.You have always been impressed by the speed of the elevators in your apartment building. You wonder about the maximum acceleration for these elevators during normal operation, so you decide to measure it by using your bathroom scale. While the elevator is at rest on the ground floor,you get in, put down your scale, and stand on it. The scale reads 50 kg. You continue standing on the scale when the elevator goes up, carefully watching the reading. During the trip to the 10th floor, the greatest scale reading was

Answer:

See explanation

Explanation:

From the formula;

0.693/t1/2 = 2.303/t log (Ao/A)

t1/2 = half life of Sodium-24

Ao = initial activity of Sodium-24

A= activity of Sodium-24 at time = t

So,

0.693/15 = 2.303/15 log (800/A)

0.0462 = 0.1535  log (800/A)

0.0462/0.1535 =  log (800/A)

0.3 = log (800/A)

Antilog(0.3) =  (800/A)

1.995 =  (800/A)

A = 800/1.995

A = 401 Bq

ii) 0.693/15 = 2.303/30 log (800/A)

0.0462 = 0.0768 log (800/A)

0.0462/0.0768 =  log (800/A)

0.6 =  log (800/A)

Antilog (0.6) =  (800/A)

3.98 =  (800/A)

A = 800/3.98

A = 201 Bq

iii)

0.693/15 = 2.303/45 log (800/A)

0.0462 = 0.0512  log (800/A)

0.0462/0.0512  =  log (800/A)

0.9 = log (800/A)

Antilog (0.9) =  (800/A)

7.94 = (800/A)

A = 800/7.94

A= 100.8 Bq

iv)

0.693/15 = 2.303/60 log (800/A)

0.0462 = 0.038 log (800/A)

0.0462/0.038 = log (800/A)

1.216 = log (800/A)

Antilog(1.216) = (800/A)

16.44 = (800/A)

A = 800/16.44

A = 48.66 Bq

Answer:

69.7 cm

Explanation:

What is the distance (in cm) between the 2nd order bright fringe and the 2nd dark fringe?

For a diffraction grating, dsinθ = mλ where d = grating spacing = 1/(3000 lines per cm) = 1/3000 × 100 m per line = 1/300000 m = 1/3 × 10⁻⁵ m, m = order of fringe and λ = wavelength of light = 550 nm = 550 × 10⁻⁹ m.

Also, tanθ = x/D where x = distance of nth order fringe from central maximum and D = distance of screen from grating = 115 cm = 1.15 m

Now sinθ = d/mλ, Since θ is small, sinθ ≅ tanθ

So, d/mλ = x/D for a second order bright fringe, m = 2.

So, d/2λ = x/D

x = dD/2λ

So, x =

For a dark fringe, we have

d/(m + 1/2)λ = x'/D where x' is the distance of the fringe from the central maximum.

For a second-order dark fringe, m = 2. So,

d/(2 + 1/2)λ = x'/D

d/(5/2)λ = x'/D

2d/5λ = x'/D

x' = 2dD/5λ

So, the distance (in cm) between the 2nd order bright fringe and the 2nd dark fringe is x" = dD/2λ - 2dD/5λ

x" = dD/10λ

Substituting the values of the variables into the equation, we have

x"= 1/3 × 10⁻⁵ m × 1.15 m/(10 × 550 × 10⁻⁹ m)

x" = 1.15/165 × 10² m

x" = 0.00697 × 10² m

x" = 0.697 m

x" = 69.7 cm

The distance (in cm) between the 2nd order bright fringe and the 2nd dark fringe 69.7 cm

Diffraction of light occurs when a light wave passes by a corner or through an opening or slit that is physically the approximate size of, or even smaller than that light's wavelength

For a diffraction grating, dsinθ = mλ

where,

d = grating spacing = 1/(3000 lines per cm) = 1/3000 × 100 m per line = 1/300000 m = 1/3 × 10⁻⁵ m,

m = order of fringe  

λ = wavelength of light = 550 nm = 550 × 10⁻⁹ m.

Also,   [tex]tan\theta=\dfrac{x}{D}[/tex]t   where

x = distance of nth order fringe from central maximum  

D = distance of screen from grating = 115 cm = 1.15 m

Now   [tex]Sin\theta =\dfrac{d}{m\lambda}[/tex] ,  Since θ is small, sinθ ≅ tanθ

So, [tex]\dfrac{d}{m\lambda}=\dfrac{x}{D}[/tex]

for a second order bright fringe, m = 2.

