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4 years ago

When energy (heat) is added to matter the molecules (particles) move __________ ?

Physics

2 answers:

Artyom0805 [142]4 years ago

7 0

They live faster or they vibrate

frozen [14]4 years ago

4 0

The molecules will start to move faster because when something it gets as I say pumped so it wants to run so it’s moving faster

Molecules move faster

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Two billiard balls of equal mass move at right angles and meet at the origin of an xy coordinate system. Initially ball A is mov

frez [133]

Answer:

Speed of ball A after collision is 3.7 m/s

Speed of ball B after collision is 2 m/s

Direction of ball A after collision is towards positive x axis

Total momentum after collision is m×4·21 kgm/s

Total kinetic energy after collision is m×8·85 J

Explanation:

<h3>If we consider two balls as a system as there is no external force initial momentum of the system must be equal to the final momentum of the system</h3>

Let the mass of each ball be m kg

v $_{1}$ be the velocity of ball A along positive x axis

v $_{2}$ be the velocity of ball A along positive y axis

u be the velocity of ball B along positive y axis

Conservation of momentum along x axis

m×3·7 = m× v $_{1}$

∴ v $_{1}$ = 3.7 m/s along positive x axis

Conservation of momentum along y axis

m×2 = m×u + m× v $_{2}$

2 = u + v $_{2}$ → equation 1

<h3>Assuming that there is no permanent deformation between the balls we can say that it is an elastic collision</h3><h3>And for an elastic collision, coefficient of restitution = 1</h3>

∴ relative velocity of approach = relative velocity of separation

-2 = v $_{2}$ - u → equation 2

By adding both equations 1 and 2 we get

v $_{2}$ = 0

∴ u = 2 m/s along positive y axis

Kinetic energy before collision and after collision remains constant because it is an elastic collision

Kinetic energy = (m×2² + m×3·7²)÷2

= 8·85×m J

Total momentum = m×√(2² + 3·7²)

= m× 4·21 kgm/s

3 0

4 years ago

Two equipotential surfaces surround a +1.70 x 10-8-C point charge. How far is the 120-V surface from the 54.0-V surface?

UNO [17]

Answer:

1.55 m

Explanation:

The potential produced by a point charge, is inversely proportional to the distance from the charge to the point where the potential is being calculated, as follows:

$V =\frac{k*q}{r}$

As it only depends from the distance r, we can conclude that if the potential is the same for any point to a distance r from the point charge, the equipotencial surface must be a sphere of radius r.

Replacing q = +1.7*10⁻⁸ C, and k = 9*10⁹ N*m²/C², and V, by 120 V and 54 V, we can find the distance from the charge, to the points where we are calculating the potential V, as follows:

$r1 =\frac{k*q}{V1} = \frac{9e9 N*m2/C2*1.7e-8C}{120 V} = 1.28 m$

$r2 =\frac{k*q}{V2} = \frac{9e9 N*m2/C2*1.7e-8C}{54V} = 2.83 m$

The distance between both points, is just the difference between the radius of both spheres, as follows:

r₂ - r₁ = 1.55 m

5 0

3 years ago

A hockey goalie is standing on ice. Another player fires a puck (m = 0.170 kg) at the goalie with a velocity of +41.0 m/s. (a) I

viktelen [127]

Answer:

2991.42 N

Explanation:

For this problem, we'll use the equations: momentum= mass x velocity and impulse = change in momentum, and impulse=force x time.

initial momentum; p1 = 0.17 x 41 = 6.97 kg.m/s

final momentum; p2 = 0, because final velocity is 0 m/s

Thus,

impulse = p1 - p2= 6.97 - 0 = 6.97 kg.m/s

Finally, impulse= Force x time,

Thus, Force = Impulse/time

Force= 6.97/ (2.33 x 10^(-3)) = 2991.42 N

4 0

4 years ago

A wire carrying a 30.0-A current passes between the poles of a strong magnet that is perpendicular to its field and experiences

Zanzabum

Answer:

1.79 T

Explanation:

Applying,

F = BILsin∅................ Equation 1

Where F = Force, B = magnetic field, I = current flowing through the wire, L = length of the wire, ∅ = angle between the magntic field and the force

make B the subject of the equation

B = F/ILsin∅............. Equation 2

From the question,

Given: F = 2.15 N, I = 30 A, L = 4.00 cm = 0.04 m, ∅ = 90° (perpendicular to the field)

Substitute these values into equation 2

B = 2.15/(30×0.04×sin90°)

B = 2.15/1.2

B = 1.79 T

Hence the average field strength is 1.79 T

8 0

3 years ago

A ferry boat is 4.0 m wide and 6.0 m long. When a truck pulls onto it,

d1i1m1o1n [39]

Should the question say “what is the weight of the truck”? :)
I can then answer it. There is no information to find the weight of the boat ;)

6 0

3 years ago