At one instant, a 17.0-kg sled is moving over a horizontal surface of snow at 4.10 m/s. After 6.15 s has elapsed, the sled stops. Use a momentum approach to find the magnitude of the average friction force acting on the sled while it was moving.

Answers

Answer 1
Answer:

Answer:

force = 11.33 kg-m/s^(2)

Explanation:

given data:

sled mass = 17.0 kg

inital velocity (U) = 4.10 m/s

elapsed time (T) 6.15 s

final velocity (V) = 0

final momentum P2 = 0

Initial momentum of sledge is

P_(1)=mU

P_(1)= 17.0 * 4.10 = 69.7 kg- m/s

from newton second law of motion

F=(\Delta P)/(\Delta t)

F = (P_(1)-P_(2))/(T)

Kgm/s^2

F = (69.7-0)/(6.15)= 11.33[tex]kg-m/s^(2)[/tex]


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Someone help Match the definition to the term.

1.

the noun that follows the verb and answers the questions

whom or what

linking verb

2. a word that expresses action

sentence

3. who or what the sentence is about

predicate noun

4.

a noun that indicates to or for whom or what something is

done

predicate adjective

5. follows a linking verb and renames the subject

direct object

subject

6. follows a linking verb and describes a subject

verb that joins a subject and a predicate noun or predicate

7.

adjective

indirect object

8. expresses a complete thought

verb

Answers

Answer:

Explanation:

No 1 is linked to the option provided at no 5.

A direct object is the noun that follows the verb and answers the question.

No 2 is linked to the option provided at no 8.

A verb is a word that expresses action.

No 3 is linked to the option provided at no 5.

The subject is what or whom the sentence is about.

No 5 is linked to the option provided at no 3.

A predicate noun follows a linking verb, and renames the subject.

No 6 is linked to the option provided at no 4.

A predicate adjective follows a linking verb, and describes a subject.

No 7 is linked to the option provided at no 1.

A linking verb joins a subject and a predicate.

No 8 is linked to the option provided at no 2.

A sentence expresses a complete thought....

Final answer:

The terms like direct object, verb, subject, indirect object, predicate noun, predicate adjective, and sentence are explained in the context of English grammar and matched to their definitions.

Explanation:

Here are the correct matches for the terms:

  1. The noun that follows the verb and answers the questions whom or what is typically referred to as the direct object.
  2. A word that expresses action is a verb.
  3. Who or what the sentence is about is known as the subject.
  4. A noun that indicates to or for whom or what something is done is an indirect object.
  5. Follows a linking verb and renames the subject is a predicate noun.
  6. Follows a linking verb and describes a subject is a predicate adjective.
  7. Expresses a complete thought is the definition for a sentence in the context of English grammar.

Learn more about English Grammar Terms here:

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Points A, B, and C are at the corners of an equilateral triangle of side 8 m. Equal positive charges of 4 mu or micro CC are at A and B. (a) What is the potential at point C? 8.990 kV * [2.5 points] 2 attempt(s) made (maximum allowed for credit = 5) [after that, multiply credit by 0.5 up to 10 attempts] 8.990 OK (b) How much work is required to bring a positive charge of 5 mu or micro CC from infinity to point C if the other charges are held fixed? .04495 J * [2.5 points] 1 attempt(s) made (maximum allowed for credit = 5) [after that, multiply credit by 0.5 up to 10 attempts] .04495 OK (c) Answer parts (a) and (b) if the charge at B is replaced by a charge of -4 mu or micro CC. Vc= kV [2.5 points] 0 attempt(s) made (maximum allowed for credit = 5) [after that, multiply credit by 0.5 up to 10 attempts] W =

Answers

Answer:

a) 8.99*10³ V  b) 4.5*10⁻² J c) 0 d) 0

Explanation:

a)

  • The electrostatic potential V, is the work done per unit charge, by the electrostatic force, producing a displacement d from infinity (assumed to be the reference zero level).
  • For a point charge, it can be expressed as follows:

        V =(k*q)/(d)

  • As the electrostatic force is linear with the charge (it is raised to first power), we can apply superposition principle.
  • This means that the total potential at a given point, is just the sum of the individual potentials due to the different charges, as if the others were not there.
  • In our case, due to symmetry, the potential, at any corner of the triangle, is just the double of the potential due to the charge located at  any other corner, as follows:

