An electric field E⃗ =5.00×105ı^N/C causes the point charge in the figure to hang at an angle. What is θ?

Answer:

d = 6.32 m

Explanation:

Given that,

The mass of a puck, m = 2 kg

It is pushed straight north with a constant force of 5N for 1.50 s and then let go.

We need to find the distance covered by the puck when move from rest in 2.25 s.

We know that,

F = ma

[tex]a=\dfrac{F}{m}\\\\a=\dfrac{5}{2}\\\\a=2.5\ m/s^2[/tex]

Let d is the distance moved in 2.25 s. Using second equation of motion,

[tex]d=ut+\dfrac{1}{2}at^2\\\\d=0+\dfrac{1}{2}\times 2.5\times (2.25)^2\\\\d=6.32\ m[/tex]

So, it will move 6.32 m from rest in 2.25 seconds.

Answer:

-4/7

Explanation:

Given the following

A=4i-10j and B= 7i+5j

A+ bB = 4i-10j + (7i+5j)b

A+ bB =  4i-10j + 7ib+5jb

A+ bB =

The vector along the x-axis is expressed as i + 0j

If the vector A+ bB is pointing in the direction of the x-axis then;

[tex]A+ bB * \frac{i+0j}{|i+0j|} = 0 \\ (4+7b)i-(10-5b)j* \frac{i+0j}{\sqrt{1^2+0^2} } = 0\\(4+7b)i-(10-5b)j *(i+0j) = 0\\4+7b-0 =0\\7b=-4\\b = -4/7[/tex]

Hence the value of b is -4/7

The value of [tex]\beta[/tex] such that [tex]\vec C = \vec A + \beta \cdot \vec B = c\,\hat{i}[/tex] is 2.

According to the statement, we have following system of vectorial equations:

[tex]\vec A = 4\,\hat {i} - 10\,\hat{j}[/tex] (1)

[tex]\vec {B} = 7\,\hat{i} + 5\,\hat{j}[/tex] (2)

[tex]\vec C = \vec A + \beta \cdot \vec B = c\,\hat{i}[/tex] (3)

By applying (1) and (2) in (3):

[tex](4\,\hat{i}-10\,\hat{j}) + \beta\cdot (7\,\hat{i}+5\,\hat{j}) = c\,\hat{i}[/tex]

[tex](4+7\cdot \beta)\,\hat{i} +(-10+5\cdot \beta)\,\hat{j} = c\,\hat{i}[/tex]

And we get two scalar equations after analyzing each component:

[tex]4+7\cdot \beta = c[/tex] (4)

[tex]-10+5\cdot \beta = 0[/tex] (5)

We solve for [tex]\beta[/tex] in (5):

[tex]\beta = 2[/tex]

And for [tex]c[/tex] in (4):

[tex]c = 4+7\cdot (2)[/tex]

[tex]c = 18[/tex]

The value of [tex]\beta[/tex] such that [tex]\vec C = \vec A + \beta \cdot \vec B = c\,\hat{i}[/tex] is 2.

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Answer:

   x = 9.32 cm

Explanation:

For this exercise we have an applied torque and the bar is in equilibrium, which is why we use the endowment equilibrium equation

Suppose the counterclockwise turn is positive, let's set our reference frame at the left end of the bar

          - W l / 2 - W_{child} x + N₂ l = 0

             x = [tex]\frac{-W l/2 + n_2 l}{W_{child}}[/tex]             1)

now let's use the expression for translational equilibrium

         N₁ - W - W_(child) + N₂ = 0

indicate that N₂ = 4 N₁

we substitute

           N₁ - W - W_child + 4 N₁ = 0

           5 N₁ -W - W_{child} = 0

           N₁ = ( W + W_{child}) / 5

we calculate

           N₁ = (450 + 250) / 5

          N₁ = 140 N

we calculate with equation 1

           x = -250 1.50 + 4 140 3) / 140

           x = 9.32 cm

Explanation:

Hey there!

According to the question;

A person can lift mass of 60 kg on earth.

mass(m1) = 60kg

acceleration due to gravity on earth (a) = 9.8m/s²

Now;

force (f) = m.a

= 60*9.8

= 588 N

Since, there is application of same magnitude of force on moon,

mass(m) =?

acceleration due to gravity on moon (a) = 1.67m/s²

Now;

force (f) = m.a

588 = m*1.67

m = 352.09 kg

Therefore, the person who can lift the mass of 60 kg on earth can lift mass of 352 kg on moon.

