A piece of tape is pulled from a spool and lowered toward a 100-mg scrap of paper. Only when the tape comes within 8.0 mm is the electric force magnitude great enough to overcome the gravitational force exerted by Earth on the scrap and lift it.

Requried:
Determine the magnitude and direction of the electric force exerted by the tape on the paper at this distance.

The question is not complete, the value of z is not given.

Assuming the value of z = 4.0m

Answer:

the magnitude of the electric field at any point having z(4.0 m)  =

E = 5.65 N/C

Explanation:

given

σ(surface density) = 0.20 nC/m² = 0.20 × 10⁻⁹C/m²

z = 4.0 m

Recall

E =F/q (coulumb's law)

E = kQ/r²

σ = Q/A

A = 4πr²

∴ The electric field at point z =

E = σ/zε₀

E = 0.20 × 10⁻⁹C/m²/(4 × 8.85 × 10⁻¹²C²/N.m²)

E = 5.65 N/C

Answer:

1:4

Explanation:

The formula for calculating kinetic energy is:

[tex]KE=\dfrac{1}{2}mv^2[/tex]

If the mass is multiplied by 4, then, the kinetic energy must be increased by 4 as well. Since they will be travelling at the same speed when they are at the same point, the relation between KA and KB must be 1:4 or 1/4. Hope this helps!

The relation between the kinetic energies of the freely falling balls A and B is obtained as [tex]\frac{KE_{A}}{KE_{B}} =\frac{1}{4}[/tex].

The kinetic energy of an object depends on the mass and velocity with which it moves.

While under free-fall, the mass of an object does not affect the velocity with which it falls.

So, the velocities of both the balls are the same.

Let the mass of ball A is 'm'

So, the mass of ball B is '4m'

The kinetic energy of ball A is given by;

[tex]KE_{A}=\frac{1}{2} mv^2[/tex]

The kinetic energy of ball B is given by;

[tex]KE_{B}=\frac{1}{2} 4mv^2 = 2mv^2[/tex]

Therefore, the ratio of kinetic energies of A and B is,

[tex]\frac{KE_{A}}{KE_{B}} =\frac{1}{4}[/tex]

Learn more about kinetic energy here:

https://brainly.com/question/11580018

Answer:

0.087976 Nm

Explanation:

The magnetic torque (τ) on a current-carrying loop in a magnetic field is given by;

τ = NIAB sinθ     --------- (i)

Where;

N = number of turns of the loop

I = current in the loop

A = area of each of the turns

B = magnetic field

θ = angle the loop makes with the magnetic field

From the question;

N = 200

I = 4.0A

B = 0.70T

θ = 30°

A = π d² / 4        [d = diameter of the coil = 2.0cm = 0.02m]

A = π x 0.02² / 4 = 0.0003142m²         [taking π = 3.142]

Substitute these values into equation (i) as follows;

τ = 200 x 4.0 x 0.0003142 x 0.70 sin30°

τ = 200 x 4.0 x 0.0003142 x 0.70 x 0.5

τ = 200 x 4.0 x 0.0003142 x 0.70      

τ = 0.087976 Nm

Therefore, the torque on the coil is 0.087976 Nm

Answer:

b. 2 m

Explanation:

Given that:

the identical speakers are connected in phases ;

Let assume ; we have speaker A and speaker B which are = 3 meter apart

An observer stands at X = 4m in front of one speaker.

If the amplitudes are not changed, the sound he hears will be least intense if the wavelength is:                  

From above;  the distance between speaker  A and speaker B can be expressed as:

[tex]\sqrt{3^2 + 4^2 } \\ \\ = \sqrt{9+16 } \\ \\ = \sqrt{25} \\ \\ = 5 \ m[/tex]

The path length difference  will now be:

= 5 m - 4 m

= 1 m

Since , we are to determine the least intense sound; the destructive interference for that path length  will be half the wavelength; which is

= [tex]\dfrac{1}{2}*4 \ m[/tex]

= 2 m

The sound will be heard with least intensity if the wavelength is 2 m. Hence, option (b) is correct.

