A laser beam is incident at an angle of 30.2° to the vertical onto a solution of corn syrup in water. (a) If the beam is refracted to 18.82° to the vertical, what is the index of refraction of the syrup solution? (b) Suppose the light is red, with wavelength 632.8 nm in a vacuum. Find its wavelength in the solution. nm (c) Find its frequency in the solution. Hz (d) Find its speed in the solution.

Answer:

The velocity of the ball after collision is  [tex]v_{2x} = -4 m/s[/tex]

Explanation:

From the question we are told that

     The mass of the ball is  [tex]m_b = 1.00\ kg[/tex]

      The velocity of the ball is  [tex]v_{1x}= 12.0 \ m/s[/tex]

      The mass of the rectangular door is  [tex]m_d = 30 \ kg[/tex]

        The width of the door is  [tex]a = 1.00 \ m[/tex]

         The distance of impact from the hinge is  [tex]L = 75 \ cm = \frac{75}{100} = 0.75 \ m[/tex]

         The angular speed of the door is  [tex]w = 1.20 \ rad/s[/tex]

So the moment of inertia of the door is given from the question as

            [tex]I = \frac{1}{3} M a^2[/tex]

substituting values

           [tex]I = \frac{1}{3} * 30 * (1)[/tex]

           [tex]I = 10 \ kg \cdot m^2[/tex]

According to the law of angular momentum conservation

           [tex]L_i = L_f[/tex]

Where  [tex]L_i[/tex] is the initial angular momentum of the system(the door and the ball) which is mathematically represented as

           [tex]L_i = m_b * v_{1x} + Iw_i[/tex]

so  [tex]w_i[/tex] is the initial angular speed of the door which is zero

So    

                 [tex]L_i = m_b * v_{1x}[/tex]

[tex]L_f[/tex] is the final angular momentum of the system(the door and the ball) which is mathematically represented as

         [tex]L_f = I w + m_b v_{2x} * L[/tex]

So  

      [tex]m_b * v_{1x} = I w + m_b v_{2x} * L[/tex]

Substituting values

      [tex]1 * 12 * 0.75 = 10* 1.2 * v_{2x} * 0.75[/tex]

         [tex]v_{2x} = -4 m/s[/tex]

The negative sign show a reversal in the balls direction

Answer:

a) 82.8°

b)1.44 rad

c)0.23λ

Explanation:

Wave function form for each wave will be

-> wave 1:

[tex]y_{1}[/tex](x,t)=[tex]y_{m[/tex]sin( kx -  ωt)

-> wave 2:

[tex]y_{2[/tex](x,t)=[tex]y_{m[/tex]sin( kx -  ωt +φ)

The summation of above two functions is the resultant wave.

Y(x,t)=[tex]y_{m[/tex]sin( kx -  ωt)+[tex]y_{m[/tex]sin( kx -  ωt +φ)

Using the trigonometric addition formula for sin i.e

sin(A)+ sin(B) = 2 sin(A+B/2) cos(A-B/2)

Y(x,t)=2[tex]y_{m[/tex]cos(φ/2) sin(kx - ωt +φ)

When comparing to the general wave form, the amplitude is 2[tex]y_{m[/tex]cos(φ/2)m

Also, [tex]y_{m[/tex]cos(φ/2)= 1.5 [tex]y_{m[/tex]

φ/2=[tex]cos^{-1}[/tex](1.5/s)

(a) φ= 82.8°

(b)φ= 82.8π/180 =>1.44 rad

(c) one complete wave is 2π radians

Therefore, for two waves,  φ/2π= 1.44/2π => 0.23 fraction of a complete wave from each other i.e they are separated by 0.23λ

Answer:

a) 750W

b) 800W

  1018.9W

c) the circular frame

Explanation:

a) To find the electric flux you use the following formula:

[tex]\Phi_E=EAcos\theta=EA[/tex]

where you have assumed that the plane of the loop and B are perpendicular between them.

By replacing you obtain:

[tex]\Phi_E=(2*10^4N/C)(0.25m)(0.15m)=750W[/tex]

b) In order to calculate the electric flux in a square frame you first calculate the perimeter of the rectangular frame of wire:

P = 0.25m+0.25m+0.15m+0.15m=0.8m

A square with this length will have sides of 0.2m

Hence. you have that the electric flux is:

[tex]\Phi_E=(2*10^4N/C)(0.2m)^2=800W[/tex]

for a circular frame you have that the radius is:

[tex]s=2\pi r\\\\r=\frac{s}{2\pi}=\frac{0.8m}{2\pi}=0.127m[/tex]

Then, the electric flux will be:

[tex]\Phi_E=(2*10^4N/C)(\pi)(0.127m)^2=1018.9W[/tex]

c) The electric flux is maximum in the circular frame.

Answer:

Explanation:

For acceleration due to gravity

g = GM / R²  ---------------------------- ( 1 )

At height 2R , radius of orbit is 3R .

If speed of satellite be v

m v² / 3R = GMm / (3R)² ( centripetal force = gravitational force )

v² = GM / 3R -------------------------------------- ( 2 )

Dividing (1) and (2)

v² / g = R / 3

v² = gR / 3

v = √( gR / 3 )

Answer:

A) first laser

B) 0.08m

C) 0.64m

Explanation:

To find the position of the maximum you use the following formula:

[tex]y=\frac{m\lambda D}{d}[/tex]

m: order of the maximum

λ: wavelength

D: distance to the screen = 4.80m

d: distance between slits

A) for the first laser you use:

[tex]y_1=\frac{(1)(d/20)(4.80m)}{d}=0.24m\\[/tex]

for the second laser:

[tex]y_2=\frac{(1)(d/15)(4.80m)}{d}=0.32m[/tex]

hence, the first maximum of the first laser is closer to the central maximum.

