3. What is electric current?
The flow of moving electrons

electrons that move one time

Answer:

K = 0.076 J

Explanation:

The height of the target, h = 0.860  m

The mass of the steel ball, m = 0.0120 kg

Distance moved, d = 1.50 m

We need to find the kinetic energy (in joules) of the target ball just after it is struck. Let t is the time taken by the ball to reach the ground.

[tex]h=ut+\dfrac{1}{2}at^2\\\\t=\sqrt{\dfrac{2h}{g}}[/tex]

Put all the values,

[tex]t=\sqrt{\dfrac{2\times 0.860 }{9.8}} \\\\=0.418\ s[/tex]

The velocity of the ball is :

[tex]v=\dfrac{1.5}{0.418}\\\\= $$3.58\ m/s[/tex]

The kinetic energy of the ball is :

[tex]K=\dfrac{1}{2}mv^2\\\\K=\dfrac{1}{2}\times 0.0120\times 3.58^2\\\\=0.076\ J[/tex]

So, the required kinetic energy is 0.076 J.

Answer:

100J

Explanation:

Kinetic energy=1/2mv^2

Kinetic energy=(1/2 x 8)x5^2

Kinetic energy=4x25

Kinetic energy=100

100J

Answer:

I guess it is A. I am not sure

Explanation:

kinetic energy was converted to potential energy in the spring.

the answer is in the above image

Answer:

a   c

Explanation:

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Answer:

The velocity of the particle = -1.92 m/s

The speed of the particle = 5.72 m/s

Explanation:

Given equation of motion;

[tex]f(t) = 18 \ + \ \frac{48}{t} \ + \ 1[/tex]

Velocity is defined as the change in displacement with time.

[tex]V = \frac{df(t)}{dt} = -\frac{48}{t^2} \\\\at \ t = 5 \ s\\\\V = -\frac{48}{5^2} = \frac{-48}{25} = - 1.92 \ m/s[/tex]

The distance traveled by the particle in 5 s:

[tex]s = f(5) = 18 + \frac{48}{5} + 1\\\\s= 28.6 \ m[/tex]

The speed of the particle when t = 5s

[tex]Speed = \frac{28.6}{5} = 5.72 \ m/s[/tex]

Answer:

Charles law

Explanation:

Charle's law states that the volume (V) of a given gas is directly proportional to the absolute temperature (T) at a constant pressure.

That is;

: V ∝ T

: V/T = K

According to this question, the volume of a balloon is warmed up but the pressure is kept the same. Charles law will be used because it shows the relationship between the volume (V) and the temperature (heat) at a constant pressure (P).

Explanation:

.....................................

Answer:

1 second is defined as 1/86400th part of a mean solar day.

Explanation:

the answer is in the above image

Answer:

a)  [tex]d=11m[/tex]

b)  [tex]T=4.68s[/tex]

Explanation:

From the question we are told that:

Wavelength [tex]\lambda=22m[/tex]

Velocity [tex]v=4.6m/s[/tex]

a)

Generally the equation for distance between her and the wall d is mathematically given by

Since

The First Anti node distance is [tex]\frac{\lambda}{2}[/tex]

Therefore

 [tex]d= \frac{\22}{2}[/tex]

 [tex]d=11m[/tex]

b)

Generally the equation for her up-and-down motion is mathematically given by

 [tex]T=\frac{22}{4.7}[/tex]

 [tex]T=4.68s[/tex]

Answer:

29.15 N

Explanation:

Applying,

Pythagoras theorem,

a² = b²+c²................ Equation 1

Where a = resultant of the two forces, b = first force, c = second force.

From the diagram,

Given: b = 15 N, c = 25 N

Substitute these values into equation 1

a² = 15²+25²

a² = 225+625

a² = 850

a = √850

a = 29.15 N

Hence the resultant of the two forces is 29.15 N

Answer:

The surface tension is 190.2 N/m.