Given data:

The distance between the speakers is, d = 3 m.

The distance between the observer and speaker is, s = 4 m.

The amplitude of sound wave is the vertical distance from the base to peak of wave. Since sound amplitudes are not changed in the given problem. Then  the distance between speaker  A and speaker B can be expressed as:

[tex]=\sqrt{3^{2}+4^{2}}\\\\=\sqrt{25}\\\\=5\;\rm m[/tex]

And the path length difference is,

= 5 m - 4 m

= 1 m

Since , we are to determine the least intense sound; the destructive

interference for that path length  will be half the wavelength; which is

 [tex]=\dfrac{1}{2} \times s\\\\=\dfrac{1}{2} \times 4[/tex]

= 2 m

Thus, we can conclude that the sound will be heard with least intensity if the wavelength is 2 m.

Learn more about the interference here:

https://brainly.com/question/16098226

Answer:

Increase of volume (F)  = 8.01%

Explanation:

Given:

Density of fish = 1,080 kg/m³

Density of water = 1,000 kg/m³

density of air = 1.28 kg/m³

Find:

Increase of volume (F)

Computation:

1,080 kg/m³  + [F × 1.28 kg/m³ ] = (1+F) × 1,000 kg/m³  

1,080 + 1.28 F =1,000 F + 1,000

80 = 998.72 F

F = 0.0801 (Approx)

F = 8.01%  (Approx)

Answer:

a

 [tex]\theta = 0.0022 rad[/tex]

b

 [tex]I = 0.000304 I_o[/tex]

Explanation:

From the question we are told that  

   The  wavelength of the light is [tex]\lambda = 550 \ nm = 550 *10^{-9} \ m[/tex]

    The  distance of the slit separation is  [tex]d = 0.500 \ mm = 5.0 *10^{-4} \ m[/tex]

Generally the condition for two slit interference  is  

     [tex]dsin \theta = m \lambda[/tex]

Where m is the order which is given from the question as  m = 2

=>    [tex]\theta = sin ^{-1} [\frac{m \lambda}{d} ][/tex]

 substituting values  

      [tex]\theta = 0.0022 rad[/tex]

Now on the second question  

   The distance of separation of the slit is  

       [tex]d = 0.300 \ mm = 3.0 *10^{-4} \ m[/tex]

The  intensity at the  the angular position in part "a" is mathematically evaluated as

      [tex]I = I_o [\frac{sin \beta}{\beta} ]^2[/tex]

Where  [tex]\beta[/tex] is mathematically evaluated as

       [tex]\beta = \frac{\pi * d * sin(\theta )}{\lambda }[/tex]

  substituting values

     [tex]\beta = \frac{3.142 * 3*10^{-4} * sin(0.0022 )}{550 *10^{-9} }[/tex]

    [tex]\beta = 0.06581[/tex]

So the intensity is  

    [tex]I = I_o [\frac{sin (0.06581)}{0.06581} ]^2[/tex]

   [tex]I = 0.000304 I_o[/tex]

Answer:

Explanation:

The stored energy varies with the square of the electric charge stored in the capacitor. If you double the charge, the stored energy in the capacitor will quadruple or increase by a factor of 4.

Doubling the potential across a given capacitor causes the energy stored in that capacitor to reduce to :

D. Quadruple

Doubling the potential across a given capacitor causes the energy stored in that capacitor to reduce to Quadruple.

The stored energy shifts with the square of the electric charge put away within the capacitor.

In case you twofold the charge, the put away vitality within the capacitor will fourfold or increment by a calculate of 4.

Thus, the correct answer is D.