B) The difference between the first maximum:

[tex]\Delta y=y_2-y_1=0.32m-0.24m=0.08m=8cm[/tex]

hence, the distance between the first maximum is 0.08m

C) you calculate the second maximum of laser 1:

[tex]y_{m=2}=\frac{(2)(d/20)(4.80m)}{d}=0.48m[/tex]

and for the third minimum of laser 2:

[tex]y_{minimum}=\frac{(m+\frac{1}{2})(\lambda)(D)}{d}\\\\y_{m=3}=\frac{(3+\frac{1}{2})(d/15)(4.80m)}{d}=1.12m[/tex]

Finally, you take the difference:

[tex]1.12m-0.48m=0.64m[/tex]

hence, the distance is 0.64m

Answer:

5.4  × 10⁸ W/m²

Explanation:

Given that:

The Power (P) of Betelgeuse is estimated to release 3.846 × 10³¹ W

the mass of the exoplanet = 5.972 × 10²⁴ kg

radius of the earth = 1.27 × 10⁷ m

half the distance (i.e radius r ) = 7.5  × 10¹⁰ m

a) What is the intensity of Betelgeuse at the "earth’s" surface?

The Intensity of  Betelgeuse  can be determined by using the formula:

[tex]Intensity \ I = \frac{P}{4 \pi r^2}[/tex]

[tex]I = \frac{3.846*10^{31}}{4 \pi (7.5*10^{10})^2}[/tex]

I = 544097698.8 W/m²

I = 5.4  × 10⁸ W/m²

Answer:

1.97*10^14 W/m^2

Explanation:

To find the intensity of the light emitted by Betelgeuse you taken into account that the light from Betelgeuse expands spherically in space.

You use the following formula:

[tex]I=\frac{P}{A}=\frac{P}{4\pi r^2}[/tex]     (1)

I: intensity of light

r: distance from Betelgeuse to the exoplanet = (1/2)*7.5*10^10m

P: power = 100,000*3.486*10^31 W

By replacing the values of the parameters in (1) you obtain:[tex]I=\frac{100000(3.486*10^{31}W}{4\pi (0.5*7.5*10^{10}m)^2}=1.97*10^{14}\frac{W}{m^2}[/tex]

Answer:

Option A, C and D are correct statements.

Explanation:

Torque is the twisting effect of a force which causes an object to acquire angular acceleration.

The direction of the torque depends on the direction of the force on the axis. The SI unit for torque is the Newton-meter.

The net torque is the sum of the individual torques.

If the net torque on a rotatable object is zero then:

1. It will be in rotational equilibrium and the angular velocity of the body remains constant.

2. It will be in rotational equilibrium and not able to acquire angular acceleration. Thus, it will not be able to acquire angular velocity.

3. Angular momentum will be conserved.

If the forces on an object are balanced, the net force is zero.

If there net force on an object is zero, its speed and direction of motion do not change, including if it is at rest, the object will not accelerate and the velocity will remain constant.

By considering the basic fundamentals of Torque, the statements (A), (C) and (D) are correct.

The given problem is based on the concept and fundamentals of Torque. Torque is the twisting effect of a force which causes an object to acquire angular acceleration.

The direction of the torque depends on the direction of the force on the axis. The SI unit for torque is the Newton-meter.  The net torque is the sum of the individual torques.

If the net torque on a rotatable object is zero then:

Thus, we can conclude that by considering the basic fundamentals of Torque, the statements (A), (C) and (D) are correct.

Learn more about the fundamentals of Torque here:

https://brainly.com/question/19247046

Answer:

6.444 × [tex]10^{-5}[/tex] T

Explanation:

Find the given attachments

The magnitude of the magnetic field at the point [tex](x = 3 \text{ cm}, y = 2 \text{ cm})[/tex] is approximately [tex]\( 1.08 \times 10^{-5} \text{ T} \)[/tex].

The magnitude of the magnetic field [tex]\( B \)[/tex] at a distance [tex]\( r \)[/tex] from a long straight wire carrying a current [tex]\( I \)[/tex] is given by:

[tex]\[ B = \frac{\mu_0 I}{2\pi r} \][/tex]

where [tex]\( \mu_0 \)[/tex] is the permeability of free space [tex]\( \mu_0 = 4\pi \times 10^{-7} \text{ T}\cdot\text{m/A} \)[/tex].

For the horizontal wire, the distance from the wire to the point is [tex]\( y = 2 \text{ cm} = 0.02 \text{ m} \)[/tex]. Thus, the magnetic field due to the horizontal wire is:

[tex]\[ B_x = \frac{\mu_0 (4.5 \text{ A})}{2\pi (0.02 \text{ m})} \][/tex]

Next, consider the vertical section of the wire (along the y-axis). Similarly, the magnetic field due to this section at a perpendicular distance [tex]\( r \)[/tex] is given by the same formula. Here, the distance from the wire to the point is [tex]\( x = 3 \text{ cm} = 0.03 \text{ m} \)[/tex].

The magnetic field due to the vertical wire is:

[tex]\[ B_y = \frac{\mu_0 (4.5 \text{ A})}{2\pi (0.03 \text{ m})} \][/tex]

The total magnetic field [tex]\( B \)[/tex] at the point is the vector sum of [tex]\( B_x \)[/tex] and [tex]\( B_y \)[/tex]. We can use Pythagoras' theorem to find the magnitude of the resultant magnetic field:

[tex]\[ B = \sqrt{B_x^2 + B_y^2} \][/tex]

Substituting the expressions for [tex]\( B_x \)[/tex] and [tex]\( B_y \)[/tex] we get:

[tex]\[ B = \sqrt{\left(\frac{\mu_0 (4.5 \text{ A})}{2\pi (0.02 \text{ m})}\right)^2 + \left(\frac{\mu_0 (4.5 \text{ A})}{2\pi (0.03 \text{ m})}\right)^2} \][/tex]