Explanation:

Initial radius, r = 4 cm

final radius, r' = 6 cm

Work doen, W = 15 J

Let the surface tension is T.

The work  done is given by

W = Surface Tension x change in surface area

[tex]15 = T \times 4\pi^2(r'^2 - r^2)\\\\15 = T \times 4 \times 3.14\times 3.14 (0.06^2- 0.04^2)\\\\15 = T\times 0.0788\\\\T = 190.2 N/m[/tex]

Answer:

the correct answer is B

Explanation:

Let's propose the solution of the problem, for this we form a system formed by the two cars, so that the forces during the collision are internal, the momentum is conserved

instantly starts. Before the crash

         p₀ = M v +0

final instant. After the crash

        m_f = (M + M) v_f

the moment is preserved

        p₀ = p_f

        M v = 2 M v_f

        v_f = v / 2

let's look for kinetic energy

before the crash

       K₀ = ½ M v²

after the crash

       K_f = ½ 2M (v_f)²

       K_f = ½ 2M (v/2)²

       K_f = (½ M v²) ½

       K_f = K₀ / 2

therefore the correct answer is B

Answer:

  T = 9056 K

Explanation:

In the exercise they indicate that the body can be approximated by a black body, for which we can use the Wien displacement relation

                 λ T = 2,898 10⁻³

where lam is the wavelength of the maximum emission

                T = 2,898 10⁻³ /λ

let's calculate

                 T = 2,898 10⁻³ / 320 10⁻⁹

                  T = 9.056 10³ K

                  T = 9056 K

Answer:

Thus, [tex]\frac{1 rev}{min} =\frac{2\pi}{60} rad/s[/tex]

Explanation:

The angular speed is defined as the rate of change of angular velocity.

Its SI unit is rad/s and other units are rev/min or rev/s.

[tex]\frac{1 rev}{min } = \frac{1 rev}{60 sec}\\\\1 rev = 2\pi rad\\\\So\\\\\frac{1 rev}{min} = \frac{2\pi}{60} rad/s[/tex]

Answer:

2.86×10⁻¹⁸ seconds

Explanation:

Applying,

P = VI................ Equation 1

Where P = Power, V = Voltage, I = Current.

make I the subject of the equation

I = P/V................ Equation 2

From the question,

Given: P = 0.414 W, V = 1.50 V

Substitute into equation 2

I = 0.414/1.50

I = 0.276 A

Also,

Q = It............... Equation 3

Where Q = amount of charge, t = time

make t the subject of the equation

t = Q/I.................. Equation 4

From the question,

4.931020 electrons has a charge of (4.931020×1.6020×10⁻¹⁹) coulombs

Q = 7.899×10⁻¹⁹ C

Substitute these value into equation 4

t =  7.899×10⁻¹⁹/0.276

t = 2.86×10⁻¹⁸ seconds

Answer:

The answer is "Option A"

Explanation:

Energy stored in capcitor[tex]=\frac{1}{2}\ cv^2[/tex]

For point A:

[tex]C_A=\frac{\varepsilon_0 2A}{\frac{d}{2}}=\frac{4\ \varepsilon_0 A}{d}\\\\[/tex]

For point B:

[tex]C_B=\frac{\varepsilon_0 2A}{2d}=\frac{\varepsilon_0 A}{d}\\\\[/tex]

For point C:

[tex]C_c=\frac{\varepsilon_0 A}{d}\\\\[/tex]

For point D:

[tex]C_D=\frac{\varepsilon_0 A}{2 \frac{d}{2}}=\frac{\varepsilon_0 A}{d}\\\\[/tex]

For point E:  [tex]C_E=\frac{\varepsilon_0 A}{2 \times 2d}=\frac{\varepsilon_0 A}{4d}\\\\[/tex]

therefore C_A has the maximum capacitance and max energy same energy is dir proportional to C for the same J

Answer:

16.33°C

Explanation:

Applying,

Heat lost by copper = heat gained by water

cm(t₁-t₃) = c'm'(t₃-t₂).............. Equation 1

Where c = specific heat capacity of copper, m = mass of copper, c' = specific heat capacity of water, m' = mass of water, t₁ = initial temperature of copper, t₂ = initial temperature of water, t₃ = final equilibrium temperature.

