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Tema [17]
3 years ago
14

A thermometer is placed in water in order to measure the water’s temperature. What would cause the liquid in the thermometer to

rise?
A.The molecules in the water move closer together. B.The molecules in the thermometer’s liquid spread apart.C.The kinetic energy of the water molecules decreases. D.The kinetic energy of the thermometer’s liquid molecules decreases. 
Physics
1 answer:
yKpoI14uk [10]3 years ago
5 0
It would be D. The kinetic energy would rise.
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A point charge q1 = 3.0 µC is at the origin and a point charge q2 = 6.0 µC is on the x axis at x = 10 m.
UkoKoshka [18]

Answer:

a) 1.6 mN  b) -1.6 mN  c) -1.6 mN  d) 1.6 mN

Explanation:

The electrostatic force between 2 point charges, obeys the Coulomb's Law, that can be expressed as follows:

F₁₂ = k*q₁*q₂/(r₁₂)² (in magnitude)

The direction of the force, is along the  line that joins the  charges (along the x axis) and as q₁ and q₂ are of the same sign, aims away from both charges.

a) So, for the force on q₂, we have:

F₁₂ = 9*18*10⁻⁵ N = 1.6 mN (positive as it is aiming in the positive x direction)

b) The force on q1, according to Newton's 3rd Law, is just equal and opposite to the one on q2:

F₂₁ = (-9*18*10⁻⁵) N = -1.6 mN (towards the negative x direction, away from q1)

c) If q₂ were -6.0 μC, the force will be the same in magnitude, but as now both charges have different signs, they wil attract each other, so the direction of the forces will be exactly the opposite to the first case:

F₁₂ = -1.6 mN (going towards the origin, where q₁ is located)

F₂₁ =  1.6 mN (going in the positive x direction, towards q₂)

6 0
3 years ago
Read 2 more answers
football player A has a mass of 110 kg is running on the field is velocity of 2 m/s football player B has a mass of 120 kg and a
valkas [14]

Complete Question:

Football player A has a mass of 110 kg, and he is running down the field with a velocity of 2 m/s. Football player B has a mass of 120 kg and is stationary. What is the total momentum after the collision?

Answer:

Total momentum = 220 Kgm/s.

Explanation:

<u>Given the following data;</u>

For footballer A

Mass, M1 = 110kg

Velocity, V1 = 2m/s

For footballer B

Mass, M1 = 120kg

Velocity, V1 = 0m/s since he's stationary.

To find the total momentum;

Momentum can be defined as the multiplication (product) of the mass possessed by an object and its velocity. Momentum is considered to be a vector quantity because it has both magnitude and direction.

Mathematically, momentum is given by the formula;

Momentum = mass * velocity

a. To find the momentum of A;

Momentum \; A = 110 * 2

Momentum A = 220 Kgm/s.

b. To find the momentum of B;

Momentum \; B = 120 * 0

Momentum B = 0 Kgm/s.

c. To find the total momentum of the two persons;

Total \; momentum = Momentum \; A + Momentum \; B

Substituting into the equation, we have;

Total \; momentum = 220 + 0

<em>Total momentum = 220 Kgm/s. </em>

7 0
3 years ago
Consider a skateboarder who starts from rest at the top of ramp that is inclined at an angle of 18.0 ∘ to the horizontal.
kotegsom [21]

Answer:

Explanation:

v= u + at

v is final velocity , u is initial velocity . a is acceleration and t is time

Initial velocity u = 0 . Putting the given values in the equation

v = 0 + g sin 18 x 3.5

= 10.6 m /s

3 0
3 years ago
In the physics lab, a block of mass M slides down a frictionless incline from a height of 35cm. At the bottom of the incline it
bogdanovich [222]

Solution :

Given :

M = 0.35 kg

$m=\frac{M}{2}=0.175 \ kg$

Total mechanical energy = constant

or $K.E._{top}+P.E._{top} = K.E._{bottom}+P.E._{bottom}$

But $K.E._{top} = 0$ and $P.E._{bottom} = 0$

Therefore, potential energy at the top = kinetic energy at the bottom

$\Rightarrow mgh = \frac{1}{2}mv^2$

$\Rightarrow v = \sqrt{2gh}$

      $=\sqrt{2 \times 9.8 \times 0.35}$      (h = 35 cm = 0.35 m)

      = 2.62 m/s

It is the velocity of M just before collision of 'm' at the bottom.

We know that in elastic collision velocity after collision is given by :

$v_1=\frac{m_1-m_2}{m_1+m_2}v_1+ \frac{2m_2v_2}{m_1+m_2}$

here, $m_1=M, m_2 = m, v_1 = 2.62 m/s, v_2 = 0$

∴ $v_1=\frac{0.35-0.175}{0.5250}+\frac{2 \times 0.175 \times 0}{0.525}

      $=\frac{0.175}{0.525}+0$

     = 0.33 m/s

Therefore, velocity after the collision of mass M = 0.33 m/s

 

3 0
2 years ago
What features were the last to form on the moon?
kap26 [50]
<span>Rayed craters-</span><span> were the last features to form on the moon.</span>
3 0
2 years ago
Read 2 more answers
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