URGENT : Which of the following is the most stable isotope? Explain.


URGENT : Which Of The Following Is The Most Stable Isotope? Explain.

Answers

Answer 1

Answer:

Plutonium–238

Explanation:

The stability of isotopes is largely dependent on their half-life.

Half life of an isotope is the time taken for the initial mass of the isotope to be halfed or we can say that the half-life of an isotope is the time taken for the mass of the isotope to become half the initial mass.

From the above definition, we discovered that if the time taken for the mass of the isotope to become half its initial mass is long, then the isotope must be very stable. On the other hand, if the time taken to become half its initial mass is short, then the isotope is unstable because.

We can thus say that, the longer the half-life the more stable the isotope and the shorter the half-life, the less stable the isotope will be.

Considering the table given in the question above and with the ideas we have obtained from the explanation above, we can see that Plutonium–238 has the longest half-life. Therefore Plutonium–238 will be more stable.


Related Questions

A conducting bar with mass m and length L slides over horizontal rails that are connected to a voltage source. The voltage source maintains a constant current I in the rails and bar, and a constant, uniform, vertical magnetic field B fills the region between the rails . Find the magnitude and direction of the net force on the conducting bar. Ignore friction, air resistance and electrical resistance :
A. v2m / ILB to yhe right
B. 3v2m /2 ILB to yhe left
C. 5v2m/ 2ILB to the right
D. v2m / 2ILB to the left

Answers

Answer:

F = ILB

Explanation:

To find the net force on the conducting bar you take into account the following expression:

[tex]\vec{F}=I( \vec{L}X \vec{B})[/tex]

I: current in the conducting bar

L: length of the bar

B: magnitude of the magnetic field

In this case the direction of the magnetic field and the motion of the bar are perpendicular between them. The direction of the bar is + i, and the magnetic field poits upward + k. The cross product of these vector give us the direction of the net force:

+i X +k = +j

The direction of the force is to the right and its magnitude is F = ILB

A tuning fork is held over a resonance tube, and resonance occurs when the surface of the water is 12 cm below the top of the tube. Resonance occurs again when the water is 34 cm below the top of the tube. If the air temperature is 23 degrees Celsius, find the frequency of the tuning fork.

Answers

Answer:

786 Hz

Explanation:

Recall, the speed of sound is

v = 332 + 0.6t

Where t = 23°

v = 332 + 0.6(23)

v = 332 + 13.8

v = 345.8 m

Also, it is known that distance between two consecutive resonance length is half of the wavelength.

L2 - L1 = λ/2

34 - 12 = λ/2

λ/2 = 22

λ = 44 cm

Finally, remember that also

Frequency = speed/ wavelength

Frequency = 345.8/0.4

Frequency = 786 Hz

Therefore, the frequency of the tuning fork is 786 Hz

Countries create quotas and tariffs to increase the volume of trade with their neighbors.

Answers

Oooooo, that statement is not true.  Countries create quotas and tariffs to LIMIT the volume of trade with other countries, including their neighbors.

Answer:

False

Explanation:

I took the text :)

HELPP MEE
Which image illustrates the desired interaction of a sound wave with
soundproofing material in a recording studio?

Answers

Soundproofing material is required for blocking sound during some works like recording voice in the studio. Image D represents the interaction of a sound wave with soundproofing material in a recording studio.

What is the basis of soundproofing?

Soundproofing is done by absorbing the sound. A very much used material for this is a dense foam.

Foam and like materials absorbs sound and it travels directly into the soft surface resulting in soundproofing.

Thus, the correct option is C, as the D image is showing the absorption.

For more details regarding soundproofing, visit:

https://brainly.com/question/8980142

#SPJ2

Answer: C.D

Explanation:...

A plastic rod of length d = 1.5 m lies along the x-axis with its midpoint at the origin. The rod carries a uniform linear charge density λ = 2.5 nC/m. The point P is located on the positive y-axis at a distance y0 = 15 cm from the origin. The z-axis points out of the screen. Integrate your correct choice in part (b) over the length of the rod and choose the correct expression for the y-component of the electric field at point P.

