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3 years ago

2 Here are some numbers: 22 23 25 27 29 ܕ ST From the list, write down all of the prime numbers .

Physics

1 answer:

lapo4ka [179]3 years ago

5 0

Answer:

only 23 and 29 are prime numbers from that list

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Suppose the voltage source for a series RL-circuit were given as V0sin(ωt) instead of V0cos(ωt). Write an expression for the cur

Gala2k [10]

Answer:

Explanation:

This is an RL circuit, therefore:

Impedance; z = $\mathbf{\sqrt{R^2+L^2}}$

$\mathbf{z = \sqrt{R^2+(Lw)^2}}$

Current amplitude

$\mathbf{I_o = \dfrac{V_o}{z}} \\ \\ \mathbf{I_o = \dfrac{V_o}{\sqrt{R^2+L^2\omega ^2}}}$

Given that:

$V_o = 1.9 \ V \\ \\ \omega= 51 \ rad/s\\\\ R = 21 \Omega \\ \\ L = 0.52 H$

∴

$I_o= \dfrac{1.9}{\sqrt{21^2+(0.52\times 51)^2}}$

$\mathbf{I_o= 0.0562} \\ \\ \mathbf{I_o = 56.2 \ mA}$

Phase constant :

$tan \ \phi = \dfrac{L \omega}{R } \\ \\ tan \ \phi = \dfrac{0.52 \times 51}{21} \\ \\ tan \phi = 1.263$

$\text{Phase constant : }\phi = tan^{-1} (1.263) \\ \\ \phi = 51.6^0\\ \\\text{To radians} \phi = 51.6 \times \dfrac{\pi}{180} \\ \\ \phi = 0.287 \pi \\ \\ \mathbf{\phi = 0.9 \ rad}$

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3 years ago

A force of 660 n stretches a certain spring a distance of 0.300 m. what is the potential energy of the spring when a 70.0 kg mas

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A force of 660 n stretches a certain spring a distance of 0.300 m. what is the potential energy of the spring when a 70.0 kg mass hangs vertically from it?

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4 years ago

Find the location of the object at times 0,1,2,. 1,000. Save the x-coordinates of the object at each time in a 1 ×1001 row vecto

xeze [42]

The x-coordinates of the object at each time in a 1 ×1001 row vector named a11 using coordinate.

Coordinates are distances or angles, represented by numbers, that uniquely perceive factors on surfaces of dimensions (second) or in space of 3 dimensions.

Coordinates are hard and fast values that help to expose the exact position of a factor within the coordinated aircraft. A coordinate aircraft is a 2nd plane that's shaped by the intersection of perpendicular traces known as the x-axis and y-axis.

clear all

close all

alpha=-0.003;

w=0.05;

A=[1-alpha,-w;w,1-alpha];

A_inv=inv(A);

x0=[1;-1];

ans1=[1];

ans2=[-1];

for i=1:1000

x0=A_inv*x0;

ans1(i+1)=x0(1);

ans2(i+1)=x0(2);

end

ans1

Learn more about coordinate here:-brainly.com/question/17206319

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2 years ago

Once the genes are copied, where do they go?

Anna [14]

By copying their genomes, they retain the tool kit and at the same time generate a garage full of spare parts. Gene duplication can provide the raw material for expression changes to occur, and polyploidy itself can trigger epigenetic changes

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3 years ago

A 0.0550-kg ice cube at −30.0°C is placed in 0.400 kg of 35.0°C water in a very well-insulated container. What is the final temp

KatRina [158]

Answer:

19.34°C

Explanation:

When the ice cube is placed in the water, heat will be transferred from the hot water to it such that the heat gained (Q₁) by the ice is equal to the heat lost(Q₂) by the hot water and a final equilibrium temperature is reached between the melted ice and the cooling/cooled hot water. i.e

Q₁ = -Q₂ ----------------------(i)

{A} Q₁ is the heat gained by the ice and it is given by the sum of ;

(i) the heat required to raise the temperature of the ice from -30°C to 0°C. This is given by [m₁ x c₁ x ΔT]

<em>Where;</em>

m₁ = mass of ice = 0.0550kg

c₁ = a constant called specific heat capacity of ice = 2108J/kg°C

ΔT₁ = change in the temperature of ice as it melts from -30°C to 0°C = [0 - (-30)]°C = [0 + 30]°C = 30°C

(ii) and the heat required to melt the ice completely - This is called the heat of fusion. This is given by [m₁ x L₁]

