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Nookie1986 [14]
3 years ago
9

Consult Multiple Concept Example 10 in preparation for this problem. Traveling at a speed of 15.4 m/s, the driver of an automobi

le suddenly locks the wheels by slamming on the brakes. The coefficient of kinetic friction between the tires and the road is 0.680. What is the speed of the automobile after 1.11 s have elapsed? Ignore the effects of air resistance.
Physics
1 answer:
ella [17]3 years ago
7 0

Answer: 8.07 m/s

Explanation:

Applying the definition of acceleration, as the rate of change of velocity, we can extract the value of the velocity v at a given time t, as follows:

v = v₀ + a t (1)

As we already know v₀ and t, our  single unknown is the acceleration a.

Once the friver slammed on the brakes, there exists an external force, due to friction, which is the only external force acting in the horizontal direction, producing a deceleration.

This friction force, is equal to the product of  the coeficient of  kinetic friction, times the normal force.

In this case, the normal force is numerically equal to the gravity force, which we call weight.

This force opposes to the direction of the initial velocity, as it always opposes to the relative movement between both surfaces in contact.

Applying Newton's 2nd Law, we get:

Fnet = ma ⇒ Ff = m a ⇒ -μk . m . g = m.a

Solving for a :

a = -μk . g = -0.68 . 9.8 m/s² = -6.66 m/s²

Replacing in (1)

v = 15.4 m/s +(-6.66) m/s². 1.11 s = 8.07 m/s

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Five different forces act on an object. Is it possible for the net force on the object to be zero?
bazaltina [42]
No because there must be an even # if their is an even amount one of the forces isn’t being cancelled
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3 years ago
Two charged particles separated by a distance of = 3 and experienced electrostatic forces of = 60 . What would be this force if
klemol [59]

Answer: 539.4 N

Explanation:

Let's begin by explaining that Coulomb's Law establishes the following:  

"The electrostatic force F_{E} between two point charges q_{1} and q_{2} is proportional to the product of the charges and inversely proportional to the square of the distance d that separates them, and has the direction of the line that joins them"

What is written above is expressed mathematically as follows:

F_{E}= K\frac{q_{1}.q_{2}}{d^{2}} (1)

Where:

F_{E}=60 N  is the electrostatic force

K=8.99(10)^{9} Nm^{2}/C^{2} is the Coulomb's constant  

q_{1} and q_{2} are the electric charges

d=3 m is the separation distance between the charges  

Then:

60 N= 8.99(10)^{9} Nm^{2}/C^{2}\frac{q_{1}.q_{2}}{(3 m)^{2}} (2)

Isolating q_{1} and q_{2}:

q_{1}q_{2}=6(10)^{-8} C^{2} (3)

Now, if we keep the same charges but we decrease the distance to d_{1}=1 m, (1) is rewritten as:

F_{E}=8.99(10)^{9} Nm^{2}/C^{2}\frac{6(10)^{-8} C^{2}}{(1 m)^{2}} (4)

Then, the new electrostatic force will be:

F_{E}= 539.4 N (5) As we can see, the electrostatic force is increased when we decrease the distance between the charges.

4 0
3 years ago
Calculate the east component of a resultant 32.5 m/s, 35.0° east of north.
ValentinkaMS [17]

Answer:

East component is: 18.64 m/s

Explanation:

If the resultant is 32.5 m/s directed 35 degrees east of north, then we use the sin(35) projection to find the east component of the velocity:

East component = 32.5 m/s * sin(35) = 18.64 m/s

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3 years ago
Which of the following is NOT considered electromagnetic radiation? x-rays radio waves cosmic rays ultraviolet light
OlgaM077 [116]

Answer:

Cosmic ray is not considered electromagnetic radiation.

Explanation:

Electromagnetic radiation refers to

  • It is Radiation that has both electric and magnetic fields and travels in waves.
  • Electromagnetic radiation can vary in strength from low energy to high energy.

X rays are the rays produced when a negatively charged electrode is heated by electricity and electrons are released, thereby producing energy. It is a type of radiation called EM waves.

Radio wave are wave from the portion of the electromagnetic spectrum at lower frequencies than microwaves. It is an EM wave.

Ultraviolet wave Invisible rays that are part of the energy that comes from the sun. It is an example of EM wave.

Cosmic ray is a high-speed particle either an atomic nucleus or an electron that travels through space.

Cosmic ray is not an electromagnetic wave.

Hence

Cosmic ray is not considered electromagnetic radiation

learn more about electromagnetic radiation here:

<u>brainly.com/question/13695751</u>

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4 0
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I WILL MARK BRAINLIEST!!ASAP!!! Wet Lab - Coulomb's Law lab from edge!!
snow_tiger [21]

Answer:

h

Explanation:

Coulomb's law, or Coulomb's inverse-square law, is an experimental law[1] of physics that quantifies the amount of force between two stationary, electrically charged particles. The electric force between charged bodies at rest is conventionally called electrostatic force or Coulomb force.[2] The law was first discovered in 1785 by French physicist Charles-Augustin de Coulomb, hence the name. Coulomb's law was essential to the development of the theory of electromagnetism, maybe even its starting point,[1] as it made it possible to discuss the quantity of electric charge in a meaningful way.[3]

The law states that the magnitude of the electrostatic force of attraction or repulsion between two point charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them,[4]

{\displaystyle F=k_{\text{e}}{\frac {q_{1}q_{2}}{r^{2}}}}{\displaystyle F=k_{\text{e}}{\frac {q_{1}q_{2}}{r^{2}}}}

Here, ke is Coulomb's constant (ke ≈ 8.988×109 N⋅m2⋅C−2),[1] q1 and q2 are the signed magnitudes of the charges, and the scalar r is the distance between the charges.

The force is along the straight line joining the two charges. If the charges have the same sign, the electrostatic force between them is repulsive; if they have different signs, the force between them is attractive.

Being an inverse-square law, the law is analogous to Isaac Newton's inverse-square law of universal gravitation, but gravitational forces are always attractive, while electrostatic forces can be attractive or repulsive.[2] Coulomb's law can be used to derive Gauss's law, and vice versa. In the case of a single stationary point charge, the two laws are equivalent, expressing the same physical law in different ways.[5] The law has been tested extensively, and observations have upheld the law on the scale from 10−16 m to 108 m.[5]

7 0
3 years ago
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