Answers

Answer:

Explanation:

Let the plastic rod extends from - L to + L .

consider a small length of dx on the rod on the positive x axis at distance x . charge on it =  λ dx where  λ is linear charge density .

It will create a field at point P on y -axis . Distance of point P

= √ x² + .15²

electric field at P due to small charged length

dE = k λ dx x  / (x² + .15² )

Its component along Y - axis

= dE cosθ where θ is angle between direction of field dE and y axis

= dE x .15 / √ x² + .15²

=  k λ dx  .15 / (x² + .15² )³/²

If we consider the same strip along the x axis at the same position  on negative x axis , same result will be found . It is to be noted that the component of field in perpendicular to y axis will cancel out each other . Now for electric field due to whole rod at point p , we shall have to integrate the above expression from - L to + L

E = ∫  k λ  .15  / (x² + .15² )³/² dx

=  k λ  x L / .15 √( L² / 4 + .15² )

b) The length of the rod:

[tex]E = \int\limits dx . k \lambda .15 / (x^2 + .15^2 )^{1/2} dx\\\\E= \frac{k \lambda * L}{0.15} \sqrt{( L^2 / 4 + .15^2 )[/tex]

Given:

d = 1.5 mλ = 2.5 nC/m

Let the plastic rod extends from - L to + L .Consider a small length of dx on the rod on the positive x axis at distance x . charge on it = λ dx where λ is linear charge density .It will create a field at point P on y -axis.

Distance of point P =[tex]\sqrt{x^2 + 0.15^2}[/tex]

How to calculate Electric Field?

E.F at P due to small charged length[tex]dE = \frac{ k \lambda x.dx}{(x^2 + .15^2 )}[/tex]

Its component along Y - axis = dE cosθ where θ is angle between direction of field dE and y axis

[tex]= \frac{dE x .15 }{\sqrt{x^2 + .15^2} }\\\\= \frac{k \lambda dx .15}{(x^2 + .15^2 )^{1/2}}[/tex]

If we consider the same strip along the x axis at the same position on negative x axis , same result will be found . We can say that the component of field in perpendicular to y axis will cancel out each other.

Now for electric field due to whole rod at point p , we shall have to integrate the above expression from - L to + L

[tex]E = \int\limits dx . k \lambda .15 / (x^2 + .15^2 )^{1/2} dx\\\\E= \frac{k \lambda * L}{0.15} \sqrt{( L^2 / 4 + .15^2 )}[/tex]

Find more information about Electric field here:

brainly.com/question/14372859

In Physics lab, a lab team places a cart on one of the horizontal, linear tracks with a fan attached to it. The cart is positioned at one end of the track, and the fan is turned on. Starting from rest, the cart takes 4.34 s to travel a distance of 1.62 m. The mass of the cart plus fan is 354 g. Assume that the cart travels with constant acceleration.
A) What is the net force exerted on the cart-fan combination?B) Mass is added to the cart until the total mass of the cart-fan combination is 762 g, and the experiment is repeated. How long does it take for the cart, starting from rest, to travel 1.62 m now?

Answers

Answer:

A. F = 0.06 N

B. t = 6.37 s

Explanation:

A)

First we need to find the constant acceleration of the cart. For this purpose, we use 2nd equation of motion:

s = (Vi)(t) + (0.5)at²

where,

s = distance traveled = 1.62 m

Vi = 0 m/s   (Since, it starts from rest)

t = Time Taken = 4.34 s

a = acceleration = ?

Therefore,

1.62 m = (0 m/s)(4.34 s) + (0.5)(a)(4.34 s)²

1.62 m/9.4178 s² = a

a = 0.172 m/s²

Now, from Newton's Second law, we know that:

F = ma

where,

F = Net Force of the combination = ?

m = Mass pf combination = 354 g = 0.354 kg

Therefore,

F = (0.354 kg)(0.172 m/s²)

F = 0.06 N

B)

Now, for the same force, but changed mass = 762 g = 0.762 kg, we have the acceleration to be:

F = ma

a = F/m

a = 0.06 N/0.762 kg

a = 0.08 m/s²

Now, using  2nd equation of motion:

s = (Vi)(t) + (0.5)at²

1.62 m = (0 m/s)(t) + (0.5)(0.08 m/s²)t²

t² = 1.62 m/(0.04 m/s²)

t = √40.54 s²

t = 6.37 s

Question
20
what would be the advantages if your body had magnetic properties science subject​

Answers

Answer:

Some of the advantages if our body had magnetic properties are as follows:

Magnetic properties can have health benefits such as recovering quickly from a stroke, resolving bladder problems, and reducing blood pressure.Brain will be able to control more activities of the nervous system and other organs of the body using magnetic power.Heart will have many benefits of magnetic properties and able to provide more energy to the entire body through the circulation of blood.Magnetic properties in body will be able to maintain the production of melatonin that controls the sleep patterns.Magnetic properties will be able to kill cancer causing cells.

Hence, magnetic properties are somehow beneficial for humans.  

A Michelson interferometer operating at a 400 nm wavelength has a 3.95-cm-long glass cell in one arm. To begin, the air is pumped out of the cell and mirror M2 is adjusted to produce a bright spot at the center of the interference pattern. Then a valve is opened and air is slowly admitted into the cell. The index of refraction of air at 1.00 atmatm pressure is 1.00028.

Required:
How many bright-dark-bright fringe shifts are observed as the cell fills with air?

Answers

Answer:

55.3

Explanation:

The computation of the number of bright-dark-bright fringe shifts observed is shown below:

[tex]\triangle m = \frac{2d}{\lambda} (n - 1)[/tex]

where

d = [tex]3.95 \times 10^{-2}m[/tex]

[tex]\lambda = 400 \times 10^{-9}m[/tex]

n = 1.00028

Now placing these values to the above formula

So, the  number of bright-dark-bright fringe shifts observed is

[tex]= \frac{2 \times3.95 \times 10^{-2}m}{400 \times 10^{-9}m} (1.00028 - 1)[/tex]

= 55.3

We simply applied the above formula so that the number of bright dark bright fringe shifts could come

A parallel-plate capacitor has square plates that are 7.20 cm on each side and 3.40 mm apart. The space between the plates is completely filled with two square slabs of dielectric, each 7.20 cm on a side and 1.70 mm thick. One slab is Pyrex glass and the other slab is polystyrene. If the potential difference between the plates is 96.0 V, find how much electrical energy (in nJ) can be stored in this capacitor.

Answers

Answer:

U = 218 nJ

Explanation:

We are given;

Spacing between the plates; d = 3.4 mm = 3.4 × 10^(-3) m

Voltage across the capacitor; V = 96 V

Dimension of the square plates is 7.2cm x 7.2cm.

So, Area = 7.2 × 7.2 = 51.84 cm² = 51.84 × 10^(-4) m²

Permittivity of free space; ε_o = 8.85 × 10^(-12) C²/N.m²

From relative permeability table;

Dielectric constant of Pyrex; k1 = 5.6

Dielectric constant of polystyrene; k2 = 2.56

Now, formula for capacitance of a capacitor with Dielectric is;

C = kC_o

Where, C_o = ε_o(A/d)

Since there are 2 capacitors, d will now be d/2 = (3.4 × 10^(-3))/2 m = 1.7 × 10^(-3)

Since we have 2 capacitor, thus ;

C1 = k1*ε_o*(A/d)

C1 = (5.6 × 8.85 × 10^(-12) × (51.84 × 10^(-4))/(1.7 × 10^(-3))

C1 = 1.51 × 10^(-10) F

Similarly;

C2 = (2.56 × 8.85 × 10^(-12) × (51.84 × 10^(-4))/(1.7 × 10^(-3))

C2 = 0.691 × 10^(-10) F

For capacitors in series, formula for total capacitance(Cs) is;

