The advantage of a hydraulic lever is A : it transforms a small force acting over a large distance into a large force acting over a small distance. B : it transforms a small force acting over a small distance into a large force acting over a large distance. C : it allows you to exert a larger force with less work. D : it transforms a large force acting over a large distance into a small force acting over a small distance. E : it transforms a large force acting over a small distance into a small force acting over a large distance.

Answer:

A. 30.38°

B 5.04N

Explanation:

Using

F= ILBsin theta

2 .55N= 8.4Ax 0.5mx 1.2T x sintheta

Theta = 30.38°

B. If theta is 90°

Then

F= 8.4Ax 0.5mx 1.2x sin 90°

F= 5.04N

Answer:

The speed of the proton is 4059.39 m/s

Explanation:

The centripetal force on the particle is given by;

[tex]F = \frac{mv^2}{r}[/tex]

The magnetic force on the particle is given by;

[tex]F = qvB[/tex]

The centripetal force on the particle must equal the magnetic force on the particle, for the particle to remain in the circular path.

[tex]\frac{mv^2}{r} = qvB\\\\r = \frac{mv^2}{qvB} \\\\r = \frac{mv}{qB}[/tex]

where;

r is the radius of the circular path moved by both electron and proton;

⇒For electron;

[tex]r = \frac{(9.1*10^{-31})(7.45*10^6)}{(1.602*10^{-19})(1.1*10^{-5})}\\\\r = 3.847 \ m[/tex]

⇒For proton

The speed of the proton is given by;

[tex]r = \frac{mv}{qB}\\\\mv = qBr\\\\v = \frac{qBr}{m} \\\\v = \frac{(1.602*10^{-19})(1.1*10^{-5})(3.847)}{1.67*10^{-27}} \\\\v = 4059.39 \ m/s[/tex]

Therefore, the speed of the proton is 4059.39 m/s

Answer:

Upper plate Q/3

Lower plate 2Q/3

Explanation:

See attached file

B. A chemical change

Explanation:

I'm guessing ?

Answer:

Explanation:

Using the Continuity equation

v X A = v' xA'

so if A is 1/2of A' then A velocity must be 2 times the A'

after-contraction v = 2 x 5.0m/s = 10m/s

Using the Bernoulli equation

p₁ + ½ρv₁² + ρgh₁ = p₂ + ½ρv₂² + ρgh₂

, the "h" terms cancel

3.5 x 10^ 5Pa + ½ x 1000kg/m³x (5.0m/s)² = p₂ + ½ x 1000kg/m³ x (10m/s)²

p₂ = 342500pa

Answer:

a) 5.5×10^17 Hz

b) visible light

Explanation:

Since the wavelength of the electromagnetic radiation must be about the size of the about itself, this implies that;

λ= 5.5 × 10^-10 m

Since;

c= λ f and c= 3×10^8 ms-1

f= c/λ

f= 3×10^8/5.5 × 10^-10

f= 5.5×10^17 Hz

The electromagnetic wave is visible light

Answer:

Will be equal to alpha x r; less than UsN

Answer:

covalent

Explanation:

covalent bonds share electrons

Answer:

[tex]v_B/v_A=\sqrt{3}[/tex]

Explanation:

Consider the two kinematic equations for velocity and position of an object falling due to the action of gravity:

[tex]v=-g\,t\\ \\position=-\frac{1}{2} g\,t^2[/tex]

Therefore, if we consider [tex]t_A[/tex] the time for the object to reach point A, and [tex]t_B[/tex] the time for it to reach point B, then:

[tex]v_A=-g\,t_A\\v_B=-g\,t_B\\\frac{v_B}{v_A}= \frac{-g\,t_B}{-g\,t_A} =\frac{t_B}{t_A}[/tex]

Let's work in a similar way with the two different positions at those different times, and for which we have some information;

[tex]x_A=-x=-\frac{1}{2}\, g\,t_A^2\\x_B=-3\,x=-\frac{1}{2}\, g\,t_B^2\\ \\\frac{x_B}{x_A} =\frac{t_B^2}{t_A^2} \\\frac{t_B^2}{t_A^2}=\frac{-3\,x}{-x} \\\frac{t_B^2}{t_A^2}=3\\(\frac{t_B}{t_A})^2=3[/tex]

Notice that this quotient is exactly the square of the quotient of velocities we are looking for, therefore:

[tex](\frac{t_B}{t_A})^2=3\\(\frac{v_B}{v_A})^2=3\\ \frac{v_B}{v_A}=\sqrt{3}[/tex]

Answer:

Vrel= 0.75c

Explanation:

See attached file

Answer:

The movement of energy and matter in a system differs from one system to another. On the other hand, in open system both the matter and energy move into and out of the system. Therefore, matter and energy in a system naturally change over time will decrease in entropy.

