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3 years ago

14

Which statement bet explains the relationship between the electric force between two charged objects and the distance between th

em

1 answer:

Nataliya [291]3 years ago

4 0

Coulomb's Law: Force = k x q1x q2 divided distance square
where k=9x10^9 , q1 and q2 are the charge
So if you distance is halved, your force is stronger by 4 times
and if you distance is doubled, your force is 1/4
Ask me again if you aren't clear :)

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At an accident scene on a level road, investigators measure a car's skid mark to be 84 m long. It was a rainy day and the coeffi

Flura [38]

The given data is incomplete. The complete question is as follows.

At an accident scene on a level road, investigators measure a car's skid mark to be 84 m long. It was a rainy day and the coefficient of friction was estimated to be 0.36. Use these data to determine the speed of the car when the driver slammed on (and locked) the brakes. (why does the car's mass not matter?)

Explanation:

Let us assume that v is the final velocity and u is the initial velocity of the car. Let s be the skid marks and $\mu$ be the friction coefficient and m be the mass of car.

Hence, the given data is as follows.

v = 0, s = 84 m, $\mu$ = 0.36

According to Newton's law of second motion the expression for acceleration is as follows.

F = ma

$-\mu N$ = ma

$-\mu mg$ = ma

a = $-\mu g$

Also,

$v^{2} = u^{2} + 2as$

$(0)^{2} = u^{2} + 2(-\mu g)s$

$u^{2} = 2(\mu g)s$

= $\sqrt{2(0.36)(9.81 m/s^{2})(84 m)}$

= 24.36 m/s

Thus, we can conclude that the speed of the car when the driver slammed on (and locked) the brakes is 24.36 m/s.

4 0

3 years ago

The diagram shows the top view of a 65-kg student at point A on an amusement park ride. The ride spins the student in a horizont

Elza [17]

Answer:

1923 N

Explanation:

From the question given above, the following data were obtained:

Mass (m) = 65 Kg

Radius (r) = 2.5 m

Velocity (v) = 8.6 m/s

Centripetal force (F) =?

The centripetal force, F, can be obtained by using the following formula:

F = mv²/r

F = 65 × 8.6² / 2.5

F = 65 × 73.96 / 2.5

F = 4807.4 / 2.5

F = 1922.96 ≈ 1923 N

Thus, the magnitude of the centripetal's force acting on the student is approximately 1923 N

3 0

3 years ago

A physics professor wants to perform a lecture demonstration of Young's double-slit experiment for her class using the 633-nm li

Alexeev081 [22]

Note: if the professor wants the distance between the m = 0 and m = 1 maxima to be 25 cm

Answer:

d = 1.0128×10⁻⁵m

Explanation:

given:

length L = 4.0m

maximum distance between m = 0 and m = 1 , y = 25cm = 0.25m

wavelength λ = 633nm = 633×10⁻⁹m

note:

dsinθ = mλ (constructive interference)

where d is slit seperation, θ is angle of seperation , m is order of interference , and λ is wavelength

for small angle

sinθ ≈ tanθ

$d (\frac{y}{L}) =$ mλ

$d (\frac{y}{L}) = (1)(633nm)$

$d(\frac{0.25}{4} ) = (1)(633nm)$

d = 1.0128×10⁻⁵m

6 0

3 years ago

An engineering team has come to the stage in the engineering design

SIZIF [17.4K]

Answer: D

Explanation: read the answers above and compare to your test

6 0

3 years ago

A circuit containing an inductor and a capacitor in series is designed to have a resonant frequency of 4511 Hz. If the inductor

OLEGan [10]

Answer:

(e) $6.835\times 10^{-7}F$

Explanation:

At resonance we know that $X_l=X_C$

That is $\omega L=\frac{1}{\omega C}$

$\omega ^2=\frac{1}{LC}$

$\omega =\frac{1}{\sqrt{LC}}$

$f=\frac{1}{2\pi \sqrt{LC}}$

We have given resonance frequency f =4511 Hz and inductance L=1.82 mH

So $4511=\frac{1}{2\pi \sqrt{LC}}$

$LC=\frac{1}{4\pi ^2\times 4511^2}$

$LC=1.244\times 10^{-9}$

$C=\frac{1.244\times 10^{-9}}{1.82\times 10^{-3}}=0.6835\times 10^{-6}=6.835\times 10^{-7}F$

So option e is the correct answer

3 0

3 years ago