A fish appears to be 2.00 m below the surface of a pond when viewed almost directly above by a fisherman. What is the actual depth of the fish

Given that,

Temperature = 10⁸ K

We need to calculate the peak wavelength in the blackbody spectrum

Using formula of peak wavelength

[tex]peak\ wavelength = \dfrac{2.898\times10^{-3}}{T}[/tex]

Where, T= temperature

Put the value into the formula

[tex]peak\ wavelength = \dfrac{2.898\times10^{-3}}{10^{8}}[/tex]

[tex]peak\ wavelength = 2.90\times10^{-11}\ m[/tex]

[tex]peak\ wavelength = 290\ nm[/tex]

This range of wavelength is ultraviolet.

Hence, The peak wavelength in the blackbody spectrum is 290 nm and the range of wavelength is ultraviolet electromagnetic spectrum .

Answer:

Explanation:

Formula used for calculating power P = current * voltage

P = IV

From ohms law, V = IR where R is the resistance. Substituting V = IR into the formula for calculating power, we will have;

P = IV

P =(V/R)V

P = V²/R

Given parameters

Power rating of the bulb P = 100 Watts

Source voltage V = 110V

Required

Resistance of the bulb R

Substituting the given parameters into the formula for calculating power to get Resistance R;

P = V²/R

100 = 110²/R

R = 110²/100

R = 110 * 110/100

R = 12100/100

R = 121 ohms

Hence, the resistance of this bulb is 121 ohms

Answer

76200meters

Explanation:

we know that 1km=1000meters

to convert km into meters we we divide km by meters

=76.2/1000

=76200meters

Answer:

0.03143 Gy

Explanation:

Mass of the human body = 70 kg

Mass of potassium in the human body = 140 g

chemical atomic mass of potassium = 39.1

From avogadros number, we know that 1 atomic mass of an element contains 6.023 × 10^(23) atoms

Thus,

140g of potassium will contain;

(140 × 6.023 × 10^(23))/(39.1) = 2.1566 × 10^(24) atoms

We are told that the natural abundance of one of the 40K isotopes is 0.012%.

Thus;

Number of atoms of this isotope = 0.012% × 6.023 × 10^(23) = 7.2276 × 10^(19) K-40 atoms

Formula for activity of K-40 is given as;

Activity = (0.693 × number of K-40 atoms)/half life

Activity = (0.693 × 7.2276 × 10^(19))/1300000000

Activity = 3.85 × 10^(10)

We are told that each decay deposits 1.0 MeV of energy into the body.

Thus;

Total energy absorbed by the body in a year = 3.85 × 10^(10) × 1 × 365 = 1405.25 × 10^(10) MeV

Now, 1 MeV = 1.602 × 10^(-13) joules

Thus;

Total energy absorbed by the body in a year = 1405.25 × 10^(10) × 1.602 × 10^(-13) = 2.25 J

1 Gy = 1 J/kg

Thus;

Yearly dose = 2.25/70 = 0.03143 Gy

Answer:

it is going to D. all of these are resistors

Answer:

b

Explanation:

Answer:

A.   number of neutrons of Magnesium Mg = 13

B.   The average mass of Mg = 22.29 amu

C.   the magnesium composition on Mars is not the same as that on Earth.

Explanation:

Isotopes are atoms with the same atomic number but different mass number. This is due to the difference in mass of the neutrons.

The atomic number of Magnesium Mg = 12

The atomic number of an element is the number of protons present in the atomic nucleus of the element

i.e Atomic number = number of protons = 12

The mass number of an element is the sum of the protons and neutrons in the atomic nucleus of the element.

Mass number = number of protons + number of neutrons

Given that the mass number of Mg = 25

Then;

25 = 12 + number of neutrons

25 - 12 = number of neutrons

13 = number of neutrons

number of neutrons of Magnesium Mg = 13

B. What is the average atomic mass of magnesium in these rocks?

The average atomic mass of an element which exhibit isotopy is the average mass of its various isotopes as they occur naturally in any quantity of the element.