Explanation:

Answer:

Decrease in entropy

Explanation:

Various systems which exist in nature possess energy and matter that move through these system continuously. The movement of energy and matter in a system differs from one system to another.

In a closed system for example, only energy flows in and out of the system while matter does not enter or leave the system.

On the other hand, in open system both the matter and energy move into and out of the system.

Answer:

A)     K / K₀ = 4   b)     v / v₀ = 4

Explanation:

A) For this exercise we can use the conservation of mechanical energy

in the problem it indicates that the displacement was doubled (x = 2xo)

starting point. At the position of maximum displacement

      Em₀ = Ke = ½ k (2x₀)²

final point. In the equilibrium position

      [tex]Em_{f}[/tex] = K = ½ m v²

        Em₀ = Em_{f}

        ½ k 4 x₀² = K

        (½ K x₀²) = K₀

         K = 4 K₀

          K / K₀ = 4

B) the speed value

          ½ k 4 x₀² = ½ m v²

          v = 4 (k / m) x₀

if we call

           v₀ = k / m x₀

          v = 4 v₀

         v / v₀ = 4

Explanation:

Consider a loop of wire, which has an area of [tex]A=14 \mathrm{cm}^{2}[/tex] and [tex]N=250[/tex] turns, it is initially placed perpendicularly in the earth magnetic field. Then it is rotated from this position to a position where its plane is parallel to the field as shown in the following figure in [tex]\Delta t=0.030[/tex] s. Given that the earth's magnetic field at the position of the loop is [tex]B=5.0 \times 10^{-5} \mathrm{T}[/tex], the flux through the loop before it is rotated is,

[tex]\Phi_{B, i} &=B A \cos \left(\phi_{i}\right)=B A \cos \left(0^{\circ}\right[/tex]

[tex]=\left(5.0 \times 10^{-5} \mathrm{T}\right)\left(14 \times 10^{-4} \mathrm{m}^{2}\right)(1)[/tex]

[tex]=7.0 \times 10^{-8} \mathrm{Wb}[/tex]

[tex]\quad\left[\Phi_{B, i}=7.0 \times 10^{-8} \mathrm{Wb}\right[/tex]

after it is rotated, the angle between the area and the magnetic field is [tex]\phi=90^{\circ}[/tex] thus,

[tex]\Phi_{B, f}=B A \cos \left(\phi_{f}\right)=B A \cos \left(90^{\circ}\right)=0[/tex]

[tex]\qquad \Phi_{B, f}=0[/tex]

(b) The average magnitude of the emf induced in the coil equals the change in the flux divided by the time of this change, and multiplied by the number of turns, that is,

[tex]{\left|\mathcal{E}_{\mathrm{av}}\right|=N\left|\frac{\Phi_{B, f}-\Phi_{B, i}}{\Delta t}\right|}{=} & \frac{1.40 \times 10^{-5} \mathrm{Wb}}{0.030 \mathrm{s}}[/tex]

[tex]& 3.6 \times 10^{-4} \mathrm{V}=0.36 \mathrm{mV}[/tex]

[tex]\mathbb{E}=0.36 \mathrm{mV}[/tex]

(a) The initial and final flux through the coil is 1.75 × 10⁻⁵ Wb and 0 Wb

(b) The induced EMF in the coil is 0.583 mV

Given that the coil has N = 250 turns

and an area of A = 14cm² = 1.4×10⁻³m².

It is rotated for a time period of Δt = 0.030s such that it is parallel with the earth's magnetic field that is B = 5×10⁻⁵T

(a) The flux passing through the coil is given by:

Ф = NBAcosθ

where θ is the angle between area vector and the magnetic field

The area vector is perpendicular to the plane of the coil.

So, initially, θ = 0°, as area vector and earth's magnetic field both are perpendicular to the plane of the coil

So the initial flux is:

Φ = NABcos0° = NAB

Ф = 250×1.4×10⁻³×5×10⁻⁵ Wb

Ф = 1.75 × 10⁻⁵ Wb

Finally, θ = 90°, and since cos90°, the final flux through the coil is 0

(b) The EMF induced is given by:

E = -ΔФ/Δt

E = -(0 - 1.75 × 10⁻⁵)/0.030

E = 0.583 × 10⁻³ V

E = 0.583 mV

Learn more about magnetic flux:

https://brainly.com/question/15359941?referrer=searchResults

Answer:

f. 80 and 90

Explanation:

