A baseball travels 50 meters in 4 seconds what is the average velocity of the baseball?

g a small smetal sphere, carrying a net charge is held stationarry. what is the speed are 0.4 m apart

Veseljchak [2.6K]

Answer:

The speed of q₂ is $4\sqrt{10}\ m/s$

Explanation:

Given that,

Distance = 0.4 m apart

Suppose, A small metal sphere, carrying a net charge q₁ = −2μC, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q₂ = −8μC and mass 1.50g, is projected toward q₁. When the two spheres are 0.800m apart, q₂ is moving toward q₁ with speed 20m/s.

We need to calculate the speed of q₂

Using conservation of energy

$E_{i}=E_{f}$

$\dfrac{1}{2}mv_{i}^2+\dfrac{kq_{1}q_{2}}{r_{i}}=\dfrac{kq_{1}q_{2}}{r_{f}}+\dfrac{1}{2}mv_{f}^2$

$\dfrac{1}{2}m(v_{i}^2-v_{f}^2)=kq_{1}q_{2}(\dfrac{1}{r_{f}}-\dfrac{1}{r_{i}})$

Put the value into the formula

$\dfrac{1}{2}\times1.5\times10^{-3}(20^2-v_{f}^2)=9\times10^{9}\times-2\times10^{-6}\times-8\times10^{-6}(\dfrac{1}{(0.4)}-\dfrac{1}{(0.8)})$

$0.00075(400-v_{f}^2)=0.18
$

$400-v_{f}^2=\dfrac{0.18}{0.00075}$

$-v_{f}^2=240-400$

$v_{f}^2=160$

$v_{f}=4\sqrt{10}\ m/s$

Hence, The speed of q₂ is $4\sqrt{10}\ m/s$

7 0

2 years ago

In the system illustrated by the diagram, the magnetic field is increasing. In which

Black_prince [1.1K]

The emf will be induced in anti-clockwise direction.

<u>Explanation</u>

Lenz's law tells us the direction us the direction that the current will flow. It states that the direction is always such that it will oppose the change in flux which produced it. This means that any magnetic field produced by an induced current will be in opposite direction to the change in the original field.

To find the direction of emf, Stretch the forefinger, middle finger and the thumb of the right hand mutually perpendicular to each other. If the force finger points in the direction of the magnetic field, the thumb gives the direction of the motion of the conductor then the middle finger gives the direction of the induced current.

3 0

3 years ago

n electromagnetic wave in vacuum has an electric field amplitude of 611 V/m. Calculate the amplitude of the corresponding magnet

enot [183]

Answer:

The corresponding magnetic field is

Explanation:

From the question we are told that

The electric field amplitude is $E_o = 611\ V/m$

Generally the magnetic field amplitude is mathematically represented as

$B_o = \frac{E_o }{c }$

Where c is the speed of light with a constant value

$c = 3.0 *0^{8} \ m/s$

So

$B_o = \frac{611 }{3.0*10^{8}}$

$B_o = 2.0 4 *10^{-6} \ Vm^{-2} s$

Since 1 T is equivalent to $V m^{-2} \cdot s$

$B_o = 2.0 4 *10^{-6} \ T$

6 0

3 years ago

3 Points

AleksAgata [21]

Answer:

B.C. D. G.

Explanation:

A vector quantity, has both magnitude and direction. A tip to remember is if you can add a direction to it! You wouldnt say 30 pounds north, but you would say 30 mph north.

<em>I hope this helped! Comment if you have any questions! :)</em>

3 0

3 years ago

Uma partícula com carga “Q”, no vácuo, gera um potencial elétrico de 600 volts a uma distância de 6,0 m, determine o valor dessa

garik1379 [7]

Answer:

q = 400 nC

the correct answer is b

Explanation:

The expression for the electric potential of a point charge is

V = k q / r

they ask us for the electrical charge

q = V r / k

let's calculate

Q = 600 6.0 / 9 10⁹

Q = 4 10⁻⁷ C

let's reduce to nC

Q = 4 10⁻⁷ C (10⁹ nC / 1C)

q = 4 10² nC = 400 nC

the correct answer is b

Traslate

La expresión para el potencial eléctrico de una carga puntual es

V = k q/r

nos piden la carga eléctrica

q= V r /k

calculemos

Q= 600 6,0 / 9 10⁹

Q= 4 10⁻⁷ C

reduzcamos a nC

Q = 4 10⁻⁷ C(10⁹ nC/1C )

q = 4 10² nC = 400 nC

la respuesta correcta es b

8 0

2 years ago