Therefore the average atomic mass of magnesium can be calculated as:

= [tex]\mathtt{\dfrac{(23.9872 \times 79.70) + ( 24.9886 \times 10.13) + (25.9846 \times 10.17) }{79.7 + 10.13 +10.17}}[/tex]

= [tex]\mathtt{\dfrac{(1911.77984) + ( 53.134518) + (264.263382) }{100}}[/tex]

= [tex]\mathtt{\dfrac{2229.17774 }{100}}[/tex]

The average mass of Mg = 22.29 amu

C. Is the magnesium composition on Mars the same as that on Earth? Explain.

The average atomic weight of magnesium on Earth is said to be 24.305 amu while that of Mars is 22.29 amu.

There difference in the average atomic weight result into difference in their composition. Therefore,the magnesium composition on Mars is not the same as that on Earth.

Answer:

Explanation:

Using the Continuity equation

v X A = v' xA'

so if A is 1/2of A' then A velocity must be 2 times the A'

after-contraction v = 2 x 5.0m/s = 10m/s

Using the Bernoulli equation

p₁ + ½ρv₁² + ρgh₁ = p₂ + ½ρv₂² + ρgh₂

, the "h" terms cancel

3.5 x 10^ 5Pa + ½ x 1000kg/m³x (5.0m/s)² = p₂ + ½ x 1000kg/m³ x (10m/s)²

p₂ = 342500pa

Answer:

1.34 mm

Explanation:

A double slit experiment is conducted with a light which has a wavelength of 620 nm

The fringes are separated 2.3 mm apart

The light is changed to a wavelength length of 360 nm

Let x represent the fringe spacing as a result of the change in wavelength

Therefore,the fringe spacing can be calculated as follows

2.3mm/x= 620nm/360nm

Multiply both sides

x × 620= 2.3×360

620x= 828

x= 828/620

x= 1.34 mm

Answer:

a)   Δt = 24.96 s , b)  τ = 0.078 N m

Explanation:

This is a rotational kinematics exercise

        θ = w₀ t - ½ α t²

Let's reduce the magnitudes the SI system

       θ = 60 rev (2π rad / 1 rev) = 376.99 rad

       w₀ = 7.0 rot / s (2π rad / 1 rpt) = 43.98 rad / s

      α = (w₀ t - θ) 2 / t²

let's calculate the annular acceleration

      α = (43.98 10 - 376.99) 2/10²

      α = 1,258 rad / s²

Let's find the time it takes to reach zero angular velocity (w = 0)

        w = w₀ - alf t

         t = (w₀ - 0) / α

         t = 43.98 / 1.258

         t = 34.96 s

this is the total time, the time remaining is

         Δt = t-10

         Δt = 24.96 s

To find the braking torque, we use Newton's law for angular motion

        τ = I α

the moment of inertia of a circular ring is

       I = M r²

we substitute

         τ = M r² α

we calculate

        τ = 0.625  0.315²  1.258

        τ = 0.078 N m

The total time taken by the wheel to come to rest is 25.18 s and the magnitude of the frictional torque is 25.18 N-m.

Given data:

The initial angular speed of wheel is, [tex]\omega = 7.0 \;\rm rps[/tex]   (rps means rotation per second).

The time interval is, t' = 10.0 s.

The number of rotations made by wheel is, n = 60.0.

The mass of bike wheel is, m = 0.625 kg.

The radius of wheel is, r = 0.315 m.

The problem is based on rotational kinematics. So, apply the second rotational equation of motion as,

[tex]\theta = \omega t-\dfrac{1}{2} \alpha t'^{2}[/tex]

Here, [tex]\theta[/tex] is the angular displacement, and its value is,

[tex]\theta =2\pi \times 60\\\\\theta = 376.99 \;\rm rad[/tex]

And, angular speed is,

[tex]\omega = 2\pi n\\\omega = 2\pi \times 7\\\omega = 43.98 \;\rm rad/s[/tex]

Solving as,

[tex]376.99 = 43.98 \times 10-\dfrac{1}{2} \alpha \times 10^{2}\\\\\alpha = 1.25 \;\rm rad/s^{2}[/tex]

Apply the first rotational equation of motion to obtain the value of time to reach zero final velocity.

[tex]\omega' = \omega - \alpha t\\\\0 = 43.98 - 1.25 \times t\\\\t = 35.18 \;\rm s[/tex]

Then total time is,

T = t - t'

T = 35.18 - 10

T = 25.18 s

Now, use the standard formula to obtain the value of braking torque as,

[tex]T = m r^{2} \alpha\\\\T = 0.625 \times (0.315)^{2} \times 1.25\\\\T = 0.0775 \;\rm Nm[/tex]

Thus, we can conclude that the total time taken by the wheel to come to rest is 25.18 s and the magnitude of the frictional torque is 25.18 N-m.

