How do you measure potential and kinetic energy?

Answer: the furthest from the end prying out the nail

Answer:

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

the furthest from the end prying out the nail

Explanation:

Answer:

a)1984.5nm

b)523mm

Explanation:

A)A destructive interference can be explained as when the phase shifting between the waves is analysed by the path lenght difference

θ=(m+0.5)λ where m= 1,2.3....

Where given from the question the 4th dark Fringe which will take place at m= 3

θ=7/2y

Where y= 567nm

= 7/2(567)=1984.5nm

But

B)tan θ ≈ y/d

And sinθ = mλ/d

y=mλd when m= 1 which is the first bright we have

Then y=(1× 567.D)/d

But the distance from Central to the 4th dark Fringe is 1.83cm then

y= 7λD/2d= 1.83cm

D/d=(2)×(1.83×10^-2)/(7×567×10^-9)

=92221.5

y= (567×10^-9)× (92221.5)

=0.00523m

Therefore, the distance between the first and center is y1-y0= 523mm

Answer:gamma ray

Explanation:

It depends on what stage of the mission you're talking about.

==>  While it's sitting on the pad before launch, the forces on the rocket are balanced, so there's no net force on it.

==>  When the engines ignite, their thrust (d) is greater than the force of gravity.  So the net force on the rocket is upward, and the spacecraft accelerates upward.

==>  After the engines shut down, the net force acting on the rocket is due to Gravity (c).

. . . If the rocket has enough vertical speed, it escapes the Earth completely, and just keeps going.  

. . . If it has enough horizontal speed, it enters Earth orbit.  

. . . If it doesn't have enough vertical or horizontal speed, it falls back to Earth.    

A rocket will preserve to speed up so long as there's a resultant pressure upwards resulting from the thrust of the rocket engine.

A rocket launches whilst the pressure of thrust pushing it upwards is greater than the burden force because of gravity downwards. This unbalanced pressure reasons a rocket to accelerate upwards. A rocket will maintain to hurry up so long as there's a resultant force upwards resulting from the thrust of the rocket engine.

What's the net pressure of unbalanced?

If the forces on an item are balanced, the net pressure is zero. If the forces are unbalanced forces, the results do not cancel each difference. Any time the forces acting on an object are unbalanced, the net pressure is not 0, and the movement of the item modifications.

Learn more about the thrust of the rocket engine. here:  https://brainly.com/question/10716695

#SPJ2

The ball's horizontal position in the air is

[tex]x=\left(27.5\dfrac{\rm m}{\rm s}\right)\cos38.5^\circ t[/tex]

It hits the wall when [tex]x=17.5\,\mathrm m[/tex], which happens at

[tex]17.5\,\mathrm m=\left(27.5\dfrac{\rm m}{\rm s}\right)\cos38.5^\circ t\implies t\approx0.813\,\mathrm s[/tex]

Meanwhile, the ball's vertical position is

[tex]y=\left(27.5\dfrac{\rm m}{\rm s}\right)\sin38.5^\circ t-\dfrac g2t^2[/tex]

where [tex]g[/tex] is the acceleration due to gravity, 9.80 m/s^2.

At the time the ball hits the wall, its vertical position (relative to its initial position) is

[tex]y=\left(27.5\dfrac{\rm m}{\rm s}\right)\sin38.5^\circ(0.813\,\mathrm s)-\dfrac g2(0.813\,\mathrm s)^2\approx\boxed{10.7\,\mathrm m}[/tex]

Answer:

this raise the temperature is x = 3

Explanation:

Heat capacity is the relationship between heat and temperature change

          C = Q / ΔT

if the heat in the system is given by the change in energy and we carry this differential formulas

          [tex]c_{v}[/tex] = dE / dT

In this problem we are told that the energy of thermal radiation is

        E = A V T⁴

Let's look for the specific heat

        c_{v} = AV 4 T³

the power to which this raise the temperature is x = 3

Answer:

We can think the water stream as a solid object that is fired.

The distance between the fireperson and the building is 50m. (i consider that the position of the fireperson is our position = 0)

The angle is 30 above the horizontal. (yo wrote 300, but this has no sense because 300° implies that he is pointing to the ground).

