A sphere has a mass of 8200 kg and a radius of 90 cm and floats with nearly zero friction on a thin layer of pressurized water. Suppose you pushes on the sphere tangent to its surface with a steady force of F = 75N and that the pressured water provides a frictionless support.

Required:
How long will it take her to rotate the sphere one time, starting from rest?

Answer:

the two beams must travel paths that differ by one-half of a wavelength.

Complete Question

The uniform purely axial magnetic induction required by the experiment in a volume large enough to accommodate the Lorentz Tube is produced by the Helmholtz Coils. What is the magnetic induction due to a coil current 1.5 Ampere? Convert the result in the still popular non-SI unit Gauss (1 Tesla = 10^4 Gauss).

B = N*mue*I/(2*r)

# of loops = 140

radius of the coil = 0.14m

Answer:

 The magnetic induction is [tex]B = 2.639 \ Gauss[/tex]

Explanation:

From the question we are told that

     The coil current is  [tex]I = 1.5 \ A[/tex]

     The number of loops is  [tex]N = 140[/tex]

The magnetic field due to the current is mathematically represented as

           [tex]B = \mu_o * N * I[/tex]

[tex]\mu_o[/tex] is the permeability of free space with value  [tex]\mu_o = 4\pi * 10^{-7} N/A^2[/tex]

substituting value

           [tex]B = 4\pi * 10^{-7} * 140 * 1.5[/tex]

           [tex]B = 2.639*19^{-4} \ T[/tex]

From question

        (1 Tesla = [tex]10^4 \ Gauss[/tex]).

=>      [tex]B = 2.693 *10^{-4} *10^4 = 2.63 \ Gauss[/tex]

=>      [tex]B = 2.639 \ Gauss[/tex]

Answer:

A)   E = 0N/C

B)   0i + 0^^j

C)   F = 0N

D)   0^i  + 0^j

Explanation:

You assume that the rings are in the zy plane but in different positions.

Furthermore, you can consider that the origin of coordinates is at the midway between the rings.

A) In order to calculate the magnitude of the electric field at the middle of the rings, you take into account that the electric field produced by each ring at the origin is opposite to each other and parallel to the x axis.

You use the following formula for the electric field produced by a charge ring at a perpendicular distance of r:

[tex]E=k\frac{rQ}{(r+R^2)^{3/2}}[/tex]               (1)

k: Coulomb's constant = 8.98*10^9Nm^2/C

Q: charge of the ring

r: perpendicular distance to the center of the ring

R: radius of the ring

You use the equation (1) to calculate the net electric field at the midpoint between the rings:

[tex]E=k\frac{rQ}{(r^2+R^2)^{3/2}}-k\frac{rQ}{(r^2+R^2)^{3/2}}=0\frac{N}{C}[/tex]

The electric field produced by each ring has the same magnitude but opposite direction, then, the net electric field is zero.

B) The direction of the electric field is 0^i + 0^j

C) The magnitude of the force on a proton at the midpoint between the rings is:

[tex]F=qE=q(0N/C)=0N[/tex]

D) The direction of the force is 0^i + 0^j

Part A: The magnitude of the electric field generated at the midpoint between two rings is equal that is 0 N/C.

Part B: The direction of the electric field at the midpoint is opposite.

Part C: The magnitude of the force generated on a proton at the midpoint between two rings is equal that is 0 N.

Part D: The direction of the force on a proton at the midpoint is opposite.

An electric field is defined as the region that surrounds electrically charged particles and exerts a force on all other charged particles within the region, either attracting or repelling them.

Given that diameter of the ring is 10 m and they are 25 m apart from each other. The charge on the left ring is -25nC and on the right ring is 25nC. The electric field can be given as below.

[tex]E = \dfrac {kQ}{(r+R)^2}\\ [/tex]

Where Q is the charge, r is the radius of the ring, R is the mid-point distance and k is the constant.

Part A

The electric field at the mid-point will be the sum of the electric field generated by both the rings. Substituting the values in the above equation,

[tex]E = \dfrac {8.9\times 10^9\times 25}{(10 +12.5)^2}+\dfrac {8.9\times 10^9\times (-25)}{(10 +12.5)^2}[/tex]

[tex]E = \dfrac {222.5\times 10^9}{506.25} - \dfrac {222.5\times 10^9}{506.25}\\ [/tex]

[tex]E = 0\;\rm N/C[/tex]

Hence we can conclude that both the rings generate the electric field with the same magnitude but they are opposite in direction.

