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Digiron [165]
3 years ago
14

{WORTH BRAINLIEST WITH AN EXPLANATION} Janine is a police officer patrolling a street. She drives 2.2 km east, then 4.4 km west,

then 1.7 km east. If east is the positive direction, which of the following equations are accurate? (select all that apply)

Physics
1 answer:
icang [17]3 years ago
5 0

Answer:

Options (3) and (6)

Explanation:

Janine drives 2.2 km East.

Since, distances measured in the East are positive,

Displacement = 2.2 km

Then she drives 4.4 km west

Displacement = -4.4 km

Followed by 1.7 km in the East

Displacement = 1.7 km

Total displacement = 2.2 - 4.4 + 1.7 = -0.5 km

\overrightarrow{d}=-0.5 km

Total distance covered by Janine = 2.2 + 4.4 + 1.7 = 8.3 km

d = 8.3 km

Therefore, Options (3) and (6) will be the answer.

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(#1). (D).

(#2). (C).

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3 years ago
Question 1
svetlana [45]

300

Explanation:

100 x 3 =300 simple and easy

7 0
2 years ago
A 1.50-mm-diameter glass sphere has a charge of + 1.60 nC. What speed does an electron need to orbit the sphere 1.60 mm above th
saveliy_v [14]

Answer :

Velocity will be 3.28\times 10^{-11}m/sec

Explanation:

We have given glass surface has a diameter of 1.5 mm

And charge q = 1.60 nC

Radius of electrons orbit r = height of electron above surface + radius of sphere  = =1.6+\frac{1.5}{2}=2.35mm = 0.00235m

Force on electron is given by F=\frac{1}{4\pi \epsilon _0}\frac{qe}{r^2}, here q is charge on sphere and e is charge on electron

F=\frac{1}{4\pi \epsilon _0}\frac{qe}{r^2}=\frac{kqe}{r^2}=\frac{9\times 10^9\times 1.6\times 10^{-9}\times 1.6\times 10^{-19}}{0.00235^2}=4.172\times 10^{-13}N

This force work as centripetal force

So F=\frac{mv^2}{r}

4.172\times 10^{-13}=\frac{9.11\times 10^{-31}v^2}{0.00235}

v = =0.0328\times 10^{-9}=3.28\times 10^{-11}m/sec

   

6 0
3 years ago
Define investigation to show its scientific meaning.
Murrr4er [49]

Answer:

the action of investigating something or someone; formal or systematic examination or research.

Explanation:

This definition is provided by Oxford Languages

7 0
3 years ago
An AM radio station broadcasts isotropically (equally in all directions) with an average power of 3.40 kW. A receiving antenna 6
lara [203]

To solve the problem we will apply the concepts related to the Intensity as a function of the power and the area, as well as the electric field as a function of the current, the speed of light and the permeability in free space, as shown below.

The intensity of the wave at the receiver is

I = \frac{P_{avg}}{A}

I = \frac{P_{avg}}{4\pi r^2}

I = \frac{3.4*10^3}{4\pi(4*1609.34)^2} \rightarrow 1mile = 1609.3m

I = 6.529*10^{-6}W/m^2

The amplitude of electric field at the receiver is

I = \frac{E_{max}^2}{2\mu_0 c}

E_{max}= \sqrt{2I\mu_0 c}

The amplitude of induced emf by this signal between the ends of the receiving antenna is

\epsilon_{max} = E_{max} d

\epsilon_{max} = \sqrt{2I \mu_0 cd}

Here,

I = Current

\mu_0 = Permeability at free space

c = Light speed

d = Distance

Replacing,

\epsilon_{max} = \sqrt{2(6.529*10^{-6})(4\pi*10^{-7})(3*10^{8})(60.0*10^{-2})}

\epsilon_{max} = 0.05434V

Thus, the amplitude of induced emf by this signal between the ends of the receiving antenna is 0.0543V

6 0
4 years ago
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