The initial speed of the stream is 40m/s.

First, using the fact that:

x = R*cos(θ)

y = R*sin(θ)

in this case R = 40m/s and θ = 30°

We can use the above relation to find the components of the velocity:

Vx = 40m/s*cos(30°) = 34.64m/s

Vy = 20m/s.

First step:

We want to find the time needed to the stream to hit the buildin.

The horizontal speed is 34.64m/s and the distance to the wall is 50m

So we want that:

34.64m/s*t = 50m

t = 50m/(34.64m/s) = 1.44 seconds.

Now we need to calculate the height of the stream at t = 1.44s

Second step:

The only force acting on the water is the gravitational one, so the acceleration of the stream is:

a(t) = -g.

g = -9.8m/s^2

For the speed, we integrate over time and we get:

v(t) = -g*t + v0

where v0 is the initial speed: v0 = 20m/s.

The velocity equation is:

v(t) = -g*t + 20m/s.

For the position, we integrate again over time:

p(t) = -(1/2)*g*t^2 + 20m/s*t + p0

p0 is the initial height of the stream, this data is not known.

Now, the height at the time t = 1.44s is

p(1.44s) = -5.9m/s^2*(1.44s)^2 + 20m/s*1.44s + po

             = 16.57m + p0

So the height at wich the stream hits the building is 16.57 meters above the initial height of the fire hose.

Answer:

Explanation:

A) 131 dB = 10*log(I / 1e-12W/m²)

where I is the intensity at 2.6 m away.

13.1 = log(I / 1e-12W/m²

1.25e13= I / 1e-12W/m²

I = 1.25 x10^1W/m²

power = intensity * area

P = I * A = 12.5W/m² * 4π(2.6m)² =1061 W ◄

B) 86 dB = 10*log(I / 1e-12W/m²)

8.6 = log(I / 1e-12W/m²)

3.98e8 = I / 1e-12W/m²

I = 3.98e-4 W/m²

area A = P / I = 1061W / 3.98e-4W/m² = 2.66e6 m²

A = 4πr²

2.66e6 m² = 4πr²

r = 14.5m ◄

Answer:

The room with the lower temperature

Explanation:

Using

PV=nRT

Since both the rooms same volume and are connected, so they will have same pressure

PV=nRT=constant

nT=Constant/R=constant

If T is more n has to be less

Thus, lower the temperature, more the number molecules.

Answer:

d' = 75.1 cm

Explanation:

It is given that,

The actual depth of a shallow pool is, d = 1 m

We need to find the apparent depth of the water in the pool. Let it is equal to d'.

We know that the refractive index is also defined as the ratio of real depth to the apparent depth. Let the refractive index of water is 1.33. So,

[tex]n=\dfrac{d}{d'}\\\\d'=\dfrac{d}{n}\\\\d'=\dfrac{1\ m}{1.33}\\\\d'=0.751\ m[/tex]

or

d' = 75.1 cm

So, the apparent depth is 75.1 cm.

Answer:

4 / 7

Explanation:

1/total resistance = 1/1 + 1/2 + 1/4

= 1¾

total resistance = 1 ÷ 1¾

= 4/7

Answer:

a,)3.042s

b)4.173s

c)3.281s

Explanation:

For a some pendulum the period in seconds T can be calculated using below formula

T=2π√(L/G)

Where L = length of pendulum in meters

G = gravitational acceleration = 9.8 m/s²

Then we are told to calculate

(a) What is the period of small oscillations for this pendulum if it is located in an elevator accelerating upward at 3.00 m/s2?

Since oscillations for this pendulum is located in the elevator that is accelerating upward at 3.00 then

use G = 9.8 + 3.0 = 12.8 m/s²

Period T=2π√(L/G)

T= 2π√(3/12.8)

T=3.042s

b) (b) What is the period of small oscillations for this pendulum if it is located in an elevator accelerating downward at 3.00 m/s2?

G = 9.8 – 3.0 = 6.8 m/s²

T= 2π√(3/6.8)

T=4.173s

C)(c) What is the period of this pendulum if it is placed in a truck that is accelerating horizontally at 3.00 m/s2?

