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tester [92]
2 years ago
7

A 1000-kg car is driving toward the north along a straight horizontal road at a speed of 20.0 m/s. The driver applies the brakes

and the car comes to a rest uniformly in a distance of 120 m. What are the magnitude and direction of the net force applied to the car to bring it to rest
Physics
1 answer:
OlgaM077 [116]2 years ago
8 0

Answer:

Explanation:

mass of car, m = 1000 kg

initial velocity, u = 20 m/s

final velocity, v = 0 m/s

distance, s = 120 m

Let a be the acceleration of motion

use third equation of motion

v² = u² + 2 as

0 = 20 x 20 + 2 x a x 120

a = - 1.67 m/s²

Let F be the force

Force, F  mass x acceleration

F = - 1000 x 1.67

F = - 1666.67 N

The direction of force is towards south and the magnitude of force is 1666.67 N.

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A ray of light crosses a boundary between two transparent materials. The medium the ray enters has a larger index of refraction.
Margarita [4]

Answer:

True, True, False, False, False, False.

Explanation:

The refraction index of a material is given by the formula n=c/v, where c is the speed of light in vacuum and v the speed of light in the material. If a ray of light crosses a boundary between two transparent materials and the medium the ray enters has a larger index of refraction it means that in this new medium the speed of light is smaller than on the other one, and then its wavelength is also reduced since f must remain the same (and \lambda=v/f), otherwise there is a discontinuity on number of vibrations per second, which cannot happen. So we know that:

1) The wavelength of the light decreases as it enters into the medium with the greater index of refraction. True.

2) The frequency of the light remains constant as it transitions between materials. True.

3) The speed of the light remains constant as it transitions between materials. False.

4) The speed of the light increases as it enters the medium with the greater index of refraction. False.

5) The frequency of the light decreases as it enters into the medium with the greater index of refraction. False.

6) The wavelength of the light remains constant as it transitions between materials.  False.

7 0
2 years ago
A ray of yellow light ( f = 5.09 × 1014 hz) travels at a speed of 2.04 × 108 meters per second in
SashulF [63]
Velocity = fλ

where f is frequency in Hz, and λ is wavelength in meters.

<span>2.04 * 10⁸ m/s =  5.09 * 10¹⁴  Hz   *  λ </span>

<span>(2.04 * 10⁸ m/s) / (5.09 * 10¹⁴  Hz ) = λ </span>

<span>4.007*10⁻⁷  m =  λ </span>

<span>The wavelength of the yellow light = 4.007*10⁻⁷  m<span> </span></span>
6 0
2 years ago
What is the best approximate value for the elastic potential energy (EPE) of the spring elongated by 3.0 meters?
DIA [1.3K]

The elastic potential energy of the spring is 6.8 J

Explanation:

The elastic potential energy of a compressed/stretched spring is given by the equation:

E=\frac{1}{2}kx^2

where

k is the spring constant

x is the elongation of the spring

The spring constant of the spring in this problem can be found by keeping in mind the relationship between force (F) and elongation (x) (Hooke's law):

F=kx

By looking at the graph and comparing it with the formula, we realize that the slope of the force-elongation graph corresponds to the spring constant. Therefore in this case,

k=\frac{15.0-0}{10.0-0}=1.5 N/m

Therefore when the spring has a elongation of x=3.0 m, its potential energy is

E=\frac{1}{2}(1.5)(3.0)^2=6.8 J

Learn more about potential energy:

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

3 0
2 years ago
) a charge of 6.15 mc is placed at each corner of a square 0.100 m on a side. determine the magnitude and direction of the force
Nana76 [90]
Because charges are positioned on a square the force acting on one charge is the same as the force acting on all others. 
We will use superposition principle. This means that force acting on the charge is the sum of individual forces. I have attached the sketch that you should take a look at.
We will break down forces on their x and y components:
F_x=F_3+F_2cos(45^{\circ})
F_y=F_1+F_2sin(45^{\circ})
Let's figure out each component:
F_1=\frac{1}{4\pi \epsilon}\frac{q^2}{a^2}\\&#10;F_3=\frac{1}{4\pi \epsilon}\frac{q^2}{a^2}\\
F_2=\frac{1}{4\pi \epsilon}\frac{q^2}{(\sqrt{2}a)^2}
Total force acting on the charge would be:
F=\sqrt{F_x^2+F_y^2}
We need to calculate forces along x and y axis first( I will assume you meant micro coulombs, because otherwise we get forces that are huge).
F_x=F_3+F_2cos(45^{\circ})=\frac{1}{4\pi \epsilon}\frac{q^2}{a^2}+\frac{1} {4\pi \epsilon}\frac{q^2}{(\sqrt{2}a)^2}\cdot\cos(45)=46N
F_y=\frac{1}{4\pi \epsilon}\frac{q^2}{(\sqrt{2}a)^2}\cdot sin(45)+\frac{1}{4\pi \epsilon}\frac{q^2}{a^2}=46N
Now we can find the total force acting on a single charge:
F=\sqrt{F_x^2+F_y^2}=\sqrt{46^2+46^2}=65N
As said before, intensity of the force acting on charges is the same for all of them.

5 0
2 years ago
The alpha particles leave visible tracks in the cloud chamber because
julia-pushkina [17]
<h2>Answer: Ionization </h2>

The inner atmosphere of a <u>cloud chamber</u> is composed of an easily ionizable gas, this means that little energy is required to extract an electron from an atom. <u>This gas is maintained in the supercooling state, so that a minimum disturbance is enough to condense it</u> in the same way as the water is frozen.

<h2>Then, when a charged particle with enough energy interacts with this gas, it <u>ionizes</u> it. </h2>

This is how alpha particles are able to ionize some atoms of the gas contained inside the chamber when they cross the cloud chamber.

These ionized atoms increase the surface tension of the gas around it allowing it to immediately congregate and condense, making it easily distinguishable inside the chamber like a <u>small cloud</u>. In this way, it is perfectly observable the path the individual particles have traveled, simply by observing the cloud traces left in the condensed gas.

6 0
2 years ago
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