Que es la friccion <br> Cual es la primera ley de newton

Help please!!!!!!!!! I will mark brainliest!!!

suter [353]

Answer:

solving for: velocity

equation: velocity = distance / time

substitution: velocity = 1425 km / 12.5 hrs

answer: 114 km/hr

5 0

3 years ago

Light does bend in a gravitational field. Why is this bending not taken into consideration by surveyors who use laser beams as s

Goshia [24]

Answer:

Because the effect is not big enough to be noticeable.

Explanation:

The light is bent by gravitational fields, but the bend is not that big unless we are talking about objects with a massive amount of mass. To be noticed the bent, you need to stay far away from the object that causes the blend and the object also needs to be far away from the source of the light. For example, you can observe the blend in the light of a far-away star when the light travels close to the sun to reach earth, and the deflection will be around 1.75 arc-seconds. The deflection occurs also with light beams on the earth but the effect is too small to be taken into consideration.

8 0

3 years ago

A boy who exerts a 300-N force on the ice of a skating rink is pulled by his friend with a force of 75 N, causing the boy to acc

JulsSmile [24]

Answer:

Choice a. $70\; \rm N$ , assuming that the skating rink is level.

Explanation:

<h3>Net force in the horizontal direction</h3>

There are two horizontal forces acting on the boy:

The pull of his friend, and
Frictions.

The boy should be moving in the direction of the pull of his friend. The frictions on this boy should oppose that motion. Therefore, the frictions on the boy would be in the opposite direction of the pull of his friend.

The net force in the horizontal direction should then be the difference between the pull of the friend, and the friction on this boy.

$\text{Net force, horizontal} = 75\; \rm N - 5\; \rm N = 70\; \rm N$ .

<h3>Net force in the vertical direction</h3>

The net force on this boy should be zero in the vertical direction. Consider Newton's Second Law of motion. The net force on an object is proportional to its acceleration. In this question, the net force on this boy in the vertical direction should be proportional to the vertical acceleration of this boy.

However, because (by assumption) the ice rink is level, the boy has no motion in the vertical direction. His vertical acceleration will be zero. As a result, the net force on him should also be zero in the vertical direction.

<h3>Net force</h3>

Therefore, the (combined) net force on this boy would be:

$\sqrt{(70\; \rm N)^2 + (0\; \rm N)^2} = 70\; \rm N$ .

4 0

3 years ago

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How much kinetic energy does a proton gain if it is accelerated, with no friction, through a potential difference of 1.00 V? The

ra1l [238]

Answer:

If energy is conserved, then the sum of the potential energy and the kinetic energy is a constant.

Assuming the proton starts from rest, so it's kineitc energy is zero, but it has a potential energy, PE equal to:

PE = qV

where q =1.6 x 10^-19 C

and V = 1.00 V

Assuming the proton no longer experiences the potential energy and it is all converted to kinetic energy then:

PE* = 0,

KE* = 1/(2mv^2)

Now since

PE + KE = Total energy =PE* + KE*

Therefore,

qV + 0 = 0 + 1/2mv^2

Or

KE = qV = 1.6 10^-19 J

4 0

3 years ago

Ccording to coulomb's law, which pair of charged particles has the lowest potential energy? according to coulomb's law, which pa

Sladkaya [172]

Coulombs law says that the force between any two charges depends on the amount of charges and distance between them. This force is directly proportional to the magnitude of the two charges and inversely proportional to the distance between them.

$F=k\frac{|q_1| |q_2|}{r^2}$

where $q_1\hspace{1mm}and\hspace{1mm}q_2$ are charges, $r$ is the distance between them and k is the coulomb constant.

case 1:

$q_1=-e\\ q_2=+3e\\ r=100pm\\ \Rightarrow F=k\frac{|-e||3e|}{(100pm)^2}=3ke^2\times10^8$

case 2

$q_1=-e\\ q_2=+2e\\ r=100pm\\ \Rightarrow F=k\frac{|-e||2e|}{(100pm)^2}=2ke^2\times10^8$

case 3:

$q_1=-e\\ q_2=+e\\ r=100pm\\ \Rightarrow F=k\frac{|-e||e|}{(200pm)^2}=0.25ke^2\times10^8$

Comparing the 3 cases:

The maximum potential force according to coulombs law is between -1 charge and +3 charge separated by a distance of 100 pm.

3 0

3 years ago

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Que es la friccion Cual es la primera ley de newton