In an adiabatic process:
a. the energy absorbed as heat equals the work done by the systemon its environment
b. the energy absorbed as heat equals the work done by theenvironment on the system
c. the work done by the environment on the system equals the changein internal energy

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:

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:

Net force = 2.3686 × 10^(20) N

Explanation:

To solve this, we have to find the force of the earth acting on the moon and the force of the sun acting on the moon and find the difference.

Now, from standards;

Mass of earth;M_e = 5.98 × 10^(24) kg

Mass of moon;M_m = 7.36 × 10^(22) kg

Mass of sun;M_s = 1.99 × 10^(30) kg

Distance between the sun and earth;d_se = 1.5 × 10^(11) m

Distance between moon and earth;d_em = 3.84 × 10^(8) m

Distance between sun and moon;d_sm = (1.5 × 10^(11)) - (3.84 × 10^(8)) = 1496.96 × 10^(8) m

Gravitational constant;G = 6.67 × 10^(-11) Nm²/kg²

Now formula for gravitational force between the earth and the moon is;

F_em = (G × M_e × M_m)/(d_em)²

Plugging in relevant values, we have;

F_em = (6.67 × 10^(-11) × 5.98 × 10^(24) × 7.36 × 10^(22))/(3.84 × 10^(8))²

F_em = 1.9909 × 10^(20) N

Similarly, formula for gravitational force between the sun and moon is;

F_sm = (G × M_s × M_m)/(d_sm)²

Plugging in relevant values, we have;

F_se = (6.67 × 10^(-11) × 1.99 × 10^(30) ×

7.36 × 10^(22))/(1496.96 × 10^(8))²

F_se = 4.3595 × 10^(20) N

Thus, net force = F_se - F_em

Net force = (4.3595 × 10^(20) N) - (1.9909 × 10^(20) N) = 2.3686 × 10^(20) N

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:

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:

B. 1.6 x 10⁻³ m

Explanation:

The magnetic field at the center of the loop is given by;

[tex]B = \frac{\mu_o I }{2R}[/tex]

Where;

μ₀ is the permeability of free space

I is the current in the loop

R is the radius of the circular loop

B is the magnetic field

Given;

diameter of the loop = 1cm

radius of the loop, r = 0.5 cm = 0.005 m

magnetic field, B = 2.5mT = 2.5 x 10⁻³ T

The current in the loop is calculated as;

[tex]I = \frac{2BR}{\mu_o} \\\\I = \frac{2*2.5*10^{-3}*0.005}{4\pi*10^{-7}} \\\\I = 19.89 \ A[/tex]

The magnetic at a distance from the long straight wire is calculated as;

[tex]B = \frac{\mu_o I}{2\pi d}[/tex]

where;

d is the distance from the wire;

[tex]d = \frac{\mu_o I}{2\pi B} \\\\d = \frac{4\pi *10^{-7} * 19.89}{2\pi *2.5*10^{-3}} \\\\d = 1.6 *10^{-3} \ m[/tex]

Therefore, the distance from the wire where the magnetic field is 2.5 mT is 1.6 x 10⁻³ m.

B. 1.6 x 10⁻³ m

This question involves the concepts of the magnetic field due to a loop and a  current-carrying wire and current.

A long straight wire carrying the same current as the loop of wire has a magnetic field of 2.5 mT at a distance of b "B. 1.5 x 10⁻³ m".

The magnetic field at the center of a loop of wire is given by the following formula:

[tex]B=\frac{\mu_o I}{2r}[/tex]

where,

B = Magnetic Field = 2.5 mT = 2.5 x 10⁻³ T

μ₀ = permeability of free space = 4π x 10⁻⁷ N/A²

I = current = ?

r = radius = diameter/2 = 1 cm/2 = 0.5 cm = 0.005 m

Therefore,

[tex]I = \frac{(2.5\ x\ 10^{-3}\ T)(2)(0.005\ m)}{4\pi\ x\ 10^{-7}\ N/A^2}[/tex]

I = 19.9 A

Now, the magnetic field at a distance from the straight wire is given by the following formula:

[tex]B=\frac{\mu_o I}{2\pi R}[/tex]

where,

R = distance from wire = ?

Therefore,

[tex]R = \frac{(4\pi \ x \ 10^{-7}\ N/A^2)(19.9\ A)}{2\pi(2.5\ x\ 10^{-3}\ T)}[/tex]

R = 1.6 x 10⁻³ m

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Answer:

C) radio waves, microwaves, infrared, visible, ultraviolet, x-rays, gamma rays

Explanation:

radio waves have lowest  energy , lowest  frequency and highest  wavelength

gamma rays  have highest  energy , highest  frequency and least  wavelength

Answer: C

Explanation:

Answer

123.5 x 10 ^-3 radian

Explanation:

Given the Width of slit a = 1 x 10⁻³ / 200

a = 5x 10⁻⁶ m .

angle at which first order peak is formed

= λ / a (where λ is wavelength and a is width of slit)

given λ = 617.3 x 10⁻⁹ m

a = 5 x 10⁻⁶

θ = 617.3 x 10⁻⁹ / 5 x 10⁻⁶

= 123.5x 10⁻³ radian .

first order peak is formed at an angle of 123.5 x 10⁻³ radian .

