A package is dropped from a helicopter moving upward at 1.5 m/s. If it takes 16.0 s before the package strikes the ground, how high above the ground was the package when it was released if air resistance is negligible?

Answer:

A, C

Explanation:

Since the frequency is inversely proportional to the length of a string, then I want to increase the frequency of the lowest

A. Replacing the string with a thicker string.

Thicker strings have more density. The more density the string has, the lower the sound.

Mathematically, we can see the proportionality (direct and inverse) by looking at those formulas for Frequency and Speed, when combined:

For:

[tex]f=\frac{v}{\lambda}[/tex]

[tex]f=\frac{v}{\lambda}*\sqrt{\frac{T}{D} }[/tex]

See above, how density (D) and [tex](\lambda)[/tex] wave length are inversely proportional.

C. Doubling the length of the string.

Because the length of the string is inversely proportional to the frequency.

The longer the string, the lower the frequency.

So, if we double string, we'll hear lower sounds in any string instrument

--

In short,  for A, and C  We can justify both since length and density are inversely proportional to the Frequency, we need longer or thicker string.

Answer:

v = 16 m/s

Explanation:

It is given that,

Acceleration of a particle along x -axis is [tex]2.5\ m/s^2[/tex]

At t = 0s, its velocity is 6 m/s

We need to find the velocity at t = 4 s

It means that the initial velocity of the particle is 6 m/s

Let v is the velocity at t = 4 s

So,

v = u + at

[tex]v=6+2.5\times 4\\\\v=16\ m/s[/tex]

So, the velocity at t = 4 s is 16 m/s.

Answer:

v = 16 m/s

Explanation:

It is given that,

Acceleration of a particle along x -axis is  

At t = 0s, its velocity is 6 m/s

We need to find the velocity at t = 4 s

It means that the initial velocity of the particle is 6 m/s

Let v is the velocity at t = 4 s

So,

v = u + at

So, the velocity at t = 4 s is 16 m/s.

q = 40 C

e = 1.6×10^-19 C

n = ?

n = q/e

n = 40/1.6×10^-19 C

= 2.6×10^20

Complete Question

 Now if you look at the equation for acceleration  given in the question i.e

        [tex]a = a_1 + F * m[/tex]

We see that evaluating it in terms of dimension it is  incorrect  instead the equation should be  

       [tex]a = a_1 + \frac{F}{m}[/tex]

So in the solution below we will be making use of [tex]a = a_1 + \frac{F}{m}[/tex]

Answer:

The values of a is [tex]a = 4.714 \ m/s^2[/tex]

Explanation:

From the question we are told that

    The  expression for the acceleration is  [tex]a = a_1 + \frac{F}{m}[/tex]

     The value of  [tex]a_ 1 = 3.0 \ m/s^2[/tex]

     The values of  [tex]F = 12.0 \ kg \cdot m/s^2[/tex]  

      The values of m is  [tex]m = 7.0 \ kg[/tex]

substituting values

        [tex]a = 3 + \frac{12}{7}[/tex]

       [tex]a = 4.714 \ m/s^2[/tex]

Answer:

Explanation:

Kinetic energy at the height = 1/2 m v²

= 1/2 x 750 x 20²

= 150000 J

Its potential energy = mgh

= 750 x 9.8 x 5

=36750 J

Total energy = 186750 J

Its total kinetic energy will be equal to 186750 J , according to conservation of mechanical energy

If v be its velocity at the bottom

1/2 m v² = 186750

v = √498

= 22.31 m /s

Answer:

Length of a tube = 1.574 m

Explanation:

Given:

Fundamental frequency (f1) = 108 Hz

First overtone (f2) = 216 Hz

Speed of sound (v) = 340 m/s

Find:

Length of a tube

Computation:

We know that,

f = v / λ

f = nv / 2L  [n = number 1,2,3]

So,

f1 = 1(340) / 2L

f1 = 170 / L

L = 170 / 108 = 1.574 m

f2 = 2(340) / 2L

L = 340 / 216

L = 1.574 m

Answer:

B: I2>I1

Explanation:

See attached file

Answer:

v = 6t² − 2t^³/₂

s = 2t³ − ⅘t^⁵/₂ + 15

Explanation:

a = 12t − 3t^½

Integrate to find velocity.

v = ∫ a dt

v = ∫ (12t − 3t^½) dt

v = 6t² − 2t^³/₂ + C

Use initial condition to find C.