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Answer:Light Amplification by Stimulated Emission of Radiation. It is able to convert light or electrical energy into focused high energy beam to treat some sickness and diseases.

Explanation:

Answer:

Light amplification by stimulated emission of radiation

Answer:

The uncertainty in momentum is 5.25x 10^25Jsm

Explanation:

We know that

h bar = h/2π

So

1.05x 10^34=h/2pπ

h=1.05x 10^ 34(2π)=6.597x 10^-34Js

dp=(6.597x10^-34/4pπ)/(1x10^-10)

=5.25x10^-25 Jsm

Answer:

The highest percentage of change corresponds to the thinnest rod, the correct answer is a

Explanation:

For this exercise we are asked to change the length of the bar by the action of a force applied along its length, in this case we focus on the expression of longitudinal elasticity

               F / A = Y ΔL/L

where F / A is the force per unit length, ΔL / L is the fraction of the change in length, and Y is Young's modulus.

In this case the bars are made of the same material by which Young's modulus is the same for all

              ΔL / L = (F / A) / Y

the area of ​​the bar is the area of ​​a circle

               A = π r² = π d² / 4

               A = π / 4 d²

we substitute

              ΔL / L = (F / Y) 4 /πd²

changing length

               ΔL = (F / Y 4 /π) L / d²

The amount between paracentesis are all constant in this exercise, let's look for the longitudinal change

a) values ​​given d and 3L

               ΔL = cte 3L / d²

               ΔL = cte L /d²  3

To find the percentage, we must divide the change in magnitude by its value and multiply by 100.

                ΔL/L % = [(F /Y  4/π 1/d²) 3L ] / 3L 100

                ΔL/L  % = cte 100%

b) 3d and L value, we repeat the same process as in part a

               ΔL = cte L / 9d²

               ΔL = cte L / d² 1/9

               ΔL / L% = cte 100/9

               ΔL / L% = cte 11%

c) 2d and 2L value

               ΔL = (cte L / d ½ )/ 2L

               ΔL/L% = cte 100/4

               ΔL/L% = cte 25%

d) value 4d and L

               ΔL = cte L / d² 1/16

                ΔL/L % = cte 100/16

                ΔL/L % = cte 6.25%

The highest percentage of change corresponds to the thinnest rod, the correct answer is a

Answer:

the pressure after contraction is 2×10^5 Pa

the speed after contraction is 15m/s

Explanation:

We were given Pressure P to be 3.5 x 10^5 that is Flowing with speed of 5.0 m/s,

For us to calculate pressure we need to calculate the area first as;

Let initial Area = A₁

And Final area A₂

We were told that in a horizontal pipe it contracts to 1/3 its former area. Which means

A₂= A₁/3.................

V₁ is the speed

the pressure and speed of the water after the contraction can be calculated using equation of continuity below

A₂V₂ = A₁V₁

But

If we substitute given value in the expresion we have

V₂ = (3A *5)/A

V₂ = 15m/s

Therefore, the speed after contraction is 15m/s

Now we can calculate the pressure using

Bernoulli's equation

p₁ + ½ρv₁² + ρgh₁ = p₂ + ½ρv₂² + ρgh₂

But we know that the pipe is horizontal, then "h" terms cancel out then

p₁ + ½ρv₁² = p₂ + ½ρv₂²

Making P₂ subject of formula we have

p₂ = 0.5ρ( V ₁² - v₂² ) + P₁

P₂=. 0.5 × 1000 (5² -15² ) + 3*10^5

=2×10^5 Pa

Therefore, the pressure after contraction is 2×10^5 Pa

(a)  the final speed of the water after contraction is 15 m/s.

(b) The final pressure of the water after contraction is 2.5 x 10⁵ Pa.