1 x 10⁻¹² W/m² sound intensity falls within 0 sound level

1 x 10⁻¹¹ W/m² sound intensity falls within 10 sound level

1 x 10⁻¹⁰ W/m² sound intensity falls within 20 sound level

1 x 10⁻⁹ W/m² sound intensity falls within 30 sound level

1 x 10⁻⁸ W/m² sound intensity falls within 40 sound level

1 x 10⁻⁷ W/m² sound intensity falls within 50 sound level

1 x 10⁻⁶ W/m² sound intensity falls within 60 sound level

1 x 10⁻⁵ W/m² sound intensity falls within 70 sound level

1 x 10⁻⁴ W/m² sound intensity falls within 80 sound level

1 x 10⁻³ W/m² sound intensity falls within 90 sound level

Given sound intensity (3.66 x 10⁻⁴ W/m²) falls with 1 x 10⁻⁴ W/m² of intensity which is within 80 and 90 sound level.

f. 80 and 90

Answer:

-50N

Explanation:

F=ma=m(Vf-Vi)/t

F=(10)(-10)/(2)=-50N

So the force acting on the block is -50N, where the negative sign simply tells us that the force is opposite to the direction of movement.

Explanation:

(a) Given that,

Area of a parallel plate capacitor, [tex]A=1.8\ cm^2=1.8\times 10^{-4}\ m^2[/tex]

The separation between the plates of a capacitor, [tex]d=0.01\ mm = 10^{-5}\ m[/tex]

The dielectric constant of, k = 2.1

When a dielectric constant is inserted between parallel plate capacitor, the capacitance is given by :

[tex]C=\dfrac{k\epsilon_o A}{d}[/tex]

Putting all the values we get :

[tex]C=\dfrac{2.1\times 8.85\times 10^{-12}\times 1.8\times 10^{-4}}{0.01\times 10^{-3}}\\\\C=3.345\times 10^{-10}\ F\\\\C=334.5\ pF[/tex]

(b) We know that the Teflon has dielectric strength of 60 MV/m, [tex]E=60\times 10^6\ V/m[/tex]

The voltage difference between the plates at this critical voltage is given by :

[tex]V=Ed\\\\V=60\times 10^6\times 0.01\times 10^{-3} \\\\V=600\ V[/tex]

or

V = 0.6 kV

We have that the Capacitance and potential difference is mathematically given as



Question Parameters:

having a plate area of 1.80 cm2 and a plate separation of 0.010 0 mm

having a plate area of 1.80 cm2 and a plate separation of 0.010 0 mm.

a)

Generally the equation for the Capacitance  is mathematically given as

[tex]C=\frac{ke_0A}{d}\\\\Therefore\\\\C=\frac{2.1*1.80e-4*8.85e12}{0.01e-3}\\\\[/tex]

C=334.68pF

b)

Generally the equation for the Capacitance  is mathematically given as

[tex]Vmax=\frac{Q}{C}[/tex]

Where

Q is the charge on the plates, and hence not given

Therefore, maximum potential difference is

[tex]Vmax=\frac{Q}{334.68pF}[/tex]

For more information on potential difference visit

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Answer:

Explanation:

To solve for the power consumed by the trains motor we have to employ the formula for power which is

Power= current * voltage

Given that

voltage V= 800 V

current I= 2130 A

Substituting in the formula for power we have

Power= 2130*800=  1704000 watt

Power = 1704 kW

This is the amount of energy consumed, transferred or converted per unit of time

Hence the power consumed  by the trains motor is 1704 kW

Answer:

1.125×10⁻⁹ J

Explanation:

Applying,

E = 1/2CV²................... Equation 1

Where E = Energy stored in the capacitor, C = capacitance of the capacitor, V = Voltage of the battery.

Given; C = 1.0 nF,  = 1.0×10⁻⁹ F, V = 1.5 V

Substitute into equation 1

E = 1/2(1.0×10⁻⁹×1.5²)

E = 1.125×10⁻⁹ J

Hence the energy stored by the capacitor is 1.125×10⁻⁹ J

Answer:

b) True. potencial diferencie does not depend on orientation

Explanation:

In this exercise we are asked to show which statements are true.

The expression the potential with respect to earth or the electric field with respect to earth refers to the potential or electric charge of the planet that is assumed to be very large and does not change in value during work.

It does not refer to the height of the system.

We can now review the claims

a) False. Potential not to be refers to height

b) True. Does not depend on orientation

c) False The potential does not refer to the altitude but to the Earth's charge

Answer

76200meters

Explanation:

we know that 1km=1000meters

to convert km into meters we we divide km by meters

=76.2/1000

=76200meters

Answer:

-0.73mA

Explanation:

Using amphere's Law

ε =−dΦB/ dt

=−(2.6T)·(7.30·10−4 m2)/ 1.00 s

=−1.9 mV

Using ohms law

ε=V =IR

I = ε/ R =−1.9mV/ 2.60Ω =−0.73mA

Answer:

a. to the west.