        V = (2*q*k)/(d) = (2*8.99e9N*m2/C2*4e-6C)/(8m) =\n \n V= 8.99e3 V

  • The potential at point C is 8.99*10³ V

b)

  • The work required to bring a positive charge of 5μC from infinity to the point C, is just the product of the potential at this point times the charge, as follows:

        W = V * q = 8.99e3 V* 5e-6C = 4.5e-2 J

  • The work needed is 0.045 J.

c)

  • If we replace one of the charges creating the potential at the point  C, by one of the same magnitude, but opposite sign, we will have the following equation:

       V = (8.99e9N*m2/C2*(4e-6C))/(8m)  + ((8.99e9N*m2/C2*(-4e-6C))/(8m)) = 0

  • This means that the potential due to both charges is 0, at point C.

d)

  • If the potential at point C is 0, assuming that at infinity V=0 also, we conclude that there is no work required to bring the charge of 5μC from infinity to the point C, as no potential difference exists between both points.

Assume that the speed of light in a vacuum has the hypothetical value of 18.0 m/s. A car is moving at a constant speed of 14.0 m/s along a straight road. A home owner sitting on his porch sees the car pass between two telephone poles in 8.89 s. How much time does the driver of the car measure for his trip between the poles?

Answers

Answer:

Observed time, t = 5.58 s  

Explanation:

Given that,

Speed of light in a vacuum has the hypothetical value of, c = 18 m/s

Speed of car, v = 14 m/s along a straight road.

A home owner sitting on his porch sees the car pass between two telephone poles in 8.89 s.

We need to find the time the driver of the car measure for his trip between the poles. The relation between real and observed time is given by :

T=\frac{t}{\sqrt{1-(v^2)/(c^2)} }

t is observed time.

t=T* \sqrt{1-(v^2)/(c^2)} \n\nt=8.89* \sqrt{1-(14^2)/(18^2)} \n\nt=5.58\ s

So, the time observed by the driver of the car measure for his trip between the poles is 5.58 seconds.

The planet uranus is tilted nearly on its side so that its axis or rotation is only 8 degress abway from its orbit plane. if you lived at latitude 45 degrees on uranus for what fraction of the uranian year would answer?

Answers

The rotation of Uranus, like that of Venus, is retrograde and its axis of rotation is inclined almost ninety degrees above the plane of its orbit. During its orbital period of 84 years one of the poles is permanently illuminated by the Sun while the other remains in the shade. Exactly its rotation period is equivalent to 17 hours and 14 Earth minutes and its translation period is equivalent to 84 years, 7 days and 9 Earth hours.

Only a narrow band around the equator experiences a rapid cycle of day and night, but with the Sun very low on the horizon as in the polar regions of the Earth. On the other side of the orbit of Uranus, the orientation of the poles in the direction of the Sun is inverse. Each pole receives about 42 years of uninterrupted sunlight, followed by 42 years of darkness. Therefore an observer at latitude of 45 degrees in Uranus will probably experience a long winter night that is equivalent to one third of the year uranium.

Dr. John Paul Stapp was a U.S. Air Force officer who studied the effects of extreme acceleration on the human body. On December 10, 1954, Stapp rode a rocket sled, accelerating from rest to a top speed of 282 m/s (1015 km/h) in 5.2 s and was brought jarringly back to rest in only 1 s. Calculate his (a) magnitude of acceleration in his direction of motion and (b) magnitude of acceleration opposite to his direction of motion. Express each in multiples of g (9.80 m/s2) by taking its ratio to the acceleration of gravity. g g

Answers

Answer:

    a = 5.53 g ,   a = -15g

Explanation:

This is an exercise in kinematics.

a) Let's look for the acceleration

         as part of rest v₀ = 0

          v = v₀ + a t

           a = v / t

           a = 282 / 5.2

          a = 54.23 m / s²

in relation to the acceleration of gravity

          a / g = 54.23 / 9.8

          a = 5.53 g

b) let's look at the acceleration to stop

         va = 0

         0 = v₀ -2 a y

         a = vi / y

         a = 282/2 1

         a = 141 m /s²

         a / G = 141 / 9.8

          a = -15g

Two students have the same velocity during a race. Colin has a mass of 80 kg while Kara has a mass of 80 kg. If Kara doubled her speed how does her new momentum compare to Colin’s?

Answers

Kara has twice the momentum as Colin