Part B

The electric field at the mid-point is 0 N/C. In the vector form, the electric field can be given as below.

[tex]E = 0i+0j[/tex]

The vector form shows that the electric field at the mid-point between the two rings has the same magnitude but is opposite in direction.

Part C

The force can be given as below.

[tex]F = qE[/tex]

[tex]F = 0 \;\rm N[/tex]

If the electric field at the mid-point is zero, then the force at the mid-point will be zero.

Part D

The vector form of the force at the midpoint is given below.

[tex]F = 0i+0j[/tex]

Hence we can conclude that at the midpoint of two rings, the electric field generates an equal force on the proton but in opposite direction. Hence the net force will be zero.

To know more about the electric field, follow the link given below.

https://brainly.com/question/4440057.

Given info

d = 0.000250 meters = distance between slits

L = 302 cm = 0.302 meters = distance from slits to screen

[tex]\theta_8 = 1.12^{\circ}[/tex] = angle to 8th max (note how m = 8 since we're comparing this to the form [tex]\theta_m[/tex])

[tex]x_n = x_5 = 3.33 \text{ cm} = 0.0333 \text{ meters}[/tex] (n = 5 as we're dealing with the 5th minimum )

---------------

Method 1

[tex]d\sin(\theta_m) = m\lambda\\\\0.000250\sin(\theta_8) = 8\lambda\\\\8\lambda = 0.000250\sin(1.12^{\circ})\\\\\lambda = \frac{0.000250\sin(1.12^{\circ})}{8}\\\\\lambda \approx 0.000 000 61082633\\\\\lambda \approx 6.1082633 \times 10^{-7} \text{meters}\\\\ \lambda \approx 6.11 \times 10^{-7} \text{ meters}\\\\ \lambda \approx 611 \text{ nm}[/tex]

Make sure your calculator is in degree mode.

-----------------

Method 2

[tex]\Delta x = \frac{\lambda*L*m}{d}\\\\L*\tan(\theta_m) = \frac{\lambda*L*m}{d}\\\\\tan(\theta_m) = \frac{\lambda*m}{d}\\\\\tan(\theta_8) = \frac{\lambda*8}{0.000250}\\\\\tan(1.12^{\circ}) = \frac{\lambda*8}{0.000250}\\\\\lambda = \frac{1}{8}*0.000250*\tan(1.12^{\circ})\\\\\lambda \approx 0.00000061094306 \text{ meters}\\\\\lambda \approx 6.1094306 \times 10^{-7} \text{ meters}\\\\\lambda \approx 611 \text{ nm}\\\\[/tex]

-----------------

Method 3

[tex]\frac{d*x_n}{L} = \left(n-\frac{1}{2}\right)\lambda\\\\\frac{0.000250*3.33}{302.0} = \left(5-\frac{1}{2}\right)\lambda\\\\0.00000275662251 \approx \frac{9}{2}\lambda\\\\\frac{9}{2}\lambda \approx 0.00000275662251\\\\\lambda \approx \frac{2}{9}*0.00000275662251\\\\\lambda \approx 0.00000061258279 \text{ meters}\\\\\lambda \approx 6.1258279 \times 10^{-7} \text{ meters}\\\\\lambda \approx 6.13 \times 10^{-7} \text{ meters}\\\\\lambda \approx 613 \text{ nm}\\\\[/tex]

There is a slight discrepancy (the first two results were 611 nm while this is roughly 613 nm) which could be a result of rounding error, but I'm not entirely sure.

Answer:

150 miles

Explanation:

5 times 30 is 150

Answer:

(a) Distance between deer and car = 5 m

(b) Vmax = 21.92 m/s

Explanation:

a.

First we calculate distance covered during response time:

s₁ = vt   --------- equation 1

where,

s₁ = distance covered during response time = ?

v = speed of car = 20 m/s

t = response time = 0.5 s

Therefore,

s₁ = (20 m/s)(0.5 s)

s₁ = 10 m

Now, we calculate the distance covered by the car during deceleration. Using 3rd equation of motion:

2as₂ = Vf² - Vi²

s₂ = (Vf² - Vi²)/2a ------ eqation 2

where,

a = deceleration = - 10 m/s²

s₂ = Distance covered during deceleration = ?