1/Cs = (1/C1) + (1/C2)

Simplifying this, we have;

Cs = (C1*C2)/(C1 + C2)

Plugging in the relevant values ;

Cs = (1.51 × 10^(-10)*0.691 × 10^(-10))/((1.51 × 10^(-10)) + (0.691 × 10^(-10)))

Cs = 0.474 × 10^(-10) F

The formula for energy stored in a capacitor with 2 Dielectrics is given as;

U = ½Cs*V²

So,

U = ½ × 0.474 × 10^(-10) × 96²

U = 2.18 × 10^(-7) J = 218 × 10^(-9) = 218 nJ

How can socialism
impact populations?

Answers

Answer:

it represents a fundamental difference. (more info below)

Explanation:

Production is incessantly developing and expanding in socialist countries, and employment is guaranteed for the entire productive population. Consequently, the relative overpopulation problem has been eliminated. This represents the fundamental difference between socialism's demographic law and capitalism's law.

hope this helped!

It represents a fundamental difference by gaining friends and losing friends or gaining jobs and losing jobs etc

A rocket rises vertically, from rest, with an acceleration of 5.0 m/s2 until it runs out of fuel at an altitude of 960 m . After this point, its acceleration is that of gravity, downward.
(A) What is the velocity of the rocket when it runs out of fuel?
(B) How long does it take to reach this point?
(C) What maximum altitude does the rocket reach?
(D) How much time (total) does it take to reach maximum altitude?
(E) With what velocity does it strike the Earth? () How long (total) is it in the air?
a) 70.427m/s
b) 22 m
c) 1027.8m
d) 29.179 s
e) 142m/s
f ) 43.654s

Answers

Answer:

a) 98 m/s

b) 19.6 s

c)  1449.8 m

d)  29.6 s

e)  168.6 m/s

f)  46.8 s

Explanation:

Given that

Acceleration of the rocket, a = 5 m/s²

Altitude of the rocket, s = 960 m

a)

Using the equation of motion

v² = u² + 2as, considering that the initial velocity, u is 0. Then

v² = 2as

v = √2as

v = √(2 * 5 * 960)

v = √9600

v = 98 m/s

b)

Using the equation of motion

S = ut + ½at², considering that initial velocity, u = 0. So that

S = ½at²

t² = 2s/a

t² = (2 * 960) / 5

t² = 1920 / 5

t² = 384

t = √384 = 19.6 s

c)

Using the equation of motion

v² = u² + 2as, where u = 98 m/s, a = -9.8 m/s², so that

0 = 98² + 2(-9.8) * s

9600 = 19.6s

s = 9600/19.6

s = 489.8 m

The maximum altitude now is

960 m + 489.8 m = 1449.8 m

d)

Using the equation of motion

v = u + at, where initial velocity, u = 98 m, a = -9.8 m/s. So that

0 = 98 +(-9.8 * t)

98 = 9.8t

t = 98/9.8

t = 10 s

Total time then is, 10 + 19.6 = 29.6 s

e) using the equation of motion

v² = u² + 2as, where initial velocity, u = o, acceleration a = 9.8 m/s, and s = 1449.8 m. So that,

v² = 0 + 2 * 9.8 * 1449.8

v² = 28416.08

v = √28416.08

v = 168.6 m/s

f) using the equation of motion

S = ut + ½at², where s = 1449.8 m and a = 9.8 m/s

1449.8 = 0 + ½ * 9.8 * t²

2899.6 = 9.8t²

t² = 2899.6/9.8

t² = 295.88

t = √295.88

t = 17.2 s

total time in air then is, 17.2 + 29.6 = 46.8 s

A meter stick hurtles through space at a speed of 0.95c relative to you, with its length perpendicular to the direction of motion. You measure its length to be equal to:_______
a. 0 m.
b. 0.05 m.
c. 0.95 m.
d. 1.00 m.
e. 1.05 m.

Answers

Answer:

d. 1.00 m

Explanation:

In 1905, Einstein proposed special theory of relativity of light.

