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Daniel [21]
3 years ago
14

Mrs. Smith walks from her house 11.2 m to the grocer. Then she walks from the grocer 5.7 m to the pet store. Then she walks 32.7

m from the pet store to her daughter's house. Mrs. Smith's house is directly 23.5 m away from her daughter's house. If the trip took 17 minutes, what was the average velocity of Mrs. Smith?
Physics
1 answer:
Liono4ka [1.6K]3 years ago
8 0

Average velocity is defined as the ratio of displacement and time

Here we know that the total distance is the sum of all given distances but we need the displacement in order to find the velocity

So here displacement is straight line distance or direct distance from her home to her daughter home

It is given as 23.5 m

and it took total time of 17 minutes

so average velocity is

v = \frac{displacement}{time}

v = \frac{23.5}{17\times 60}

v = 0.023 m/s

So average velocity of Mrs Smith is 0.023 m/s

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A nonconducting sphere of diameter 10.0 cm carries charge distributed uniformly inside with charge density of +5.50 µC/m3 . A pr
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Answer:

t = 2.58*10^-6 s

Explanation:

For a nonconducting sphere you have that the value of the electric field, depends of the region:

rR:\\\\E=k\frac{Q}{r^2}

k: Coulomb's constant = 8.98*10^9 Nm^2/C^2

R: radius of the sphere = 10.0/2 = 5.0cm=0.005m

In this case you can assume that the proton is in the region for r > R. Furthermore you use the secon Newton law in order to find the acceleration of the proton produced by the force:

F=m_pa\\\\qE=m_pa\\\\k\frac{qQ}{r^2}=m_pa\\\\a=k\frac{qQ}{m_pr^2}

Due to the proton is just outside the surface you can use r=R and calculate the acceleration. Also, you take into account the charge density of the sphere in order to compute the total charge:

Q=\rho V=(5.5*10^{-6}C/m^3)(\frac{4}{3}\pi(0.05m)^3)=2.87*10^{-9}C\\\\a=(8.98*10^9Nm^2/C^2)\frac{(1.6*10^{-19}C)(2.87*10^{-9}C)}{(1.67*10^{-27}kg)(0.05m)^2}=9.87*10^{11}\frac{m}{s^2}

with this values of a you can use the following formula:

a=\frac{v-v_o}{t}\\\\t=\frac{v-v_o}{a}=\frac{2550*10^3m/s-0m/s}{9.87*10^{11}m/s^2}=2.58*10^{-6}s

hence, the time that the proton takes to reach a speed of 2550km is 2.58*10^-6 s

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An office window has dimensions 3.1 m by 2.1 m. As a result of the passage of a storm, the outside air pressure drops to 0.954 a
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Answer:

Net forces which pushes the window is 30342.78 N.

Explanation:

Given:

Dimension of the office window.

Length of the window = 3.1 m

Width of the window = 2.1 m

Area of the window = (3.1\times 2.1) = 6.51\ m^2

Difference in air pressure = Inside pressure - Outside pressure

                                           = (1.0-0.954) atm = 0.046 atm

Conversion of the pressure in its SI unit.

⇒  1 atm = 101325 Pa

⇒ 0.046 atm = 0.046\times 101325 =4660.95 Pa

We have to find the net force.

We know,

⇒ Pressure = Force/Area

⇒ Pressure=\frac{Force }{Area}

⇒ Force =Pressure\times Area

⇒ Plugging the values.

⇒ Force =4660.95\times 6.51

⇒ Force=30342.78 Newton (N)

So,

The net forces which pushes the window is 30342.78 N.

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