Explanation:

Answer:

Explanation:

The general decay equation is given as [tex]N = N_0e^{-\lambda t} \\\\[/tex], then;

[tex]\dfrac{N}{N_0} = e^{-\lambda t} \\[/tex] where;

[tex]N/N_0[/tex] is the fraction of the radioactive substance present = 1/16

[tex]\lambda[/tex] is the decay constant

t is the time taken for decay to occur = 8,450s

Before we can find the half life of the material, we need to get the decay constant first.

Substituting the given values into the formula above, we will have;

[tex]\frac{1}{16} = e^{-\lambda(8450)} \\\\Taking\ ln\ of \both \ sides\\\\ln(\frac{1}{16} ) = ln(e^{-\lambda(8450)}) \\\\\\ln (\frac{1}{16} ) = -8450 \lambda\\\\\lambda = \frac{-2.7726}{-8450}\\ \\\lambda = 0.000328[/tex]

Half life f the material is expressed as [tex]t_{1/2} = \frac{0.693}{\lambda}[/tex]

[tex]t_{1/2} = \frac{0.693}{0.000328}[/tex]

[tex]t_{1/2} = 2,112.8 secs[/tex]

Hence, the half life of the material is approximately 2113 seconds

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:

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

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:

1.8/0.61 =2.95 ft

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Answer:

Knowledge of vectors is important because many quantities used in physics are vectors. If you try to add together vector quantities without taking into account their direction you'll get results that are incorrect.

Explanation:

An example of the importance of vector addition could be the following:

Two cars are involved in a collision. At the time of the collision car A was travelling at 40 mph, car B was travelling at 60 mph. Until I tell you in which directions the cars were travelling you don't know how serious the collision was.

The cars could have been travelling in the same direction, in which case car B crashed into the back of car A, and the relative velocity between them was 20 mph. Or the cars could have been travelling in opposite directions, in which case it was a head on collision with a relative velocity between the cars of 100 mph!

Answer:

Explanation:

B. There is a counterclockwise induced current in the loop

Explanation:

This in line with the right hand grip rule,

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

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

1. Which example best describes a restoring force?

B) the force applied to restore a spring to its original length

2. A spring is compressed, resulting in its displacement to the right. What happens to the spring when it is released?

C) The spring exerts a restoring force to the left and returns to its equilibrium position.

3. A 2-N force is applied to a spring, and there is displacement of 0.4 m. How much would the spring be displaced if a 5-N force was applied?

D) 1 m

4. Hooke’s law is described mathematically using the formula Fsp=−kx. Which statement is correct about the spring force, Fsp?

D)It is a vector quantity.

5. What happens to the displacement vector when the spring constant has a higher value and the applied force remains constant?

A) It decreases in magnatude.

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:

it is going to D. all of these are resistors

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:

1.34 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: 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:

Answer A : Fusion followed by beta decay (electron emission)

Explanation:

Notice that you want the Atomic number to increase, that is the number of protons in a nucleus. So if all four cases given experience the same fusion of nuclei, the only one that net increases the number of protons in the last stage, is the reaction that undergoes a beta decay (with emission of an electron) thus leaving a positive imbalance of positive charge (proton generated in the beta decay of a neutron).

Therefore, answer A is the correct one.

Answer:

A : Fusion followed by beta decay (electron emission)

Explanation:

Ap3x

Answer:

The  value [tex]y = 0.0349 \ m[/tex]

Explanation:

From the question we are told that

   The  distance of the screen is  [tex]D = 2.30 \ m[/tex]

   The  width of the slit is  [tex]d = 0.0368 \ nm = 0.0368 *10^{-3} \ m[/tex]

   The  wavelength is  [tex]\lambda = 558 \ nm = 558 *10^{-9} \ m[/tex]

The  width of the central diffraction peak is  mathematically represented as

        [tex]k = 2 * y[/tex]

Where  y is the distance from the center to the high peak which  is mathematically represented as

       [tex]y = \frac{\lambda * D }{d }[/tex]

substituting values

      [tex]y = \frac{ 558 *10^{-8} * 2.30 }{0.0368 *10^{-3} }[/tex]

      [tex]y = 0.0349 \ m[/tex]

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:

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.

Answer:

0.09N, attractive

Explanation:

It can be deducted from the question that the currents are arranged in parallel settings, then it is obvious that the force on each of the wire will be attractive toward the other wire.

the magnitude of force can be determined by using below formula;

F2 = (μ₀/2π)(I₁I₂/d)I₂

μ₀ = constant = 4π × 10^-7 H/m,

I₁, I₂ = currents= 30A

L = the length o the wire=30m

d = distance between these two wires= 0.06m

Since the current are arranged in the same direction, they exhibit attractive force on each other.

Then plugging the values Into the formula above we have

F₂ = (4π × 10^-7 T.m/A)/2π) × ((30A)²/ 0.06m)× 30 m

= 0.09 N, attractive

Therefore, the magnitude and direction of the force is 0.09 N, attractive

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 direction of the force will be towards the east

Explanation:

From the question we are told that  

    The direction of the  downward

Generally according to Fleming's right-hand rule(

          Thumb -  direction of force

           Middle finger -  direction of current

           Index finger -  direction of the magnetic field

) and the fact that the earth magnetic field acts  from south to north with respect to the four cardinal points then the direction of the  force will be toward the east with respect to the four cardinal point on the earth