From the question,

Given: m = 50 kg, t₁ = 140°C, m' = 90 L = 90 kg, t₂ = 10°C

Constant: c = 385  J/kg°C, c' = 4200J/kg°C

Substitute these values into equation 1

50(385)(140-t₃) = 90(4200)(t₃-10)

(140-t₃) = 378000(t₃-10)/19250

(140-t₃) = 19.64(t₃-10)

140-t₃ = 19.64t₃-196.6

19.64t₃+t₃ = 196.4+140

20.64t₃ = 336,4

t₃ = 336.4/20.6

t₃ = 16.33°C

We call the acceleration of an object moving in uniform circular motion— resulting from a net external force—the centripetal ...

Answer:

Flux is 21 Nm^2/C.

Explanation:

Electric field, E = 6 N/C along X axis

Electric filed vector, E = 6 i N/C

Area, A = 4 square meter

Area vector

[tex]\overrightarrow{A} = 4 (cos30 \widehat{i} + sin 30 \widehat{j})\\\\\overrightarrow{A} = 3.5 \widehat{i} + 2 \widehat{j}\\[/tex]

The flux is given by

[tex]\phi= \overrightarrow{E}.\overrightarrow{A}\\\\\phi = 6 \widehat{i} . \left (3.5 \widehat{i} + 2 \widehat{j} \right )\\\\\phi = 21 Nm^2/C[/tex]

Answer:

48.9 is the answer I think !

Answer:

28.4

Explanation:

Answer:

T=0,25s

Explanation

5s>20vong  1s>4vong

omega= 8pi

omega=2pi/T

Explanation:

No. From your own reference frame, nothing will change. Everything will look the same to you even if you're traveling at 99% speed of light. However, an outside observer will see you shrink to about 14% of your original length along the direction of your motion according to the Lorentz contraction predicted by special relativity:

[tex]L=L_{0} \sqrt{1 - \frac{v^2 }{c^2 }}[/tex]

Answer:

They investigated Hooke's law.

Explanation:

[tex]{ \tt{force \: \alpha \: extension }} \\ { \boxed{ \bf{F = kx}}} \\ x \: is \: extension \\ F \: is \: force \\ k \: is \: constant : k = 1[/tex]

Answer:

The correct answer is "4000 J".

Explanation:

Given that,

Force,

= 200 N

Displacement,

= 20 m

Now,

The work done will be:

⇒ [tex]Work=Force\times displacement[/tex]

By putting the values, we get

              [tex]=200\times 20[/tex]

              [tex]=4000 \ J[/tex]

Answer:

Option b, cosine.

Explanation:

Below you can see an image that illustrates this situation.

Remember that for a triangle rectangle with a given angle θ, we have:

Cos(θ) = (adjacent cathetus)/(hypotenuse)

In the image, you can see a vector of magnitude M, and the angle θ defined between the vector and the positive y-axis.

In this case, the y-component is the adjacent cathetus and the hypotenuse is the magnitude of the vector.

Then we will have:

Cos(θ) = (adjacent cathetus)/(hypotenuse) = y/M

solving that for y, we get:

y = Cos(θ)*M

Then the y-component is the vector's magnitude multiplied by the cosine of the angle between the vector and the y-axis.

The correct option is b.

Answer:

(b) cosine

Explanation:

In a 2-dimensional Cartesian coordinate system, a vector has a x-component and/or a y-component. To get these components, the magnitude of the vector is resolved with respect to the x-axis and the y-axis by multiplying it (the magnitude) by some trigonometric function with respect to the angle between the vector and the x or y axis.