Net acceleration is

g'= √(g² + a²)

=√(9² + 3²)

Then period is

T=2π√(3/11)

T=3.281s

Answer:

The magnetic field at the center of the solenoid is 2.1  × 10⁻³ T

Explanation:

The magnetic field B at the center of the solenoid is given by

B = μ₀ni where μ₀ = permeability of free space = 4π × 10⁻⁷H/m, n = number of turns per unit length of the solenoid = 1100 turns per meter and i = current in the solenoid = 1.5 A.

So B = μ₀ni

= 4π × 10⁻⁷H/m × 1100 × 1.5 A

= 4π × 10⁻⁷H/m × 1650 A-turns/m

= 20734.5 × 10⁻⁷T  

= 2.07345 × 10⁻³ T

≅ 2.1  × 10⁻³ T

So the magnetic field at the center of the solenoid is 2.1  × 10⁻³ T

Answer:

A) y = 3.56 mm

B) y = 3.56 mm

Explanation:

A) The distance from the central maximum to the first bright region can be found using Young's double-slit equation:

[tex] y = \frac{m\lambda L}{d} [/tex]

Where:

λ: is the wavelength = 546.1 nm

m: is first bright region = 1

L: is the distance between the screen and the plane of the parallel slits = 1.50 m

d: is the separation between the slits = 0.230 mm

[tex] y = \frac{m\lambda L}{d} = \frac{1*546.1 \cdot 10^{-9} m*1.50 m}{0.230 \cdot 10^{-3} m} = 3.56 \cdot 10^{-3} m [/tex]  

B) The distance between the first and second dark bands is:

[tex] \Delta y = \frac{\Delta m*\lambda L}{d} [/tex]

Where:

[tex] \Delta m = m_{2} - m_{1} = 2 - 1 = 1 [/tex]

[tex] \Delta y = \frac{1*546.1 \cdot 10^{-9} m*1.50 m}{0.230 \cdot 10^{-3} m} = 3.56 \cdot 10^{-3} m [/tex]      

I hope it helps you!

Answer:

(a) 1.47 x 10⁴ V/m

(b) 1.28 x 10⁻⁷C/m²

(c) 3.9 x 10⁻¹²F

(d) 9.75 x 10⁻¹¹C

Explanation:

(a) For a parallel plate capacitor, the electric field E between the plates is given by;

E = V / d               -----------(i)

Where;

V = potential difference applied to the plates

d = distance between these plates

From the question;

V = 25.0V

d = 1.70mm = 0.0017m

Substitute these values into equation (i) as follows;

E = 25.0 / 0.0017

E = 1.47 x 10⁴ V/m

(c) The capacitance of the capacitor is given by

C = Aε₀ / d

Where

C = capacitance

A = Area of the plates = 7.60cm² = 0.00076m²

ε₀ = permittivity of free space =  8.85 x 10⁻¹²F/m

d = 1.70mm = 0.0017m

C = 0.00076 x  8.85 x 10⁻¹² / 0.0017

C = 3.9 x 10⁻¹²F

(d) The charge, Q, on each plate can be found as follows;

Q = C V

Q =  3.9 x 10⁻¹² x 25.0

Q = 9.75 x 10⁻¹¹C

Now since we have found other quantities, it is way easier to find the surface charge density.

(b) The surface charge density, σ, is the ratio of the charge Q on each plate to the area A of the plates. i.e

σ = Q / A

σ = 9.75 x 10⁻¹¹ /  0.00076

σ = 1.28 x 10⁻⁷C/m²

Answer:

South and West

Explanation:

Those people are pushing the hardest. It will move south faster than it moves west.

