If two opposing forces are equal, then the net force is 0 N.<br><br> true or false?

The mass of a golf ball is 45.9 g . if it leaves the tee with a speed of 62.0 m/s , what is its corresponding wavelength?

inn [45]

The wavelength of the golf ball is <u>2.328×10⁻³⁴m.</u>

All moving particles with mass have a matter wave associated with it. These matter waves are called deBroglie waves.

The deBroglie wavelength λ of a particle is given by,

$\lambda=\frac{h}{mv}$

Here, h is the Planck's constant, m is the mass of the ball and v is its velocity.

Calculate the deBroglie wavelength of the moving golf ball by substituting 6.626×10⁻³⁴J s for h, 45.9×10⁻³kg for m and 62.0 m/s for v.

$\lambda=\frac{h}{mv}\\ =\frac{6.626*10^-^3^4Js}{(45.9*10^-^3kg)(62.0m/s)} \\ =2.328*10^-^3^4m$

The wavelength of the golf ball is <u>2.328×10⁻³⁴m.</u>

4 0

4 years ago

A. Draw the electric field lines around a negative charge.

Alborosie

<h2>a. Answer:</h2>

We use Electric field lines for visualizing electric fields, so this helps us to see the problem more real. So an electric field line is an imaginary line or curve drawn through a region of space such that the tangent at any point comes from the direction of the electric-field vector at that point. The electric field lines around a negative charge is shown in the First figure below.

<h2>b. Answer:</h2>

Electric forces can be found by using the Coulomb Law's that states <em>that The magnitude of the electric force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. </em>This can be expressed as follows:

$F=k\frac{\left | q_{1}q_{2} \right |}{r^2} \\ \\ Where: \\ \\ k=9\times 10^9Nm^2/c^2 \\ \\ q_{1}=0.00150 C \ and \ q_{2}=0.00240 C \\ \\ r=0.900 m$

Then:

$F=9\times 10^9\frac{\left | 0.00150 \times 0.00240 \right |}{(0.900)^2} \\ \\ \therefore \boxed{F=40000N}$

This force is repulsive because the two charges are positive and recall that two positive charges or two negative charges repel each other while a positive charge and a negative charge attract each other.

<h2>C. Answer:</h2>

From the statement, we have two charged objects. Let's say that this charges are:

$q_{1} \ and \ q_{2}$

If the amount of charge on one of the objects is tripled, let's say this is the charge $q_{2}$ , then the new charge is:

$q_{N}=3q_{2}$

In the formula of Coulomb:

$F=k\frac{\left | q_{1}q_{N} \right |}{r^2} \\ \\ \therefore F=k\frac{\left | q_{1}(3q_{2}) \right |}{r^2} \\ \\ \therefore \boxed{F=3k\frac{\left | q_{1}q_{2} \right |}{r^2}}$

<em>The conclusion is that if the amount of charge on one of the objects is tripled, the electric force between two charged objects is also tripled</em>

<h2>d. Answer:</h2>

Let's use the Coulomb's Law again to solve this problem. We want to know how the electric force between two charged objects changes if the charges are moved closer together:

$F=k\frac{\left | q_{1}q_{N} \right |}{r^2}$

<em>By saying that the charges are moved closer together, we want to express that r becomes smaller. Since r is in the denominator, this implies that the electric force between these two charged objects becomes greater.</em>

<h2>e. Answer:</h2>

From the figure, we can see a metal sphere on a stand. There we have both positive and negative charges. We can say that the positive charge of this sphere is +10q and the negative and the negative charge is -10q. Since the electric charge is conserved, then the algebraic sum of all the electric charges in any closed system is constant. In conclusion, <em>the sphere has no net charge.</em>

<h2>f. Answer:</h2>

Here we want to know how the negative charges in the same sphere are redistributed when a positively charged rod is brought near it. Therefore, positive charge on rod repels positive charges on the sphere, creating zones of negative and positive charge as indicated in the second Figure.

7 0

3 years ago

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Help ASAP <br> Just answer the first question for me please!

Usimov [2.4K]

Answer:

Im pretty sure its b

4 0

3 years ago

A projectile is fired horizontally from a height of 10 m above level ground. The projectile lands a horizontal distance of 15 m

Degger [83]

Answer:

a)t = 1,43 s

b) V = 10,49 m/s

c) V₀ₓ = 10,49 m/s ; V₀y = 14,01 m/s

d) Vf = 17,5 m/s

Explanation:

According to the problem statement

V₀ = V₀ₓ and V₀y = 0

And at the end of the movement t = ? the distance y = 10 m

Therefore as

h = V₀y - (1/2)*g*t²

Vertical distance y = h = 10 = V₀y*t - 0,5 (-9,8)*t²

10 = 4,9*t²

t² = 10/4,9 ⇒ t² = 2,04 s

t = 1,43 s

a) 1,43 s is the time of movement

b) V₀ = V₀ₓ V₀y = 0 and V₀ₓ = Vₓ ( constant )

Just before touching the ground, the horizontal distance is

hd = 15 = Vₓ * t

Then 15 /1,43 = Vₓ = V₀ₓ

Vₓ = 10,49 m/s

Then initial speed is V = 10,49 m/s since V₀y = 0

Vf² = Vₓ² + Vy²

Vyf = V₀y - g*t

Vyf = 0 - 9,8 *1,43

Vyf = - 14,01 m/s

And finally the speed when the projectile strike the ground is:

Vf² = Vₓ² + Vy²

Vf = √ (10,49)² + (14,01)²

Vf = 17,50 m/s

3 0

3 years ago

Two students stand at rest, facing each other on frictionless skates. They then start tossing a heavy ball back and forth betwee

Rzqust [24]

<span>The students will move away from each other as explained by Newton’s Third Law which states that “To every action, there is an equal and opposite reaction.” If the students are on frictionless skates, their action of throwing the ball will propel them in the opposite direction. Once the motion is initiated, they continue to move away from each other unless stopped by another force.</span><span />

8 0

3 years ago

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If two opposing forces are equal, then the net force is 0 N. true or false?