A projectile is shot directly away from Earth's surface. Neglect the rotation of the Earth. What multiple of Earth's radius RE gives the radial distance (from the Earth's center) the projectile reaches if (a) its initial speed is 0.785 of the escape speed from Earth and (b) its initial kinetic energy is 0.785 of the kinetic energy required to escape Earth

0.46Ω

The electromotive force (E) in the circuit is related to the terminal voltage(V), of the circuit and the internal resistance (r) of the battery as follows;

E = V + Ir                      --------------------(a)

Where;

I = current flowing through the circuit

But;

V = I x Rₓ                    ---------------------(b)

Where;

Rₓ = effective or total resistance in the circuit.

First, let's calculate the effective resistance in the circuit:

The effective resistance (Rₓ) in the circuit is the one due to the resistances in the two lightbulbs.

Let;

R₁ = resistance in the first bulb

R₂ = resistance in the second bulb

Since the two bulbs are both rated at 4.0W ( at 12.0V), their resistance values (R₁ and R₂) are the same and will be given by the power formula;

P = [tex]\frac{V^{2} }{R}[/tex]

=> R = [tex]\frac{V^{2} }{P}[/tex]             -------------------(ii)

Where;

P = Power of the bulb

V = voltage across the bulb

R = resistance of the bulb

To get R₁, equation (ii) can be written as;

R₁ = [tex]\frac{V^{2} }{P}[/tex]    --------------------------------(iii)

Where;

V = 12.0V

P = 4.0W

Substitute these values into equation (iii) as follows;

R₁ = [tex]\frac{12.0^{2} }{4}[/tex]

R₁ = [tex]\frac{144}{4}[/tex]

R₁ = 36Ω

Following the same approach, to get R₂, equation (ii) can be written as;

R₂ = [tex]\frac{V^{2} }{P}[/tex]    --------------------------------(iv)

Where;

V = 12.0V

P = 4.0W

Substitute these values into equation (iv) as follows;

R₂ = [tex]\frac{12.0^{2} }{4}[/tex]

R₂ = [tex]\frac{144}{4}[/tex]

R₂ = 36Ω

Now, since the bulbs are connected in parallel, the effective resistance (Rₓ) is given by;

[tex]\frac{1}{R_{X} }[/tex] = [tex]\frac{1}{R_1}[/tex] + [tex]\frac{1}{R_2}[/tex]       -----------------(v)

Substitute the values of R₁ and R₂ into equation (v) as follows;

[tex]\frac{1}{R_X}[/tex] = [tex]\frac{1}{36}[/tex] + [tex]\frac{1}{36}[/tex]

[tex]\frac{1}{R_X}[/tex] = [tex]\frac{2}{36}[/tex]

Rₓ = [tex]\frac{36}{2}[/tex]

Rₓ = 18Ω

The effective resistance (Rₓ) is therefore, 18Ω

Now calculate the current I, flowing in the circuit:

Substitute the values of V = 11.7V and Rₓ = 18Ω into equation (b) as follows;

11.7 = I x 18

I = [tex]\frac{11.7}{18}[/tex]

I = 0.65A

Now calculate the battery's internal resistance:

Substitute the values of E = 12.0, V = 11.7V and I = 0.65A  into equation (a) as follows;

12.0 = 11.7 + 0.65r

0.65r = 12.0 - 11.7

0.65r = 0.3

r = [tex]\frac{0.3}{0.65}[/tex]

r = 0.46Ω

Therefore, the internal resistance of the battery is 0.46Ω

Answer:

[tex]R_i_n_t=0.45 \Omega[/tex]

Explanation:

Internal resistance is a concept that helps model the electrical consequences of the complex chemical reactions that occur within a battery. When a charge is applied to a battery, the internal resistance can be calculated using the following equation:

[tex]R_i_n_t=(\frac{V_N_L}{V_F_L} -1)R_L[/tex]

Where:

[tex]V_F_L=Load\hspace{3}voltage=11.7V\\V_N_L= O pen\hspace{3}circuit\hspace{3}voltage=12V\\R_L=Load\hspace{3}resistance[/tex]

As you can see, we don't know the exactly value of the [tex]R_L[/tex]. However we can calculated that value using the next simple operations:

The problem tell us that the power of each lightbulb is 4.0 W at 12.0 V, hence let's calculated the power at 11.7V using Cross-multiplication:

[tex]\frac{12}{11.7} =\frac{4}{P}[/tex]

Solving for [tex]P[/tex] :

[tex]P=\frac{11.7*4}{12} =3.9W[/tex]

Now, the electric power is given by:

[tex]P=\frac{V^2}{R_b}[/tex]

Where:

[tex]R_b=Resistance\hspace{3}of\hspace{3}each\hspace{3}lightbulb[/tex]

So:

[tex]R_b=\frac{V^2}{P} =\frac{11.7^2}{3.9} =35.1\Omega[/tex]

Now, because of the lightbulbs are connected in parallel the equivalent resistance is given by:

[tex]\frac{1}{R_L} =\frac{1}{R_b} +\frac{1}{R_b} =\frac{2}{R_b} \\\\ R_L= \frac{R_b}{2} =\frac{35.1}{2}=17.55\Omega[/tex]

Finally, now we have all the data, let's replace it into the internal resistance equation:

[tex]R_i_n_t=(\frac{12}{11.7} -1)17.55=0.45\Omega[/tex]

Answer:

a)  Vki = 0.4 J  

b) v = 0.98 m/s

c)  final position = 0.11745 m from initial position

Explanation:

Given:-

- The mass of block, m = 500 g

- The spring constant, k = 80 N/m

- The coefficient of the surface, u = 0.20

- The inclination of the block, θ = 30°

- The block is initially compressed, xi = 10 cm

- The block is initially at rest, vi = 0 m/s

Find:-

(a) How much potential energy was stored in the block-spring-support system when the block was just released?