Plugging in the value of [tex]\( \mu_0 \)[/tex] and simplifying, we find:

[tex]\[ B = \sqrt{\left(\frac{4\pi \times 10^{-7} \text{ T}\cdot\text{m/A} (4.5 \text{ A})}{2\pi (0.02 \text{ m})}\right)^2 + \left(\frac{4\pi \times 10^{-7} \text{ T}\cdot\text{m/A} (4.5 \text{ A})}{2\pi (0.03 \text{ m})}\right)^2} \][/tex]

[tex]\[ B = \sqrt{\left(\frac{18 \times 10^{-7} \text{ T}\cdot\text{m}}{0.02 \text{ m}}\right)^2 + \left(\frac{18 \times 10^{-7} \text{ T}\cdot\text{m}}{0.03 \text{ m}}\right)^2} \][/tex]

[tex]\[ B = \sqrt{\left(9 \times 10^{-6} \text{ T}\right)^2 + \left(6 \times 10^{-6} \text{ T}\right)^2} \][/tex]

[tex]\[ B = \sqrt{81 \times 10^{-12} \text{ T}^2 + 36 \times 10^{-12} \text{ T}^2} \][/tex]

[tex]\[ B = \sqrt{117 \times 10^{-12} \text{ T}^2} \][/tex]

[tex]\[ B \approx 1.08 \times 10^{-5} \text{ T} \][/tex]

Answer:

a) 0.83H

b) 3.22/N Henry

Explanation:

Given two inductors L1 = 1.31 H and L2 = 2.24 H connected in parallel, their equivalent inductance derivative is similar to that of resistance in parallel.

Derivation:

If the voltage across an inductor

VL = IXL

I is the current

XL is the inductive reactance

XL = 2πfL

VL = I(2πfL)

L is the inductance.

From the formula, I = V/2πfL

Given two inductors in parallel, different current will flow through them.

I1 = V/2πfL1 (current in L1)

I2 = V/2πfL2 (current in L2)

Total current I = I1+I2

I = V/2πfL1 + V/2πfL2

I = V/2πf{1/L1+1/L2}

V/2πfL = V/2πf{1/L1+1/L2}

1/L = 1/L1+1/L2 (equivalent inductance in parallel)

Given L1 = 1.31 H and L2 = 2.24

1/L = 1/1.31 + 1/2.24

1/L = 0.763 + 0.446

1/L = 1.209

L = 1/1.209

L = 0.83H

The equivalent inductance is 0.83H

b) Given similar inductors L = 3.22H in parallel, the equivalent inductance will be:

1/L = 1/3.22+1/3.22+1/3.22+1/3.22+1/3.22

+1/3.22+1/3.22+1/3.22+1/3.22+1/3.22

1/L = 10/3.22 (since that all have the same denominator)

L = 3.22/10

If N = 10, the generalization of 10 similar inductors in parallel will be:

L = 3.22/N Henry

Answer:

Explanation:

moment of inertia of the door M = 1 /3 m ( l² + b² + d² )

= 1 / 3 x 7.5 x ( 2.3² + 1² + .03² )

= 1 / 3 x 7.5 x 6.2909

I = 15.7272.

moment of inertia of the door + small toy

15.7272 + .12 x .5²

= 15.7272 + .03

= 15.7572

Applying conservation of angular momentum law

mvr = M ω

m is mass of the toy thrown with velocity v at distance r from the axis , M is moment of inertia of door + toy and ω is angular velocity ith which door opens.

.12 x v x .5 =  15.7572 x .19

v = 49.89 m /s .

Answer:

a  

The tension in the string is  [tex]T = 0.85 N[/tex]

 b

 The new balance reading is  [tex]M_b = 885.86 g[/tex]

Explanation:

From the question we are told that

    The mass of the beaker of water is  [tex]m = 875 .0g[/tex]

     The diameter of the copper ball is  [tex]d = 2.75 cm = \frac{2.75}{100} = 0.0275 m[/tex]

There are two forces acting on the copper ball

   The first is the Buoyant force of the water pushing it up which is mathematically represented as

                     [tex]F = \rho V g[/tex]

Where [tex]\rho[/tex] is the density of water which has value of  [tex]\rho = 1000 kg/m^3[/tex]

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

          [tex]V[/tex] is the volume of water displaced by the copper ball  which is mathematically evaluated as

                             [tex]V = \frac{4}{3} \pi r^3[/tex]

The radius r is  [tex]r = \frac{d}{3} = \frac{0.0275}{2} = 0.01375 m[/tex]

Substituting value  

                        [tex]V = \frac{4}{3} * 3.142 * (0.01375)^3[/tex]

                            [tex]V = 1.08 9 *10^{-5 } m^3[/tex]  

   Substituting for  F

              [tex]F = 1000 * 1.089 *10^{-5} * 9.8[/tex]

               [tex]F = 0.1067 N[/tex]      

     The second force is the weight of the copper ball which is mathematically represented as

       [tex]W_c = mg[/tex]

Now m is the mass which can be mathematically evaluated as

        [tex]m = \rho_c * V[/tex]

Where  is the density of copper with  value of  [tex]\rho_c = 8960 kg /m^3[/tex]

So      [tex]m = 8960 * 1.089 *10^{-5}[/tex]

         [tex]m = 0.0976[/tex]

So the weight of copper is  

             [tex]W_c = 0.09756 * 9.8[/tex]

            [tex]W_c = 0.956 N[/tex]

Now the tension the string would be mathematically evaluated as

            [tex]T = W_c - F[/tex]

So        [tex]T = 0.956 - 0.1067[/tex]

           [tex]T = 0.85 N[/tex]

From this value we that the string is holding only 0.85 N of the copper weight thus (0.956 - 0.85 = 0.1065 N ) is being held by the balance

Now the mass equivalent of this weight is mathematically evaluated as

             [tex]m_z = \frac{1.0645 }{9.8 }[/tex]

             [tex]m_z = 0.01086 kg[/tex]

Converting to grams  

                     [tex]m_z = 10.86 g[/tex]

So the new balance reading is  

                  [tex]M_b = 875.0 +10.86[/tex]

                  [tex]M_b = 885.86 g[/tex]

Answer:

Yes, the design has a hope of working.