Vf = Final Velocity = 0 m/s (since car finally stops)

Vi = Initial Velocity = 20 m/s

Therefore,

s₂ = [(0 m/s)² - (20 m/s)²]/2(-10 m/s²)

s₂ = (400 m²/s²)/(20 m/s²)

s₂ = 20 m

thus, the total distance covered by the car before coming to rest is given as:

s = s₁ + s₂

s = 10 m + 20 m

s = 30 m

Now, the distance between deer and car, when it comes to rest, can be calculated as:

Distance between deer and car = 35 m - s = 35 m - 30 m

Distance between deer and car = 5 m

b.

Since, the distance covered by the car in total must be equal to 35 m at maximum. Therefore,

s₁ + s₂ = 35 m

using equation 1 and equation 2 from previous part:

Vi t + (Vf² - Vi²)/2a = 35 m

Vi(0.5 s) + [(0 m/s)² - Vi²]/2(-10 m/s²) = 35 m

0.5 Vi + 0.05 Vi² = 35

0.05 Vi² + 0.5 Vi - 35 = 0

solving this quadratic equation, we get:

Vi = - 31.92 m/s  (OR)  Vi = 21.92 m/s

For maximum velocity:

Vmax = 21.92 m/s

Answer:

The astronaut and the satellite are 53.718 m apart.

Explanation:

Given;

mass of spacewalking astronaut, = 88 kg

mass of satellite, = 645 kg

force exerts by the satellite, F = 110N

time for this action, t = 0.45 s

Determine the acceleration of the satellite after the push

F = ma

a = F / m

a = 110 / 645

a = 0.171 m/s²

Determine the final velocity of the satellite;

v = u + at

where;

u is the initial velocity of the satellite = 0

v = 0 + 0.171 x 0.45

v = 0.077 m/s

Determine the displacement of the satellite after 1.4 m

d₁ = vt

d₁ = 0.077 x (1.4 x 60)

d₁ = 6.468 m

According to Newton's third law of motion, action and reaction are equal and opposite;

Determine the backward acceleration of the astronaut after the push;

F = ma

a = F / m

a = 110 / 88

a = 1.25 m/s²

Determine the final velocity of the astronaut

v = u + at

The initial velocity of the astronaut = 0

v = 1.25 x 0.45

v = 0.5625 m/s

Determine the displacement of the astronaut after 1.4 min

d₂ = vt

d₂ = 0.5625 x (1.4 x 60)

d₂ = 47.25 m

Finally, determine the total separation between the astronaut and the satellite;

total separation = d₁ + d₂

total separation = 6.468 m + 47.25 m

total separation = 53.718 m

Therefore, the astronaut and the satellite are 53.718 m apart.

Answer:

The maximum speed of the mass is 4.437 m/s.

Explanation:

Given;

length of pendulum, L = 1.62 m

attached mass, m = 117 g

angle of inclination, θ = 38°

This mass was raised to a height of

h = 1.62 - cos38° = 1.0043 m

Apply the principle of conservation of mechanical energy

PE = KE

mgh = ¹/₂mv²

v  = √(2gh)

v = √(2 * 9.8 * 1.0043)

v = 4.437 m/s.

Therefore, the maximum speed of the mass is 4.437 m/s.

Answer:

A) Force = 2303.925 N in the negative x-direction

B) F ≈ 143998.28 N

C) Ratio = 62.5

Explanation:

A) Since the brakes are the only thing making the van to come to a stop, then first of all, we will calculate the force (in a component along the direction of motion of the car) that the brakes will apply on the van.

Let's find the deceleration using Newton's law of motion formula;

v² = u² + 2as

where;

v = final velocity,

u = initial velocity,

s = displacement

a = acceleration

We are given;

u = 87.5 km/h = 24.3056 m/s

s = 125 m

v = 0 m/s

Thus;

0 = (24.3056)² + 2a(125)

- (24.3056)²= 250a

a = - 24.3056²/250

a = - 2.363 m/s²

Now, force = mass × acceleration

We are given mass = 975 kg

Thus;

Force = 975 x (-2.363)

Force = 2303.925 N in the negative x-direction

B) formula for kinetic energy is

KE = ½mv²

KE = ½(975)(24.3056)²

= 287996.568288 J

Work done on impact = F x 2

Thus;