(b) Determine the speed of the block when it crosses the point when the spring is neither compressed nor stretched.

(c) Determine the position of the block where it just comes to rest on its way up the incline.

Solution:-

- The potential energy initially stored in a spring "Vki" with constant "k" which has undergone a displacement of "x" is given by:

                            Vki = 0.5*k*xi^2

                            Vki = 0.5*80*0.1^2

                            Vki = 0.4 J  

- The block is released from rest when the energy stored in the spring is dispensed in three forms of energy.

- There would be an increase in the potential energy " Ep " of the block as it moves up.

- There would be an increase in kinetic energy for some time " Ek "

- There would be loss of total energy due to fictitious force ( Ff) the work done against friction will dissipate energy.

- The increase in potential energy "Ep" at displacement xo = 0 from mean position is given by:

                         ΔEp = m*g*Δh ... change in vertical height h

                         ΔEp = m*g*( xi - xo )*sin ( θ )

                         ΔEp = 0.5*9.81*(0.1)*sin ( 30 )

                         ΔEp = 0.24525 J

- The increase in kinetic energy "Ek" at displacement xi = 10 cm, vi = 0 m/s from initial position to mean position at xo = 0 ,its velocity is "vf" is given by:

                         ΔEk = 0.5*m*( vf^2 - vi^2 ) ... change in velocity

                        ΔEk = 0.5*m*( vf )^2

- There would be loss of total energy due to fictitious force ( Ff ) the work done against friction will dissipate energy. First apply the equilibrium conditions on the block normal to slope and determine the contact force ( Nc ):

                        Nc - m*g*cos ( θ ) = 0

                        Nc = m*g*cos ( θ )

- The friction force ( Ff ) is given by:

                        Ff = Nc*u

                        Ff = u*m*g*cos ( θ )

- The work done by block against friction is given by:

                        Wf = -Ff*( xi - xo )

                        Wf = -u*m*g*xi*cos ( θ )

                        Wf = -0.2*0.5*9.81*0.1*cos ( 30 )  

                        Wf = -0.08495 J

- We can now express the work done principle for the block:

                        Vki = ΔEp + ΔEk + Wf

                        ΔEk = - ΔEp - Wf + Vki

                        0.5*m*( vf )^2 = -0.24525 + 0.08495 +0.4

                        vf^2 = 4*(-0.24525 + 0.08495 +0.4 )

                        vf = √0.9588

                        vf = 0.98 m/s

- We will denote the extension of spring at top most position from mean position as "x".

- From mean position xo = 0 m. The block will further move up the slope and expense all its kinetic energy "Ek" in the form of gain in potential energy and gain in elastic potential energy "Vk" and work is done against friction.

               Vki = Ep2 + Wf + Vk2

              0.4 = mg*x*sin ( 30 ) + 0.24525 + 0.08495 + u*m*g*x*cos ( θ ) + 0.5*k*x^2

               0.5*80*x^2 + x*(0.5*9.81*sin(30) + 0.2*0.5*9.81*cos ( 30 ) ) + 0.3302-0.4 =0

               40x^2 + 3.30207x - 0.0698 = 0

Solve the quadratic equation:

               x = -0.1 m (10 cm) - initial compression at rest ;

               x = 0.01745 m

- So the extension in spring at the rest position is x = 0.01745 m. The position of the next resting point is:

                final position = xi + x

                final position = 0.1 + 0.01745

                final position = 0.11745 m from initial position.

                x =  

(a) The stored PE will be "0.40 J".

(b) The block's speed will be "0.48 m/s".

(c) The block's position will be at "12 cm".

According to the question,

Mass, m = 500 h

Spring constant = 80 N/m

Coefficient of friction = 0.20

Inclined angle = 30°

(a) We know the formula,

Spring P.E = [tex]\frac{1}{2}[/tex] kx²

By substituting the values,

                 = [tex]\frac{1}{2}[/tex] × 80 × (0.1)²

                 = 0.40 J

(b) We know that,

→ [tex]P.E_s[/tex] = ΔKE + Work done by friction + Δ[tex]PE_g[/tex]

or,

          = [tex]\frac{1}{2}[/tex] mv² + [tex]\mu_R[/tex] mgx Cosθ + mgx Sinθ

By substituting the values,

 0.40 = [tex]\frac{1}{2}[/tex] × 0.5 × v² + 0.2 × 0.5 × 9.8 × Cos30° × 0.1 + 0.5 × 9.8 × 0.1 ×  Sin30°

          = 0.48 m/s

(c) By using the above formula,

→ 0.40 = [tex]\mu_R[/tex] mgl Cosθ + mgl Sinθ

           = 0.2 × 0.5 × 9.8 × Cos30° × l + 0.5 × 9.8 × l × Sin30°

         l = 0.12 m or,

           = 12 cm

Thus the above responses are correct.