Answer:

0.002699 m or 2.699 mm

Explanation:

y = Fringe distance

d= Distance between slits = 0.310mm

L = Screen distance = 4.40m

λ= Wavelength

Given from question

λ₁= 660 nm = 6.6 x 10^-9 m

λ₂= 470 nm = 4.7 x 10^-9 m

d = 0.340 mm = 3.4 x 10^-3 m

L = 4.40 m

In the case of constructive interference, we use below formula

y/L = mλ/d

For first order wavelength

(y₁/4.40) =(1×660x10⁻⁹)/(0.310*10⁻³)

y₁= (0.310*10⁻³)×(4.40)/(0.310*10⁻³)

y₁=0.00937m

(y2/4.40) =(1×470x10⁻⁹)/(0.310*10⁻³)

y2= =(1×470x10⁻⁹)×(4.40)/(0.310*10⁻³)

y2=0.00667m

distance between the fringes is given by (y₁ -y2)

=0.00937-0.00667=0.002699m

Therefore, distance on the screen between the first-order bright fringes for the two wavelengths is 0.002699 m or 2.699 mm

i

CHECK COMPLETE QUESTION BELOW

inductor is connected to the terminals of a battery that has an emf of 12.0 VV and negligible internal resistance. The current is 4.96 mAmA at 0.800 msms after the connection is completed. After a long time the current is 6.60 mAmA.

Part A)What is the resistance RR of the inductor

PART B) what is inductance L of the conductor

Answer:

A)R=1818.18 ohms

B)L=1.0446H

Explanation:

We were given inductor L with resistance R , there is a connection between the battery and the inductor with Emf of 12V, we can see that the circuit is equivalent to a simple RL circuit.

There is current of 4.96mA at 0.8ms, at the end of the connection the current increase to 6.60mA,

.

a)A)What is the resistance RR of the inductor?

The current flowing into RL circuit can be calculated using below expresion

i=ε/R[1-e⁻(R/L)t]

at t=∞ there is maximum current

i(max)= ε/R

Where ε emf of the battery

R is the resistance

R=ε/i(max)

= 12V/(6.60*10⁻³A)

R=1818.18 ohms

Therefore, the resistance R=1818.18 ohms

b)what is inductance L of the conductor?

i(t=0.80ms and 4.96mA

RT/L = ⁻ln[1- 1/t(max)]

Making L subject of formula we have

L=-RT/ln[1-i/i(max)]

If we substitute the values into the above expresion we have

L= -(1818.18 )*(8.0*10⁻⁴)/ln[1-4.96/6.60)]

L=1.0446H

Therefore, the inductor L=1.0446H

Answer:

The bumper will be able to move by 0.01155m.

Explanation:

The magnitude of deceleration of the car in the front end collision.

[tex]a = \frac{F_m}{m} \\[/tex]

[tex]a = \frac{80000}{1500} \\[/tex]

[tex]a = 53.33[/tex]

This is the deceleration of the car that is generated to stop due to a front end collision.

4 km/h = 1.11 m/s

Now, the initial speed of the bumper in the relation of car, Vi = 0

Now, the initial speed of the bumper in the relation of car, Vf = 1.11 m/s

Use the below equation:

[tex]s = \frac{(Intitial \ speed)^2 – (Final \ speed)^2}{2a} \\[/tex]

[tex]s = \frac{(1.11)^2 – (0)}{2 \times 53.33} \\[/tex]

[tex]s = 0.01155 \\[/tex]

Thus, the bumper can move relative to the car is 0.01155 m .

Answer:

Explanation:

According to Flemming's left hand rule, it sates that if the first three fingers of the left hand are held mutually at right angles to one another, the fore finger will point in the direction of magnetic field, the middle finger will point in direction of current, while the thumb will point to the direction of force.

Mathematically the law is stated as

F= BIL

given data

Magnetic field B=  0.43T

Current I= 4.9 A

length of conductor L= 3.3cm to meter , 3.3/100=  0.033 m

Applying the formula the force is calculated as

F= 0.43*4.9* 0.033= 0.069 N

According to Flemming's rule the direction of all parameters are mutually perpendicular to one another, then the Force is in the X direction

Answer:

a) 0.56°C

b) 273.56 K

Explanation:

If we want to convert from Fahrenheit scale to Celcius scale we use the formula;

T(°C) = (T(°F) - 32) × 5/9

Where T(°F) = 33°F

Hence;

T(°C) = (33°F - 32) × 5/9

T°C = 0.56°C

b)

T(K) = T°C + 273

T(K) = 0.56 + 273

T(K) = 273.56 K

Answer:

The wavelength of the emitted photons 532 nm, corresponds to a visible light having GREEN color.