This theory had a number of consequences or results. One of them is called "Length Contraction".

According to this consequence, whenever an object travels at a speed comparable to the speed of light, its length decreases.

But this decrease in length is only seen in the dimension, which is parallel to the direction of motion of the body. All other dimensions of the object remains same.

In the given situation, the length of meter stick is not parallel to the direction of motion, but it is perpendicular. Hence, the length of meter stick will be same as the length of meter stick at rest. Hence, the correct option will be:

d. 1.00 m

Two hockey pucks, labeled A and B, are initially at rest on a smooth ice surface and are separated by a distance of 18.0 m . Simultaneously, each puck is given a quick push, and they begin to slide directly toward each other. Puck A moves with a speed of 3.90 m/s , and puck B moves with a speed of 4.30 m/s . What is the distance covered by puck A by the time the two pucks collide

Answers

Answer:

The distance covered by puck A before collision is  [tex]z = 8.56 \ m[/tex]

Explanation:

From the question we are told that

   The label on the two hockey pucks is  A and  B

    The distance between the  two hockey pucks is D   18.0 m

     The speed of puck A is  [tex]v_A = 3.90 \ m/s[/tex]

        The speed of puck B is  [tex]v_B = 4.30 \ m/s[/tex]

The distance covered by puck A is mathematically represented as

     [tex]z = v_A * t[/tex]

  =>  [tex]t = \frac{z}{v_A}[/tex]

 The distance covered by puck B  is  mathematically represented as

      [tex]18 - z = v_B * t[/tex]

=>   [tex]t = \frac{18 - z}{v_B}[/tex]

Since the time take before collision is the same

        [tex]\frac{18 - z}{V_B} = \frac{z}{v_A}[/tex]

substituting values

          [tex]\frac{18 -z }{4.3} = \frac{z}{3.90}[/tex]

=>      [tex]70.2 - 3.90 z = 4.3 z[/tex]

=>       [tex]z = 8.56 \ m[/tex]

Which nucleus completes the following equation?
Se+?
O A. Ga
B. P
C. 31P
D. CI

Answers

Answer:First option

Explanation:

hope it helped

Can someone help me with this question

Answers

Answer:

hypothesis , hope it helps

Explanation:

Answer:

Inference

Explanation:

Inference is something you predict after testing that's a result after an hypothesis has been made. Hypothesis is an intelligent guess based on some observed phenomena which can be subjected to further testing.

An aluminum wing on a passenger jet is 30 m long when its temperature is 27 C. At what temperature would the wing be 0.03 shorter?

Answers

Answer:2000

Explanation:

When an aluminum bar is connected between a hot reservoir at 860 K and a cold reservoir at 348 K, 2.40 kJ of energy is transferred by heat from the hot reservoir to the cold reservoir
(a) In this irreversible process, calculate the change in entropy of the hot reservoir.
_______ J/K
(b) In this irreversible process, calculate the change in entropy of the cold reservoir.
_______ J/K
(c) In this irreversible process, calculate the change in entropy of the Universe, neglecting any change in entropy of the aluminum rod.
_______ J/K
(d) Mathematically, why did the result for the Universe in part (c) have to be positive?

Answers

Answer:

a) [tex]\Delta S_{in} = 2.791\,\frac{J}{K}[/tex], b) [tex]\Delta S_{out} = 6.897\,\frac{J}{K}[/tex], c) [tex]S_{gen} = 4.106\,\frac{J}{K}[/tex], d) Due to irreversibilities due to temperature differences.

