constant G as 6.67 × 10^(-11) Nm²/kg²

The required gravitational field strength

= 6.67 × 10^(-11) × 7.3 × 10^(22) / (1738000)²

= 1.61194103 N/kg

= 1.61 N/kg (corr. to 3 sig. fig.)

Answer:

Heat loss per unit length = 642.358 W/m

The heat loss per unit length on a breezy day during 8 m/s speed is = 1760.205 W/m

Explanation:

From the information given:

Diameter D [tex]= 100 mm = 0.1 m[/tex]

Surface emissivity ε = 0.8

Temperature of steam [tex]T_s[/tex] = 150° C = 423K

Atmospheric air temperature [tex]T_{\infty} = 20^0 \ C = 293 \ K[/tex]

Velocity of wind V = 8 m/s

To calculate average film temperature:

[tex]T_f = \dfrac{T_s+T_{\infty}}{2}[/tex]

[tex]T_f = \dfrac{423+293}{2}[/tex]

[tex]T_f = \dfrac{716}{2}[/tex]

[tex]T_f = 358 \ K[/tex]

To calculate volume expansion coefficient

[tex]\beta= \dfrac{1}{T_f} \\ \\ \beta= \dfrac{1}{358} \\ \\ \beta= 2.79 \times 10^{-3} \ K^{-1}[/tex]

From the table of "Thermophysical properties of gases at atmospheric pressure" relating to 358 K of average film temperature; the following data are obtained;

Kinematic viscosity (v) = 21.7984 × 10⁻⁶ m²/s

Thermal conductivity k = 30.608 × 10⁻³ W/m.K

Thermal diffusivity ∝ = 31.244 × 10⁻⁶ m²/s

Prandtl no. Pr = 0.698

Rayleigh No. for the steam line is determined as follows:

[tex]Ra_{D} = \dfrac{g \times \beta (T_s-T_{\infty}) \times D_b^3}{\alpha\times v}[/tex]

[tex]Ra_{D} = \dfrac{9.8 \times (2.79 *10^{-3})(150-20) \times (0.1)^3}{(31.244\times 10^{-6}) \times (21.7984\times 10^{-6})}[/tex]

[tex]Ra_{D} = 5.224 \times 10^6[/tex]

The average Nusselt number is:

[tex]Nu_D = \Big \{ 0.60 + \dfrac{0.387(Ra_D)^{1/6}}{[ 1+ (0.559/Pr)^{9/16}]^{8/27}} \Big \}^2[/tex]

[tex]Nu_D = \Big \{ 0.60 + \dfrac{0.387(5.224\times 10^6)^{1/6}}{[ 1+ (0.559/0.698)^{9/16}]^{8/27}} \Big \}^2[/tex]

[tex]Nu_D = \Big \{ 0.60 + \dfrac{5.0977}{[ 1.8826]^{8/27}}\Big \}^2[/tex]

[tex]Nu_D = \Big \{ 0.60 + 4.226 \Big \}^2[/tex]

[tex]Nu_D = 23.29[/tex]

However, for the heat transfer coefficient; we have:

[tex]h_D = \dfrac{Nu_D\times k}{D_b} \\ \\ h_D = \dfrac{(23.29) \times (30.608 \times 10^{-3} )}{0.1}[/tex]

[tex]h_D = 7.129 \ Wm^2 .K[/tex]

Hence, Stefan-Boltzmann constant [tex]\sigma = 5.67 \times 10^{-8} \ W/m^2.K^4[/tex]

Now;

To determine the heat loss using the formula:

[tex]q'_b = q'_{ev} + q'_{rad} \\ \\ q'_b = h_D (\pi D_o) (T_t-T_{\infty})+\varepsilon(\pi D_b)\sigma (T_t^4-T_{\infty }^4)[/tex]

[tex]q'_b = (7.129)(\pi*0.1) (423-293) + (0.8) (\pi*0.1) (5.67 *10^{-8}) (423^4-293^4) \\ \\ q'_b = 291.153 + 351.205 \\ \\ \mathbf{q'_b = 642.258 \ W/m}[/tex]

Now; here we need to determine the Reynold no and the average Nusselt number:

[tex]Re_D = \dfrac{VD_b}{v } \\ \\ Re_D = \dfrac{8 *0.1}{21.7984 \times 10^{-6}} \\ \\ Re_D = 3.6699 \times 10^4[/tex]

However, to determine the avg. Nusselt no by using Churchill-Bernstein correlation, we have;

[tex]Nu_D = 0.3 + \dfrac{0.62 \times Re_D^{1/2}* Pr^{1/3}}{[1+(0.4/Pr)^{2/3}]^{1/4}} [1+ (\dfrac{Re_D}{282000})^{5/8}]^{4/5}[/tex]

[tex]Nu_D = 0.3 + \dfrac{0.62 \times (3.6699*10^4)^{1/2}* (0.698)^{1/3}}{[1+(0.4/0.698)^{2/3}]^{1/4}} [1+ (\dfrac{3.669*10^4}{282000})^{5/8}]^{4/5}[/tex]