2F = 287996.568288

F = 287996.568288/2

F ≈ 143998.28 N

C) Ratio = Force on car/braking force = 143998.284/2303.925 = 62.5

Complete question is;

Using energy considerations and assuming negligible air resistance, show that a rock thrown from a bridge 25.0 m above water with an initial speed of 20.0 m/s strikes the water with a final speed of 31.1 m/s, independent of the direction thrown

Answer:

It is proved that the final speed is truly 31.1 m/s

Explanation:

From energy - conservation principle;

E_i = Initial potential energy + Initial Kinetic Energy

Or

E_i = U_i + K_i

Similarly, for final energy

E_f = U_f + K_f

So, expressing the formulas for potential and kinetic energies, we now have;

E_i = (m × g × y_i) + (½ × m × v_i²)

Similarly,

E_f = (m × g × y_f) + (½ × m × v_f²)

We are given;

y_i = 25 m

y_f = 0 m

v_i = 20 m/s

v_f = 31.1 m/s

So, plugging in relevant values;

E_i = m((9.8 × 25) + (½ × 20²))

E_i = 485m

Similarly;

E_f = m((9.8 × 0) + (½ × v_f²)

E_f ≈ ½m•v_f²

From energy conservation principle, E_i = E_f.

Thus;

485m = ½m•v_f²

m will cancel out to give;

½v_f² = 485

v_f² = 485 × 2

v_f² = 970

v_f = √970

v_f ≈ 31.1 m/s

Answer:

a) -75 Hz

b) 0.11 [tex]m^{-1}[/tex]

Explanation:

a) Let us first find the frequency detected by the person on the platform.

We have to find the frequency observed by the person when the train was approaching and when the train was receding.

When the train was approaching:

[tex]f_o = \frac{v}{v - v_s} f_s[/tex]

where fo = frequency observed

fs = frequency from the source = 320 Hz

v = speed of sound = 343 m/s

vs = speed of the train = 40 m/s

Therefore:

[tex]f_o = \frac{343}{343 - 40} * 320\\\\f_o = \frac{343}{303} * 320\\\\f_o = 362 Hz[/tex]

The person on the platform heard the sound at a frequency of 362 Hz when the train was approaching.

When the train was receding:

[tex]f_o = \frac{v}{v + v_s} f_s[/tex]

[tex]f_o = \frac{343}{343 + 40} * 320\\\\f_o = \frac{343}{383} * 320\\\\f_o = 287 Hz[/tex]

The person on the platform heard the sound at a frequency of 287 Hz when the train was receding.

Therefore, the frequency change is given as:

Δf = 287 - 362 = -75 Hz

b) We can find the wavelength detected by the person on the platform as the train approaches by using the formula for speed:

[tex]v = \lambda f[/tex]

where λ = wavelength

f = frequency of the train as it approaches = 362 Hz

v = speed of train = 40 m/s

Therefore, the wavelength detected is:

40 = λ * 362

λ = 40 / 362 = 0.11 [tex]m^{-1}[/tex]

Answer:

4.7ft

Explanation:

Pls see attached file

Answer:

Explanation:

Heat absorbed by oil

= mass x specific heat x rise in temperature

= 30 x 10⁻³ x 950 x 2.2 x 10³ x ( 50-10 )

= 25.08 x 10⁵ J  

Heat absorbed by air

= 50 x 1.2 x 1.0054 x 10³ x ( 20-10 )

= 6.03 x 10⁵ J

Total heat absorbed = 31.11 x 10⁵ J

If time required = t

heat lost from room

= .35 x 10³ t

Total heat generated in time t

= 1.8 x 10³ t

Heat generated = heat used

1.8 x 10³ t =  .35 x 10³ t  + 31.11 x 10⁵

1.45 x 10³ t = 31.11 x 10⁵

t = 31.11 x 10⁵ / 1.45 x 10³

t = 2145.5 s

Answer:

the right answer is option C

Explanation:

at the horizon

The position of rainbows depends on the some factors like wavelength of light and position of sun. So, it can occur at the horizon or above the horizon at the 40 - 42 Degrees. Hence, option (A) and (C) are correct.

The given problem is based on the concepts and fundamentals of Rainbow. The rainbow is the band of seven of seven colors namely, violet, indigo, blue, green, yellow, orange and red.