So,   [tex]\dfrac{d}{2\lambda}=\dfrac{x}{D}[/tex]

[tex]x=\dfrac{dD}{2\lambda}[/tex]

For a dark fringe, we have

[tex]\dfrac{d}{(m+\dfrac{1}{2})\lambda}=\dfrac{X'}{D}[/tex]

where x' is the distance of the fringe from the central maximum.

For a second-order dark fringe, m = 2. So,

[tex]\dfrac{d}{(m+\dfrac{1}{2})\lambda}=\dfrac{X'}{D}[/tex]

[tex]\dfrac{d}{ \dfrac{5}{2}\lambda}=\dfrac{X'}{D}[/tex]

[tex]X'=\dfrac{2dD}{5\lambda}[/tex]

So, the distance (in cm) between the 2nd order bright fringe and the 2nd dark fringe is

[tex]X''=\dfrac{dD}{2\lambda}-\dfrac{2dD}{5\lambda}[/tex]

[tex]X''=\dfrac{dD}{10\lambda}[/tex]

Substituting the values of the variables into the equation, we have

[tex]x''=\dfrac{1}{3\times10^{-5}}\times \dfrac{1.15}{10\times 550\times 10^{-9}}[/tex]x

x" = 0.00697 × 10² m

x" = 0.697 m

x" = 69.7 cm

Hence the distance (in cm) between the 2nd order bright fringe and the 2nd dark fringe 69.7 cm

To know more about Diffraction follow

https://brainly.com/question/16749356

Answer:

Lol

Explanation:

Answer:

C

Explanation:

Albert Bandura emphasized the idea of Self efficacy which is the belief one has in one’s own ability to succeed.  

A person's self-efficacy relates to their confidence in their ability to carry out the behaviors required to achieve particular performance goals (Bandura, 1977, 1986, 1997).

The belief in one's capacity to exercise control over one's own motivation, behavior, and social environment is known as self-efficacy. The goals for which people strive, the amount of effort put out to obtain goals, and the possibility of achieving particular levels of behavioral performance are all influenced by these cognitive self-evaluations.

Self-efficacy beliefs, unlike conventional psychological notions, are anticipated to change according to the operating domain and the environment in which an action occurs.

Therefore, Albert Bandura emphasized the idea of Self efficacy which is the belief one has in one’s own ability to succeed.  

To learn more about Self efficacy, refer to the link:

https://brainly.com/question/28215515

#SPJ6

Is earth’s freshwater

Only about 3 percent of Earth’s water is fresh water.

Answer:

IMA = 8

Explanation:

Given the following data:

Length of plank = 160cm

Length of resistance = 20cm

To find the ideal mechanical advantage (IMA);

[tex] IMA= \frac {length \; of \; effort}{length \; of \; resistance} [/tex]

[tex] IMA= \frac {160}{20} [/tex]

IMA = 8

Answer:

    d₃ = 37,729 km,     θ=  5.1º North of West

Explanation:

This is a velocity addition problem, the easiest way to solve it is to decompose the velocities in a Cartesian system, the x-axis coincides with the West-East direction and the y-axis with the South-North direction

* first displacement is

           d₁ₓ = 11 km

* second offset is

          cos 6 = d₂ₓ / d₂

          sin 6 = d_{2y} / d₂

          d₂ₓ = d₂ cos 6

          d_{2y} = d₂ sin 6

          d₂ₓ = 6 cos 6 = 5.967 km

          d_{2y} = 6 sin 6 = 0.6272 km

* third displacement is unknown

* fourth and last displacement

          cos (-11) = d₄ₓ / d₄

          sin (-11) = d_{4y} / d₄

          d₄ₓ = d₄ cos (-11)

          d_{4y} = d₄ sin (-11)

          d₄ₓ = 21 cos (-11) = 20.61 km

          d_{4y} = 21 sin (-11) = -4.007 km

They tell us that at the end of the tour you are back on the island, so the displacement must be zero