[tex]Nu_D = (0.3 +\dfrac{105.359}{1.140}\times 1.218) \\ \\ Nu_D = 112.86[/tex]

SO, the heat transfer coefficient for forced convection is determined as follows afterward:

[tex]h_D = \dfrac{Nu_{D}* k}{D_b} \\ \ h_D = \dfrac{112.86*30.608 *10^{-3}}{0.1} \\ \\ h_D = 34.5 \ W/m^2 .K[/tex]

Finally; The heat loss per unit length on a breezy day during 8 m/s speed is:

[tex]q'b = h_D (\pi D_b) (T_s-T_{\infty}) + \varepsilon (\pi D_b) \sigma (T_s^4-T_ {\infty}^4) \\ \\ q'b = (34.5) (\pi *0.1) (423-293) + (0.8) (\pi*0.1) (5.67*10^{-8}) (423^4 - 293^4) \\ \\ = 1409 +351.205 \\ \\ \mathbf{q'b = 1760.205 \ W/m}[/tex]

Answer:

the potential energy decreases as it is converted to kinetic energy.

Explanation:

As things move, their potential energy converts to kinetic energy to power them along. When a ball rolls down the top of a ramp, all the potential energy it accumulated at the top of the ramp converts to kinetic energy to help it roll down. In other words, its potential energy decreases as its kinetic energy increases.

Answer:

If you have two batteries and they have precisely the same voltage then placing one backwards will effectively cancel out the voltages and no current will flow. However, batteries aren't like that. The slightest difference in voltages mean that current will flow.

Explanation:

Answer:

The movement of Sun and moon

Explanation:

When the sun rise.it is am and when it sets .it is pm.

1. false

2. true

I hope this helps ^-^

Explanation:

रात में घूमने वाला arthaarat निशाचर

The image of this hollow sphere and uniform rod is missing, so i have attached it.

Answer:

A) J = 0.7443 kg•m²

B) T = 1.9169 N•m CCW

C) α = 2.5754 rad/s²

D) a = 3.966 m/s²

Explanation:

A) The moment of inertia J of the contraption around the fulcrum is given by the formula;

J = Jℓ + Jr

Let's calculate Jℓ

Jℓ = [((0.34²/3) × 0.50 × 0.34)/1.78] + (0.48 × (0.34 + 0.64)²)

Jℓ = 0.4647 kg•m²

Now, let's Calculate Jr

Jr = ((1.78 - 0.34)²/3) × ((1.78 - 0.34)/1.78) × 0.50

Jr = 0.2796 kg•m²

Thus;

J = 0.4647 + 0.2796

J = 0.7443 kg•m²

(b) Using CCW as positive, Torque in Nm is calculated as;

T = Tℓ - Tr

Let's calculate Tℓ

Tℓ = [(0.48 × (0.64 + 0.34)) + (0.50 × 0.34/1.78) × 0.34/2)] × 9.81

Tℓ = 4.7739 N•m CCW

Now, let's Calculate Tr;

Tr = [(0.50 × (1.78 - 0.34)/1.78) × (1.78 - 0.34)/2)] × 9.81

Tr = 2.857 N•m CW

Thus;

T = 4.7739 - 2.857

T = 1.9169 N•m CCW

(c) The angular acceleration α of the contraption, using CCW is gotten from the formula;

α = T/J

α = 1.9169/0.7443

α = 2.5754 rad/s²

(d) The linear acceleration a of the right end of the rod, using up as positive is given by;

a = α*(1.78 - 0.34)

a = 2.5754 × 1.54

a = 3.966 m/s²

A) the moment of inertia of the contraption is 0.7443 kgm²

B) The torque about the fulcrum is 1.9169 Nm

C) Angular acceleration of the contraption is 2.5754 rad/s²

D) The linear acceleration of the contraption is 3.966 m/s²

(A) The moment of inertia I of the contraption around the fulcrum is given by :

[tex]I = [(0.34^2/3) \times 0.50 \times 0.34)/1.78 + (0.48 \times (0.34 + 0.64)^2)] + [(1.78 - 0.34)^2/3) \times (1.78 - 0.34)/1.78) \times 0.50][/tex]

I = 0.4647 + 0.2796

I = 0.7443 kgm²

(B) Using CCW as positive, Torque in Nm is given by;

T = [(0.48 × (0.64 + 0.34)) + (0.50 × 0.34/1.78) × 0.34/2)] × 9.81 - [(0.50 × (1.78 - 0.34)/1.78) × (1.78 - 0.34)/2)] × 9.81

T = 4.7739 - 2.857

T = 1.9169 Nm

(C) The angular acceleration (α) of the contraption is given by:

α = T/I

since, torque is defined as T = Iα

α = 1.9169/0.7443

α = 2.5754 rad/s²

(D) The linear acceleration (a) of the right end of the rod

a = αr

where r is the distance from the pivot

a = α × (1.78 - 0.34)

a = 2.5754 × 1.54

a = 3.966 m/s²

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Answer:

momentum=mass x velocity= 10 x 2 = 20kgm/s

Answer:

ks= 133.2 N/m

Explanation:

       [tex]\Delta U + \Delta K = 0 (1)[/tex]

       [tex]k_{s} =\frac{m*v^{2}}{\Delta x^{2}} = \frac{29 kg*(3m/s)^{2}}{(1.4m)^{2}} = 133.2 N/M (4)[/tex]

Answer:

An Oven

Explanation:

The heat is higher, so it moves faster. Shile in a freezer the particles are extremely slow!

