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3 years ago

Earth's surface receives about twice as much energy from the atmosphere than from the sun as a result of ____.

Physics

1 answer:

Oxana [17]3 years ago

5 0

I think the answer is “greenhouse effect”

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Two long, straight wires are parallel and 20 cm apart. One carries a current of 2.2 A, the other a current of 5.9 A. (a) If the

Leya [2.2K]

Answer: $1.298\times 10^{-5}\ N/m$

Explanation:

Given

Current in the first wire $I_1=2.2\ A$

Current in the second wire $I_2=5.9\ A$

wires are $20\ cm$ apart

Force per unit length between the current-carrying wires is

$\Rightarrow \dfrac{F}{l}=\dfrac{\mu_oI_1I_2}{2\pi r}$

Force exerted by the wires is the same

Put the values

$\Rightarrow \frac{F}{l}=f=\dfrac{4\pi \times 10^{-7}\times 2.2\times 5.9}{2\pi \times 0.2}=1.298\times 10^{-5}\ N/m$

This force will be repulsive in nature as the current is flowing opposite

4 0

2 years ago

(a) Consider the initial-value problem dA/dt = kA, A(0) = A0 as the model for the decay of a radioactive substance. Show that, i

murzikaleks [220]

Answer:

a) $t = -\frac{ln(2)}{k}$

b) See the proof below

$A(t) = A_o 2^{-\frac{t}{T}}$

c) $t = 3T \frac{ln(2)}{ln(2)}= 3T$

Explanation:

Part a

For this case we have the following differential equation:

$\frac{dA}{dt}= kA$

With the initial condition $A(0) = A_o$

We can rewrite the differential equation like this:

$\frac{dA}{A} =k dt$

And if we integrate both sides we got:

$ln |A|= kt + c_1$

Where $c_1$ is a constant. If we apply exponential for both sides we got:

$A = e^{kt} e^c = C e^{kt}$

Using the initial condition $A(0) = A_o$ we got:

$A_o = C$

So then our solution for the differential equation is given by:

$A(t) = A_o e^{kt}$

For the half life we know that we need to find the value of t for where we have $A(t) = \frac{1}{2} A_o$ if we use this condition we have:

$\frac{1}{2} A_o = A_o e^{kt}$

$\frac{1}{2} = e^{kt}$

Applying natural log we have this:

$ln (\frac{1}{2}) = kt$

And then the value of t would be:

$t = \frac{ln (1/2)}{k}$

And using the fact that $ln(1/2) = -ln(2)$ we have this:

$t = -\frac{ln(2)}{k}$

Part b

For this case we need to show that the solution on part a can be written as:

$A(t) = A_o 2^{-t/T}$

For this case we have the following model:

$A(t) = A_o e^{kt}$

If we replace the value of k obtained from part a we got:

$k = -\frac{ln(2)}{T}$

$A(t) = A_o e^{-\frac{ln(2)}{T} t}$

And we can rewrite this expression like this:

$A(t) = A_o e^{ln(2) (-\frac{t}{T})}$

And we can cancel the exponential with the natural log and we have this:

$A(t) = A_o 2^{-\frac{t}{T}}$

Part c

For this case we want to find the value of t when we have remaining $\frac{A_o}{8}$

So we can use the following equation:

$\frac{A_o}{8}= A_o 2^{-\frac{t}{T}}$

Simplifying we got:

$\frac{1}{8} = 2^{-\frac{t}{T}}$

We can apply natural log on both sides and we got:

$ln(\frac{1}{8}) = -\frac{t}{T} ln(2)$

And if we solve for t we got:

$t = T \frac{ln(8)}{ln(2)}$

We can rewrite this expression like this:

$t = T \frac{ln(2^3)}{ln(2)}$

Using properties of natural logs we got:

$t = 3T \frac{ln(2)}{ln(2)}= 3T$

8 0

2 years ago

A city bus travels 6 blocks east and 8 blocks north. Each block is 100 m long. If the bus travels this distance in 15 minutes, w

Eddi Din [679]

Answer:

1.56 m/s

Explanation:

For East direction, the 6 blocks is equivalent to 6*100=600 m

For West direction, 8 blocks is equivalent to 8*100=800 m

Speed is given by s=d/t where s is speed, d is distance covered and t is time taken

Total distance is 600+800=1400m

Time taken is 15 mins, converted to seconds will be 15*60=900 s

Speed, s=1400/900=1.5555555555555 m/s rounded off as 1.56 m/s

4 0

2 years ago

The angular separation of two stars is 0.1 arcseconds and you photograph them with a telescope that has an angular resolution of

gizmo_the_mogwai [7]

Answer:

Will see them as only one star

Explanation:

Solution:

- The angular resolution of a telescope means the minimum quantity that can be visualized. Since their angular separation ( 0.1 arcseconds ) is smaller than the telescope's angular resolution (1 arcseconds ), your photograph will seem to show only one star rather than two.

7 0

2 years ago

An airplane has an effective wing surface area of 17.0 m2 that is generating the lift force. In level flight the air speed over

olga_2 [115]

To solve this problem it is necessary to apply the equations given from Bernoulli's principle, which describes the behavior of a liquid moving along a streamline. Mathematically this expression can be given as,

$P_1 + \frac{1}{2}\rho*v_1^2 + P_2 + \frac{1}{2}*\rho*v_2^2=0$

Where,

$P_i =$ Pressure at each state

$\rho$ = Density

$v_i =$ Velocity

Re-organizing the expression we can get that

$P_1 - P_2 = \frac{1}{2}\rho (v_2^2 - v_1^2)$

Our values are given as

$v_1 = 40m/s$

$v_2 = 55m/s$

$\rho_{water} = 1.2kg/m^3 \rightarrow$ Normal Conditions

Replacing we have,

$P_1 -P_2 = \frac{1}{2}*1.2*(55^2-40^2)$

$P_1 - P_2 = 855Pa$

If we consider that there is a balance between the two states, the Force provided by gravity is equivalent to the Support Force, therefore

$F_l = F_g$

Here the lift force is the product between the pressure difference previously found by the effective area of the aircraft, while the Force of gravity represents the weight. There,

$F_g = W$

$F_l = (P_2-P_1)A$

Equating,

$(P_1 - P_2)*A = W$

$W = 855*17$

$W = 14535 N$

Therefore the weight of the plane is 14535N

3 0

3 years ago