Thus, we can conclude that the position of rainbows depends on the some factors like wavelength of light and position of sun. So, it can occur at the horizon or above the horizon at the 40 - 42 Degrees. Hence, option (A) and (C) are correct.

Learn more about the Rainbows here:

https://brainly.com/question/15758504

Answer:

Explanation:

a. Is the sphere charged negatively or positively?

The sphere us negatively charged. In a Millikan type experiment, there will be two forces that will be acting on the sphere which are the electric force which acts upward and also the gravity which acts downward.

Because the upper plate is positively charged, there'll what an attractive curve with an upward direction which will be felt by the negatively charged sphere.

b. What is the magnitude of the electric field intensity between the plates?

The magnitude of the electric field intensity between the plates is 18400v/m.

C. Calculate the magnitude of the charge on the latex sphere.

The magnitude of the charge on the latex sphere hae been solved and attached

d. How many excess or deficit electrons does the sphere have?

There are 5 excess electrons that the sphere has.

Check the attachment for further explanation.

Answer:

we can conclude that the component of the horizontal force and vertical force are 225.28 N and 1370 N respectively.

Coefficient of static friction = 0.26

Explanation:

Given that:

length of the ladder = 16.0 m

weight of the ladder = 520 N

angle θ = 65.0°

(a) We are to find the horizontal and vertical forces the ground exerts on the base of the ladder when an :

Force = 850 N

distance of the climber from the base of the ladder = 4.20 m

The diagrammatic illustration representing what the given information entails can be seen from the attached file below.

Let consider the Ladder being at point A with the horizontal layer of the ground.

From the whole system; the condition for the equilibrium at the point A can be computed as :

[tex]N_2 (16 \ Sin\ 65) = 850(4.2 \ \times Cos \ 65 )+ 520 (\dfrac{16}{2}) Cos \ 65[/tex]

[tex]N_2 (14.50) = 850(1.7749 )+ 520 (8) \times 0.4226[/tex]

[tex]N_2 (14.50) = 1508.665+1758.016[/tex]

[tex]N_2 (14.50) = 3266.681[/tex]

[tex]N_2 =\dfrac{ 3266.681}{14.50}[/tex]

[tex]N_2 =225.28 \ N[/tex]

[tex]N_1 = mg+F\\[/tex]

where ;

w =mg

[tex]N_1 = 520+850[/tex]

[tex]N_1 = 1370 \ N[/tex]

Therefore; we can conclude that the component of the horizontal force and vertical force are 225.28 N and 1370 N respectively.

(b) If the ladder is just on the verge of slipping when the firefighter is 9.40 m from the bottom, what is the coefficient of static friction between ladder and ground?

the coefficient of static friction between ladder and ground when the firefighter is 9.40 m from the bottom can be calculated as:

[tex]N_2 (16 \ Sin\ 65) = 850(9.4 \ \times Cos \ 65 )+ 520 (\dfrac{16}{2}) Cos \ 65[/tex]

[tex]N_2 (14.50) = 850(3.9726 )+ 520 (8) \times 0.4226[/tex]

[tex]N_2 (14.50) =3376.71+1758.016[/tex]

[tex]N_2 (14.50) =5134.726[/tex]

[tex]N_2 =\dfrac{5134.726}{14.50}[/tex]

[tex]N_2 =354.12 \ N[/tex]

Therefore; the coefficient of the static friction is;

[tex]\mu = \dfrac{f_s}{N_1}[/tex]

[tex]\mu = \dfrac{354.12}{1370}[/tex]

[tex]\mu[/tex]  = 0.26

Coefficient of static friction = 0.26

Answer:

3) Transmitted intensity of light if unpolarized light passes through a single polarizing filter = 40 W/m²

- Transmitted intensity of light if an additional polarizer is added perpendicular to the first polarizer in the setup described = 7.5 W/m²

Explanation:

Complete Question

3) What is the transmitted intensity of light if unpolarized light passes through a single polarizing filter and the initial intensity is 80 W/m²?

- What is the transmitted intensity of light if an additional polarizer is added perpendicular to the first polarizer in the setup described in Question 3 (the setup)? Show all work in your answer.

The image of this setup attached to this question as obtained from online is attached to this solution.

Solution

3) When unpolarized light passes through a single polarizer, the intensity of the light is cut in half.