Hope it helps!

Answer

E - 1/2 K x^2      potential energy of compressed spring

E = 1/2 * 4 N / m * (.5 m)^2 = 2 * .5^2 N-m = .5 N-m

Explanation:

Work cannot be increased by using a machine of some kind.

Work cannot be increased by using a machine of some kind.

A machine is any device in which the effort applied at one end overcomes a load at the other end.

Machines are generally used to perform different tasks faster.

However, a simple machine can not be used to increase the amount of work done at any time.

Force, speed and torque can all be increased using machines.

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The index of refraction of the plastic is approximately 1.461

The known values in the question are;

The thickness of the piece of plastic placed on the dot = 1.30 cm

The height to which the microscope objective is raised to bring the dot back to focus = 0.410 cm

The unknown values in the question are;

The index of refraction

Strategy;

Calculate the refractive index by making use of the apparent height and real height method for the black dot under the thick piece of plastic

[tex]\mathbf{ Refractive \ index, n = \dfrac{Real \ depth}{Apparent \ depth}}[/tex]

The real depth of the dot below the piece of plastic, d₁ = 1.30 cm

The apparent depth of the dot, d₂ = The actual depth - The height to which the microscope is raised

Therefore;

The apparent depth of the dot, d₂ = 1.30 cm - 0.410 cm = 0.89 cm

[tex]The \ refractive \ index, \ n = \dfrac{d_1}{d_2}[/tex]

Therefore, n = 1.30/0.89 ≈ 1.461

The refractive index of the plastic block, n ≈ 1.461

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The sign set after the slowdown of the bicycle will be positive for the position,  negative for velocity, and negative for acceleration.

The rate at which an object's position changes when observed from a specific point of view and when measured against a specific unit of time is known as its velocity.

According to Que, when a bicyclist moves in a straight line and slows down, then the velocity decrease as displacement is decreasing, and the acceleration also decreases only displacement increases.

Therefore, the sign set for the position is +ve, for velocity it is -ve, and for acceleration also -ve

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A biker slows down after traveling in a long, straight line at initial velocity v0. Which of the following the best \sdescribes the signs set for the initial position, initial velocity and the acceleration? Initial position Initial velocity Acceleration

A. Positive Negative Negative

B. Positive Positive Negative

C. Negative Positive Negative

D. Negative Negative Positive

E. Negative Negative Negative

Answer:

The additional load is 20g

Answer:

The speed of proton is 2.1 x 10^5 m/s .

Explanation:

potential difference, V = 234 V

let the initial speed of the proton is v.

The kinetic energy of proton is

KE = q V

[tex]0.5 mv^2 = e V \\\\0.5\times 1.67\times 10^{-27} v^2 = 1.6\times 10^{-19} \times 234\\\\v=2.1\times 10^5 m/s[/tex]

Answer:

chk picture for eqn

Explanation:

Answer:

15.88°C I am not 100% sure this is right but I am 98% sure this IS right

Answer: it would be daddy

Explanation:

Because I’m daddy

Answer:

4.73 × 10^5

Explanation:

Answer:

P = density (p) * g * h

P = 1000  kg/m^3 * 9.8 m/s^2 * 40 m = 392,000 N/m^2

since kg m / s^2 = Newtons

The average pressure is 1/2 (pressure at 0m +  pressure 80 m) for liquid of uniform density

Explanation:

Given that,

Maximum potential, V = 4. mV

Distance, d = 0.350 m

Frequency of the wave, f = 100 Hz

(a) The maximum electric field strength created is given by:

[tex]E=\dfrac{V}{d}\\\\E=\dfrac{4.1\times 10^{-3}}{0.350 }\\\\E=0.0117\ V/m[/tex]

(b) The corresponding maximum magnetic field strength in the electromagnetic wave is given by :

[tex]B=\dfrac{E}{c}\\\\B=\dfrac{0.0117}{3\times 10^8}\\\\B=3.9\times 10^{-11}\ T[/tex]

(c) The wavelength of the electromagnetic wave can be calculated as :

[tex]\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{100}\\\\=3\times 10^6\ m[/tex]

So, the wavelength of the electromagnetic wave is [tex]3\times 10^6\ m[/tex].

