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3 years ago

9

Which of the following is Not true about the noble gases

1 answer:

Alexandra [31]3 years ago

3 0

Noble gases are not highly reactive

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Two north poles will blank

soldier1979 [14.2K]

Answer:

Be pushed away from each other.

Explanation:

7 0

3 years ago

A ranger in a national park is driving at 56 km/h when a deer jumps onto the road 65 m ahead of the vehicle. After a reaction ti

vagabundo [1.1K]

Answer:

t = 1.58 s

Explanation:

given,

Speed of ranger, v = 56 km/h

v = 56 x 0.278 = 15.57 m/s

distance, d = 65 m

deceleration,a = 3 m/s²

reaction time = ?

using stopping distance formula

$d = v. t + \dfrac{v^2}{2a}$

$t = \dfrac{d}{v} -\dfrac{v}{2a}$

t is the reaction time

$t = \dfrac{65}{15.57} -\dfrac{15.57}{2\times 3}$

t = 1.58 s

hence, the reaction time of the ranger is equal to 1.58 s.

3 0

3 years ago

A box that weighs 5.00×10^2 N is sliding down a ramp at a constant speed. The angle the ramp makes with the horizontal is 25°. W

maxonik [38]

Answer:

0.466 (3 sig. fig.)

Explanation:

Frictional force acting on the box = 5.00×10^2xsin25

Normal force acting on the box = 5.00×10^2xcos25

coefficient of friction = 0.466 (3 sig. fig.)

5 0

2 years ago

Two horses are pulling a box in two different directions as shown in the below image. The image shows one 30.0 N force due north

Novay_Z [31]

Explanation:

'What is the magnitude of the force needed to stop the horses and bring the box into equilibrium?' ≈42N; according to the vectors rules.

'Where would you locate the rope to apply the force?' - in point D.

PS. zoom out the attached picture.

4 0

3 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

3 years ago