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Answer:

(a) So range of frequency [tex]f > 43.27[/tex] Hz

(b) the reactance is 89.75 Ω

Explanation:

Given:

(a)

Capacitance of a capacitor [tex]C= 23 \times 10^{-6}[/tex] F

Reactance of capacitive circuit [tex]X_{C} =[/tex] 160 Ω

From the formula of reactance,

[tex]X_{C} = \frac{1}{\omega C}[/tex]

[tex]X_{C} = \frac{1}{2\pi fC}[/tex]

   [tex]f = \frac{1}{2\pi X_{C} C }[/tex]

   [tex]f = \frac{1}{6.28 \times 160 \times 23 \times 10^{-6} }[/tex]

   [tex]f = 43.27[/tex] Hz

So range of frequency [tex]f > 43.27[/tex] Hz

(b)

Capacitance [tex]C = 41 \times 10^{-6}[/tex] F

Frequency [tex]f = 43.27[/tex] Hz

From the formula of reactance,

[tex]X_{C} = \frac{1}{2\pi fC}[/tex]

[tex]X_{C} = \frac{1}{6.28 \times 43.27 \times 41 \times 10^{-6} }[/tex]

[tex]X_{C} =[/tex] 89.75 Ω

Therefore, the reactance is 89.75 Ω

Answer:

So energy is not conserved and inelastic shock

Explanation:

In the collision between a bullet and a ballistic pendulum, characterized in that the bullet is embedded in the block, if the kinetic energy is conserved the shock is elastic and if it is not inelastic.

Let's find the kinetic energy just before the crash

         K₀ = ½ m vₓ₀²

After the crash we can use the law of conservation of energy

Starting point. Right after the crash, before starting to climb

        Em₀ = K = ½ (m + M) v₂²

Final point. At the maximum height of the pendulum

       [tex]Em_{f}[/tex] = U = (m + M) g h

Where m is the mass of the bullet and M is the mass of the pendulum

        Em₀ = Em_{f}

        ½ (m + M) v₂² = (m + M) g h

        v₂ = √ 2g h

Now we can calculate the final kinetic energy

       K_{f} = ½ (m + M) v₂²²

       K_{f} = ½ (m + M) (2gh)

The relationship between these two kinetic energies is

      K₀ / K_{f} = ½ m vₓ₀² / (½ (m + M) 2 g h)

      K₀ / K_{f} = m / (m + M)  vₓ₀² / 2 g h

We can see that in this relationship the Ko> Kf

So energy is not conserved and inelastic shock

To determine if the collision between the ball and the pendulum is elastic or inelastic, we need to calculate the kinetic energy of the system before and after the collision.

An elastic collision is one that conserves kinetic energy, while an inelastic collision does not conserve kinetic energy. To determine if the collision between the ball and the pendulum is elastic or inelastic, we need to calculate the kinetic energy of the system before and after the collision.

In this case, we can calculate the kinetic energy of the system before the collision using the value of vxo and the kinetic energy just after the collision using the experimental value of h in Eq. 9.2.

If the kinetic energy of the system is the same before and after the collision, then it is an elastic collision. If the kinetic energy decreases after the collision, then it is an inelastic collision.

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Answer:

first ball is at rest after collision and the kinetic energy of the first ball is transferred to the second ball after collision.

Explanation:

mass of each ball = m

initial velocity of first ball = u

initial velocity of second ball = 0 m/s

final velocity of second ball = initial velocity of first ball = u

Use conservation of momentum

let v is the final velocity of the first ball after collision

m x u + m x 0 = m x v + m x u

So, v = 0

It means after the collision the second ball moves with the velocity which is equal to the initial velocity of the first ball and the first ball comes at rest.

As there is no friction between the balls during the collision, the collision of two balls is perfectly elastic and thus the kinetic energy of the system is conserved.

The entire kinetic energy of the first ball is transferred to the second ball after the collision.

The entire kinetic energy of ball 1 will be totally transferred to the second ball after the collision.

According to the law of conservation of momentum, the change in momentum of a body before the collision is equal to the change in momentum after the collision.

The formula for calculating momentum is expressed as:

[tex]\rho = mv \\[/tex]

According to the conservation of momentum

[tex]m_1u_1+m_2u_2=(m_1+m_2)v[/tex]

Since ball 2 is initially at rest [tex]u_2=0[/tex]

Also if after the collision, ball 2 departs with the same velocity that ball 1 originally had, hence [tex]v=u_1[/tex]

Substituting the given parameters into the formula:

[tex]m_1u_1+m_2(0)=(m_1+m_2)u_1\\m_1u_1=m_1u_1+m_2u_2\\m_2u_2=0[/tex]

This shows that the velocity of the second ball is also zero

Due to the absence of friction between the colliding object, the collision is elastic in nature showing that the kinetic energy of the system is conserved.

The entire kinetic energy of ball 1 will be totally transferred to the second ball after the collision.

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Answer:

[tex]9.38\times 10^7 m/s[/tex]

Explanation:

We are given that

Potential ,V=25 kV=[tex]25\times 10^3 V[/tex]

Distance,r =1 cm=[tex]\frac{1}{100}=0.01 m[/tex]

1 m=100 cm

Mass of electron, m=[tex]9.1\times 10^{-31} kg[/tex]

Charge, q=[tex]1.6\times 10^{-19} C[/tex]

We have to find the final velocity of the electron.