Answer:

Regular reflection

Explanation:

Regular reflection occurs when light reflects off a very smooth surface and forms a clear image.

i hope this helps a bit.

According to the context, a sharp image is formed when light reflects from a regular reflection.

It is reflection without diffusion that obeys the laws of geometrical optics, as in mirrors.

This reflection of light happens when the angles that the two rays determine with the surface are equal.

Therefore, we can conclude that according to the context, a sharp image is formed when light reflects from a regular reflection.

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Answer:

Molecular scale. The story begins a long time ago  

when the idea that molecules are in constant motion  

was first discovered. Part of the evidence that you can  

see in everyday life was discovered by Robert Brown  

about 150 years ago when he used a microscope to  

watch how tiny dust particles move.

So how fast do molecules move? It all depends upon  

the molecule and its state: molecules in a solid state  

move slower than in a liquid state, and much slower  

than gas molecules. One estimate puts gas molecules  

in the range of 1,100 mph at room temperature. Cool  

them down to almost absolute zero and they slow  

down to less than 0.1 mph (slower than the average  

couch potato). The fact that they are always moving  

makes it a challenge to see molecules and make stuff  

out of them, but it’s a challenge that scientists  

work hard to figure out.

Explanation:

Answer:

[tex]3.52\times 10^{25}\ \text{J}[/tex]

Explanation:

G = Gravitational constant = [tex]6.674\times 10^{-11}\ \text{Nm}^2/\text{kg}^2[/tex]

M = Mass of Sun = [tex]1.989\times 10^{30}\ \text{kg}[/tex]

m = Mass of Earth = [tex]5.972\times 10^{24}\ \text{kg}[/tex]

[tex]r_i[/tex] = Initial radius of orbit = [tex]1.5\times 10^{11}\ \text{m}[/tex]

[tex]r_f[/tex] = Final radius of orbit = [tex]((1.5\times 10^{11})+2\times 10^3)\ \text{m}[/tex]

Energy required is given by

[tex]E=\dfrac{1}{2}\Delta U\\\Rightarrow E=\dfrac{GMm}{2}(\dfrac{1}{r_i}-\dfrac{1}{r_f})\\\Rightarrow E=\dfrac{6.674\times 10^{-11}\times 1.989\times 10^{30}\times 5.972\times 10^{24}}{2}(\dfrac{1}{1.5\times 10^{11}}-\dfrac{1}{(1.5\times 10^{11})+2\times 10^3})\\\Rightarrow E=3.52\times 10^{25}\ \text{J}[/tex]

The energy required would be [tex]3.52\times 10^{25}\ \text{J}[/tex].

Answer:

Q = 636.48 J

Explanation:

Given that,

The mass of gold, m = 0.052 kg

The temperature increase from 30°C to 120°C.

The specific heat of gold is 136  J/kg °C.

We need to find the heat needed to warm the gold. The formula for heat needed is given by :

[tex]Q=mc\Delta T\\\\Q=0.052\times 136\times (120-30)\\\\Q=636.48\ J[/tex]

So, 636.48 J of heat is needed to warm gold.

An inventor makes a clock using a brass rod and a heavy mass as a pendulum. when the clock gets colder then the time clock would gain time

The expansion of any material due to the variation of the temperature is known as thermal expansion. It varied differently for different materials according to their corresponding values of the coefficient of the thermal expansion.

As given in the problem statement that an inventor makes a clock using a brass rod and a heavy mass as a pendulum, when the clock gets colder then the length of the brass decreases due to thermal expansion.

The length of the pendulum gets reduced which further results in the reduction in the time period, as per the formula of the time period for the pendulum

T = 2π√(L/g)

As the length of the brass gets reduced. This means the pendulum of the clock moves faster and the clock would gain time

Thus, if a  pendulum made of a heavy mass and a brass rod is used to create a clock by an inventor. The time clock would advance in time as the clock get colder

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Answer:

8.45 m

Explanation:

From the question given above, the following data were obtained:

Mass (m) = 90 Kg

Initial velocity (u) = 13 m/s

Final velocity (v) = 0 m/s

Height (h) =?

NOTE: Acceleration due to gravity (g) = 10 m/s²

The height of the hill can be obtained as follow:

v² = u² – 2gh (since the cart is going against gravity)

0² = 13² – (2 × 10 × h)

0 = 169 – 20h

Rearrange

20h = 169

Divide both side by 20

h = 169/20

h = 8.45 m

Therefore, the height of the hill is 8.45 m

Answer:

work done= force × displacement

=800×5

=4000J

Explanation:

The amount of work done is the result of the magnitude of force applied and the displacement of the body due to the force applied. Therefore, work done is defined as the product of the applied force and the displacement of the body.