0 = 6(0)² − 2(0)^³/₂ + C

C = 0

v = 6t² − 2t^³/₂

Integrate to find position.

s = ∫ v dt

s = ∫ (6t² − 2t^³/₂) dt

s = 2t³ − ⅘t^⁵/₂ + C

Use initial condition to find C.

15 = 2(0)³ − ⅘(0)^⁵/₂ + C

15 = C

s = 2t³ − ⅘t^⁵/₂ + 15

Answer:

(a) The power wasted for 0.289 cm wire diameter is 15.93 W

(b) The power wasted for 0.417 cm wire diameter is 7.61 W

Explanation:

Given;

diameter of the wire, d = 0.289 cm = 0.00289 m

voltage of the wire, V = 120 V

Power drawn, P = 1850 W

The resistivity of the wire, ρ = 1.68 x 10⁻⁸ Ω⋅m

Area of the wire;

A = πd²/4

A = (π x 0.00289²) / 4

A = 6.561 x 10⁻⁶ m²

(a) At 26 m of this wire, the resistance of the is

R = ρL / A

R = (1.68 x 10⁻⁸  x 26) / 6.561 x 10⁻⁶

R = 0.067 Ω

Current in the wire is calculated as;

P = IV

I = P / V

I = 1850 / 120

I = 15.417 A

Power wasted = I²R

Power wasted = (15.417²)(0.067)

Power wasted = 15.93 W

(b) when a diameter of 0.417 cm is used instead;

d = 0.417 cm = 0.00417 m

A = πd²/4

A = (π x 0.00417²) / 4

A = 1.366 x 10⁻⁵ m²

Resistance of the wire at 26 m length of wire and  1.366 x 10⁻⁵ m² area;

R = ρL / A

R = (1.68 x 10⁻⁸  x 26) / 1.366 x 10⁻⁵

R = 0.032 Ω

Power wasted = I²R

Power wasted = (15.417²)(0.032)

Power wasted = 7.61 W

Answer:

The speed is  [tex]v =10.27 *10^{7} \ m/s[/tex]

Explanation:

From the question we are told that

      The  voltage  is  [tex]V = 30 kV = 30*10^{3} V[/tex]

      The  initial velocity of the electron is  [tex]u = 0 \ m/s[/tex]

Generally according to the law of energy conservation

    Electric potential Energy  =  Kinetic energy of the electron

So  

      [tex]PE = KE[/tex]

Where  

      [tex]KE = \frac{1}{2} * m* v^2[/tex]

Here  m is the mass of the electron with a value of  [tex]m = 9.11 *10^{-31} \ kg[/tex]

     and  

         [tex]PE = e * V[/tex]

      Here  e is the charge on the electron with a value  [tex]e = 1.60 *10^{-19} \ C[/tex]

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

=>      [tex]v = \sqrt{ \frac{2 * e * V}{m} }[/tex]

substituting values  

           [tex]v = \sqrt{ \frac{2 * (1.60*10^{-19}) * 30*10^{3}}{9.11 *10^{-31}} }[/tex]

          [tex]v =10.27 *10^{7} \ m/s[/tex]

Answer:

The new voltage between the plates of the capacitor is 18 V

Explanation:

The charge on parallel plate capacitor is calculated as;

q = CV

Where;

V is the battery voltage

C is the capacitance of the capacitor, calculated as;

[tex]C = \frac{\epsilon _0A}{d} \\\\q =CV = (\frac{\epsilon _0A}{d})V = \frac{\epsilon _0A V}{d}[/tex]

[tex]q = \frac{\epsilon _0A V}{d}[/tex]

where;

ε₀ is permittivity of free space

A is the area of the capacitor

d is the space between the parallel plate capacitors

If only the space between the capacitors is doubled and every other parameter is kept constant, the new voltage will be calculated as;

[tex]q = \frac{\epsilon _0A V}{d} \\\\\frac{\epsilon _0A V}{d} = \frac{\epsilon _0A V}{d} \\\\\frac{V_1}{d_1} = \frac{V_2}{d_2} \\\\V_2 = \frac{V_1d_2}{d_1} \\\\(d_2 = 2d_1)\\\\V_2 = \frac{V_1*2d_1}{d_1} \\\\(V_1 = 9V)\\\\V_2 = \frac{9*2d_1}{d_1} \\\\V_2 = 9*2\\\\V_2 = 18 \ V[/tex]

Therefore, the new voltage between the plates of the capacitor is 18 V

Answer:

The average current will be "17.69 nA".

Explanation:

The given values are:

Charge,

q = 9.2 pC

Time,

t = 0.52ms

The equivalent circuit of the cell surface is provided by:

⇒  [tex]i_{avg}=\frac{charge}{t}[/tex]

Or,

⇒  [tex]i_{avg}=\frac{q}{t}[/tex]

On substituting the given values, we get

⇒         [tex]=\frac{9.2\times 10^{-12}}{0.52\times 10^{-3}}[/tex]

⇒         [tex]=17.69^{-9}[/tex]

⇒         [tex]=17.69 \ nA[/tex]

Answer:

a) 1.94 m/s

b) 120 m

Explanation:

Convert km/h to m/s:

6 km/h = 1.67 m/s

a) The final speed is found with Pythagorean theorem:

v = √((1.67 m/s)² + (1 m/s)²)

v = 1.94 m/s

b) The time it takes the boat to cross the river is:

t = (200 m) / (1.67 m/s)

t = 120 s

The displacement in the direction of the current is:

x = (1 m/s) (120 s)

x = 120 m

Answer:

Fx = 2.5 N

Fy = 5 N

|F| = 5.59 N

Explanation:

Given:-

- The mass of puck, m = 4.0 kg

- The initial velocity of puck, u = 3.00 i m/s

- The final velocity of puck, v = ( 8.00 i + 10.00 j ) m/s

- The time interval for the duration of force, Δt = 8 seconds

Find:-

the components of the force and (b) its magnitude.

Solution:-

- We will set up a coordinate system ( x - y ) plane. With unit vectors i and j along x and y axes respectively.

- To model the situation we will seek help from Newton's second law of motion. Defined by the rate of change of linear momentum of the system.

                                [tex]F_net = \frac{m*( v - u ) }{dt}[/tex]

Where,

                   Fnet: The net force that acts on the puck-rocket system

- Here we will assume that the mass of rocket is negligible compared to the mass of the puck. The only force ( F ) acting on the puck is due to the thrust produced of the rocket. The dry and air frictions are both neglected for the analysis.

- We will apply the newton's second law of motion in component forms. And determine the components of force F, as  ( Fx ) and ( Fy ) as follows:

                         [tex]F_x = \frac{m* ( v_x - u_x)}{dt} \\\\F_x = \frac{4* ( 8 - 3)}{8} \\\\F_x = 2.5 N\\\\F_y = \frac{m* ( v_y - u_y)}{dt} \\\\F_y = \frac{4* ( 10 - 0)}{8} \\\\F_y = 5 N\\\\[/tex]

- We will apply the Pythagorean theorem and determine the magnitude of the thrust force produced by the rocket with which the puck accelerated:

                          [tex]| F | = \sqrt{( F_x)^2 + ( F_y)^2} \\\\| F | = \sqrt{( 2.5)^2 + ( 5)^2} \\\\| F | = \sqrt{31.25} \\\\| F | = 5.590[/tex]            

Answer: the magnitude of the thrust force is F = 5.59 N

Answer:

a) The average friction force exerted on the toboggan is 653.125 newtons, b) The rough region reduced the kinetic energy of the toboggan in 92.889 %, c) The speed of the toboggan is reduced in 73.333 %.

Explanation:

a) Given the existence of non-conservative forces (friction between toboggan and ground), the motion must be modelled by means of the Principle of Energy Conservation and the Work-Energy Theorem, since toboggan decrease its speed (associated with  due to the action of friction. Changes in gravitational potential energy can be neglected due to the inclination of the ground. Then:

[tex]K_{1} = K_{2} + W_{f}[/tex]

Where:

[tex]K_{1}[/tex], [tex]K_{2}[/tex] are the initial and final translational kinetic energies of the tobbogan, measured in joules.

[tex]W_{f}[/tex] - Dissipated work due to friction, measured in joules.

By applying definitions of translation kinetic energy and work, the expression described above is now expanded and simplified:

[tex]f\cdot \Delta s = \frac{1}{2}\cdot m \cdot (v_{1}^{2}-v_{2}^{2})[/tex]

Where:

[tex]f[/tex] - Friction force, measured in newtons.

[tex]\Delta s[/tex] - Distance travelled by the toboggan in the rough region, measured in meters.

[tex]m[/tex] - Mass of the toboggan, measured in kilograms.

[tex]v_{1}[/tex], [tex]v_{2}[/tex] - Initial and final speed of the toboggan, measured in meters per second.

The friction force is cleared:

[tex]f = \frac{m\cdot (v_{1}^{2}-v_{2}^{2})}{2\cdot \Delta s}[/tex]

If [tex]m = 375\,kg[/tex], [tex]v_{1} = 4.50\,\frac{m}{s}[/tex], [tex]v_{2} = 1.20\,\frac{m}{s}[/tex] and [tex]\Delta s = 5.40 \,m[/tex], then:

[tex]f = \frac{(375\,kg)\cdot \left[\left(4.50\,\frac{m}{s} \right)^{2}-\left(1.20\,\frac{m}{s}\right)^{2}\right]}{2\cdot (5.40\,m)}[/tex]

[tex]f = 653.125\,N[/tex]

The average friction force exerted on the toboggan is 653.125 newtons.

b) The percentage lost by the kinetic energy of the tobbogan due to friction is given by the following expression, which is expanded and simplified afterwards:

[tex]\% K_{loss} = \frac{K_{1}-K_{2}}{K_{1}}\times 100\,\%[/tex]

[tex]\% K_{loss} = \left(1-\frac{K_{2}}{K_{1}} \right)\times 100\,\%[/tex]

[tex]\% K_{loss} = \left(1-\frac{\frac{1}{2}\cdot m \cdot v_{2}^{2}}{\frac{1}{2}\cdot m \cdot v_{1}^{2}} \right)\times 100\,\%[/tex]

[tex]\% K_{loss} = \left(1-\frac{v_{2}^{2}}{v_{1}^{2}} \right)\times 100\,\%[/tex]

[tex]\%K_{loss} = \left[1-\left(\frac{v_{2}}{v_{1}}\right)^{2} \right]\times 100\,\%[/tex]

If [tex]v_{1} = 4.50\,\frac{m}{s}[/tex] and [tex]v_{2} = 1.20\,\frac{m}{s}[/tex], then:

[tex]\%K_{loss} = \left[1-\left(\frac{1.20\,\frac{m}{s} }{4.50\,\frac{m}{s} }\right)^{2} \right]\times 100\,\%[/tex]

[tex]\%K_{loss} = 92.889\,\%[/tex]

The rough region reduced the kinetic energy of the toboggan in 92.889 %.

c) The percentage lost by the speed of the tobbogan due to friction is given by the following expression:

[tex]\% v_{loss} = \frac{v_{1}-v_{2}}{v_{1}}\times 100\,\%[/tex]

[tex]\% v_{loss} = \left(1-\frac{v_{2}}{v_{1}} \right)\times 100\,\%[/tex]

If [tex]v_{1} = 4.50\,\frac{m}{s}[/tex] and [tex]v_{2} = 1.20\,\frac{m}{s}[/tex], then:

[tex]\% v_{loss} = \left(1-\frac{1.20\,\frac{m}{s} }{4.50\,\frac{m}{s} } \right)\times 100\,\%[/tex]

[tex]\%v_{loss} = 73.333\,\%[/tex]

The speed of the toboggan is reduced in 73.333 %.

The average frictional force exerted on the toboggan by the rough surface is 661.5 N.

The percentage of the toboggan kinetic energy reduction is 7.11%.

The percentage of the toboggan speed reduction is 26.67%.

The given parameters;

The normal force on the toboggan is calculated as follows;

Fₙ = mg

Fₙ = 375 x 9.8 = 3675 N

The acceleration of the toboggan is calculated as follows;

[tex]v^2 = u^2 + 2as\\\\2as = v^2 - u^2\\\\a = \frac{v^2 - u^2 }{2s} \\\\a = \frac{(1.2)^2 - (4.5)^2 }{2(5.4)}\\\\a = -1.74 \ m/s^2[/tex]

The coefficient of friction is calculated as follows;

[tex]\mu_k = \frac{a}{g} \\\\\mu_k = \frac{1.74}{9.8} \\\\\mu_k = 0.18[/tex]

The average frictional force exerted on the toboggan by the rough surface;

[tex]F_k = \mu_k F_n\\\\F_k = 0.18 \times 3675\\\\F_k = 661.5 \ N[/tex]

The initial kinetic energy of the toboggan is calculated as follows;

[tex]K.E_i = \frac{1}{2} mu^2\\\\K.E_i = \frac{1}{2} \times 375\times 4.5^2\\\\K.E_i = 3,796.88 \ J[/tex]

The final kinetic energy of the toboggan is calculated as follows;

[tex]K.E_f = \frac{1}{2} mv^2\\\\K.E_f = \frac{1}{2} \times 375\times 1.2^2\\\\K.E_f = 270 \ J[/tex]

The percentage of the toboggan kinetic energy reduction is calculated as follows;

[tex]\frac{K.E_f}{K.E_i} \times 100\% = \frac{270}{3796.88} \times 100\% = 7.11 \%[/tex]

The percentage of the toboggan speed reduction is calculated as follows;

[tex]\frac{1.2}{4.5} \times 100\% = 26.67 \%[/tex]

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Answer:

The  tension on the clotheslines is  [tex]T = 8.83 \ N[/tex]

Explanation:

The  diagram illustrating this  question is  shown on the first uploaded image

From the question we are told that  

    The distance between the two poles is  [tex]d = 12 \ m[/tex]

     The mass tie to the middle of the clotheslines [tex]m = 1 \ kg[/tex]

     The length at which the clotheslines sags is  [tex]l = 4 \ m[/tex]

Generally the weight due to gravity at the middle of the  clotheslines is mathematically represented as

          [tex]W = mg[/tex]

let the angle which the tension on the  clotheslines makes with the horizontal be  [tex]\theta[/tex] which mathematically evaluated using the SOHCAHTOA as follows

        [tex]Tan \theta = \frac{ 4}{6}[/tex]

=>     [tex]\theta = tan^{-1}[\frac{4}{6} ][/tex]

=>     [tex]\theta = 33.70^o[/tex]

   So the vertical component of this  tension is  mathematically represented a  

      [tex]T_y = 2* Tsin \theta[/tex]

Now at equilibrium the  net horizontal force is  zero which implies that

          [tex]T_y - mg = 0[/tex]

=>       [tex]T sin \theta - mg = 0[/tex]

substituting values

          [tex]T = \frac{m*g}{sin (\theta )}[/tex]

substituting values

           [tex]T = \frac{1 *9.8}{2 * sin (33.70 )}[/tex]

           [tex]T = 8.83 \ N[/tex]

Answer:

Do u have a picture of the graph?