Hence, if the initial intensity of unpolarized light is I₀ = 80 W/m²

The intensity of the light rays thay pass through the first single polarizer = I₁ = (I₀/2) = (80/2) = 40 W/m²

- According to Malus' law, the intensity of transmitted light through a polarizer is related to the intensity of the incident light and the angle at which the polarizer is placed with respect to the major axis of the polarizer before the current polarizer of concern.

I₂ = I₁ cos² θ

where

I₂ = intensity of light that passes through the second polarizer = ?

I₁ = Intensity of light from the first polarizer that is incident upon the second polarizer = 40 W/m²

θ = angle between the major axis of the first and second polarizer = 30°

I₂ = 40 (cos² 30°) = 40 (0.8660)² = 30 W/m²

In the same vein, the intensity of light that passes through the third/additional polarizer is related to the intensity of light that passes through the second polarizer and is incident upon this third/additional polarizer through

I₃ = I₂ cos² θ

I₃ = intensity of light that passes through the third/additional polarizer = ?

I₂ = Intensity of light from the second polarizer that is incident upon the third/additional polarizer = 30 W/m²

θ = angle between the major axis of the second and third/additional polarizer = 60° (although, it is 90° with respect to the first polarizer, it is the angle it makes with the major axis of the second polarizer, 60°, that matters)

I₃ = 30 (cos² 60°) = 30 (0.5)² = 7.5 W/m²

Hope this Helps!!!

Answer:

The distance between the two charges is =4.4mm

Answer:

The speed will be "3.4×10⁴ m/s²".

Explanation:

The given values are:

Angular speed,

w = 7200 rpm

i.e.,

  = [tex]7200 \times \frac{2 \pi}{60}[/tex]

  = [tex]753.6 \ rad/s[/tex]

Speed from the center,

r = 6.0 cm

As we know,

⇒  Linear speed, [tex]v=wr[/tex]

On putting the estimated values, we get

                               [tex]=753.6\times 0.06[/tex]

                               [tex]=45.216 \ m[/tex]

Now,

Acceleration on disk will be:

⇒  [tex]a=\frac{v^2}{r}[/tex]

       [tex]=34074 \ m/s^2[/tex]

       [tex]=3.4\times 10^4 \ m/s^2[/tex]

Answer:

B = 4.1*10^-3 T = 4.1mT

Explanation:

In order to calculate the strength of the magnetic field, you use the following formula for the magnetic flux trough a surface:

[tex]\Phi_B=S\cdot B=SBcos\alpha[/tex]        (1)

ФB: magnetic flux trough the circular surface = 6.80*10^-5 T.m^2

S: surface area of the circular plate = π.r^2

r: radius of the circular plate = 8.50cm = 0.085m

B: magnitude of the magnetic field = ?

α: angle between the direction of the magnetic field and the normal to the surface area of the circular plate = 43.0°

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

[tex]B=\frac{\Phi_B}{Scos\alpha}=\frac{6.80*10^{-5}T.m^2}{(\pi (0.085m)^2)cos(43.0\°)}\\\\B=4.1*10^{-3}T=4.1mT[/tex]

The strength of the magntetic field is 4.1mT

Answer:

The main points of difference between a citizen and alien are: (a) A citizen is a permanent resident of a state, while an alien is a temporary resident, who comes for a specific duration of time as a tourist or on diplomatic assignment. ... Aliens do not possess such rights in the state where they reside temporarily

Explanation:

Answer:

I think it is the 4th answer choice

Explanation:

Heat is thermal energy that flows in the direction of high temp to low temp, and internal energy is the "energy contained in a system", and thermal energy is a part of that.

Answer:

he correct answers are a, b

Explanation:

In the two-slit interference phenomenon, the expression for interference is

          d sin θ= m λ                       constructive interference

          d sin θ = (m + ½) λ             destructive interference

in general this phenomenon occurs for small angles, for which we can write

           tanθ = y / L

           tan te = sin tea / cos tea = sin tea

           sin θ = y / La

un

derestimate the first two equations.

Let's do the calculation for constructive interference

         d y / L = m λ

the distance between maximum clos is and

         y = (me / d) λ

this is the position of each maximum, the distance between two consecutive maximums

         y₂-y₁ = (L   2/d) λ - (L 1 / d) λ₁          y₂ -y₁ = L / d λ

examining this equation if the wavelength decreases the value of y also decreases

the same calculation for destructive interference

         d y / L = (m + ½) κ

         y = [(m + ½) L / d] λ

again when it decreases the decrease the distance

the correct answers are a, b

Answer:

Explanation:

The current through the  resistor is 0.83 A

.

