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lesantik [10]
4 years ago
6

True or False: Solids always have a higher density than liquids and gases.

Physics
2 answers:
Hunter-Best [27]4 years ago
7 0

Answer:

<h3>False</h3>

<h3> I hoped this helpful for you....</h3>

Thank you ☺️☺️

adoni [48]4 years ago
3 0

Answer:

True.

Explanation:

This is because the solid particles are very close to each other due to strong intermolecular force attraction between them.

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Find the work performed when the given force f f is applied to an object, whose resulting motion is represented by the displacem
Ahat [919]

work = force x distance

fd foot pounds

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3 years ago
These types of electromagnetic waves are listed in order of decreasing
viktelen [127]
The answer you’re looking for is
D. Gamma rays
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4 0
3 years ago
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Three charges lie along the x axis. The positive charge q1 = 15 μC is at x= 2.0 m and the positive charge q2 = 6.0μC is at the o
Delicious77 [7]

Answer:

x = 0.775m

Explanation:

Conceptual analysis

In the attached figure we see the locations of the charges. We place the charge q₃ at a distance x from the origin. The forces F₂₃ and F₁₃ are attractive forces because the charges have an opposite sign, and these forces must be equal so that the net force on the charge q₃ is zero.

We apply Coulomb's law to calculate the electrical forces on q₃:

F_{23} = \frac{k*q_{2}*q_{3}}{x^2} (Electric force of q₂ over q₃)

F_{13} = \frac{k*q_{1}*q_{3}}{(2-x)^2} (Electric force of q₁ over q₃)

Known data

q₁ = 15 μC = 15*10⁻⁶ C

q₂ = 6 μC = 6*10⁻⁶ C

Problem development

F₂₃ = F₁₃

\frac{k*q_{2}*q_{3}}{x^2} = \frac{k*q_{1}*q_{3}}{(2-x)^2} (We cancel k and q₃)

\frac{q_2}{x^2}=\frac{q_1}{(2-x)^2}

q₂(2-x)² = q₁x²

6×10⁻⁶(2-x)² = 15×10⁻⁶(x)² (We cancel 10⁻⁶)

6(2-x)² = 15(x)²

6(4-4x+x²) = 15x²

24 - 24x + 6x² = 15x²

9x² + 24x - 24 = 0

The solution of the quadratic equation is:

x₁ = 0.775m

x₂ = -3.44m

x₁ meets the conditions for the forces to cancel in q₃

x₂ does not meet the conditions because the forces would remain in the same direction and would not cancel

The negative charge q₃ must be placed on x = 0.775 so that the net force is equal to zero.

6 0
3 years ago
Read 2 more answers
A dentist’s drill starts from rest. After 4.28 s of constant angular acceleration it turns at a rate of 28940 rev/min. Find the
mylen [45]

Answer:

Angular acceleration, is 708.07\ rad/s^2

Explanation:

Given that,

Initial speed of the drill, \omega_i=0

After 4.28 s of constant angular acceleration it turns at a rate of 28940 rev/min, final angular speed, \omega_f=28940\ rev/min=3030.58\ rad/s

We need to find the drill’s angular acceleration. It is given by the rate of change of angular velocity.

\alpha =\dfrac{\omega_f}{t}\\\\\alpha =\dfrac{3030.58\ rad/s}{4.28\ s}\\\\\alpha =708.07\ rad/s^2

So, the drill's angular acceleration is 708.07\ rad/s^2.

4 0
3 years ago
A bird is flying east at 5.2 kilometers/hour relative to the air. There's a crosswind blowing at 3.1 kilometers/hour toward the
alexandr402 [8]

Answer:

r=6.05km/hr

z=59.1 degree to the horizontal

Explanation:

A bird is flying east at 5.2 kilometers/hour relative to the air. There's a crosswind blowing at 3.1 kilometers/hour toward the south relative to the ground. What is the bird's velocity relative to the ground? State your answer to one decimal place

can be solved using pythagoras theorem

r^2=o^2+a^2

r^2=5.2^2+3.1^2

r^2=36.65

r=6.1km/hr is te birds velocity relative to the ground

tanz=5.2/3.1

z=tan^-1(5,2/3.1)

z=59.1 degree to the horizontal

7 0
3 years ago
Read 2 more answers
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