Speed of electron,[tex]v=\sqrt{\frac{2qV}{m}}[/tex]

Using the formula

[tex]v=\sqrt{\frac{2\times 1.6\times 10^{-19}\times 25\times 10^3}{9.1\times 10^{-31}}[/tex]

v=[tex]9.38\times 10^7 m/s[/tex]

Hence, the final velocity of the electron=[tex]9.38\times 10^7 m/s[/tex]

Complete Question:

An object oscillates back and forth on the end of a spring.

Which of the following statements are true at some time during the course of the motion? Check all that apply. Check all that apply. The object can have zero acceleration and, simultaneously, nonzero velocity. The object can have zero velocity and, simultaneously, nonzero acceleration. The object can have zero velocity and, simultaneously, zero acceleration. The object can have nonzero velocity and nonzero acceleration simultaneously.

Answer:

a) the object can have zero acceleration and, simultaneously, nonzero velocity

b) the object can have zero velocity and, simultaneously, nonzero acceleration

d) the object can have nonzero velocity and nonzero acceleration simultaneously

Explanation:

For an object oscillating back and forth on the end of a spring, when the object swings to the extremes, it momentarily stops and the velocity becomes zero. Rather than being zero, acceleration is maximum at these points because the net force at these extremes is maximum. This justifies option B

At the middle, which is the equilibrium position, the net force is zero because the object is motionless, and hence the acceleration is zero. At this point, the velocity is maximum and not zero. This justifies option A

When the object is swinging and it is neither at the middle nor at the extremes, both acceleration and velocity are not zero. This justifies option D

It is not possible for both the acceleration and the velocity of the swinging object to be simultaneously zero. Option C is wrong

An object can have zero acceleration and non-zero velocity or have non-zero velocity and acceleration simultaneously at some point during its motion. However, an object can also have zero velocity and zero acceleration at all times during its motion.

Both the statements "The object can have zero acceleration and, simultaneously, nonzero velocity" and "The object can have nonzero velocity and nonzero acceleration simultaneously" are true at some time during the course of motion.

For example, when an object is thrown upwards, at the highest point of its trajectory, its velocity becomes zero, but it still experiences the acceleration due to gravity, which is nonzero. This satisfies the first statement. Additionally, when an object moves in a circular path at a constant speed, it has a nonzero velocity and nonzero acceleration directed towards the center of the circle. This satisfies the second statement.

The statement "The object can have zero velocity and, simultaneously, zero acceleration." is true at all times during the course of motion. For instance, when an object is at rest, both its velocity and acceleration are zero.

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Answer:

the magnitude of the electric force on the projectile is 0.0335N

Explanation:

time of flight t = 2·V·sinθ/g

= (2 * 6.0m/s * sin35º) / 9.8m/s²

= 0.702 s

The body travels for this much time and cover horizontal displacement x from the point of lunch

So, use kinematic equation for horizontal motion

horizontal displacement

x = Vcosθ*t + ½at²

2.9 m = 6.0m/s * cos35º * 0.702s + ½a * (0.702s)²

a = -2.23 m/s²

This is the horizontal acceleration of the object.

Since the object is subject to only electric force in horizontal direction, this acceleration is due to electric force only

Therefore,the magnitude of the electric force on the projectile will be

F = m*|a|

= 0.015kg * 2.23m/s²

= 0.0335 N

Thus, the magnitude of the electric force on the projectile is 0.0335N

Answer:

Magnitude of electric force = 0.03345 N

Explanation:

We are given;

Mass; m = 15g = 0.015kg

Angle above horizontal; θ = 35°

Speed; v = 6 m/s

Horizontal displacement; d = 2.9m

Now formula for time of flight is given as;

time of flight; t = (2Vsinθ)/g

Thus, plugging in values, we have

t = (2 x 6.0 x sin35)/9.8

t = (12 x 0.5736)/9.8

t = 0.7024 s

Now, let's find the acceleration

The formula for horizontal displacement is given by;

d = (Vcosθ)t + (1/2)at²

Plugging in the relevant values ;

2.9 = [6(cos35) x 0.7024] + (1/2)a(0.7024)²

2.9 = (4.2144 x 0.8192) + (0.2467)a

2.9 = 3.45 + (0.2467)a

(0.2467)a = 2.9 - 3.45

a = -0.55/0.2467

a = -2.23 m/s²

Since we are looking for the magnitude of the electric force, we will take the absolute value of a. Thus, a = 2.23 m/s²

We know that F = ma

Thus,Force = 0.015kg x 2.23m/s² =

= 0.03345 N

Answer:

[tex]1.6026299569\times 10^{-11}\ m[/tex]

Explanation:

Grating constant

[tex]d=\dfrac{1}{6200}=0.000161\ cm=0.000161\times 10^{-2}\ m[/tex]

Number of slits

[tex]N=3.14\times 6200=19468[/tex]

Order

[tex]m=\dfrac{d}{\lambda}\\\Rightarrow m=\dfrac{0.000161\times 10^{-2}}{624\times 10^{-9}}\\\Rightarrow m\approx 2[/tex]

At m = 1

[tex]\Delta\lambda=\dfrac{\lambda}{mN}\\\Rightarrow \Delta\lambda=\dfrac{624\times 10^{-9}}{1\times 19468}\\\Rightarrow \Delta\lambda=3.2052599137\times 10^{-11}\ m[/tex]

At m = 2

[tex]\Delta\lambda=\dfrac{\lambda}{mN}\\\Rightarrow \Delta\lambda=\dfrac{624\times 10^{-9}}{2\times 19468}\\\Rightarrow \Delta\lambda=1.6026299569\times 10^{-11}\ m[/tex]

The wavelengths can be close by [tex]1.6026299569\times 10^{-11}\ m[/tex]

Answer:

The minimum wall thickness Tmin required for the spherical tank is 65.90mm

Explanation:

Solution

Recall that,

Tmin = The minimum wall thickness =PD/2бp

where D = diameter of 8.0 m

Internal pressure = 1.62 MPa

Then

The yield strength = 295MPa/3.0 = 98.33

thus,

PD/2бp = 1.62 * 8000/ 2 *98.33

= 12960/196.66 = 65.90

Therefore the wall thickness Tmin required for the spherical tank is 65.90mm

Answer:

the coefficient of lift is 3.5561

the coefficient of drag is 2.3153

Explanation:

the solution is in the attached Word file

Answer:

2.59 T

Explanation:

Parameters given:

Current flowing through the wire, I = 29 A

Angle between the magnetic field and wire, θ = 90°

Magnetic force, F = 2.25 N

Length of wire, L = 3 cm = 0.03 m

The magnetic force, F, is related to the magnetic field, B, by the equation below:

F = I * L * B * sinθ

Inputting the given parameters:

2.25 = 29 * 0.03 * B * sin90

2.25 = 0.87 * B

=> B = 2.25/0.87

B = 2.59 T

The magnetic field strength between the poles is 2.59 T

Answer:

Average induced emf will be equal to 0.00156 volt

Explanation:

We have given diameter of the wire d = 16.1 cm

So radius [tex]r=\frac{16.1}{2}=8.05cm=0.08m[/tex]

Cross sectional area of the wire [tex]A=\pi r^2=3.14\times 0.08^2=0.02m^2[/tex]

Magnetic field is changing from 0.53 T to 0.19 T

So change in magnetic field dB = 0.19-0.53 = -0.34 T

Time taken to change in magnetic field dt = 0.23 sec

Induced emf is given by [tex]e=\frac{-d\Phi }{dt}=-A\frac{dB}{dt}=-0.02\times \frac{0.34}{0.23}=0.00156volt[/tex]

So average induced emf will be equal to 0.00156 volt

Answer: 6.29*10^-12 N

Explanation:

given,

Potential difference of the electron, v = 1950 V

Magnetic field of the electron, B = 1.50 T

If the electron is accelerated through 19500 V from rest its Potential Energy has to be

converted to Kinetic Energy. This allows us solve for the velocity.

PE = Vq

PE = 1950 * 1.6*10^-19

PE = 3.12*10^-16 J

Also, PE = 1/2mv²

3.12*10^-16 = 1/2mv²

v = 2.62*10^7 m/s

to get F(max), we use,

F(max) = qvB

F(max) = 1.6*10^-19 * 2.62*10^7 * 1.5

F(max) = 6.29*10^-12 N

6.3 x 10⁻¹²N

As stated by Lorentz Force law, the magnitude of a magnetic force, F, can be expressed in terms of a fixed amount of charge, q, which is moving at a constant velocity, v, in a uniform magnetic field, B, as follows;

F =  qvB sin θ              ------------(i)

Where;

θ = angle between the velocity and the magnetic field vectors

When the electron passes through a potential difference, V, it is made to accelerate as it gains some potential energy ([tex]P_{E}[/tex]) which is then converted to kinetic energy ([tex]K_{E}[/tex]) as it moves. i.e

[tex]P_{E}[/tex] = [tex]K_{E}[/tex]                 ----------------(ii)

But;

[tex]P_{E}[/tex] = qV

And;

[tex]K_{E}[/tex] = [tex]\frac{1}{2}[/tex] x m x v²

Therefore substitute these into equation (ii) as follows;

qV = [tex]\frac{1}{2}[/tex] x m x v²

Make v subject of the formula;

2qV = mv²

v² = [tex]\frac{2qV}{m}[/tex]

v = [tex]\sqrt{\frac{2qV}{m} }[/tex]      ---------------(iii)

From the question;

q = 1.6 x 10⁻¹⁹C       (charge on an electron)

V = 1.95 x 10³V

m = 9.1 x 10⁻³¹kg

Substitute these values into equation (iii) as follows;

v = [tex]\sqrt{\frac{2*1.6*10^{-19} * 1.95*10^{3}}{9.1*10^{-31}} }[/tex]

v =  [tex]\sqrt{\frac{6.24*10^{-16}}{9.1*10^{-31}} }[/tex]

v = 2.63 x 10⁷m/s

Now, from equation (i), the magnitude of the magnetic force will be maximum when the angle between the velocity and the magnetic field is 90°. i.e when θ = 90°

Substitute the values of θ, q v and B = 1.50T into equation (i) as follows;

F =  qvB sin θ

F = 1.6 x 10⁻¹⁹ x 2.63 x 10⁷ x 1.50 x sin 90°

F = 6.3 x 10⁻¹²N

Therefore the maximum magnitude of the magnetic force this particle can experience is 6.3 x 10⁻¹²N

Answer:

Explanation:

mass of rocket m = .5 kg

height of wall h = 40 m

initial horizontal velocity u = .5 m /s

horizontal acceleration = force / mass

= 20 / .5

a = 40 m /s²

Let rocket falls or covers 40 m vertically downwards in time t

h = 1/2 gt² , initial vertical velocity = 0

40 = 1/2 x 9.8 x t²

t = 2.8566 s

During this period it will cover horizontal distance with initial velocity of .5 m /s and acceleration a = 40m /s²

horizontal distance = ut + 1/2 at²

= .5 x 2.8566 + .5 x 40 x 2.8566²

= 1.4283 + 163.2033

= 164.63 m .