Explanation:

Given;

energy of the emitted photons, E = 3.74 x 10⁻¹⁹ J

speed of light, c = 3 x 10⁸ m/s

Planck's constant, h = 6.63 x 10⁻³⁴ J.s

The wavelength of the emitted light will be calculated by applying energy of photons;

[tex]E = hf[/tex]

where;

E is the energy emitted light

h is Planck's constant

f is frequency of the emitted photon

But f = c / λ

where;

λ is the wavelength of the emitted photons

[tex]E = \frac{hc}{\lambda} \\\\\lambda = \frac{hc}{E} \\\\\lambda = \frac{6.63*10^{-34} *3*10^{8}}{3.74*10^{-19}} \\\\\lambda = 5.318 *10^{-7} \ m\\\\\lambda = 531.8 *10^{-9} \ m\\\\\lambda = 531.8 \ nm[/tex]

λ ≅ 532 nm

the wavelength of the emitted photons is 532 nm.

Therefore, the wavelength of the emitted photons 532 nm, corresponds to a visible light having GREEN color.

Answer:

The induced emf is [tex]\epsilon = 0.0280 \ V[/tex]

Explanation:

From the question we are told

    The diameter of the loop is  [tex]d = 6.7 cm = 0.067 \ m[/tex]

    The magnetic field is  [tex]B = 1.27 \ T[/tex]

    The time taken is  [tex]dt = 0.16 \ s[/tex]

Generally the induced emf is mathematically represented as

          [tex]\epsilon = - N * \frac{\Delta \phi}{dt}[/tex]

Where  N =  1 given that it is only a circular loop

            [tex]\Delta \phi = \Delta B * A[/tex]

Where  [tex]\Delta B = B_f - B_i[/tex]

   where [tex]B_i[/tex] is  1.27 T  given that the loop of wire was initially in the magnetic field

    and  [tex]B_f[/tex] is  0 T given that the loop was removed from the magnetic field

Now the area of the of the loop is evaluated as

          [tex]A = \pi r^2[/tex]

Where r is the radius which is mathematically represented as

       [tex]r = \frac{d}{2}[/tex]

substituting values

       [tex]r = \frac{0.067}{2}[/tex]

        [tex]r = 0.0335 \ m[/tex]

So

         [tex]A = 3.142 * (0.0335)^2[/tex]

          [tex]A = 0.00353 \ m^2[/tex]

So

      [tex]\Delta \phi = (0- 127)* (0.00353)[/tex]

      [tex]\Delta \phi = -0.00448 Weber[/tex]

=>    [tex]\epsilon = - 1 * \frac{-0.00448}{0.16}[/tex]

=>   [tex]\epsilon = 0.0280 \ V[/tex]

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:

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:

65.3

Explanation:

1 foot = 12 inches

Sammy is 5 feet tall.

5 feet = ? inches

Multiply the feet value by 12 to find in inches.

5 × 12

= 60

Add 5.3 inches to 60 inches.

60 + 5.3

= 65.3

Answer:

It will be 》》》》1.664716m

Answer:

2,66

Explanation:

The refractive index= real depth/ apparent depth

real depth = refractive index * apparent depth

Let's assume index for water is 1.33

real depth = 2*1,33 = 2,66

Answer:

 λ = 605.80 nm

Explanation:

These double-slit experiments the equation for constructive interference is

          d sin θ = m λ

where d is the distance between the slits, λ the wavelength of light and m an integer that determines the order of interference.

In this case, the distance between the slits is d = 1.11 mm = 1.11 10⁻³ m, the distance to the screen is L = 8.63 m, the range number m = 10 and ay = 4.71 cm

Let's use trigonometry to find the angle

         tan θ = y / L

as the angles are very small

          tan θ = sin θ / cos θ = sin θ

we substitute

         sin θ = y / L

we substitute in the first equation

         d y / L = m λ          

          λ = d y / m L

let's calculate

           λ = 1.11 10⁻³ 4.71 10⁻²/ (10 8.63)

           λ = 6.05805 10⁻⁷ m

let's reduce to nm

          λ = 6.05805 10⁻⁷ m (10⁹ nm / 1m)

          λ = 605.80 nm

Answer:

4 x 10¹⁵

Explanation: