2 Name the device in a car that we use to change the speed of the car

What is this question formula a=V²-U² /2S?

erik [133]

Answer:

sorry I don't really know about that question.

Explanation:

7 0

3 years ago

Imagine a universe in which, like in ours, there are two kinds of charges (positive and negative), with the like charges repelli

GuDViN [60]

Answer:

the static charge is not always distributed on the surface of the conductor, there are also charges in the volume but of lesser magnitude

Explanation:

In this hypothetical system the electric force is of type

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

in this case the force decays to zero much faster,

if we call Fo the force of Coulomb's law

F₀ = $k \frac{q_1 q_2 }{r^2}$

assuming the constant k is the same

the relationship between the two forces is

F / F₀ = 1 / r

F = F₀ / r

when analyzing this expression the force decays much faster to zero.

In an electric conductor, charges of the same sign may not feel any repulsive force from other charges that are at a medium distance, so there is a probability that some charges are distributed in the volume of the material, this does not happen with coulomb's law

Consequently, the static charge is not always distributed on the surface of the conductor, there are also charges in the volume but of lesser magnitude

5 0

3 years ago

A technical machinist is asked to build a cubical steel tank that will hold 130l of water. calculate in meters the smallest poss

disa [49]

Answer:0.507m

To answer this question, you need to know how to calculate a cube volume. Cube volume is calculated by multiplying the length^3, make its unit become cubic meter.
In this case we know the volume is 130L and asked how long is the length in meter. Then you need to convert the liter into m3. The calculation would be:
V=R^3
130L= R^3
R= 3<span>√(130m3/1000)
R= 5.065797/10 m
R= 0.506579m
Rounded up to 0.507m</span>

3 0

3 years ago

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John, paul, and george are standing in a strawberry field. paul is 14.0 m due west of john. george is 36.0 m from paul, in a dir

sdas [7]

Position of paul with respect to john is given as

14 m due west of john

$r_{pj} = r_p - r_j = -14\hat i$

position of George with respect to Paul is given as 36 m in direction 37 degree south of east

$r_{GP} = r_G - r_p = 36cos37\hat i - 36 sin37\hat j$

now we need to find the position of George with respect to John

$r_{GJ} = r_G - r_j[\tex]now for the above equation we can add the two equations[tex]r_{Gj} = -14\hat i + 36 cos37\hat i - 36sin37\hat j$

$r_{Gj} = 14.75\hat i - 21.67\hat j$

so the magnitude is given as

$r = \sqrt{14.75^2 + 21.67^2} = 26.2 m$

and direction is given as

$\theta = tan^{-1}\frac{21.67}{14.75}= 55.75 degree$

<em>so it is 26.2 m at an angle 55.75 degree South of east</em>

7 0

3 years ago

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A rugby player sits on a scrum machine that weighs 200 Newtons. Given that the coefficient of static friction is 0.67, the coeff

Trava [24]

a. 850 N is the minimum force needed to get the machine/player system moving, which means this is the maximum magnitude of static friction between the system and the surface they stand on.

By Newton's second law, at the moment right before the system starts to move,

• net horizontal force

∑ F[h] = F[push] - F[s. friction] = 0

• net vertical force

∑ F[v] = F[normal] - F[weight] = 0

and we have

F[s. friction] = µ[s] F[normal]

It follows that

F[weight] = F[normal] = (850 N) / (0.67) = 1268.66 N

where F[weight] is the combined weight of the player and machine. We're given the machine's weight is 200 N, so the player weighs 1068.66 N and hence has a mass of

(1068.66 N) / g ≈ 110 kg

b. To keep the system moving at a constant speed, the second-law equations from part (a) change only slightly to

∑ F[h] = F[push] - F[k. friction] = 0

∑ F[v] = F[normal] - F[weight] = 0

so that

F[k. friction] = µ[k] F[normal] = 0.56 (1268.66 N) = 710.45 N

and so the minimum force needed to keep the system moving is

F[push] = 710.45 N ≈ 710 N

4 0

2 years ago