A dull face should yield more force.

Explanation:

According to the de broglie theory, and the wave-particle duality, waves can be assumed and made to behave as a matter and vice versa. Under these circumstances, the impinging wave can act as a particle (photon) colliding with a relatively stationary body, yielding some of its momentum to the relatively stationary body. With this, the momentum of the photons can be used to propel the ship by its solar sail.

Making the face dull should yield more force in the sense that the incident photon is not reflected and all of its momentum is yielded to the sail compared to when it is reflected (bouncing off) off the sail. This will be a case of inelestic collision between the sail and the photon.

Answer:

22.4 m/s

Explanation:

To find the speed of the crate you take into account the momentum conservation. The momentum before Jack and Zack jump must be equal to the momentum after. Then you have:

[tex]p_b=p_a\\\\[/tex]

Furthermore, you take into account that the momentum before, when crater, Jack and Zack are at rest, the momentum is zero. Hence:

[tex]p_b=(m_Z+m_J+m_c)v=0\\\\m_Z=45kg\\\\m_J=70kg\\\\m_c=15kg\\\\p_a=m_Zv_Z+m_Jv_J+m_cv_c[/tex]

but you know what are the velocities of Zack and Jack after they jump.

[tex]p_a=p_b=0\\\\-v_c=\frac{m_Zv_Z+m_Jv_J}{m_c}\\\\-v_c=\frac{(70kg)(4m/s)+(45kg)(4m/s)}{15kg}=-22.4\frac{m}{s}[/tex]

the minus sign means that the velocity of the crater is in an opposite direction of the spee of Jack and Zack.

hence, the speed of the crater is 22/4m/s

The final speed of the crate when Jack and Zack jump off simultaneously in the same direction is calculated to be approximately 3.54 m/s. The negative sign indicates opposite direction to the jump.

To solve for the final speed of the crate, we will use the principle of conservation of momentum. Since the system is initially at rest, the total initial momentum is zero.

Let's denote the mass of Jack as mJ, the mass of Zack as mZ, and the mass of the crate as mC. Also, let the velocity of Jack, Zack, and the crate relative to the ground be VJ, VZ, and VC respectively.

[tex]mJ = 70 kg[/tex]

[tex]mZ = 45 kg[/tex]

[tex]mC = 15 kg[/tex]

Since Jack and Zack both push off with a speed of 4 m/s relative to the crate:

VJ = VC + 4 m/s

VZ = VC + 4 m/s

Using conservation of momentum:

Initial momentum = Final momentum

[tex]0 = mJ(VC + 4) + mZ(VC + 4) + mCVC[/tex]

[tex]0 = 70(VC + 4) + 45(VC + 4) + 15VC[/tex]

Combine and solve for VC:

[tex]0 = 70VC + 280 + 45VC + 180 + 15VC[/tex]

[tex]0 = 130VC + 460[/tex]

[tex]-460 = 130VC[/tex]

[tex]VC = -460 / 130[/tex]

[tex]VC = -3.54 m/s[/tex]

The negative sign indicates the crate moves in the opposite direction to their jump. Thus, the final speed of the crate is 3.54 m/s.

Answer:

a) L = 2.10x10⁴⁰ kg*m²/s

b) τ = 1.12x10²⁴ N.m

Explanation:

a) The angular momentum (L) of the pulsar can be calculated using the following equation:

[tex] L = I \omega [/tex]

Where:

I: inertia momentum

ω: angular velocity

First we need to calculate ω and I. The angular velocity can be calculated as follows:

[tex] \omega = \frac{2 \pi}{T} [/tex]

Where:

T: is the period = 33.5x10⁻³ s

[tex] \omega = \frac{2 \pi}{T} = \frac{2 \pi}{33.5 \cdot 10^{-3} s} = 187.56 rad/s [/tex]

The inertia moment of the pulsar can be calculated using the following relation:

[tex] I = \frac{2}{5}mr^{2} [/tex]

Where:

m: is the mass of the pulsar = 2.8x10³⁰ kg

r: is the radius = 10.0 km

[tex] I = \frac{2}{5}mr^{2} = \frac{2}{5}2.8\cdot 10^{30} kg*(10\cdot 10^{3} m)^{2} = 1.12 \cdot 10^{38} kg*m^{2} [/tex]

Now, the  angular momentum of the pulsar is:

[tex] L = I \omega = 1.12 \cdot 10^{38} kg*m^{2}*187.56 rad/s = 2.10 \cdot 10^{40} kg*m^{2}*s^{-1} [/tex]

b) If the angular velocity decreases at a rate of 10⁻¹⁴ rad/s², the torque of the pulsar is:

[tex] \tau = I*\alpha [/tex]

Where:

α: is the angular acceleration = 10⁻¹⁴ rad/s²

[tex]\tau = I*\alpha = 1.12 \cdot 10^{38} kg*m^{2} * 10^{-14} rad*s^{-2} = 1.12 \cdot 10^{24} N.m[/tex]

I hope it helps you!

Final answer:

To find the angular momentum of the pulsar, we can use the formula L = Iω, where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.

Explanation:

To find the angular momentum of the pulsar, we can use the formula L = Iω, where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.

The moment of inertia for a solid sphere can be calculated using the formula I = (2/5)mr², where m is the mass and r is the radius.

Using the given radius and mass of the pulsar, we can calculate the moment of inertia.

Once we have the moment of inertia, we can use the given period of the pulsar to calculate the angular velocity.