Explanation:

I can solve it with refraction

Answer:

t = 0.31s

Explanation:

In order to calculate the time that the object takes to travel from the point with its maximum speed to the point with the maximum acceleration, you first use the following formulas, for the maximum speed and the maximum acceleration:

[tex]v_{max}=\omega A\\\\a_{max}=\omega^2A[/tex]

A: amplitude

v_max = 1.38m/s

a_max = 6.83m/s^2

w: angular frequency

From the previous equations you can obtain the angular frequency w.

You divide vmax and amax, and solve for w:

[tex]\frac{v_{max}}{a_{max}}=\frac{\omega A}{\omega^2 A}=\frac{1}{\omega}\\\\\omega=\frac{a_{max}}{v_{max}}=\frac{6.83m/s^2}{1.38m/s^2}=4.94\frac{rad}{s}[/tex]

Next, you take into account that the maximum speed is obtained when the object passes trough the equilibrium point, and the maximum acceleration for the maximum elongation, that is, the amplitude. In such a trajectory the time is T/4 being T the period.

You calculate the period  by using the information about the angular frequency:

[tex]T=\frac{2\pi}{\omega}=\frac{2\pi}{4.94rad/s}=1.26s[/tex]

Then the required time is:

[tex]t=\frac{T}{4}=\frac{1.26s}{4}=0.31s[/tex]

Complete Question

A proton of mass m​p​​= 1.67×10​−27​​ kg and a charge of q​p​​= 1.60×10​−19​​ C is moving through vacuum at a constant velocity of 10,000 m/s directly to the east when it enters a region of uniform electric field that points to the south with a magnitude of E = 3.62e+3 N/C . The region of uniform electric field is 5 mm wide in the east-west direction. How far (in meters) will the proton have been deflected towards the south by the time it exits the region of uniform electric field. You may neglect the effects of friction and gravity, and assume that the electric field is zero outside the specified region. Answer is to be in units of meters

Answer:

    [tex]s = 0.039 \ m[/tex]

Explanation:

From the question we are told that

    The  mass of the proton is  [tex]m = 1.67 *10^{-27} \ g[/tex]

    The charge of on the proton is [tex]q = 1.60 *10^{-19} \ C[/tex]

      The speed of the proton is  [tex]v = 10000 \ m/s[/tex]

     The magnitude of the electric field is  [tex]E = 3.62*10^{3 } \ N/C[/tex]

       The width covered by the electric field    [tex]d = 5mm = 5 *10^{-3} \ m[/tex]      

Generally the acceleration of the proton due to the electric toward the south  (at the point where the force on the proton is equal to the electric force due to the electric field) is  mathematically represented as

       [tex]a = \frac{q* E}{m}[/tex]  

Substituting values

       [tex]a = \frac{1.60*10^{-19 } * 3.26 *10^{3}}{ 1.67*10^{-27}}[/tex]

      [tex]a = 3.12*10^{11} \ m/s^2[/tex]

Generally the time it will take the proton to cross the electric field is  mathematically represented as

      [tex]t = \frac{d}{v}[/tex]

Substituting values

      [tex]t = \frac{5 *10^{-3}}{10000}[/tex]

     [tex]t = 5 *10^{-7} \ s[/tex]

Generally the the distance covered by the proton toward the south is  

       [tex]s = ut + \frac{1}{2} * a*t^2[/tex]

   Here  u = 0  m/s  this  because before the proton entered the electric field region the it velocity towards the south is  zero

     So

       [tex]s = \frac{1}{2} * a*t^2[/tex]

Substituting values

      [tex]s = \frac{1}{2} * 3.12 *10^{11}*(5 *10^{-7})^2[/tex]

      [tex]s = 0.039 \ m[/tex]

Answer:

W = 14700 J

Explanation:

This is an exercise on Newton's second law.