Learn more about "energy":

https://brainly.com/question/1932868?referrer=searchResults

Answer:

The workdone is [tex]W_v = - 20 \ J[/tex]

Explanation:

From the question we are told that

    The mass of the object is [tex]m = 2.5 \ kg[/tex]

     The speed of fall is [tex]v = 2.5 \ m/s[/tex]

     The depth of fall is  [tex]d = 80\ cm = 0.8 \ m[/tex]

Generally according to the work energy theorem

      [tex]W = \frac{1}{2} mv_2^2 - \frac{1}{2} mv_1^2[/tex]

Now here given the that the velocity is  constant  i.e  [tex]v_1 = v_2 = v[/tex] then

We have that

    [tex]W = \frac{1}{2} mv^2 - \frac{1}{2} mv^2 = 0 \ J[/tex]  

So in terms of workdone by the potential energy of the object and that of the viscous liquid we have

       [tex]W = W_v - W_p[/tex]

Where  [tex]W_v[/tex] is workdone by viscous liquid

             [tex]W_p[/tex] is the workdone by the object which is mathematically represented as

            [tex]W_p = mgd[/tex]

So  

       [tex]0 = W_v + mgd[/tex]

=>    [tex]W_v = - m * g * d[/tex]

substituting values

       [tex]W_v = - (2.5 * 9.8 * 0.8)[/tex]

      [tex]W_v = - 20 \ J[/tex]

2Q

When a capacitor carries some certain charge, the energy stored in the capacitor is its electric potential energy E. The magnitude of this potential energy is given by;

E  = [tex]\frac{1}{2}qV[/tex]            ------------(i)

Where;

q = charge between the plates of the capacitor

V = potential difference between the plates of the capacitor

From the question;

q = Q

E = Ep

Therefore, equation (i) becomes;

Ep = [tex]\frac{1}{2} QV[/tex]              ----------------(ii)

Make V subject of the formula in equation (ii)

V = [tex]\frac{2E_{p}}{Q}[/tex]

Now, when the energy is doubled i.e E = 2Ep, equation (i) becomes;

2Ep = [tex]\frac{1}{2}qV[/tex]

Substitute the value of V into the equation above;

2Ep = [tex]\frac{1}{2}[/tex]([tex]q *\frac{2E_{p}}{Q}[/tex])

Solve for q;

[tex]2E_{p}[/tex] = [tex]\frac{2qE_p}{2Q}[/tex]

[tex]2E_{p}[/tex] = [tex]\frac{qE_p}{Q}[/tex]

[tex]q = 2Q[/tex]

Therefore, the charge, when the energy stored is twice the originally stored energy, is twice the original charge. i.e 2Q

Answer:

8.7 m/s^2

82.15°

Explanation:

Given:-

- The initial speed of the car, vi = 25 m/s

- The radius of track, r = 129 m

- Car makes a circular " quarter turn "

- The constant tangential acceleration, at = 1.2 m/s^2

Solution:-

- We will solve the problem using rotational kinematics. Determine the initial angular velocity of car ( wi ) as follows:

                          [tex]w_i = \frac{v_i}{r} \\\\w_i = \frac{25}{129}\\\\w_i = 0.19379 \frac{rad}{s}[/tex]

- Now use the constant tangential acceleration ( at ) and determine the constant angular acceleration ( α ) for the rotational motion as follows:

                           at = r*α

                           α = ( 1.2 / 129 )

                           α = 0.00930 rad/s^2

- We know that the angular displacement from the initial entry to the exit of the turn is quarter of a turn. The angular displacement would be ( θ = π/2 ).