To find the torque on the pulsar, we can use the formula τ = Iα, where τ is the torque, I is the moment of inertia, and α is the angular acceleration.

The angular acceleration can be calculated using the given rate at which the angular velocity is decreasing.

Using these formulas and the given values, we can find the angular momentum and torque of the pulsar.

Answer:

[tex]\dfrac{F_2}{F_1}=\dfrac{1}{4}[/tex]

Explanation:

G = Gravitational constant

r = Distance between Earth and object

M = Mass of Earth

m = Mass of object

Centripetal force on the space station

[tex]F_1=\dfrac{GMm}{r^2}[/tex]

Centripetal force on the satellite

[tex]F_2=\dfrac{GMm}{(2r)^2}\\\Rightarrow F_2=\dfrac{GMm}{4r^2}[/tex]

From the question the required ratio is

[tex]\dfrac{F_2}{F_1}=\dfrac{\dfrac{GMm}{4r^2}}{\dfrac{GMm}{r^2}}\\\Rightarrow \dfrac{F_2}{F_1}=\dfrac{1}{4}[/tex]

The ratio is [tex]\dfrac{F_2}{F_1}=\dfrac{1}{4}[/tex]

Complete Question

A power supply has an open-circuit voltage of 40.0 V and an internal resistance of 2.00 [tex]\Omega[/tex]. It is used to charge two storage batteries connected in series, each having an emf of 6.00 V and internal resistance of 0.300[tex]\Omega[/tex] . If the charging current is to be 4.00 A, (a) what additional resistance should be added in series? At what rate does the internal energy increase in (b) the supply, (c) in the batteries, and (d) in the added series resistance? (e) At what rate does the chemical energy increase in the batteries?

Answer:

a

The additional resistance is [tex]R_z = 4.4 \Omega[/tex]

b

The rate at which internal energy increase at the supply is [tex]Z_1 = 32 W[/tex]

c

The rate at which internal energy increase in the battery  is  [tex]Z_1 = 32 W[/tex]

d

The rate at which internal energy increase in the added series resistance is  [tex]Z_3 = 70.4 W[/tex]

e

the increase rate of the chemically energy in the battery is [tex]C = 48 W[/tex]

Explanation:

From the question we are told that

    The  open circuit voltage is  [tex]V = 40.0V[/tex]

     The internal resistance is [tex]R = 2 \Omega[/tex]

     The emf of each battery is [tex]e = 6.00 V[/tex]

      The internal resistance of the battery is  [tex]r = 0.300V[/tex]

      The  charging current is  [tex]I = 4.00 \ A[/tex]

Let assume the the additional resistance to to added to the circuit is  [tex]R_z[/tex]

 So this implies that

        The total resistance in the circuit is

                              [tex]R_T = R + 2r +R_z[/tex]

Substituting values

                             [tex]R_T = 2.6 +R_z[/tex]

And  the difference in potential in the circuit is  

                         [tex]E = V -2e[/tex]

                 =>   [tex]E = 40 - (2 * 6)[/tex]

                        [tex]E = 28 V[/tex]

Now according to ohm's law

            [tex]I = \frac{E}{R_T}[/tex]

Substituting values

           [tex]4 = \frac{28}{R_z + 2.6}[/tex]        

Making [tex]R_z[/tex] the subject of the formula

So    [tex]R_z = \frac{28 - 10.4}{4}[/tex]

           [tex]R_z = 4.4 \Omega[/tex]

The  increase rate of   internal energy at the supply is mathematically represented as

        [tex]Z_1 = I^2 R[/tex]

Substituting values

     [tex]Z_1 = 4^2 * 2[/tex]

     [tex]Z_1 = 32 W[/tex]

The  increase rate of   internal energy at the batteries  is mathematically represented as

         [tex]Z_2 = I^2 r[/tex]

Substituting values

         [tex]Z_2 = 4^2 * 2 * 0.3[/tex]

         [tex]Z_2 = 9.6 \ W[/tex]

The  increase rate of  internal energy at the added  series resistance  is mathematically represented as

        [tex]Z_3 = I^2 R_z[/tex]

Substituting values

       [tex]Z_3 = 4^2 * 4.4[/tex]

      [tex]Z_3 = 70.4 W[/tex]

Generally the increase rate of the chemically energy in the battery is  mathematically represented as

         [tex]C = 2 * e * I[/tex]

Substituting values

       [tex]C = 2 * 6 * 4[/tex]

      [tex]C = 48 W[/tex]

Answer:

a. A input = 0.001669 m²

b. W= 4.193775 KJ

c. h output = 0.60357 cm

d. n = 74.55638 strokes

e. 4.1937 N = 4.1937 N

Explanation:

Hydraulic jacks lift loads using the force created by the pressure in the cylinder chamber by applying small effort. It works on Pascal's principles which explains that the pressure at a certain level and through a mass of fluid at rest is the same in all the directions.

Parameters given:

Diameter of the output piston,   d =  0.23 m

Mass of the car,  m= 950 kg

Force applied at the input piston,  f = 375 N

Height, h  = 0.45 m

(a) Finding the area of the input piston:

First, we use Pascal's principle to find the area

(f ÷ A output)  ÷ (f ÷ A input)

Where A= area

            g = 9.81

          A output = πd² ÷ 4

(f ÷ (πd² ÷ 4)) = (f ÷ A input)

[(950 x 9.81)  ÷ ((3.14 x 0.23²) ÷ 4) ] = 375 ÷ A input

9319.5 ÷ 0.0415 = 375 ÷ A input

A input = 0.001669 m²

(b) Finding the work done in lifting the car 45 cm

Work done, W = force, f x distance ( which in this case is height, h)

= (950 x 9.81) x 0.45

W= 4.193775 KJ

(c) Finding how the car move up for each stroke if the input piston moves 15 cm in each stroke.

