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3 years ago

Cat walks 100 m East, then turns around and walks 25 m west. What is the cats displacement?

Physics

1 answer:

kvasek [131]3 years ago

8 0

Answer:

Explanation:

If a cat walk 100m East, this means that it is walking in the positive x direction, the distance will therefore be +100m

If it turns around and walks 25 m west, the direction of movement is in the negative x direction i.e -25m

Taking the sum;

Displacement = +100m - 25m

Displacement of the cat = 75m

hence the cats displacement is 75m

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Help asap please I will give you 5stars

boyakko [2]

Explanation:

In the parallel combination, the equivalent resistance is given by :

$\dfrac{1}{R}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+....$

4. When three 150 ohms resistors are connected in parallel, the equivalent is given by :

$\dfrac{1}{R}=\dfrac{1}{150}+\dfrac{1}{150}+\dfrac{1}{150}\\\\R=50\ \Omega$

5. Three resistors of 20 ohms, 40 ohms and 100 ohms are connected in parallel, So,

$\dfrac{1}{R}=\dfrac{1}{20}+\dfrac{1}{40}+\dfrac{1}{100}\\\\=11.76\ \Omega$

Hence, this is the required solution.

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3 years ago

A 75-g bullet is fired from a rifle having a barrel 0.540 m long. choose the origin to be at the location where the bullet begin

lyudmila [28]

Part a) The work done by the gas on the bullet is the integral of the force in dx, where x is the distance covered by the bullet inside the barrel with respect to the origin:
$W= \int\limits^{0.540m}_{0} {F} \, dx = \int\limits^{0.540m}_{0} {(16000+10000x-26000x^2)} \, dx =$
$=16000x+10000 \frac{x^2}{2} - 26000 \frac{x^3}{3}$
By substituting the length of the barrel, L=0.540 m, we find the total work done by the gas on the bullet:
$W=16000(0.540m)+10000 \frac{(0.540m)^2}{2} - 26000 \frac{(0.540m)^3}{3} =$
$=8733 J=8.73 kJ$

part b) The resolution of the problem is the same, we just have to use the new length of the barrel (L=0.95 m) inside the final formula, and we find the new value of the work:
$W=16000(0.95m)+10000 \frac{(0.95m)^2}{2} - 26000 \frac{(0.95m)^3}{3} =$
$=12280 J=12.28 kJ$

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2 years ago

A tree falls in a forest. How many years must pass before the 14C activity in 1.03 g of the tree's carbon drops to 1.02 decay pe

Illusion [34]

Answer:

t = 5.59x10⁴ y

Explanation:

To calculate the time for the ¹⁴C drops to 1.02 decays/h, we need to use the next equation:

$A_{t} = A_{0}\cdot e^{- \lambda t}$ (1)

<em>where $A_{t}$ : is the number of decays with time, A₀: is the initial activity, λ: is the decay constant and t: is the time.</em>

To find A₀ we can use the following equation:

$A_{0} = N_{0} \lambda$ (2)

<em>where N₀: is the initial number of particles of ¹⁴C in the 1.03g of the trees carbon </em>

From equation (2), the N₀ of the ¹⁴C in the trees carbon can be calculated as follows:

$N_{0} = \frac{m_{T} \cdot N_{A} \cdot abundance}{m_{^{12}C}}$

<em>where $m_{T}$ : is the tree's carbon mass, $N_{A}$ : is the Avogadro's number and $m_{^{12}C}$ : is the ¹²C mass. </em>

$N_{0} = \frac{1.03g \cdot 6.022\cdot 10^{23} \cdot 1.3\cdot 10^{-12}}{12} = 6.72 \cdot 10^{10} atoms ^{14}C$

Similarly, from equation (2) λ is:

$\lambda = \frac{Ln(2)}{t_{1/2}}$

<em>where t 1/2: is the half-life of ¹⁴C= 5700 years </em>

$\lambda = \frac{Ln(2)}{5700y} = 1.22 \cdot 10^{-4} y^{-1}$

So, the initial activity A₀ is:

$A_{0} = 6.72 \cdot 10^{10} \cdot 1.22 \cdot 10^{-4} = 8.20 \cdot 10^{6} decays/y$

Finally, we can calculate the time from equation (1):

$t = - \frac{Ln(A_{t}/A_{0})}{\lambda} = - \frac {Ln(\frac{1.02decays \cdot 24h \cdot 365d}{1h\cdot 1d \cdot 1y \cdot 8.20 \cdot 10^{6} decays/y})}{1.22 \cdot 10^{-4} y^{-1}} = 5.59 \cdot 10^{4} y$

I hope it helps you!

4 0

3 years ago

If R is the total resistance for a parallel circuit with two resistors of resistance r1 and r2, then . Find the resistance, r1,

Goshia [24]

For a parallel circuit with two resistors, the total resistance is calculated from the expression:

1/R = 1/R1 + 1/R2

We are given the total resistance, R, which is 20 ohms and R2 which is 75 ohms. We calculate R1 as follows:

1/20 = 1/R1 + 1/75
1/R1 = 11/300
R1 = 27.27 ohms

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3 years ago

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astra-53 [7]

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3 years ago

Cat walks 100 m ￼ East, then turns around and walks 25 m west. What is the cats displacement￼￼?

Cat walks 100 m East, then turns around and walks 25 m west. What is the cats displacement?