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Ivenika [448]
3 years ago
11

A car is traveling at 30 m/s in a straight line. The drive applies the brakes for 3 seconds and the car slows down to 12 m/s. Wh

at is the acceleration of the car?
Physics
1 answer:
jeyben [28]3 years ago
3 0

Answer:

The answer to your question is: -6m/s²

Explanation:

Data

vo = 30 m/s

vf = 12 m/s

t = 3 s

a = ?

Formula

vf = vo + at

a = (vt - vo) / t

Process

                a = (12 - 30) / 3                substitution

                a = -18 / 3                         simplify

                a = -6 m/s²                        result, is negative because the car

                                                          is slowing down.

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I NEED HELP PLEASE, THANKS! :)
Zina [86]

Answer:

charge C = greatest net force

charge B = the smallest net force

ratio  = 9 : 1

Explanation:

we know that in Electrostatic Forces, when 2 charges are at same sign then they repel each other and if they are different signed charges then they attract each other

so as per Coulomb's formula of Electrostatic Forces

F = \frac{k\ q_1\ q_2}{r^2}     .....................1

and here k is 9 × 10^9 N.m²/c² and we consider each charge at distance d

so two charge force at A to B is

F1 = \frac{k\ q^2}{d^2}

and force between charges at A to C, at 2d distance

F1 = \frac{k\ q^2}{(2d)^2}  =  \frac{k\ q^2}{4d^2}

force between charges at A to D,  3d distance

F1 = \frac{k\ q^2}{(3d)^2}  = \frac{k\ q^2}{9d^2}  

so

Charge a It receives force to the left from b and c and to the right from d

so at a will be

F(a)  = -F1 - F2 + F3             ....................2

put here value

F(a) = -\frac{k\ Q^2}{d^2}-\frac{k\ Q^2}{4d^2}+\frac{k\ Q^2}{9d^2}

solve it

F(a) = \frac{k\ q^2}{d^2}(-1-\frac{1}{4}+\frac{1}{9})  

F(a) = -\frac{41}{36}\ F1   = 1.13 F1  

and

Charge b It  receives force to the right from a and d and to the left from c

F(b) = F1 - F1 + F2            ....................3

F(b)  =  \frac{k\ q^2}{d^2}-\frac{k\ q^2}{d^2}+\frac{k\ q^2}{4d^2}    

F(b)  = \frac{1}{4} \ F1    =  0.25 F1

and

Charge c It receives forces to the right from all charges.

F(c) = F2 + F 1 + F 1      ....................4

F(c) = \frac{k\ q^2}{4d^2}+\frac{k\ q^2}{d^2}+\frac{k\ q^2}{d^2}      

F(c) =  \frac{9}{4} \ F1   = 2.25 F1

and

Charge d It receives forces to the left from all charges

F(d) = - F3 - F2 -F 1      ....................5

F(d) = -\frac{k\ q^2}{9d^2}-\frac{k\ q^2}{4d^2}-\frac{k\ q^2}{d^2}  

so

F(d) = -\frac{49}{36} \ F1    = 1.36 F1

and

now we get here ratio of the greatest to the smallest net force that is

ratio = \frac{2.25}{0.25}

 ratio  = 9 : 1

5 0
3 years ago
Metallic bonds are responsible for many properties of metals, such as conductivity. Why is this possible? (1 point)
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Answer:

The bonds can shift because valence electrons are held loosely and more freely

Explanation:

Please give brainliest if you can,have a good day<3 :)

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a force of 50 newtons pulls a rope attached to a 150 newton sled across a horizontal surface at a constant velocity of 5 meters
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Answer:

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infrared radiation from young stars can pass through the heavy dust clouds surrounding them, allowing astronomers here on earth
Mashcka [7]

The wavelength of the infrared radiation is  λ = 3.174×10^{-5}m.

<h3>What is infrared radiation?</h3>

An infrared telescope is tuned to detect infrared radiation with a frequency of 9.45 THz.

We know that,

1 THz = 10¹² Hz

So,

f = 9.45 × 10¹² Hz

We need to find the wavelength of the infrared radiation.

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λ = 3×10^{8}/9.45×10^{12}

λ = 3.174 ×10^{-5} m

The term "infrared radiation" (IR) refers to a part of the electromagnetic radiation spectrum with wavelengths between about 700 nanometers (nm) and one millimeter (mm). Longer than visible light waves but shorter than radio waves are infrared waves.

Electromagnetic radiation with wavelengths longer than those of visible light is known as infrared, also known as infrared light. Since it is undetectable to the human eye, The typical range of wavelengths considered to be infrared (IR) is from about 1 millimeter to the nominal red edge of the visible spectrum, or about 700 nanometers.

To learn more about infrared radiation from the given link:

brainly.com/question/13163856

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5 0
1 year ago
Describe two sources of earth's energy that are not produced
Anvisha [2.4K]
Here is the answer. Two sources of Earth's energy that are not produced would be Cosmic rays and Tidal Energy. Cosmic rays <span>are high-energy protons and atomic nuclei that come from outside the solar system. Whereas, tidal energy is the energy produced by both the moon (2/3) and the sun (1/3). Hope this answers your question.</span>
6 0
3 years ago
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