Answer:

c

Explanation:

Liquid plastic is frozen in a physical change that increases its volume,it can be known about the plastic that Its mass will remain the same, therefore the correct answer is option C.

Anything which has mass and occupies space is known as matter ,mainly there are four states of matter solid liquid gases, and plasma.

These different states of matter have different characteristics according to which they vary their volume and shape.

It is known about plastic that its mass will remain the same when liquid plastic is frozen, by increasing its volume.

Liquid plastic is frozen in a physical change that increases its volume,it can be known about the plastic that Its mass will remain the same,  therefore the correct answer is C.

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[tex]\textbf{A}×\textbf{B}= 68\hat{\textbf{i}} + 52\hat{\textbf{j}} - 8\hat{\textbf{k}}[/tex]

Explanation:

Given:

[tex]\textbf{A} = 5\hat{\textbf{i}} - 7\hat{\textbf{j}} - 3\hat{\textbf{k}}[/tex]

[tex]\textbf{B} = -4\hat{\textbf{i}} + 4\hat{\textbf{j}} - 8\hat{\textbf{k}}[/tex]

The cross product [tex]\textbf{A}×\textbf{B}[/tex] is given by

[tex]\textbf{A}×\textbf{B} = \left|\begin{array}{ccc}\hat{\textbf{i}} & \hat{\textbf{j}} & \hat{\textbf{k}} \\\:\:5 & -7 & -3 \\ -4 & \:\:4 & -8 \\ \end{array}\right|[/tex]

[tex]= \left|\begin{array}{cc}-7 & -3\\\:4 & -8\\ \end{array}\right|\:\hat{\textbf{i}}\:+\:\left|\begin{array}{cc}-3 & \:\:5\\-8 & -4\\ \end{array}\right|\:\hat{\textbf{j}}\:+\: \left|\begin{array}{cc}\:\:5 & -7\\-4 & \:\:4\\ \end{array}\right|\:\hat{\textbf{k}}[/tex]

[tex]= 68\hat{\textbf{i}} + 52\hat{\textbf{j}} - 8\hat{\textbf{k}}[/tex]

The comparison of the speeds and kinetic energy of the identical balls are as follows

The speed and the kinetic energy of the first and fourth ball are equal, while the speed and kinetic energy of the second and third balls are equal

The reason for the above comparison results areas follows;

Known parameters;

First ball is thrown straight up

Second ball is thrown 30° above the horizontal

Third ball it thrown 30° below the horizontal

The fourth ball is thrown straight down

Unknown:

Comparison of the speed and kinetic energy of the four balls

Method:

The kinetic energy, K.E. = (1/2) × m × v²

The velocity of the ball, v = u × sin(θ)

Where;

u = The initial velocity of the ball

θ = The reference angle

θ = 90°

∴ [tex]v_y[/tex] = u

The final velocity of the ball as it strikes the ground is v₂ = u² + 2gh

Where;

h = The height of the cliff

∴ Kinetic energy of first ball, K.E.₁ = (1/2) × m × (u₁² + 2gh)²

K.E. = (1/2) × m × ((u×sin30)² + 2·g·h)² = K.E. = (1/2) × m × ((0.5·u)² + 2·g·h)²

Kinetic energy  K.E.₂ = (1/2) × m × ((0.5·u₂)² + 2·g·h)²

K.E. = (1/2) × m × ((u×sin30)² + 2·g·h)² = K.E. = (1/2) × m × ((0.5·u)² + 2·g·h)²

Kinetic energy K.E.₃ = (1/2) × m × ((0.5·u₃)² + 2·g·h)²

Kinetic energy K.E.₄ = (1/2) × m × (u₄² + 2gh)²

Therefore, as the ball strike the ground, the speed and the kinetic energy of the first and fourth ball are equal, while the speed and kinetic energy of the second and third balls are equal

u₁ = u₄, K.E₁ = K.E.₄, u₂ = u₃, K.E₂ = K.E.₃

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Answer:

Answer:30 cm

Answer:30 cmExplanation:

Answer:30 cmExplanation:Given=ROC= 60cm

Answer:30 cmExplanation:Given=ROC= 60cmObject be placed so that the rays that came from the object to them mirror are reflected from the mirror, and, then travel parallel to each other= 30cm at focus.

