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alisha [4.7K]
3 years ago
6

Margaret walks to the store using the following path 0.630 mi west, 0.370 mi north, 0.180 mi east. assume north to be along the

+y-axis and west to be along the -x-axis. what is the magnitude of her total displacement?​
Physics
1 answer:
svp [43]3 years ago
3 0

Answer:

d = 0.583 mi

Explanation:

d = √((-0.630 + 0.180)² + 0.370²)

d = √(-0.450² + 0.370²) = 0.582580... = 0.583 mi

direction would be arctan(0.370/ -0.450) = 39.42780... = 39.4° N of W

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An air-filled parallel-plate capacitor is constructed with a plate area of 0.40 m2 and a plate separation of 0.10 mm. It is then
seropon [69]

Answer:

The charge stored  is  Q =  4.25 *10^{-7 } \ C

The energy stored is  E = 2.55*10^{-6} \ J

Explanation:

From the question we are told that

    The area of the plates is  A =  0.40 \ m^2

     The separation between the plate is d =  0.10 \ mm =0.0001 \ m

      The potential difference is  V =  12 \ V

       The permitivity of free space is  \epsilon_o  = 8.85 *10^{-12} C^2 \cdot N^{-1} \cdot m^2

         The dielectric constant of glass is K =  5.0

Generally the capacitance of this capacitor is

      C =  \frac{\epsilon_o  * A}{d}

substituting values

       C =  \frac{8.85*10^{-12}  * 0.40}{0.0001}

        C =   3.34 *10^{-8} \ C

The charge stored is mathematically evaluated as

         Q = CV

substituting values

        Q =  (3.54*10^{-8} * 12)

         Q =  4.25 *10^{-7 } \ C

The energy stored is  

         E =  0.5 *  CV^2

substituting values

         E =  0.5 *  (4.25 *10^{-7} * 12^2)

           E = 2.55*10^{-6} \ J

7 0
3 years ago
If the velocity of a body varied uniformly from 10 m s-1 to 25 m s-1
BaLLatris [955]

Answer:

3m/s2

Explanation:

a = (v-u) / t

a = (25 - 10) / 5

a = 15 / 5

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4 0
2 years ago
Figure 1.18 (Chapter 1) shows the Hoover Dam Bridge over
stiks02 [169]

Answer: 7.436 s

Explanation:

This situation is related to vertical motion, specifically free fall and can be modelled by the following equation:

y=y_{o}+V_{o} t+\frac{gt^{2}}{2}  

Where:

y= 0m is the final height of the object (when it makes splash)

y_{o}=271 m  is the initial height of the object

V_{o}=0 m/s  is the initial velocity of the object (it was dropped)

g=-9.8m/s^{2}  is the acceleration due gravity (directed downwards)

t is the time since the objecct is dropped until it makes splash

0=y_{o}+0+\frac{gt^{2}}{2}  

Clearing t:

t=\sqrt{\frac{-2y_{o}}{g}}  

t=\sqrt{\frac{-2(271 m)}{-9.8m/s^{2}}}  

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6 0
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Sergeeva-Olga [200]

Answer:

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The potential of the smaller sphere will be:

V_S = \frac{kq}{r}

The potential of the larger sphere will be:

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V_L = \frac{k(Q - q)}{4r}

Since V_S = V_L = V,

\frac{k(Q - q)}{4r} = \frac{kq}{r}

=> Q - q = 4q

=> 5q = Q

q = 0.2Q

The fraction of the charge Q that rests on the smaller sphere is 0.2

The charge of the larger sphere is:

Q - q = Q - 0.2Q = 0.8Q

∴ The fraction of the total charge Q that rests on the larger sphere is 0.8

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If you travel at 3 m/s for 12 seconds, how far did you travel?
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Answer:

V=3 m/s

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S=?

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S=36meters

So distance you travel is 36meters.

4 0
3 years ago
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