The distance from the base of the wall does the rocket land is 164.63 m .

A 500 g model rocket is resting horizontally at the top edge of a 40-m-high wall when it is accidentally bumped. The bump pushes it off the edge with a horizontal speed of 0.5 m/s

We know that

horizontal acceleration = force / mass

a = 20 / .5

a = 40 m /s²

Now

h = 1/2 gt² ,

Here initial vertical velocity = 0

So,

40 = 1/2 x 9.8 x t²

t = 2.8566 s

Now

horizontal distance = ut + 1/2 at²

= .5 x 2.8566 + .5 x 40 x 2.8566²

= 1.4283 + 163.2033

= 164.63 m .

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Answer:

B=1.89*10^{-4} T

Explanation:

First we have to calculate the electric force on the electron, and then we have to take into account that this force is equal to the force generated by the magnetic field.

The formula is

[tex]F=q_eE\\\\E=\frac{V}{d}[/tex]

q: charge of the electron = 1.6*10^{-19}C

V: potential

d: separation between plates

[tex]E=\frac{100V}{17*10^{-3}m}=5882.35N/C[/tex]

[tex]F=(1.6*10^{-19}C)(5882.35N/C)=9.4*10^{-16}N[/tex]

This force must equal the Lorentz's force

[tex]F_E=F_B\\\\F_B=qvB\\\\B=\frac{F_B}{q_ev}[/tex]

But before we have to calculate the speed of the electron by using  (me=9.1*10^{-31}kg)

[tex]E_e=\frac{1}{2}m_ev^2=q_eV\\\\v=\sqrt{\frac{2q_eV}{m_e}}=\sqrt{\frac{2(1.6*10^{-19}C)(2.8*10^3V)}{9.1*10^{-31}kg}}\\\\v=3.13*10^7\frac{m}{s}[/tex]

Hence, we have

[tex]B=\frac{9.4*10^{-16}N}{(1.6*10^{-19}C)(3.1*10^{7})\frac{m}{s}}=1.89*10^{-4}T[/tex]

hope this helps!!

Answer:

The magnetic field that is necessary is equal to 1.88x10⁻⁴ T

Explanation:

If the electron is accelerated:

[tex]e*V=\frac{1}{2} mv^{2} \\v=\sqrt{\frac{2eV}{m} }[/tex]

Where

e = 1.6x10⁻¹⁹C

V = 2.8 kV = 2800 V

m = 9.1x10⁻³¹kg

Replacing:

[tex]v=\sqrt{\frac{2*1.6x10^{-19}*2800 }{9.1x10^{-31} } } =3.13x10^{7} m/s[/tex]

When the electron is moving in straight line, the magnetic force is balanced with the electric force, thus:

V = E * d

Where V = 100 V

d = 17 mm = 0.017 m

E = V/d = 100/0.017 = 5882.35 N/C

The magnetic field that is necessary is equal to:

B = E/v = 5882.35/3.13x10⁷ = 1.88x10⁻⁴ T

Answer:

a) [tex]\Delta U_{g} = 12.945\,J[/tex], b) [tex]\Delta U_{k} = 12.945\,J[/tex], c) [tex]k = 2930.059\,\frac{N}{m}[/tex]

Explanation:

a) The change in the gravitational potential energy of the marble-Earth system is:

[tex]\Delta U_{g} = (0.06\,kg)\cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (22\,m)[/tex]

[tex]\Delta U_{g} = 12.945\,J[/tex]

b) The change in the elastic potential energy of the spring is equal to the change in the gravitational potential energy, then:

[tex]\Delta U_{k} = 12.945\,J[/tex]

c) The spring constant of the gun is:

[tex]\Delta U_{k} = \frac{1}{2} \cdot k \cdot x^{2}[/tex]

[tex]k = \frac{2\cdot \Delta U_{k}}{x^{2}}[/tex]

[tex]k = \frac{2\cdot (12.945\,J)}{(0.094\,m)^{2}}[/tex]

[tex]k = 2930.059\,\frac{N}{m}[/tex]

Answer:

1.04x[tex]10^{-3}[/tex] s

Explanation:

->The maximum current through the resistor is

[tex]I_{max}[/tex] = V/R = V/[tex]Re^{-t/RC}[/tex]= V/R×[tex]e^{0}[/tex] = V/R

Voltage 'V'=50V

Effective resistance 'R'= 25.0-Ω+ 10.0 Ω= 35.0 Ω

Therefore, [tex]I_{max}[/tex]=50/35=> 1.43 A

->The maximum charge can be determined by

Q = CV

where,

Capacitance of the capacitor 'C' = 30.0µF = 30×10-⁶F

Therefore,

Q=30×10-⁶ x 50=>1.5 x [tex]10^{-3}[/tex]

