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3 years ago

an automobile is traveling 65km/h the brakes decelerate it at a rate of -6.0 m/s^2 how long will it take to stop the car?

Physics

1 answer:

joja [24]3 years ago

6 0

Explanation:

It is given that,

Initial speed of the automobile, u = 65 km/hr =

Final speed of the automobile, v = 0

Deceleration of the automobile, $a=-6\ m/s^2$

We need to find the distance covered by the car as it comes to rest. It can be calculated using third equation of motion as :

$v^2-u^2=2ax$

$a=\dfrac{v^2-u^2}{2x}$

$a=\dfrac{0-(18.05)^2}{2\times (-6)}$

$a=27.15\ m/s^2$

So, the acceleration of the car is $27.15\ m/s^2$ . Hence, this is the required solution.

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an electric current of 285.0 ma flows 674. milliseconds. Calculate the amount of electric charge transported. Be sure your answe

yulyashka [42]

Answer:

The amount of electric charge transported = 0.192 C

Explanation:

Electric Charge: This is defined as the product of electric current and time in an electric circuit, The S.I unit of electric charge is Coulombs (C)

Q = It..................... Equation 1

Where Q = Electric charge, I = electric current, t = time.

Given: I = 285 mA, t = 674 milliseconds.

Conversion: (i) Convert from 285 mA to A = (285/1000) A = 0.285 A

(ii) convert from 674 milliseconds to seconds = (674/1000) s = 0.674 s

Substituting these values into equation 1

Q = 0.285 × 0.674

Q = 0.192 C

Therefore the amount of electric charge transported = 0.192 C

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3 years ago

Which particles contribute to the net charge and how does each change the net charge?

Licemer1 [7]

The two subatomic particles that contribute to the net charge of an ion are electrons and protons.

Atom is the smallest possible amount of matter which still retains its identity as a chemical element, now known to consist of a nucleus surrounded by electrons.

The atom is made up of three components called subatomic particles as follows;

Protons
Electrons
Neutrons

The proton is the positively charged subatomic particle forming part of the nucleus of an atomwhile the electron is the subatomic particle having a negative charge and orbiting the nucleus.

This suggests that the two subatomic particles that contribute to the net charge of an ion are electrons and protons. That is;

Net charge = protons - electrons

Learn more about subatomic particles at:brainly.com/question/13303285

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2 years ago

An earth scientist who studies hydrosphere

ipn [44]

Oceanographer? I think that is what it is.

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3 years ago

Calculate, for the judge, how fast you were going in miles per hour when you ran the red light because it appeared Doppler-shift

sammy [17]

Answer:

The doppler effect equation is:

$f' = \frac{v +v0}{v - vs}*f$

In the equation we have frequencies, but then we have the wavelengths of the lights, remember the relation:

v = f*λ

then:

f = v/λ

and v is the speed of light, then:

f = c/λ

where:

f' is the observed frequency, in this case, is equal to f = (3*10^17nm/s)/550 nm

f is the real frequency, in this case, is (3*10^17nm/s)/650 nm

vs is the speed of the source, in this case, the source is not moving, then vs = 0 m/s.

v is the speed of the wave, in this case, is equal to the speed of light, v = 3*10^8 m/s

v0 is your speed, this is what we want to find.

Replacing those quantities in the equation, we get:

(3*10^17nm/s)/550 = (3*10^8 m/s + v0)/(3*10^8 m/s)*(3*10^17nm/s)/650 nm

(650nm)/(550nm) = (3*10^8 m/s + v0)/(3*10^8 m/s)

1.182*(3*10^8 m/s) = (3*10^8 m/s + v0)

1.182*(3*10^8 m/s) - (3*10^8 m/s) = v0 = 54,600,000 m/s

So your speed was 54,600,000 m/s, which is a lot.

6 0

3 years ago

A +1.0 nC charge is at x = 0 cm, a -1.0 nC charge is at x = 1.0 cm and a 4.0 nC at x= 2 cm. What is the electric potential energ

lesantik [10]

Answer:

- 2.7 x 10^-6 J

Explanation:

q1 = 1 nC at x = 0 cm

q2 = - 1 nC at x = 1 cm

q3 = 4 nC at x = 2 cm

The formula for the potential energy between the two charges is given by

$U=\frac{Kq_{1}q_{2}}{r}$

where r be the distance between the two charges

By use of superposition principle, the total energy of the system is given by

$U = U_{1,2}+U_{2,3}+U_{3,1}$

$U=\frac{Kq_{1}q_{2}}{0.01}+\frac{Kq_{2}q_{3}}{0.01}+\frac{Kq_{3}q_{1}}{0.02}$

$U=-\frac{9\times10^{9}\times 1\times10^{-9}\times 1\times10^{-9}}}{0.01}-\frac{9\times10^{9}\times 1\times10^{-9}\times 4\times10^{-9}}}{0.01}+-\frac{9\times10^{9}\times 1\times10^{-9}\times 4\times10^{-9}}}{0.02}$

U = - 2.7 x 10^-6 J

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3 years ago