To solve it we must fix a coordinate system, the most common is an axis parallel to the ramp and the other perpendicular axis, we write Newton's second law

Y Axis . Perpendicular to the ramp

       N - Wy = 0

X axis. Parallel to the ramp, we assume it is positive when the ramp is going up

        F - Wx = m a

in this case F = 1470 N and it is parallel to the plane.

Work is defined by

      W = F .d

boldface indicates vectors

      W = F d cos θ

     let's calculate

      W = 1470 10 cos 0

       W = 14700 J

Answer:

6 N

Explanation:

From the laws of friction

F = ¶R = 0.3 × 20 = 6 N

The force of friction opposing the block's motion is 6 N.

The given parameters;

The force of friction which opposes the motion of the block is obtained by applying Newton's second law of motion.

F = ma

Fₓ = μF

Substitute the given parameters to calculate the frictional force on the object.

Fₓ = 0.3 x 20

Fₓ = 6 N

Thus, the force of friction opposing the block's motion is 6 N.

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Answer:

2.20°

Explanation:

For the central bright spot, we will use the constructive pattern for a double slit interference,

[tex]m\times w = d \times Sin\beta[/tex]

where w indicates the wavelength

and [tex]\beta[/tex] indicates the angle between the bright spot and center line.

now we will use the given values,

1 × 500 × 10^-9 = 1.3 × 10^-5 × Sin [tex]\beta[/tex]

Solving for [tex]\beta[/tex],

[tex]\beta[/tex] = 2.204° ~ 2.20°

Therefore the correct answer is 2.20°

Answer:

An asteroid that has an orbital period of 3 years will have an orbital with a semi-major axis of about  2 years.

Explanation:

Given;

orbital period of 3 years, P = 3 years

To calculate the years of an orbital with a semi-major axis, we apply Kepler's third law.

Kepler's third law;

P² = a³

where;

P is the orbital period

a is the orbital semi-major axis

(3)² = a³

9 = a³

a = [tex]a = \sqrt[3]{9} \\\\a = 2.08 \ years[/tex]

Therefore, An asteroid that has an orbital period of 3 years will have an orbital with a semi-major axis of about  2 years.

Answer:

B) The wavelength of both transverse and longitudinal waves is measured parallel to the direction of the travel of the wave

Answer:

B) The wavelength of both transverse and longitudinal waves is measured parallel to the direction of the travel of the wave.

Explanation: hope this helps ;)

Answer:

 f = 6.66 cm

Explanation:

For this exercise we will use the constructor equation

         1 / f = 1 / p + 1 / q

where f is the focal length, p is the distance to the object and q is the distance to the image

the expression for magnification is

          m = h '/ h = - q / p

with this we have a system of two equations with two unknowns, in the problem they give us the distance to the image q = 10 cm and a magnification of m = -0.5

        -0.5 = - q / p

         p = - q / 0.5

         p = - 10 / 0.5

         p = 20 cm

now we can with the other equation look for the focal length

         1 / f = 1/20 + 1/10

         1 / f = 0.15

          f = 6.66 cm

Answer: Net electrostatic force on C is 24.2×[tex]10^{-2}[/tex] Newtons.

Explanation: Coulomb's Law is used to determine Electrostatic Force. Its formula is:

F = k.[tex]\frac{q_{0}.q_{1}}{r^{2}}[/tex]

where:

k is electrostatic constant (k = 8.987×[tex]10^{9}[/tex] Nm²/C²);

q is the charge of the object in Coulumb;

r is the distance between charges;

The net force is the sum of all the forces acting on C, so:

Force B on C:

They are both positive, so there is a relpusive force acting between them on the y-axis.