- Now we will use the third rotational kinematic equation of motion to determine the angular velocity at the exit of the turn (wf) as follows:

                            [tex]w_f^2 = w_i^2 + 2\alpha*theta\\\\w_f = \sqrt{0.19379^2 + 0.00930\pi } \\\\w_f = 0.25840 \frac{rad}{s}[/tex]

- We will use the evaluated final velocity ( wf ) and determine the corresponding velocity ( vf ) as follows:

                            [tex]v_f = r*w_f\\\\v_f = 129*0.2584\\\\v_f = 33.33380 \frac{x}{y}[/tex]

- Now use the formulation to determine the centripetal acceleration ( ac ) at this point as follows:

                            [tex]a_c = \frac{v_f^2}{r} \\\\a_c = \frac{33.3338^2}{129} \\\\a_c = 8.6135 \frac{m}{s^2}[/tex]

- To determine the magnitude of acceleration we will use find the resultant of the constant tangential acceleration ( at ) and the calculated centripetal acceleration at the exit of turn ( ac ) as follows:

                             [tex]|a| = \sqrt{a^2_t + a_c^2} \\\\|a| = \sqrt{1.2^2 + 8.6135^2} \\\\|a| = 8.7 \frac{m}{s^2}[/tex]

- To determine the angle between the velocity vector and the acceleration vector. We need to recall that the velocity vector only has one component and always tangential to the curved path. Hence, the velocity vector is parallel to the tangential acceleration vector ( at ). We can use the tangential acceleration ( at ) component of acceleration ( a ) and the centripetal acceleration ( ac ) component of the acceleration and apply trigonometric ratio as follows:

                          [tex]q = arctan \frac{a_c}{a_t} = arctan \frac{8.7}{1.2} \\\\q = 82.15 ^.[/tex] 

Answer: The angle ( q ) between acceleration vector ( a ) and the velocity vector ( v ) at the exit of the turn is 82.15° .

Answer:

The average angular acceleration is 0.05 radians per square second.

Explanation:

Let suppose that Ferris wheel accelerates at constant rate, the angular acceleration as a function of change in angular position and the squared final and initial angular velocities can be clear from the following expression:

[tex]\omega^{2} = \omega_{o}^{2} + 2 \cdot \alpha\cdot (\theta-\theta_{o})[/tex]

Where:

[tex]\omega_{o}[/tex], [tex]\omega[/tex] - Initial and final angular velocities, measured in radians per second.

[tex]\alpha[/tex] - Angular acceleration, measured in radians per square second.

[tex]\theta_{o}[/tex], [tex]\theta[/tex] - Initial and final angular position, measured in radians.

Then,

[tex]\alpha = \frac{\omega^{2}-\omega_{o}^{2}}{2\cdot (\theta-\theta_{o})}[/tex]

Given that [tex]\omega_{o} = 0\,\frac{rad}{s}[/tex], [tex]\omega = 0.70\,\frac{rad}{s}[/tex] and [tex]\theta-\theta_{o} = 4.9\,rad[/tex], the angular acceleration is:

[tex]\alpha = \frac{\left(0.70\,\frac{rad}{s} \right)^{2}-\left(0\,\frac{rad}{s} \right)^{2}}{2\cdot \left(4.9\,rad\right)}[/tex]

[tex]\alpha = 0.05\,\frac{rad}{s^{2}}[/tex]

Now, the time needed to accelerate the Ferris wheel uniformly is described by this kinematic equation:

[tex]\omega = \omega_{o} + \alpha \cdot t[/tex]

Where [tex]t[/tex] is the time measured in seconds.

The time is cleared and obtain after replacing every value:

[tex]t = \frac{\omega-\omega_{o}}{\alpha}[/tex]

If [tex]\omega_{o} = 0\,\frac{rad}{s}[/tex],  [tex]\omega = 0.70\,\frac{rad}{s}[/tex] and [tex]\alpha = 0.05\,\frac{rad}{s^{2}}[/tex], the required time is:

[tex]t = \frac{0.70\,\frac{rad}{s} - 0\,\frac{rad}{s} }{0.05\,\frac{rad}{s^{2}} }[/tex]

[tex]t = 14\,s[/tex]

Average angular acceleration is obtained by dividing the difference between final and initial angular velocities by the time found in the previous step. That is:

[tex]\bar \alpha = \frac{\omega-\omega_{o}}{t}[/tex]

If [tex]\omega_{o} = 0\,\frac{rad}{s}[/tex],  [tex]\omega = 0.70\,\frac{rad}{s}[/tex] and [tex]t = 14\,s[/tex], the average angular acceleration is:

[tex]\bar \alpha = \frac{0.70\,\frac{rad}{s} - 0\,\frac{rad}{s} }{14\,s}[/tex]

[tex]\bar \alpha = 0.05\,\frac{rad}{s^{2}}[/tex]

The average angular acceleration is 0.05 radians per square second.