The given parameters;

Let the initial area of the pipe = A₁

Apply the continuity equation to determine the final speed of the water after contraction as follows;

[tex]A_1 V_1 = A_2 V_2\\\\V_2 = \frac{A_1V_1}{A_2} \\\\V_2 = \frac{A_1 \times 5}{\frac{1}{3} A_1 } \\\\V_2 = 15 \ m/s[/tex]

The final pressure of the water after contraction is determined by applying Bernoulli's equation for horizontal pipe;

[tex]P_1 + \frac{1}{2} \rho V_1^2= P_2 + \frac{1}{2} \rho V_2^2\\\\P_2 = \frac{1}{2} \rho (V_1^2 - V_2^2) + P_1\\\\P_2 = \frac{1}{2} \times 1000(5^2 - 15^2) + 3.5 \times 10^5\\\\P_2 = 2.5 \times 10^5 \ Pa[/tex]

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Answer:

Increase in frequency of the laser

Explanation:

Because An increase in frequency will result in more lines per centimeter and a smaller distance between each consecutive line. And a decrease in distance between each gratin

Answer: The properties of magnets are used to make electricity. Moving magnetic fields pull and push electrons. Moving a magnet around a coil of wire, or moving a coil of wire around a magnet, pushes the electrons in the wire and creates an electrical current.

Answer:

t = 23.6 min

Explanation:

First we need to find the mass of air in the room:

m = ρV

where,

m = mass of air in the room = ?

ρ = density of air at room temperature = 1.2041 kg/m³

V = Volume of room = 16 m x 16.5 m x 12.3 m = 3247.2 m³

Therefore,

m = (1.2041 kg/m³)(3247.2 m³)

m = 3909.95 kg

Now, we find the amount of energy consumed to heat the room:

E = m C ΔT

where,

E = Energy consumed = ?

C = Specific Heat of air at room temperature = 1 KJ/kg.⁰C

ΔT = Change in temperature = 21 °C - 13.5 °C = 7.5 °C

Therefore,

E = (3909.95 kg)(1 KJ/kg.°C)(7.5 °C)

E = 29324.62 KJ

Now, the time period can be calculated as:

P = E/t

t = E/P

where,

t = Time needed = ?

P = Power of heater = 20.7 KW

Therefore,

t = 29324.62 KJ/20.7 KW

t = (1416.65 s)(1 min/60 s)

t = 23.6 min

Answer:

metal conducts heat better than wood.

Explanation:

Metals are generally good conductors of heat, and they usually conduct heat at a relatively rapid rate. Inside the room with a uniform temperature, a metal when touched will rapidly conduct the heat from your hand, leaving your hand with a cooler feeling. Wood on the other hand is a poor heat conductor, so the heat is not conducted from your hand fast enough to cool it up to the point that your hand feels cool.

Answer:

34°

Using the relation

θᶜ = sin^-1(n₂/n₁),

where n1= the refractive index of light is propagating from a medium

And n2 = refractive index of medium into which light is entering

So we know that

refractive index of diamond at 589nm = 2.41= n₁

refractive index of ethanol at 589nm and 20°C = 1.36= n₂

Thus. θᶜ = sin^-1(1.361/2.417) = 0.58radians = 34°

Explanation:

Answer:

attached below is the free body diagram of the missing  illustration

Initial kinetic energy of the electron = 3 eV

Explanation:

The conclusion that can be drawn about the kinetic energy of the electron is

[tex]E_{e} = E_{3} - E_{1}[/tex]

E[tex]_{e}[/tex] = initial kinetic energy of the electron

E[tex]_{1}[/tex] = -4 eV

E[tex]_{3}[/tex] = -1 eV

insert the values into the equation above

[tex]E_{e}[/tex] = -1 -(-4)  eV

   = -1 + 4 = 3 eV

Answer:

His angular velocity will increase.

Explanation:

According to the conservation of rotational momentum, the initial angular momentum of a system must be equal to the final angular momentum of the system.

The angular momentum of a system = [tex]I[/tex]'ω'

where

[tex]I[/tex]' is the initial rotational inertia

ω' is the initial angular velocity

the rotational inertia = [tex]mr'^{2}[/tex]

where m is the mass of the system

and r' is the initial radius of rotation

Note that the professor does not change his position about the axis of rotation, so we are working relative to the dumbbells.

we can see that with the mass of the dumbbells remaining constant, if we reduce the radius of rotation of the dumbbells to r, the rotational inertia will reduce to [tex]I[/tex].

From

[tex]I[/tex]'ω' = [tex]I[/tex]ω

since [tex]I[/tex] is now reduced, ω will be greater than ω'

therefore, the angular velocity increases.