Part b

The current through  resistor is 0.53 A

.

Part c

The current through  resistor is 0.30 A

Answer:

F' = F/4

Thus, the magnitude of electrostatic force will become one-fourth.

Explanation:

The magnitude of force applied by each charge on one another can be given by Coulomb's Law:

F = kq₁q₂/r²   -------------- equation 1

where,

F = Force applied by charges

k = Coulomb's Constant

q₁ = magnitude of first charge

q₂ = magnitude of 2nd charge

r = distance between the charges

Now, in the final state the charges on both spheres are halved. Therefore,

q₁' = q₁/2

q₂' = q₂/2

Hence, the new force will be:

F' = kq₁'q₂'/r²

F' = k(q₁/2)(q₂/2)/r²

F' = (kq₁q₂/r²)(1/4)

using equation 1:

F' = F/4

Thus, the magnitude of electrostatic force will become one-fourth.

The magnitude of the electrostatic force will be F' = F/4

Here we used Coulomb's Law:

F = kq₁q₂/r²   -------------- equation 1

Here

F = Force applied by charges

k = Coulomb's Constant

q₁ = magnitude of first charge

q₂ = magnitude of 2nd charge

r = distance between the charges

Now

q₁' = q₁/2

q₂' = q₂/2

So, the new force should be

F' = kq₁'q₂'/r²

F' = k(q₁/2)(q₂/2)/r²

F' = (kq₁q₂/r²)(1/4)

So,

F' = F/4

Learn more about force here: https://brainly.com/question/14282312

Answer:

the image is virtual and erect and the lens divergent; therefore the correct answer is C

Explanation:

In a thin lens the magnification given by

      m = h '/ h = - q / p

where h ’is the height of the image, h is the height of the object, q is the distance to the image and p is the distance to the object.

It indicates that the object is straight and is placed at a distance p> f

analyze the situation tells us that the magnification is positive so the distance to the image must be negative, that is, that the image is on the same side as the object.

Consequently the lens must be divergent

The magnification value is

          0.4 = h ’/ h

          h ’= 0.4 h

therefore the erect images

therefore the image is virtual and erect and the lens divergent; therefore the correct answer is C

Answer:

θ₀ = 84.78° (OR) 5.22°

Explanation:

This situation can be treated as projectile motion. The parameters of this projectile motion are:

R = Range of Projectile = 150 m

V₀ = Launch Speed of Projectile = 90 m/s

g = 9.8 m/s²

θ₀ = Launch angle (OR) Angle of Elevation = ?

The formula for range of a projectile is given as:

R = V₀² Sin 2θ₀/g

Sin 2θ₀ = Rg/V₀²

Sin 2θ₀ = (150 m)(9.8 m/s²)/(90 m/s)²

2θ₀ = Sin⁻¹ (0.18)

θ₀ = 10.45°/2

θ₀ = 5.22°

Also, we know that for the same launch velocity the range will be same for complementary angles. Therefore, another possible value of angle is:

θ₀ = 90° - 5.22°

θ₀ = 84.78°

Answer:

a) a = + 1,758 10¹¹ m / s ,  b)  = √ (2 1,758 10¹¹ E x)

Explanation:

a) A charged particle is an electric field undergoing force given by the expression

          F = qE

where q is the charge of the paticle and E electric field.

In this case we are told that the particle is positron

         q = + 1.6 10⁻¹⁹ C

let's calculate the force

         F = + 1.6 10⁻¹⁹ E

we write the positive sign, to show that the particle accelerates in the same direction of the electric field

let's write Newton's second law to find the acceleration

    F = ma

     a = F / m

     a = + 1.6 10-19 / 9.1 10-31 E

     a = + 1,758 10¹¹ m / s

b) the velocity of the particle starting from rest

       v² = v₀² + 2 a x

       v = √ (2 1,758 10¹¹ E x)

Answer:

3.09 m/s²

Explanation:

Given:

Δx = -105 m

v₀ = -27.7 m/s

v = -10.9 m/s

Find: a

v² = v₀² + 2aΔx

(-10.9 m/s)² = (-27.7 m/s)² + 2a (-105 m)

a = 3.09 m/s²