X axis

           x = d₁ₓ + d₂ₓ + d₃ₓ + d₄ₓ

           0 = 11 +5.967 + d₃ₓ + 20.61

           d₃ₓ = -11 - 5.967 - 20.61

           d₃ₓ = -37.577 km

Y axis  

          y = d_{1y} + d_{2y} + d_{3y} + d_{4y}

          0 = 0 + 0.6272 + d_{3y} -4.007

          d_{3y} = 4.007 - 0.6272

          d_{3y} = 3.3798 km

This distance can be given in the form of module and angle

Let's use the Pythagorean theorem for the module

           d₃ = [tex]\sqrt{d_{3x}^2 + d_{3y}^2}[/tex]

           d₃ = [tex]\sqrt{37.577^2 + 3.3798^2}[/tex]

           d₃ = 37,729 km

Let's use trigonometry for the angle

            tan θ = d_{3y} / d₃ₓ

            θ = tan⁻¹ [tex]\frac{d_{3y}}{d_{3x}}[/tex]

            θ = tan-1 (-3.3798 / 37.577)

            θ = 5.1º

Since the y coordinate is positive and the x coordinate is negative, this angle is in the second quadrant, so the direction given in the form of cardinal coordinates is

            θ=  5.1º North of West

Answer:

the kinetic energy gained by the toy is 6J.

Explanation:

Given;

net applied to the toy by dog, F = 2 N

distance moved by the toy, d = 3 m

Apply the principle of work-energy theorem to determine the kinetic energy gained by the toy.

ΔK.E = W

         = F x d

         = 2 x 3

         = 6 J

Therefore, the kinetic energy gained by the toy is 6J.

Answer:

The force must be applied during 8 seconds to reach trhe same final speed.

Explanation:

By Impulse Theorem, a change in the magnitude of linear momentum of a system with constant mass can be done by applying a force during a given time. That is:

[tex]m\cdot (v_{f}-v_{o}) = F \cdot \Delta t[/tex]

Where:

[tex]m[/tex] - Mass, measured in kilograms.

[tex]v_{o}[/tex], [tex]v_{f}[/tex] - Initial and final speed, measured in meters per second.

[tex]F[/tex] - Net external foce, measured in newtons.

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

We can eliminate mass and speeds by constructing the following relationship:

[tex]F_{1}\cdot \Delta t_{1} = F_{2}\cdot \Delta t_{2}[/tex] (2)

If we know that [tex]F_{1} = F[/tex], [tex]\Delta t_{1} = 4\,s[/tex] and [tex]F_{2} = \frac{F}{2}[/tex], then the time is:

[tex]4\cdot F = \frac{F\cdot \Delta t_{2}}{2}[/tex]

[tex]\Delta t_{2} = 8\,s[/tex]

The force must be applied during 8 seconds to reach trhe same final speed.

Answer:

the altitude of these satellites above the surface of the earth;

35790 km ( SI units )

22243.63 miles ( U.S. customary unit )

Explanation:

Given the data in the question;

Time taken by the satellite to complete on revolution is 23.934 hours

= 23.934 × 60 × 60 = 86162.4 seconds

now, let h represent altitude, r represent Orbit radius, v represent Orbit speed.

we know that

v² = GM/r

= gR²/r

= (9.81m/s² × ( 6.37 × 10⁶ m)²) / r

v = 19.95 × 10⁶ / √r   ---------- let this be equation

also;

time t = 2πr/v

86162.4 s = 2πr/v -------- let this be equation 2

a) Determine the altitude of these satellites above the surface of the earth in both SI and U.S. customary units

we substitute v in equation into equation 2

so

86162.4 s = 2πr / (19.95 × 10⁶ / √r)

r = 42.16 × 10⁶ m

so, altitude h = r - R

h = 42.16 × 10⁶ m - 6.37 × 10⁶ m

h = 35790000 m

convert to kilometer

h = 35790000 / 1000

h = 35790 km

Convert to miles

h = 35790 / 1.609

h = 22243.63 miles

Therefore, the altitude of these satellites above the surface of the earth;

35790 km ( SI units )

22243.63 miles ( U.S. customary unit )

I'm assuming the question is supposed to be like this:

How  does the Orbital velocity of a satellite depends on the mass of the satellite?

Answer

It is independent of mass of satellite.

Answer:

Kinetic energy increases and potential energy decrease when velocity of an object increase.

Answer:

Are you sure it was soccer ball? Or meine hearts

Explanation:

Answer:

I think it’s c

Explanation:

Answer:

See explanation

Explanation:

Since the original count rate is 600 Bq,

i) after 1 half life, the count rate decreases to 1/2 of 600 Bq = 300 Bq

ii) after 2 half lives, the count rate decreases to 1/4 of 600 = 150 Bq

iii) after 3 half lives, the count rate decreases to 1/6 of 600 = 100 Bq

iv) after 4 half lives, the count rate decreases to 1/8 of 600 = 75 Bq

Answer:

T = 0.375 s, f = 2.66 Hz and ω = 16.71 rad/s

Explanation:

Given that,

The mass of a harmonic oscillator, m = 0.5 kg

The force constant of the spring, k = 140 N/m

The frequency of a harmonic oscillator is given by :

[tex]f=\dfrac{1}{2\pi }\sqrt{\dfrac{k}{m}}[/tex]

Substitute all the values,

[tex]f=\dfrac{1}{2\pi }\sqrt{\dfrac{140}{0.5}} \\\\f=2.66\ Hz[/tex]

Time period is given by :

[tex]T=\dfrac{1}{f}\\\\T=\dfrac{1}{2.66}\\\\T=0.375\ s[/tex]

The angular frequency is given by :

[tex]\omega=2\pi f\\\\\omega=2\pi \times 2.66\\\\\omega=16.71\ rad/s[/tex]

Hence, this is the required solution.

Answer:

The electric potential is the amount of work energy needed to move a unit of electric charge from a reference point to the specific point in an electric field with negligible acceleration of the test charge to avoid producing kinetic energy or radiation by test charge.

SI unit: volt

Other units: statvolt

In SI base units: V = kg⋅m^2⋅A^−1⋅s^−3

Dimension: M L^2 T^−3 I^−1

Explanation:

Hope it is helpful....

Explanation:

common symbol - V

SI unit - volt

other unit - statvolt

Answer:

Change in momentum is zero.

Explanation:

The following data were obtained from the question:

Mass (m) = 5000 kg

Time (t) = 1 h

Net force (F) = 0

Change in momentum =?

Force = Rate of change of momentum

0 = change in momentum

Change in momentum = 0

We can see from the above illustration that the net force is zero. Thus, the change in momentum is also zero.

Answer:

pretty sure A is correct, as they are pushing in opposite directions, hence canceling each other

Explanation:

Answer:

A and D are correct as they reduce GHG emissions while maintaining people's standard of living.

Answer:

Ig A and D

Explanation:

as :

a : Solar powered cars-  redices emissions, relies on renewable sources and maintains current living.

d: a : WIND TURBINES -  redices emissions, relies on renewable sources and is the closest to maintaining current living.

I hope im  right !!!

Answer:

Mass = 1133.33 kg (Approx.)

Explanation:

Given:

Momentum = 2.04 x 10⁴ kg[m/s]

Velocity = 18 m/s

Find:

Mass

Computation:

Mass = Momentum / Velocity

Mass = [2.04 x 10⁴] / 18

Mass = 1133.33 kg (Approx.)

Explanation:

Given that,

Two particles, an electron and a proton, are initially at rest in a uniform electric field of magnitude 570 N/C.

We need to find their speeds after 47.6 ns.

For electron,

The electric force is given by :

[tex]F=qE\\\\F=1.6\times 10^{-19}\times 570\\\\=9.12\times 10^{-17}\ N[/tex]

Let a be the acceleration of the electron. So,

F = ma

m is mass of electron

[tex]a=\dfrac{F}{m}\\\\a=\dfrac{9.12\times 10^{-17}}{9.1\times 10^{-31}}\\\\a=10^{14}\ m/s^2[/tex]

Let v be the final velocity of the electron. So,

v = u +at

u = 0 (at rest)

So,

[tex]v=10^{14}\times 47.6\times 10^{-9}\\\\v=4.76\times 10^6\ m/s[/tex]

For proton,

Acceleration,

[tex]a=\dfrac{9.12\times 10^{-17}}{1.67\times 10^{-27}}\\\\=5.46\times 10^{10}\ m/s^2[/tex]

Now final velocity of the proton is given by :

[tex]v=5.46\times 10^{10}\times 47.6\times 10^{-9}\\\\v=2598.96\ m/s[/tex]

Hence, this is the required solution.

Answer:

Uhh 2 one

Explanation