[tex]F_{BC} = 8,987.10^{9}.\frac{4.35.10^{-3}.9.67.10^{-4}}{(6.14.10^{2})^{2}}[/tex]

[tex]F_{BC} = 10.03.10^{-2}[/tex] N

Force D on C:

There is an atractive force between them on the x-axis.

[tex]F_{CD} = 8.987.10^{9}.\frac{9.67.10^{-4}.1.92.10^{-3}}{(1.42.10^{3})^{2}}[/tex]

[tex]F_{CD} = 13.64.10^{-4}[/tex] N

Force A on C:

First, find the distance between objects:

The distance is a diagonal line that divides the rectangle into a right triangle. Distance is square of the hypotenuse .

[tex]r^{2} = (6.14.10^2)^{2} + (1.42.10^{3})^{2}[/tex]

[tex]r^{2} = 37.72.10^{4}[/tex]

and hypotenuse: r = [tex]6.14.10^2[/tex]m

There is an atractive force between charges, but there are components of the force in x- and y-axis. So, because of that, force will be:

[tex]F_{CA} = F_{CA}[/tex].sinα + [tex]F_{CA}.[/tex]cosα

[tex]F_{CA} = 8.987.10^{9}.\frac{3.12.10^{-3}.9.67.10^{-4}}{37.72.10^{4}}[/tex]

[tex]F_{CA} = 7.2.10^{-2}[/tex]

The trigonometric relations is taken from the rectangle:

sinα = [tex]\frac{6.14.10^{2}}{6.14.10^{2}}[/tex]

cosα = [tex]\frac{1.42.10^{3}}{6.14.10^{2}}[/tex]

[tex]F_{CA}.[/tex]cosα = [tex]7.2.10^{-2}(\frac{1.42.10^{3}}{6.14.10^{2}})[/tex] = 0.17

[tex]F_{CA}.[/tex]sinα = [tex]7.2.10^{-2}.(\frac{6.14.10^{2}}{6.14.10^{2}} )[/tex] = 0.072

[tex]F_{CA} =[/tex] 0.17î + 0.072^j

Now, sum up all the terms in its respective axis:

X: [tex]13.64.10^{-4} + 0.17 =[/tex] 0.1714

Y: [tex]10.03.10^{-2} + 7.2.10^{-2}[/tex] = 0.1723

These forms another right triangle, whose hypotenuse is the net electrostatic force:

[tex]F_{net} = \sqrt{(0.1714)^{2} + (0.1723)^2}[/tex]

[tex]F_{net} = 24.3.10^{-2}[/tex] N

The net electrostatic force acting on C has magnitude [tex]F_{net} = 24.3.10^{-2}[/tex] N.

Answer:

The two methods will yield different results as one is subject to experimental errors that us the Archimedes method of measurement, the the density measurement method will be more accurate

Explanation:

This is because the density method using the calculated volume will huve room for less errors that's occur in practical method i.e Archimedes method due to human error

Answer:

B = 0.025T

Explanation:

In order to calculate the strength of the magnetic field at the center of the solenoid, you use the following formula:

[tex]B=\frac{\mu N i}{L}[/tex]         (1)

μ: magnetic permeability of vacuum = 4π*10^-7 T/A

N: turns of the solenoid = 500

i: current = 4.0A

L: length of the solenoid = 0.10m

You replace the values of the parameters in the equation (1):

[tex]B=\frac{(4\pi*10^{-7}T/A)(500)(4.0A)}{0.10m}=0.025T[/tex]

The strength of the magnetic field at the center of the solenoid = 0.025T

Answer:

Magnetic field strength at the center is 2.51x10^-2T

Explanation:

Pls see attached file for step by step calculation