Answer:

Gas 1

Explanation:

Answer:

8.167

Explanation:

Answer:

The arrangement with the greatest resistance is the light bulb of option C. 4 W, 4.5 V

Explanation:

The equation for electric power is

power P = IV

also,  I = V/R,

substituting into the equation, we have

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

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

a)  [tex]R = \frac{1.5^{2} }{0.8}[/tex] = 2.8 Ω

b) [tex]R = \frac{3^{2} }{6}[/tex] = 1.5 Ω

c) [tex]R = \frac{4.5^{2} }{4}[/tex] 5.06 Ω

d) [tex]R = \frac{6^{2} }{8}[/tex] = 4.5 Ω

from the calculations, one can see that the lightbulb with te greates resistance is

C. 4 W, 4.5 V

Answer:

8 Hz

Explanation:

Given that

Standing wave at one end is 24 Hz

Standing wave at the other end is 32 Hz.

Then the frequency of the standing wave mode of a string having a length, l, is usually given as

f(m) = m(v/2L), where in this case, m could be 1. 2. 3. 4 etc

Also, another formula is given as

f(m) = m.f(1), where f(1) is the fundamental frequency..

Thus, we could say that

f(m+1) - f(m) = (m + 1).f(1) - m.f(1) = f(1)

And as such,

f(1) = 32 - 24

f(1) = 8 Hz

Then, the fundamental frequency needed is 8 Hz

Explanation:

KE = q

½ mv² = mCΔT

ΔT = v² / (2C)

ΔT = (200 m/s)² / (2 × 236 J/kg/°C)

ΔT = 84.7°C

This question involves the concepts of the law of conservation of energy.

The temperature change of the bullet is "84.38°C".

According to the law of conservation of energy, total energy of the system must remain constant. Therefore, in this situation.

[tex]Kinetic\ energy\ of\ bullet\ before\ impact=heat\ absorbed\ in\ bullet\\\\\frac{1}{2}mv^2=mC\Delta T\\\\\Delta T = \frac{v^2}{2C}[/tex]

where,

Therefore,

[tex]\Delta T = \frac{(200\ m/s)^2}{2(237\ J/kg.^oC)}[/tex]

ΔT = 84.38°C

Learn more about the law of conservation of energy here:

https://brainly.com/question/20971995

#SPJ2

Answer:

B and D

Explanation:

Because

R= resistivity xlenght/ Area

Where R= resistance

Explanation:

The Earth has a solid inner core

Answer:

The maximum number of bright spot is [tex]n_{max} =5001[/tex]

Explanation:

From the question we are told that

     The  slit distance is [tex]d = 1 \ mm = 0.001 \ m[/tex]

      The  wavelength is  [tex]\lambda = 400 \ nm = 400*10^{-9 } \ m[/tex]

Generally the condition for interference is  

        [tex]n * \lambda = d * sin \theta[/tex]

Where n is the number of fringe(bright spots) for the number of bright spots to be maximum  [tex]\theta = 90[/tex]

=>     [tex]sin( 90 )= 1[/tex]

So

     [tex]n = \frac{d }{\lambda }[/tex]

substituting values

     [tex]n = \frac{ 1 *10^{-3} }{ 400 *10^{-9} }[/tex]

     [tex]n = 2500[/tex]

given there are two sides when it comes to the double slit apparatus which implies that the fringe would appear on two sides so the maximum number of bright spots is mathematically evaluated as

        [tex]n_{max} = 2 * n + 1[/tex]