In order to find that when does the maximum current occur, the time taken given the quantity of charge and the electric current is:

t= Q / I=>  1.5 x [tex]10^{-3}[/tex]/ 1.43

t= 1.04x[tex]10^{-3}[/tex] s

Answer: C. Chemical Weathering

Explanation:

Answer:

Chemical Weathering

Explanation:

Answer:

[tex]\frac {f_i}{\sqrt 2}[/tex]

Explanation:

Frequency is given by [tex]\frac {1}{2\pi}\sqrt{{\frac {k}{m}}[/tex]

Let the initial frequency be denoted by [tex]f_i[/tex]

[tex]f_i=\frac {1}{2\pi}\sqrt{{\frac {k}{M}}[/tex]

When M is 2M then

[tex]f_f=\frac {1}{2\pi}\sqrt{{\frac {k}{2M}}=\frac {f_i}{\sqrt 2}[/tex]

Therefore, the final frequency is

[tex]\frac {f_i}{\sqrt 2}[/tex]

Answer: [tex]1.8(10)^{-4} rad[/tex]

Explanation:

This problem is related to the Rayleigh Criterion, which provides the following formula to find the acuity or limit of resolution of an optic system with circular aperture (the eye in this case):

[tex]\theta=1.22\frac{\lambda}{D}[/tex]

Where:

[tex]\theta[/tex] is the angle of resolution (related to the acuity)

[tex]\lambda=554 nm=554(10)^{-9} m[/tex] is the wavelength of the light

[tex]D=3.75 mm=3.75(10)^{-3} m[/tex] is the diameter of the pupil

Solving:

[tex]\theta=1.22 \frac{554(10)^{-9} m}{3.75(10)^{-3} m}[/tex]

[tex]\theta=1.8(10)^{-4} rad[/tex]

The human eye's acuity is limited by light's diffraction through the pupil, which determines the minimum angle, Δθ, between two resolvable points. This is calculated using the formula Δθ = 1.22 (λ / D), where λ is the light's wavelength and D the diameter of the pupil.

The resolution of human visual acuity is related to the diffraction of light by the eye's pupil. This diffraction effect can be worked out using a formula that quantifies resolving power (Δθ), when the diameter of the aperture (in this case, the pupil size) and the wavelength of incident light are known.

In your example, a 3.75 mm diameter pupil encountering light of 554 nm wavelength can identify two points of light provided they subtend an angle (Δθ) at the pupil calculated by using the formula: Δθ = 1.22 (λ / D). For all light's range of visible wavelengths, the angle is extremely tiny - on the order of tens of thousands of a degree. This shows that the eye’s ability to distinguish two separate points of light is limited by the diffraction of light by the eye’s pupil.

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Answer:

0.48 W

Explanation:

Given that:

the mass of the steel ball = 0.0367 kg

C = 0.270 N

g (acceleration due to gravity) = 9.8

Now;

At Terminal Velocity Weight is balance by drag force

mg =Cv

Making v the subject of the formula:we have:

[tex]v = \frac {mg}{C}[/tex]

[tex]v = \frac {0.0367*9.8}{0.270}[/tex]

v = 1.332 m/s

Thus, the Rate at which gravitational energy is converted to thermal energy once the ball reaches terminal velocity is:

P=Fv

P = mgv

[tex]P = (0.0367*9.8)*1.332[/tex]

P=0.48 W

Final answer:

To find the rate at which gravitational energy is converted to thermal energy at terminal velocity, equate the drag force to the gravitational force on the ball and solve for power as the product of force and velocity.

Explanation:

The question asks for the rate at which gravitational energy is converted to thermal energy once the ball reaches terminal velocity in a vat of corn syrup. To find this rate, we must recognize that at terminal velocity, all the gravitational force acting on the ball (minus any buoyant force, which is ignored here) is converted into thermal energy due to viscous drag. The viscous drag force [tex]F_{drag}[/tex] is given by [tex]F_{drag[/tex] = -Cv, where C is the drag coefficient, and v is the velocity of the ball. The gravitational force acting on the ball can be calculated using [tex]F_{gravity[/tex] = mg, where m is the mass of the ball, and g is the acceleration due to gravity. At terminal velocity, [tex]F_{drag[/tex] = [tex]F_{gravity[/tex], hence the power (rate of energy conversion) P can be found as P = Fv = mgv. Assuming standard gravity (9.81 m/s2), the mass of the ball (0.0367 kg), and solving for v using [tex]F_{drag[/tex], one can determine the rate of gravitational energy conversion to thermal energy.

Answer:

It has no effect on the amplitude.

Explanation:

When the sandbag is dropped, then the cart is at its maximum speed. Dropping the sand bag does not affect the speed instantly, this is because the energy remains within the system after the bag as been dropped. The cart will always return to its equilibrium point with the same amount of kinetic energy, as a result the same maximum speed is maintained.

Answer:

It decreases the amplitude.

Explanation:

If it is located at the end, its kinetic energy will be equal to zero, so the system as a whole would only have potential energy, which is defined as:

Ep = (1/2) * k * A²

Where A is the amplitude of the oscillation.