W output = W input

F x h output = F x h input

= (950 x 9.81) x h output = 375 x 0.15

h output =  0.0060357 m

h output = 0.60357 cm

(d) Finding the number of strokes that are required to jack the car up 45 cm

n = h ÷ h output

n = 45 ÷ 0.60357 cm

n = 74.55638 strokes

(e) How the energy is conserved

W output = W input

F x h output = F x h input x n

(950 x 9.81) x 0.45 = 375  x  0.15  x 74.55638

4.1937 N = 4.1937 N

The area of the input piston will be 0.00167 square meters, the work done in lifting the car will be 4193.78 J, the height moved by car in each stroke will be 0.6035 cm, the number of strokes required to lift the car by 45 cm will be 74.57, and the energy conservation is justified.

Given information:

Mass of the car is [tex]m=950[/tex] kg

The lift of the car is [tex]h=45[/tex] cm.

The diameter of the output piston D is 23 cm.

The input force f is 375 N.

Let the output force be F and the input piston diameter be d.

Hydraulic lift follows Pascal's principle.

(a)

So, the area of the input piston will be calculated as,

[tex]\dfrac{F}{\frac{\pi}{4}D^2}=\dfrac{f}{\frac{\pi}{4}d^2}\\\dfrac{\pi}{4}d^2=\dfrac{f\frac{\pi}{4}D^2}{mg}\\\dfrac{\pi}{4}d^2=A=0.00167\rm\;m^2[/tex]

(b)

The work done in lifting the car will be,

[tex]W=mgh\\W=950\times 9.81\times0.45\\W=4193.78\rm\;J[/tex]

(c)

The input piston moves 15 cm in each stroke.

The height h' moved by car in each stroke will be calculated as,

[tex]mgh'=f\times 0.15\\h'=0.006035\rm\;m=0.6035\rm\;cm[/tex]

(d)

The number of strokes required to lift the car by 45 cm will be,

[tex]n=\dfrac{h}{h'}\\n=\dfrac{45}{0.6035}\\n=74.57[/tex]

(e)

The energy is conserved when output work is equal to input work.

So, the energy conservation can be verified as,

[tex]W_i=W_o\\f\times0.15\times n=mgh\\375\times 0.15\times74.57=950\times 9.81\times0.45\\4194.28\approx4194.1[/tex]

The input and output works are approximately equal. The value isn't exactly equal because of the rounding-off.

So, energy conservation is justified.

Therefore, the area of the input piston will be 0.00167 square meters, the work done in lifting the car will be 4193.78 J, the height moved by car in each stroke will be 0.6035 cm, the number of strokes required to lift the car by 45 cm will be 74.57, and the energy conservation is justified.

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Answer:

a)6.34 x  [tex]10^{7}[/tex]W/m²

b)1.37 x [tex]10^{3[/tex] W/m²

c) see explanation.

Explanation:

a)The relation of intensity'I' of the radiation and area 'A' is given by:

I= P/A

where P= power of sunlight i.e 3.9 x [tex]10^{26}[/tex] J

and the area of the sun is given by,

A= 4π[tex]R_{sun}[/tex] => 4π[tex](\frac{1.4*10^{9} }{2} )^{2}[/tex]

A=6.15 x [tex]10^{18}[/tex]m²

[tex]I_{sun}[/tex] =  3.9 x [tex]10^{26}[/tex] / 6.15 x [tex]10^{18}[/tex] => 6.34 x  [tex]10^{7}[/tex]W/m²

b) First determine the are of the sphere in order to determine intensity at the surface of the virtual space

A= 4π[tex]R[/tex]

Now R= 1.5 x [tex]10^{11[/tex]m

A=  4π x 1.5 x [tex]10^{11[/tex] =>2.83 x [tex]10^{23[/tex] m²

The power that each square meter of Earths surface receives

[tex]I_{earth[/tex] =   3.9 x [tex]10^{26}[/tex]/2.83 x [tex]10^{23[/tex] =>1.37 x [tex]10^{3[/tex] W/m²

c) in part (b), by assuming the shape of the wavefront of the light emitted by the Sun is a spherical shape so each point has the same distance from the source i.e sun on the wavefront.

The power per square meter reaching Earth's surface from the Sun can be estimated by considering the power output of the Sun and the distance between the Sun and Earth. The maximum power that reaches Earth's surface is approximately 1.30 kW/m². The assumptions made in this estimation include the perfect absorption of sunlight by the Earth's surface and the average radius of the Earth.

The power per square meter reaching Earth's surface from the Sun can be estimated by considering the power output of the Sun and the distance between the Sun and Earth. The power output of the Sun is given as 4.00 × 10^26 W. To estimate the power reaching Earth's surface, we need to take into account that part of the Sun's radiation is absorbed and reflected by the atmosphere. The maximum power that reaches Earth's surface is approximately 1.30 kW/m².

To estimate the power received by 1 m² of Earth's surface, we can use the fact that the area of the Earth facing the Sun is πR², where R is the radius of the Earth. Assuming the Earth is a perfect sphere, we can take the average radius of the Earth as Re ≈ 6,371 km. Therefore, the power received by 1 m² of Earth's surface can be calculated as (1.30 kW/m²) × (πR²) / (4πRe²), where Re is the radius of the Earth.

In part (b), the assumptions made include the perfect absorption of sunlight by the Earth's surface, the average radius of the Earth, and the assumption that the Earth is a perfect sphere.

Answer:

[tex]F = 2.55*10^{-4} \ N[/tex]

F = [tex]3.808 * 10^{-6} \ N[/tex]

Explanation:

The magnetic field due to the first wire can be written as:

[tex]B_1 = \frac{\mu _o I_1}{2 \pi (2 d)}[/tex]

[tex]B_1 = \frac{\mu _o I_1}{4 \pi d}[/tex]

The magnetic field due to the second wire is as follows:

[tex]B_2 = \frac{\mu _o I_2}{2 \pi d}[/tex]

The net magnetic field B = [tex]B_1 +B_2[/tex]

[tex]B = \frac{\mu _o I_1}{4 \pi d} +\frac{\mu _o I_2}{2 \pi d} \\ \\ B = \frac{ \mu_o }{2 \pi d}(\frac{I_1}{2}+ I_2)[/tex]

Also; the magnetic force on wire segment  l = 2.8 m

[tex]F = I_1lB = \frac{ \mu_o }{2 \pi d}(\frac{I_1}{2}+ I_2)[/tex]

Replacing all the values ; we have :

[tex]F = \frac{ 4 \pi * 10^{-7}*6.8*2.8}{2 \pi * 10*10^{-2}}(\frac{6.8}{2}+ 3.3)[/tex]

[tex]F = 2.55*10^{-4} \ N[/tex]

b)

Magnetic field due to the first wire is :

[tex]B_1 = \frac{\mu _o I_1}{4 \pi d}[/tex]

[tex]B_1 = \frac{4.0*10^{-7}*6.8}{4 \pi *10*10^{-2}}[/tex]

[tex]B_1 = 6.8*10^{-6} \ T[/tex]

Magnetic field due to the second wire is :

[tex]B_2 = \frac{\mu _o I_2}{2 \pi d} \\ \\ B_2 = \frac{4*10^{-7}*3.3}{2 \pi *10*10^{-2}}[/tex]

[tex]B_2 = 6.6*10^{-6}T[/tex]

[tex]B_1>B_2[/tex]

Net magnetic field B = [tex]B_1 - B_2[/tex]

B =  [tex](6.8*10^{-6} - 6.6*10^{-6})T[/tex]

B = [tex]2*10^{-7} \ T[/tex]

Magnetic force F = [tex]I_1lB[/tex]

F =  [tex]6.6*2.8 *2*10^{-7}[/tex]

F = [tex]3.808 * 10^{-6} \ N[/tex]

Answer:

Energy is transferred from mechanical energy to thermal energy and then back to mechanical energy ( D )

Explanation:

when conducting plate swings through a magnetic field, energy is transferred from mechanical energy to thermal energy and then back to mechanical energy and this is due to the induction of eddy currents during the process of that the conducting plate swings through the magnetic field.

An eddy current is induced in conductor when the magnetic field in which the conductor is located varies.

Answer:

Energy is transferred from mechanical energy to thermal energy. (A)

Explanation:

As the conducting plate is swinging, we know it posseses some form of mechanical energy (either K or U). Because the plate is conducting, charge is transfered, yielding energy transfer to thermal.

Answer:

A. The initial velocity of the bullet is [tex]= 300.6m/s[/tex]

B. Mechanical energy of the system before and after collision: 451.80 J, 0.9018 J

C. Percentage of K.E lost to heat is  = 99.8 %

Explanation:

From conservation of linear momentum,

[tex](m_{1}v_{1} +m_{2}v_{2})= (m_{1}+m_{2})v[/tex]

let the mass of the block be m1 and velocity = v1

let the mass of the bullet be m2 and velocity = v2

Let the final velocity of the system be v.

A. Plugging our parameters into the equation, we have:

[tex][(5 \times 0) +(0.01\times v_{2})]= 5.01 \times 0.6[/tex]

[tex]v_{2}=\frac{3.006}{0.01}= 300.6m/s[/tex]

Hence, the initial velocity of the bullet is [tex]= 300.6m/s[/tex]

B. The mechanical energies of the system exist in form of kinetic energy.

I. Kinetic energy of the system before collision:

[tex]0.5 \times 5\times 0^{2} + 0.5 \times 0.01 \times 300.6^{2}= 451.80 J[/tex]

II. Kinetic energy after collision:

[tex]0.5\times 5.01 \times 0.6^{2}= 0.9018 J[/tex]

C. Change in Mechanical Energy = [tex]451.8 - 0.9018 J= 450.9J[/tex]

[tex]\frac{450.9}{451.8} \times 100 =99.8%[/tex]

Percentage of K.E lost to heat is  = 99.8 %

Answer:

the final kinetic energy is 0.9eV

Explanation:

To find the kinetic energy of the electron just after the collision with hydrogen atoms you take into account that the energy of the electron in the hydrogen atoms are given by the expression:

[tex]E_n=\frac{-13.6eV}{n^2}[/tex]

you can assume that the shot electron excites the electron of the hydrogen atom to the first excited state, that is

[tex]E_{n_2-n_1}=-13.6eV[\frac{1}{n_2^2}-\frac{1}{n_1^2}]\\\\E_{2-1}=-13.6eV[\frac{1}{2^2}-\frac{1}{1}]=-10.2eV[/tex]

-10.2eV is the energy that the shot electron losses in the excitation of the electron of the hydrogen atom. Hence, the final kinetic energy of the shot electron after it has given -10.2eV of its energy is:

[tex]E_{k}=11.1eV-10.2eV=0.9eV[/tex]

The final kinetic energy the electron will have just after it collides with and excites the hydrogen atom is; E_k = 0.9 eV

The formula for the energy of the electron of hydrogen atom in its nth orbit is given by the formula;

E_n = -13.6eV/n²

Where n is the principal quantum number.

Thus;

For n = 1;

E_1 = -13.6/1²

E_1 = -13.6 eV

At n = 2;

E_2 = -13.6/2²

E_2 = -3.4 eV

Thus, the energy that the electron looses is;

E_2 - E_1 = -3.4 - (-13.6)

E_2 - E_1 = 10.2 eV

We are told that the kinetic energy of an electron is 11.1 eV just before it collides with a hydrogen atom.