Explanation:

An electron in a magnetic field always experience a force that tends to change its direction of motion through the magnetic field. According to Lorentz left hand rule (which is the opposite of Lorentz right hand rule for a positive charge), the left hand is used to represent the motion of an electron in a magnetic field. Hold out the left hand with the fingers held out parallel to the palm, and the thumb held at right angle to the other fingers. If the thumb represents the motion of the electron though the field, and the other fingers represent the direction of the field, then the palm will push in the direction of the force on the particle.

In this case, if we point the thumb (which shows the direction we shot the electron) to the south (towards your body), with the palm (shows the direction of the force) facing up to the roof, then the fingers (the direction of the field) will point west.

Answer:

A. largest: (4000 kg, 5 m/s; 1000 kg, 20 m/s)

medium: (2000 kg, 5 m/s; 500 kg, 20 m/s; 1000 kg, 10 m/s)

smallest: (500 kg, 10 m/s)

B. largest: (4000 kg, 5 m/s; 1000 kg, 20 m/s)

medium: (2000 kg, 5 m/s; 500 kg, 20 m/s; 1000 kg, 10 m/s)

smallest: (500 kg, 10 m/s)

C. You can't say anything about the forces required until we know about the time frames required for each one to stop. So If they all stopped in the same time interval, then the rankings are the same.

The researchers need to compare those who contracted the disease to those who did not.

Complete question is;

The place you get your hair cut has two nearly parallel mirrors 6.50 m apart. As you sit in the chair, your head is 3.00 m from the nearer mirror. Looking toward this mirror, you first see your face and then, farther away, the back of your head. (The mirrors need to be slightly nonparallel for you to be able to see the back of your head, but you can treat them as parallel in this problem.) How far away does the back of your head appear to be?

Answer:

13 m

Explanation:

We are given;

Distance between two nearly parallel mirrors; d = 6.5 m

Distance between the face and the nearer mirror; x = 3 m

Thus, the distance between the back-head and the mirror = 6.5 - 3 = 3.5m

Now, From the given values above and using the law of reflection, we can find the distance of the first reflection of the back of the head of the person in the rear mirror.

Thus;

Distance of the first reflection of the back of the head in the rear mirror from the object head is;

y' = 2y

y' = 2 × 3.5

y' = 7

The total distance of this image from the front mirror would be calculated as;

z = y' + x

z = 7 + 3

z = 10

Finally, the second reflection of this image will be 10 meters inside in the front mirror.

Thus, the total distance of the image of the back of the head in the front mirror from the person will be:

T.D = x + z

T.D = 3 + 10

T.D = 13m

Answer:

your question answer is 22°

Answer:

The  charge on the dust particle is  [tex]q_d = 6.94 *10^{-13} \ C[/tex]

Explanation:

From the question we are told that

    The length is  [tex]l = 2.0 \ m[/tex]

    The width is  [tex]w = 4.0 \ m[/tex]

   The charge is  [tex]q = -10\mu C= -10*10^{-6} \ C[/tex]

    The mass suspended in mid-air is [tex]m_a = 5.0 \mu g = 5.0 *10^{-6} \ g = 5.0 *10^{-9} \ kg[/tex]

Generally the electric field on the carpet is mathematically represented as

           [tex]E = \frac{q}{ 2 * A * \epsilon _o}[/tex]

Where [tex]\epsilon _o[/tex] is the permittivity of free space with value [tex]\epsilon_o = 8.85*10^{-12} \ \ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2[/tex]

substituting values

           [tex]E = \frac{-10*10^{-6}}{ 2 * (2 * 4 ) * 8.85*10^{-12}}[/tex]

           [tex]E = -70621.5 \ N/C[/tex]

Generally the electric force keeping the dust particle on the air  equal to the force of gravity acting on the particles

        [tex]F__{E}} = F__{G}}[/tex]

=>     [tex]q_d * E = m * g[/tex]

=>      [tex]q_d = \frac{m * g}{E}[/tex]

=>      [tex]q_d = \frac{5.0 *10^{-9} * 9.8}{70621.5}[/tex]

=>     [tex]q_d = 6.94 *10^{-13} \ C[/tex]

Answer:

The uncertainty in momentum is 5.25x 10^25Jsm

Explanation:

We know that

h bar = h/2π

So

1.05x 10^34=h/2pπ

h=1.05x 10^ 34(2π)=6.597x 10^-34Js

dp=(6.597x10^-34/4pπ)/(1x10^-10)

=5.25x10^-25 Jsm