For example, given a vector A of magnitude A which makes an angle α with the x-axis and an angle β with the y-axis, the x and y components of the vector A can be found as follows;

i. x-component is given by [tex]A_{x}[/tex]

[tex]A_{x}[/tex] = A cos α (with respect to the angle between A and the x-axis)  or

[tex]A_{x}[/tex] = A sin β (with respect to the angle between A and the y-axis)

ii. y-component is given by [tex]A_{y}[/tex]

[tex]A_{y}[/tex] = A sin α (with respect to the angle between A and the x-axis)  or

[tex]A_{y}[/tex] = A cos β (with respect to the angle between A and the y-axis)

Therefore, the y-component of a vector in a 2-dimensional Cartesian coordinate is given by the product of the magnitude of the vector and the cosine of the angle between the vector and the y-axis.

Answer:

a)   L = 4.75 103 kg m² / s,  b)  K_total = 2.57 10³ J,  

c)   L₀ = L_f =4.75 103 kg m² / s,  d)  K = 1.03 10⁴ J,  K = 1.03 10⁴ J

Explanation:

a) the angular momentum is the sum of the angular momentum of each astronaut

the distance is measured from the center of the circle r = 10/2 = 5.0 m

            L = 2m v r

            L = 2  88.0 5.40 5.0

            L = 4.75 103 kg m² / s

b) rotational kinetic energy

            K = ½ I w²

As there are two astronauts, the total energy is the sum of the energy of each no.

The moment of inertia of a point mass

             I = m r²

             I = 88 5²

             I = 2.2 10³ kg m²

the angular velocity is given by

             v = w r

             w = v / r

             w = 5.40 / 5

             w = 1.08 rad / s

the kinetic energy of the system

             K_total = 2 K

             K_total = 2 (½ I w²)

             K_total = 2.2 10³ 1.08²

             K_total = 2.57 10³ J

c, d) as astronauts are isolated in space, these speeds do not change unless there is an interaction between them, for example they approach each other, suppose they reduce their distance by half

             r = 2.5 m

             I = 88 2.5²

             I = 5.5 10² kg m²

for the change in angular velocity let us use the conservation of moment

            L₀ = L_f

            2Io wo = 2 I w

            w = Io / I wo

            w = 2.2 10³ / 5.5 10² 1.08

            w = 4.32 rad / s

linear velocity is

            v = w r

            v = 4.32 2.5

             K = 1.03 10⁴ J

the kinetic energy of the system is

            K = 5.5 10² 4.32²

            K = 1.03 10⁴ J

Answer:

See annex

Explanation:

By convention potential at ∞    V(∞ ) = 0

As the distance from the sphere decreases the potential increases up to the point d = R  ( R is the radius of the sphere. That potential remains constant while d = R and becomes 0 inside the sphere where there is not free charges and therefore the electric field is 0 and so is the potential.

I am sorry I could not make a better graph

The graph that best  illustrates the potential (relative to infinity) produced by this sphere as a function of the distance r from the center of the sphere is attached as an image below

[tex]V = \frac{KQ}{R}[/tex]

for r <= R

[tex]V = \frac{KQ}{r}[/tex]

for  r > R  

Therefore the graph will be

For more information on potentials as function of distance

https://brainly.com/question/24146175?referrer=searchResults

Answer: The weight of the object is 29.4 N

Explanation:

To calculate the weight of the object, we use the equation:

[tex]W=m\times g[/tex]

where,

m = mass of the object = 3 kg

g = acceleration due to gravity = [tex]9.8m/s^2[/tex]

Putting values in above equation, we get:

[tex]W=3kg\times 9.8m/s^2\\\\W=29.4N[/tex]

Hence, the weight of the object is 29.4 N

Answer:

1995 and 2000 , 4 trillions

Explanation:

Answer:

1995 and 2000.4 trillion

Explanation:

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