The  1  here represented the central bright spot

So  

      [tex]n_{max} = 2 * 2500 + 1[/tex]

     [tex]n_{max} =5001[/tex]      

sorry but I don't understand

The Tension of the string is going to be less when submerged in water by a value called the buoyancy force, so below in the attached file is explanation on how to calculate the apparent weight and density of the submerged object

Answer:

The  tension on the rope  is  T  =  900 N

Explanation:

From the question we are told that  

     The mass of the person on the left is  [tex]m_l = 100 \ kg[/tex]

      The force of the person on the left is  [tex]F_l = 1000 \ N[/tex]

       The mass of the person on the right  is  [tex]m_r = 70 \ kg[/tex]

       The force of the person on the right is  [tex]F_r = 830 \ N[/tex]

Generally the net force is  mathematically represented as

         [tex]F_{Net} = F_l - F_r[/tex]

substituting  values

        [tex]F_{Net} = 1000-830[/tex]

       [tex]F_{Net} = 170 \ N[/tex]

Now the acceleration net acceleration of the rope is mathematically evaluated as

        [tex]a = \frac{F_{net}}{m_I + m_r }[/tex]

substituting  values

     [tex]a = \frac{170}{100 + 70 }[/tex]

     [tex]a = 1 \ m/s ^2[/tex]

The  force [tex]m_i * a[/tex]) of the person on the left that caused the rope to accelerate by  a  is  mathematically represented as

        [tex]m_l * a = F_r -T[/tex]

Where T  is  the tension on the rope  

      substituting values

        [tex]100 * 1 = 1000 - T[/tex]

=>      T  =  900 N

Answer:

non of the above

Explanation:

Quantity of heat = mass× specific heat× change in temperature

m= 7kg c= 4.18 temp= 46-25=21°

.......H= 7×4.18×21= 614.46kJ

Answer:610 KJ

Explanation:A P E X answers

Answer:

11.7 m

Explanation:

The radius of the circle is 13 m.

The central angle of the arc is 0.9 radians

The length of an arc is given as:

L = r θ

where θ = central angle in radians = 0.9

=> L = 0.9 * 13 = 11.7 m

Length of the arc will be 11.7 m ≈ 10 m

Arc length refers to the distance between two points along a curve’s section.

Arc length = radius * theta

where

Arc length  = ? to find

given :

radius = 13 m

theta ( central angle) = 0.9 radians

Arc length = 13 m * 0.9 radians

                = 11.7 m ≈ 10 m

length of the arc will be 11.7 m ≈ 10 m

Learn more about Arc length

https://brainly.com/question/16403495?referrer=searchResults

#SPJ2

Answer:

a. E = 4.62 × 10⁻¹⁹J

b. n = 4.54 × 10¹⁹photons

c. 2.99 × 10²²photons/m²

d. 1.58 ×10¹⁸photons/seconds

e.  n = 1.896 × 10 ⁹ photons

f. the number of photons will be more if it travels slowly

Explanation:

Answer:

a) Only the first root is physically reasonable. Therefore, both stones hit the water in 2.866 seconds, b) The initial velocity of the second stone is -16.038 meters per second, c) The speed of the first stone is 30.227 meters per second and the speed of the second stone is 34.338 meters per second.

Explanation:

a) The time after the release after the release of the first stone can be get from the following kinematic formula for the first rock:

[tex]y_{1} = y_{1,o} + v_{1,o} \cdot t +\frac{1}{2}\cdot g \cdot t^{2}[/tex]

Where:

[tex]y_{1}[/tex] - Final height of the first stone, measured in meters.

[tex]y_{1,o}[/tex] - Initial height of the first stone, measured in meters.

[tex]v_{1,o}[/tex] - Initial speed of the first stone, measured in meters per second.

[tex]t[/tex] - Time, measured in seconds.

[tex]g[/tex] - Gravity constant, measured in meters per square second.