Where;

m₁ = mass of ice = 0.0550kg

L₁ = a constant called latent heat of fusion of ice = 334 x 10³J/kg

Therefore,

Q₁ = [m₁ x c₁ x ΔT₁] + [m₁ x L₁] ------------------(ii)

Substitute the values of m₁, c₁, ΔT₁ and L₁ into equation (ii) as follows;

Q₁ = [0.0550 x 2108 x 30] + [0.0550 x 334 x 10³]

Q₁ = [3478.2] + [18370]

Q₁ = 21848.2 J

{B} Q₂ is the heat lost by the hot water and is given by

Q₂ = m₂ x c₂ x ΔT₂ -----------------(iii)

Where;

m₂ = mass of water = 0.400kg

c₂ = a constant called specific heat capacity of water = 4200J/Kg°C

ΔT₂ = change in the temperature of water as it cools from 35°C to the final temperature of the hot water (T) = (T - 35)°C

Substitute these values into equation (iii) as follows;

Q₂ = 0.400 x 4200 x (T - 35)

Q₂ = 1680 x (T-35) J

{C} Now to get the final temperature, substitute the values of Q₁ and Q₂ into equation (i) as follows;

Q₁ = -Q₂

=> 21848.2 = - 1680 x (T-35)

=> 35 - T = 21848.2 / 1680

=> 35 - T = 13

=> T = 35 - 13

=> T = 22

Therefore the final temperature of the hot water is 22°C.

Now let's find the final temperature of the mixture.

The mixture contains hot water at 22°C and melted ice at 0°C

At this temperature, the heat ( $Q_{W}$ ) due to the hot water will be equal to the negative of the one ( $Q_{I}$ ) due to the melted ice.

i.e

$Q_{W}$ = - $Q_{I}$ -----------------(a)

Where;

$Q_{I}$ = $m_{I}$ x $c_{I}$ x Δ $T_{I}$ [ $m_{I}$ = mass of ice, $c_{I}$ = specific heat capacity of melted ice which is now water and Δ $T_{I}$ = change in temperature of the melted ice]

and

$Q_{W}$ = $m_{W}$ x $c_{W}$ x Δ $T_{W}$

[ $m_{W}$ = mass of water, $c_{W}$ = specific heat capacity of water and Δ $T_{W}$ = change in temperature of the water]

Substitute the values of $Q_{W}$ and $Q_{I}$ into equation (a) as follows

$m_{W}$ x $c_{W}$ x Δ $T_{W}$ = - $m_{I}$ x $c_{I}$ x Δ $T_{I}$

Note that $c_{W}$ and $c_{I}$ are the same since they are both specific heat capacities of water. Therefore, the equation above becomes;

$m_{W}$ x Δ $T_{W}$ = - $m_{I}$ x Δ $T_{I}$ -----------------------(b)

Now, let's analyse Δ $T_{W}$ and Δ $T_{I}$ . The final temperature ( $T_{F}$ ) of the two kinds of water(melted ice and cooled water) are now the same.

=> Δ $T_{W}$ = change in temperature of water = final temperature of water( $T_{F}$ ) - initial temperature of water( $T_{IW}$ )

Δ $T_{W}$ = $T_{F}$ - $T_{IW}$

Where;

$T_{IW}$ = 22°C [which is the final temperature of water before mixture]

=> Δ $T_{I}$ = change in temperature of melted ice = final temperature of water( $T_{F}$ ) - initial temperature of melted ice ( $T_{II}$ )

Δ $T_{I}$ = $T_{F}$ - $T_{II}$

$T_{II}$ = 0°C (Initial temperature of the melted ice)

Substitute these values into equation (b) as follows;

$m_{W}$ x Δ $T_{W}$ = - $m_{I}$ x Δ $T_{I}$

0.400 x ( $T_{F}$ - $T_{IW}$ ) = -0.0550 x ( $T_{F}$ - $T_{II}$ )

0.400 x ( $T_{F}$ - 22) = -0.0550 x ( $T_{F}$ - 0)

0.400 x ( $T_{F}$ - 22) = -0.0550 x ( $T_{F}$ )

0.400 $T_{F}$ - 8.8 = -0.0550 $T_{F}$

0.400 $T_{F}$ + 0.0550 $T_{F}$ = 8.8

0.455 $T_{F}$ = 8.8

$T_{F}$ = 19.34°C

Therefore, the final temperature of the mixture is 19.34°C

8 0

3 years ago