Answer:

Apparent depth (Da) = 60.15 cm (Approx)

Explanation:

Given:

Distance from fish (D) = 80 cm

Find:

Apparent depth (Da)

Computation:

We know that,

Refractive index of water (n2) = 1.33

So,

Apparent depth (Da) = D(n1/n2)

Apparent depth (Da) = 80 (1/1.33)

Apparent depth (Da) = 60.15 cm (Approx)

The apparent depth of the fish is 60 cm.

To calculate the apparent depth of the fish, we use the formula below.

Formula:

Where:

Make D' The subject of the equation.

From the question,

Given:

Substitute these values into equation 2

Hence, the apparent depth of the fish is 60 cm

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Answer:

Explanation:

If a sample emits 2000 counts per second when the detector is 1 meter from the sample, then;

2000 counts per second = 1 meter ... 1

In order to know the number of counts per second that would be observed when the detector is 3 meters from the sample, we will have;

x count per second = 3 meter ... 2

Solving the two expressions simultaneously for x we will have;

2000 counts per second = 1 meter

x counts per second = 3 meter

Cross multiply to get x

2000 * 3 = 1* x

6000 = x

This shows that 6000 counts per second would be observed when the detector is 3 meters from the sample

Answer:

11.2 Ω

Explanation:

The impedance of a circuit is given by;

Z= √R^2 +(XL-XC)^2

Since

Resistance R= 10 Ω

Inductive reactance XL= 12 Ω

Capacitive reactance XC= 7 Ω

Z= √10^2 + (12-7)^2

Z= √100 + 25

Z= √125

Z= 11.2 Ω

Answer:

1.6×10²⁰

Explanation:

An ampere is a Coulomb per second.

1 A = 1 C / s

The amount of charge after 5 seconds is:

5.0 A × 5 s = 25 C

The number of electrons is:

25 C × (1 electron / 1.6×10⁻¹⁹ C) = 1.6×10²⁰ electrons

Answer:

The  radius of the earth is [tex]r = 6365.4 \ km[/tex]

Explanation:

From the question we are told that

     The distance at  Alexandria is  [tex]d_a = 800 \ km = 800 *10^{3} \ m[/tex]

      The angle of the sun is  [tex]\theta = 7.2 ^o[/tex]

So we want to first obtain the circumference of the earth

   So let assume that the earth is  circular ([tex]360 ^o[/tex])

  Now from question we know that the sun made an angle of [tex]7.2 ^o[/tex] so with this we will obtain how many  [tex](7.2 ^o)[/tex]  are in [tex]360^o[/tex]

 i.e    [tex]N = \frac{360}{7.2}[/tex]

=>      [tex]N = 50[/tex]

     With this  value we can evaluate the circumference as

             [tex]c = 50 * 800[/tex]

              [tex]c = 40000 \ km[/tex]

Generally circumference is mathematically represented as

        [tex]c = 2\pi r[/tex]

         [tex]40000 = 2 * 3.142 * r[/tex]

=>        [tex]r = 6365.4 \ km[/tex]

Answer:

386.13 N

Explanation:

The kinetic energy of the baseball is converted into workdone in moving the glove backward( work energy theorem).

Therefore, KE of the ball

[tex]\frac{1}{2} mv^2 =\frac{1}{2}(0.145)35^2\\ = 88.81 \text{J}[/tex]

Now, workdone in moving the glove

W= Fd

where F = Force applied, d = displacement of the glove= 0.23 cm.

88.81 = F×0.23

F= 88.81/0.23 = 386.13 N

Answer:

The frictional force the road applies to the rear tire is static friction and it acts opposite to the direction in which the car is traveling.

Explanation:

This question suggests that the car is accelerating forward. Thus, the easiest way for us to know what friction is doing is for us to know what happens when we turn friction off.

Now, if there is no friction and the car is stopped, if we push down on the accelerator, it will make the front wheels to spin in a clockwise manner. This spin occurs on the frictionless surface with the rear wheels doing nothing while the car doesn't move.

Now, if we apply friction to just the front wheels, the car will accelerate forward while the back wheels would be dragging along the road and not be spinning. Thus, friction opposes the motion and as such, it must act im a direction opposite to where the car is going. This must be static friction.

The frictional force the road applies to the rear tire is static friction and it acts opposite to the direction in which the car is traveling.