Given that [tex]y_{1,o} = 47\,m[/tex], [tex]y_{1} = 0\,m[/tex], [tex]v_{1,o} = -2.12\,\frac{m}{s}[/tex] and [tex]g = -9.807\,\frac{m}{s^{2}}[/tex], the following second-order polynomial is built:

[tex]-4.984\cdot t^{2} - 2.12\cdot t + 47 = 0[/tex]

Roots of the polynomial are, respectively:

[tex]t_{1} \approx 2.866\,s[/tex] and [tex]t_{2}\approx -3.291\,s[/tex]

Only the first root is physically reasonable. Therefore, both stones hit the water in 2.866 seconds.

b) As the second stone is thrown a second later than first one, its height is represented by the following kinematic expression:

[tex]y_{2} = y_{2,o} + v_{2,o}\cdot (t-t_{o}) + \frac{1}{2}\cdot g \cdot (t-t_{o})^{2}[/tex]

[tex]y_{2}[/tex] - Final height of the second stone, measured in meters.

[tex]y_{2,o}[/tex] - Initial height of the second stone, measured in meters.

[tex]v_{2,o}[/tex] - Initial speed of the second stone, measured in meters per second.

[tex]t[/tex] - Time, measured in seconds.

[tex]t_{o}[/tex] - Initial absolute time, measured in seconds.

[tex]g[/tex] - Gravity constant, measured in meters per square second.

Given that [tex]y_{2,o} = 47\,m[/tex], [tex]y_{2} = 0\,m[/tex], [tex]t_{o} = 1\,s[/tex], [tex]t = 2.866\,s[/tex] and [tex]g = -9.807\,\frac{m}{s^{2}}[/tex], the following expression is constructed and the initial speed of the second stone is:

[tex]1.866\cdot v_{2,o}+29.926 = 0[/tex]

[tex]v_{2,o} = -16.038\,\frac{m}{s}[/tex]

The initial velocity of the second stone is -16.038 meters per second.

c) The final speed of each stone is determined by the following expressions:

First stone

[tex]v_{1} = v_{1,o} + g \cdot t[/tex]

Second stone

[tex]v_{2} = v_{2,o} + g\cdot (t-t_{o})[/tex]

Where:

[tex]v_{1,o}, v_{1}[/tex] - Initial and final velocities of the first stone, measured in meters per second.

[tex]v_{2,o}, v_{2}[/tex] - Initial and final velocities of the second stone, measured in meters per second.

If [tex]v_{1,o} = -2.12\,\frac{m}{s}[/tex] and [tex]v_{2,o} = -16.038\,\frac{m}{s}[/tex], the final speeds of both stones are:

First stone

[tex]v_{1} = -2.12\,\frac{m}{s} + \left(-9.807\,\frac{m}{s^{2}} \right)\cdot (2.866\,s)[/tex]

[tex]v_{1} = -30.227\,\frac{m}{s}[/tex]

Second stone

[tex]v_{2} = -16.038\,\frac{m}{s} + \left(-9.807\,\frac{m}{s^{2}} \right) \cdot (2.866\,s-1\,s)[/tex]

[tex]v_{2} = -34.338\,\frac{m}{s}[/tex]

The speed of the first stone is 30.227 meters per second and the speed of the second stone is 34.338 meters per second.

Answer:

(a)    vo = 24.98m/s

(b)    t = 5.09 s

Explanation:

(a) In order to calculate the the initial speed of the ball, you use the following formula:

[tex]y=y_o+v_ot-\frac{1}{2}gt^2[/tex]      (1)

y: vertical position of the ball = 2.44m

yo: initial vertical position = 0m

vo: initial speed of the ball = ?

g: gravitational acceleration = 9.8m/s²

t: time on which the ball is at 2.44m above the ground = 5.00s

You solve the equation (1) for vo and replace the values of the other parameters:

[tex]v_o=\frac{y-y_o+1/2gt^2}{t}[/tex]        

