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2 years ago

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A box with a mass of 2 kg only has four forces acting on it: One force of 16 N due East. One force of 24 N due South. One force

of 16 N due West. One force of 18 N due North. What is the magnitude of the box's acceleration, in m/s2

2 answers:

katrin [286]2 years ago

7 0

Answer:

The acceleration of the object is 3 m/s2.

Explanation:

The force on the object is acting along the four directions which are perpendicular to each other.
Since the north and south directions are opposite to each other, the magnitude of the net force along north-south direction is (24-18)=6 N.
Since the east and west directions are opposite to each other, the magnitude of the net force along east-west direction is (16-16)=0 N.
Therefore the magnitude of the net force on the object is 6 N
By Newton's second law, the force F on the object is given by the formula F=ma.
Here m= 2kg, therefore 6 =(2)a which gives a = 3 m/s2.

Learn more about Newton's law.

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julsineya [31]2 years ago

3 0

Answer:

The magnitude of acceleration is 3 m/s^2.

Explanation:

The net force along North-South direction is 18-24=-6 N.

The net force along East-West direction is 16-16=0 N

The net magnitude of force is square root of (-6)^2+0^2 which is 6 N.

By Newton's Second law,

Force F=ma

Therefore acceleration a=F/m

a=6/2

a=3 m/s^2

Learn more about Newton's second law.

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3 years ago

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The block slides on a horizontal frictionless surface. The block has a mass of 1.0kg and is pushed 5.0N at 45°. What is the magn

Firlakuza [10]

Answer:

$3.5\:\mathrm{m/s^2}$

Explanation:

Newton's 2nd law is given as $\Sigma F = ma$ .

To find the acceleration in the horizontal direction, you need the horizontal component of the force being applied.

Using trigonometry to find the horizontal component of the force:

$\cos 45^{\circ}=\frac{x}{5},\\\frac{\sqrt{2}}{{2}}=\frac{x}{5},\\x=\frac{5\sqrt{2}}{2}$

Use this horizontal component of the force to solve for for the acceleration of the object:

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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:

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

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3 years ago

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Harrizon [31]

Answer: 249 seconds

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Recall that speed is the distance covered per unit time.

Hence, speed = Distance / Time taken

Time taken = Distance / speed

Z = 628m / 2.52 m/s

Z = 249.2 seconds (Round to nearest tenth i.e 249 seconds)

Thus, it takes 249 seconds for the girl to get to her friends house

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3 years ago

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Svetllana [295]

Part 1)

Answer:

Explanation:

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so we will have

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here we know that

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so we have

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Part b)

Answer:

The time will increase.

Explanation:

As we know that on increasing the value of the resistance the the product of the resistance and capacitance will increase so the time will increase to get the above voltage.

Part c)

Answer:

The capacitor discharges through a very low resistance (the lamp filled with ionized gas), and so the discharge time constant is very short. Thus the flash is very brief.

Explanation:

Since the lamp resistance is very small so the energy across the lamp will totally lost in very short interval of time

Part d)

Answer:

Once the lamp has flashed, the stored energy in the capacitor is gone, and there is no source of charge to maintain the lamp current. The lamp "goes out", the lamp resistance increases, and the capacitor starts to recharge. It charges again and the process will repeat.

Explanation:

Since we know that the battery is connected to the given system so after whole energy of capacitor is flashed out it is again charged by the battery and the process will continue

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4 years ago