Answer:

148.5 N

Explanation:

Given that,

The mass of a bungee jumper, m = 55 kg

The downward acceleration, a = 7.1 m/s²

We need to find the net force acting on the jumper. As it is moving in downward direction, net force is :

T = m(g-a)

Put all the values,

T = 55(9.8 - 7.1)

= 148.5 N

So, the force exerted on the bungee cord is 148.5  N.

Answer:

The downward force is 148.5 N.

Explanation:

mass, m = 55 kg

downwards acceleration, a = 7.1 m/s^2

Let the force is F.

According to the newton's second law

m g - F = m a

F = m (g - a)

F = 55 (9.8 - 7.1)

F = 148.5 N

Answer:

26 km

Explanation:

Let's say our "cable" has a cross section of 1 m²

Then each meter of cable would weight 7900(9.8) = 77420 N

A Pascal is a Newton per square meter

2 x 10⁹ / 77420 = 25840 m or about 26 km or about 16 miles

Answer:

the body is subjected to a continuous rotation and the body is not in rotational equilibrium

Explanation:

For an object to have a static equilibrium, it must meet two relationships

             ∑ F = 0

             ∑ τ =0              

force acting on a body fulfills the relation of

         sum F = F - F = 0

when two forces do not move from position.

To find the torque we assume that the counterclockwise rotations are positive

        Σ τ = - F r - F r

        Στ = -2 Fr <> 0

consequently the body is subjected to a continuous rotation and the body is not in rotational equilibrium

momentum p= m x v = 15 x -4 = -60 N.s

A single force acts on a particle situated on the positive x axis. The torque about the origin is in the negative z direction. The force might be in the negative y direction. Thus, option B is correct.

To determine the force that could generate a torque in the negative z direction, we need to consider the right-hand rule for cross products. The torque vector, denoted by τ, is given by the cross product of the position vector, r, and the force vector, F:

[tex]τ = r × F[/tex]

In this case, the position vector, r, points along the positive x-axis. The negative z-direction torque indicates that the force vector must be perpendicular to both the position vector and the negative z-axis.

Using the right-hand rule, we can determine that the force vector must be in the negative y-direction, which is option B.

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Answer:

 N = 107.94 N

Explanation:

For this exercise we must use Newton's second law.

Let's set a reference system with the x-axis parallel to the ground and the y-axis vertical

X axis

        Fₓ = ma

ej and

       N -F_y - W = 0

let's use trigonometry to decompose the applied force

     cos -35 = Fₓ / F

     sin -35 = F_y / F

     Fₓ = F cos -35

     F_y = F sin -35

     Fₓ = 40.0 cos -35 = 32.766 N

     F_y = 40.0 sin -35 = -22.94 N

we substitute

     N = Fy + W

     N = 22.94 + 85

     N = 107.94 N

Answer:

  E_ {total} = [tex]\frac{Q }{L^2 \epsilon_o}[/tex]

Explanation:

In this exercise you are asked to calculate the electric field between two plates, the electric field is a vector

         E_ {total} = E₁ + E₂

         E_ {total} = 2 E

where E₁ and E₂ are the fields of each plate, we have used that for the positively charged plate the field is outgoing and for the negatively charged plate the field is incoming, therefore in the space between the plates for a test charge the two fields point in the same direction

to calculate the field created by a plate let's use Gauss's law

          Ф = ∫ E . dA = q_{int} /ε₀

As a Gaussian surface we use a cylinder with the base parallel to the plate, therefore the direction of the electric field and the normal to the surface are parallel, therefore the scalar product is reduced to the algebraic product.

           E 2A = q_{int} / ε₀

where the 2 is due to the surface has two faces

indicate that the surface has a uniform charge for which we can define a surface density

           σ = q_{int} / A

           q_{int} = σ A

we substitute

           E 2A = σ A /ε₀

           E = σ / 2ε₀  

therefore the total field is

           E_ {total} = σ /ε₀

let's substitute the density for the charge of the whole plate

           σ= Q / L²

            E_ {total} = [tex]\frac{Q }{L^2 \epsilon_o}[/tex]