[tex]v_o=\frac{2.44m-0.00m+1/2(9.8m/s^2)(5.00s)^2}{5.00s}\\\\v_o=24.98\frac{m}{s}[/tex]

The initial speed of the ball is 24.98m/s

(b) To find the time the ball takes to arrive to the ground you use the equation (1) for y = 0m (ground) and solve for t:

[tex]0=24.98t-\frac{1}{2}(9.8)t^2\\\\t=5.09s[/tex]

The time that the ball takes to arrive to the ground is 5.09s

We have that for the Question, it can be said that the speed of  ball and How much longer would it take the ball to reach the ground is

From the question we are told

A goalie kicks a soccer ball straight vertically into the air. It takes 5.00 s for the ball to reach its maximum height and come back down to the level of the crossbar. Assume the crossbar of a soccer goal is 2.44 m above the ground.

(a) How fast was the ball originally moving when it was kicked.

(b) How much longer would it take the ball to reach the ground?

a)

Generally the Newton equation for the Motion  is mathematically given as

[tex]S=ut+1/2at^2\\\\Therefore\\\\2.44=ut+1/2(9.8)(5)^2\\\\u=25.13m/s\\\\[/tex]

b)

Generally the Newton equation for the Motion  is mathematically given as

[tex]S=ut+1/2at^2\\\\Therefore\\\\t=\frac{-24}{a}\\\\t=\frac{-2*25.013}{9.81}\\\\t=5.095sec\\\\[/tex]

Therefore

[tex]X=5.095-5[/tex]

X=0.095sec

For more information on this visit

https://brainly.com/question/23379286

Answer:

Red light

Explanation:

This because All interference or diffraction patterns depend upon the wavelength of the light (or whatever wave) involved. Red light has the longest wavelength (about 700 nm)

Answer:

The discharging current is [tex]I_d = 36.8 \ A[/tex]

Explanation:

From the question we are told that  

     The radius of each circular plates is  R

     The displacement current is  [tex]I = 9.2 \ A[/tex]

      The radius of the central circular area is  [tex]\frac{R}{2}[/tex]

The discharging current is mathematically represented as

       [tex]I_d = \frac{A}{k} * I[/tex]

where A is the area of each plate which is mathematically represented as

       [tex]A = \pi R ^2[/tex]

and   k is central circular area which is mathematically represented as

     [tex]k = \pi [\frac{R}{2} ]^2[/tex]

So  

     [tex]I_d = \frac{\pi R^2 }{\pi * [ \frac{R}{2}]^2 } * I[/tex]

     [tex]I_d = \frac{\pi R^2 }{\pi * \frac{R^2}{4} } * I[/tex]

     [tex]I_d = 4 * I[/tex]

substituting values

     [tex]I_d = 4 * 9.2[/tex]

     [tex]I_d = 36.8 \ A[/tex]

Answer:

a) a = 2.383 m / s², b)   T₂ = 120,617 N , c)   T₃ = 72,957 N

Explanation:

This is an exercise of Newton's second law let's fix a horizontal frame of reference

in this case the mass of the sleds is 30, 20 10 kg from the last to the first, in the first the horizontal force is applied.

a) request the acceleration of the system

we can take the sledges together and write Newton's second law

     T = (m₁ + m₂ + m₃) a

    a = T / (m₁ + m₂ + m₃)

     a = 143 / (10 +20 +30)

     a = 2.383 m / s²

b) the tension of the cables we think through cable A between the sledges of 1 and 20 kg

on the sled of m₁ = 10 kg

          T - T₂ = m₁ a

in this case T₂ is the cable tension

           T₂ = T - m₁ a

            T₂ = 143 - 10 2,383

            T₂ = 120,617 N

c) The cable tension between the masses of 20 and 30 kg

            T₂ - T₃ = m₂ a

             T₃ = T₂ -m₂ a

             T₃ = 120,617 - 20 2,383

             T₃ = 72,957 N