According to the energy principle, said energy must remain conserved, therefore, the total energy is equal to:

Etot = (1/2) * k * A²

Since the system is in equilibrium, according to the expression of conservation of energy, the energy in the position located at the end would be equal to the energy in equilibrium. As the sandbag falls, its kinetic energy decreases, which is why the total energy of the system also decreases.

According to the total energy expression, it is directly proportional to the square of the oscillation amplitude. If the total energy is decreased, the amplitude of the oscillation would also decrease.

θ₃ is 27.12°

Explanation:

Using Snell's law:

n₁ sin θ₁ = n₂ sin θ₂

where,

n₁ = refractive index of material with incident light

n₂ = refractive index of material with refracted light

(a)

1 → air

2 → glas

3 → water

n(water) = n₃ = 1.333

From air to glass,

n₁ sin θ₁ = n₂sin θ₂

The angle of refraction 2 is the angle of incidence when the light comes from glass to water.

From glass to water:

n₁ sin θ₁ = n₃ sin θ₃

1 X sin 37.5° = 1.333 sin θ₃

sin θ₃ =  sin (37.5)/ 1.333

sin θ₃ = 0.456

θ₃ = 27.12°

Therefore, θ₃ is 27.12°

Answer:

Explanation:

1. They have different wavelengths - Because These radiations form a spectra that differs by the size of the wavelength.

2. They have different Frequencies (f) that is frequency = 1/ wavelength

(f = 1/wavelength)

3. They propagate at different speed though a non vacuum media (non vacuum media affect the speed based on the wavelength)

Answer:

A

Explanation:

Current, magnetic field and motion are mutually dependent and perpendicular to one another

Answer:

b) clockwise

Explanation:

The right hand rule states that to determine the direction of the magnetic force on a positive moving charge, point the thumb of the right hand in the direction of the potential v, the fingers in the direction of the magnetic field, B, and a perpendicular to the palm points in the direction of the force, F.

Since the south pole of the bar magnet is pointing downwards, the induced current developed in the wire repels it and moves in the clockwise direction.

Answer:

Explanation:

Balance point will be achieved as soon as the weight of the baby elephant creates torque equal to torque created by weight of woman about the pivot. torque by weight of woman

weight x distance from pivot

= 500x 5

= 2500 Nm

torque by weight of baby woman , d be distance of baby elephant from pivot at the time of balance

= 2500x d

for equilibrium

2500 d = 2500

d = 1 m

So elephant will have to walk up to 1 m close to pivot or middle point.

Final answer:

The distance between the reflected light beams can be found using reflection and refraction laws, applying Snell's law to calculate the angle of refraction, and subsequently using trigonometry to determine the light path within the glass and the shift caused by both reflections.

Explanation:

To determine the distance b between the light beam reflected from the top and the bottom surfaces of a glass plate, we need to use the laws of reflection and Snell's law for the refraction of light. The given information includes the thickness of the glass (2.8 cm), the index of refraction of the glass (1.6), and the angle of incidence (36°). When the laser beam strikes the top surface at an angle of incidence of 36°, it reflects at the same angle (law of reflection).

The angle of refraction can be found using Snell's law: n1 * sin(θ1) = n2 * sin(θ2), where n1 and n2 are the indices of refraction of the air and glass, respectively, and θ1 and θ2 are the angles of incidence and refraction, respectively. Once the angle of refraction is determined, the actual path of the beam within the glass can be calculated using trigonometry, and subsequently the shift b caused by the refraction at the two interfaces (again using trigonometry).

Answer:

ω2  =  216.47 rad/s

Explanation:

given data

radius r1 =  460 mm

radius r2 = 46 mm

ω =  32k rad/s

solution

we know here that power generated by roller that  is

power = T. ω    ..............1

power = F × r × ω

and this force of roller on cylinder is equal and opposite force apply by roller

so power transfer equal in every cylinder so

( F × r1 × ω1)  ÷ 2 = (  F × r2 × ω2 )  ÷  2    ................2

so

ω2  =  [tex]\frac{460\times 32}{34\times 2}[/tex]

ω2  =  216.47

The spring has been extended for 3.5 m

Explanation:

We have the formula,

PE =1/2 K X²

Rewrite the equation as

PE=1/2 K d²

multiply both the sides by 2/K to simplify the equation

2/k . PE= 1/2 K  d² . 2/K

√d²=√2PE/K

Cancelling the root value and now we have,

d=√2PE/k

d=√2×98 J / 16N/m

d=√12.25

d=3.5 m

The spring has been extended for 3.5 m

Final answer:

To find the extension of a spring with a constant of 16 N/m and 98 J of energy stored, we use the formula for potential energy. The spring is extended by 3.5 meters.

Explanation:

The question involves calculating the extension of a spring based on the energy stored in it and the spring's constant. The formula for the potential energy stored in a spring is given by

U = 1/2 kx², where U is the potential energy, k is the spring constant, and x is the extension of the spring from its equilibrium position. In this case, the energy (U) is 98 J and the spring constant (k) is 16 N/m.

The spring in question has an energy of 98 J stored in it.

Using the formula for potential energy stored in a spring, we have:

[tex]PE = 1/2 k x^2[/tex]

By substituting the known values, we find that the spring is extended by 0.7 m (70 cm

We rearrange the formula to solve for x:

x = √(2U/k).

Substituting the given values,

x = √(2*98/16)

= √(196/16)

= √(12.25)

= 3.5 meters. Therefore, the spring is extended by 3.5 meters.