Thus, the final kinetic energy the electron will have just after it collides with and excites the hydrogen atom is;

E_k = 11.1 - 10.2

E_k = 0.9 eV

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Given Information:  

Wavelength of the red laser = λr = 632.8 nm

Distance between bright fringes due to red laser = yr = 5 mm

Distance between bright fringes due to laser pointer = yp = 5.14 mm

Required Information:  

Wavelength of the laser pointer = λp = ?

Answer:

Wavelength of the laser pointer = λp = ?

Explanation:

The wavelength of the monochromatic light can be found using young's double slits formula,

y = Dλ/d  

y/λ = D/d

Where

λ is the wavelength

y is the distance between bright fringes.

d is the double slit separation distance

D is the distance from the slits to the screen

For the red laser,

yr/λr = D/d

For the laser pointer,

yp/λp = D/d

Equating both equations yields,

yr/λr = yp/λp

Re-arrange for λp

λp = yp*λr/yr

λp =  (5*632.8)/5.14

λp = 615.56 nm

Therefore, the wavelength of the small laser pointer is 615.56 nm.

Answer:

Constant

Accelerates

Constant

Positive

Answer:

V = 500 volts

I = 50 nA

Explanation:

We usually see 'high voltage' sign rather than 'high current' this is due to the fact that the amount of current depends upon the resistance of the body,

The relation of voltage, current and resistance is given by Ohm's law,

V = IR

I = V/R

As you can see, current and resistance are inversely related, the lower the resistance of the body, the higher will be the current flowing through the body.

What voltages are dangerous even if the skin is dry?

A dry human body has a resistance of approximately 100 kΩ and it gets reduced if the body is wet.

For a current of 5 mA, the corresponding voltage is,

V = 0.005*100,000

V = 500 volts

Therefore, for a dry human body, a voltage of 500 volts would be dangerous.

If voltage differences that are observed are about 0.5 mV, and the resistance of the skin is about 10 kΩ, what size currents are involved?

I = V/R

I = 0.0005/10,000

I = 50×10⁻⁹ A

I = 50 nA

Therefore, a current of 50 nA would be flowing through the EEG.

Answer:

Current flowing in the wire is 1680 A

Explanation:

It is given length of wire l = 3.5 m

Mass of the wire m = 0.03 kg

Magnetic field B = 0.00005 Tesla

Acceleration due to gravity [tex]g=9.8m/sec^2[/tex]

Mg force acting on the wire will be equal to Lorentz force acting on the wire.

Therefore [tex]mg=IBl[/tex]

[tex]0.03\times 9.8=I\times 0.00005\times 3.5[/tex]

[tex]I=1680A[/tex]

Therefore current flowing in the wire is 1680 A

Answer:

125atm

Explanation:

Given:

initial pressure [tex]P_{1[/tex]= 1 atm

initial volume  [tex]V_{1[/tex]= 250L

final volume  [tex]V_{2[/tex]=2L

Required:

Final Pressure  [tex]P_{2[/tex]=?

As Boyle’s Law expresses the pressure-volume relationship as

[tex]P_{1[/tex][tex]V_{1[/tex]= [tex]P_{2[/tex] [tex]V_{2[/tex]

(1)(250)= [tex]P_{2[/tex] (2)

[tex]P_{2[/tex] = 250/2

[tex]P_{2[/tex]=125atm

Therefore, it would take pressure of 125atm.

Using Boyle's Law, the pressure required to compress a gas of 250 L at 1 atm into a smaller volume of 2 L at constant temperature is calculated to be 125 atm.

The student is attempting to solve this problem using Boyle's Law, which states that the pressure and volume of a gas have an inverse relationship when temperature is held constant. In this context, the formula for Boyle's Law is P1V1 = P2V2.

Here, the initial conditions are a volume V1 of 250 L and a pressure P1 of 1.00 atm. The final volume V2 is given as 2.00 L, and we are asked to solve for the final pressure P2.

Replace the given values into the Boyle's Law equation: (1.00 atm) * (250.L) = P2 * (2.00 L). From this, we can solve for P2 which sums to: P2 = ((1.00 atm) * (250.L)) / (2.00 L) = 125 atm. The pressure required to compress the helium gas into a 2L container would be 125 atm.

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Answer:

- S(t) = 5t³+8t+3

- 19.7m approximately

Explanation:

Given the velocity if an object to be

v(t) = At² + B where t is the time in seconds

Velocity is the change in displacement of a body with respect to time.

V(t) = dS(t)/dt

Making S(t) the subject of the formula

dS(t) = v(t)dt

Integrating both sides

∫dS(t) =∫v(t)dt

S(t) = ∫(At²+B)dt

S(t) = At³/3+Bt + C... 1

In its initial position s(0) = 0

t = 0, s = 0

S(0) = A(0)³/3+B(0)+C

S(0)= C

a) The object's position function s(t) if A = 15, B = 8 and C = 3 can be gotten by substituting this value into eqn 1

S(t) = 15t³/3+8t+3

S(t) = 5t³+8t+3

b) For the distance traveled by the object between t equals 1 and t equals 2 seconds

When t = 1s

S(1) = 5(1)³/3+8(1)+3

S(1) = 5/3+8+3

S(1) = 5+24+9/3

S(1) = 38/3 m

When t = 2secs

S(2) = 5(2)³/3+8(2)+3

S(2) = 40/3+16+3

S(2) = (40+48+9)/3

S(2) = 97/3

Distance travelled between this times will be 97/3-38/3

= 59/3

= 19.7m approximately

Answer:

The answer is 735 Nanometer

Explanation:

Solution to this problem

Given that:

Solution)

The Redshift is categorized by the  difference related between  wavelength(emitted)  and wavelength (observer).

Now,

The wavelength can be computed as  :

1+z=\lambda (observer)\div \lambda (emitted)

hence,

1 + 0.05 = λ(observer) / 700 nm

Gives

= 1.05*700  = 735 nm

Therefore, The wavelength from that source for an observer would be 735 nanometer