Explanation:

a) The change in entropy of the hot reservoir is:

[tex]\Delta S_{in} = \frac{2400\,J}{860\,K}[/tex]

[tex]\Delta S_{in} = 2.791\,\frac{J}{K}[/tex]

b) The change in entropy of the cold reservoir is:

[tex]\Delta S_{out} = \frac{2400\,J}{348\,K}[/tex]

[tex]\Delta S_{out} = 6.897\,\frac{J}{K}[/tex]

c) The total change in entropy of the Universe is modelled after the Second Law of Thermodynamics. Let assume that process is steady:

[tex]\Delta S_{in} - \Delta S_{out} + S_{gen} = 0[/tex]

[tex]S_{gen} = \Delta S_{out} - \Delta S_{in}[/tex]

[tex]S_{gen} = 6.897\,\frac{J}{K} - 2.791\,\frac{J}{K}[/tex]

[tex]S_{gen} = 4.106\,\frac{J}{K}[/tex]

d) Since irreversibilities create entropy as process goes by. The main source of irreversibilities is the existence of temperature differences.

A small ball of mass m is aligned above a larger ball of mass M = 0.63kg (with a slight separation) and the two are dropped simultaneously from a height of 1.8m. If the larger ball rebounds elastically from the floor and the small ball rebounds elastically from the larger ball what value of m results in the larger ball stopping when it collides with the small ball?

Answers

thats hard. i would have to say work the problem out

The amount of friction divided by the weight of an object forms a unit less number called the

Answers

Answer:

Coefficient of friction.

Explanation:

The amount of friction divided by the weight of an object is equal to the coefficient of friction. It is a dimensional less number. It can be given by :

[tex]F=\mu N[/tex]

N is normal force.

[tex]\mu[/tex] = coefficient of friction

[tex]\mu=\dfrac{F}{N}[/tex]

Constants Canada geese migrate essentially along a north-south direction for well over a thousand kilometers in some cases, traveling at speeds up to about 100 km/h. The one goose is flying at 100 km/h relative to the air but a 44 km/h wind is blowing from west to east.
1. At what angle relative to the north-south direction should this bird head so that it will be traveling directly southward relative to the ground?2. How long will it take the bird to cover a ground distance of 450 from north to south?

Answers

Answer:

a. 63.89°  in the north-southward manner

b.  2.2 sec

Explanation:

The goose is flying at 100 km/h

Air from east to west is 44 km/h

angle relative to the north-south direction for the bird to travel south will be

cos∅ = 44/100 = 0.44

∅ = [tex]cos^{-1}[/tex]0.44 = 63.89°  in the north-southward manner

Speed south relative to the ground will be v

Tan 63.89 = v/100

2.04 = v/100

v = 2.04 x 100 = 204 km/hr

to cover a distance of 450 m from north to south at this speed time will be

t = d/v = 450/204 = 2.2 sec

Find the displacement of a simple harmonic wave of amplitude 6.44 m at t = 0.71 s. Assume that the wave number is 2.34 m-1, the angular frequency is 2.88 rad/s, and that the wave is propagating in the +x direction at x = 1.21 m.
A) 4.55 m.
B) 1.05 m.
C) 3.54 m.
D) 2.25 m.

Answers

Answer:

Letter A. [tex]y=4.55 m[/tex]

Explanation:

Let's use the wave equation:

[tex]y=Asin(kx-\omega t)[/tex]

A is the amplitude (A=6.44 m)t is the time (t=0.71 s)k is the wave number (k=2.34 1/m)ω is the angular frequency (ω=2.88 rad/s)x is the propagation of the x direction  (x=1.21 m)

Therefore the displacement y will be:

[tex]y=6.44*sin(2.34*1.21-2.88*0.71)[/tex]

[tex]y=4.55 m[/tex]

The answer is letter A.

I hope it helps you!

Answer:

Explanation:

Find the displacement of a simple harmonic wave of amplitude 6.44 m at t = 0.71 s. Assume that the wave number is 2.34 m-1, the angular frequency is 2.88 rad/s, and that the wave is propagating in the +x direction at x = 1.21 m.

Amplitude (A) of the simple harmonic wave = 6.44 m

wave number (k) of the given wave = 2.34 m-1

Angular frequency (ω) of the given wave = 2.88 rad/s

Displacement x = 1.21 m and time t = 0.71 s

Then the general equation for the displacement of the given simple harmonic wave at given x and time t is given by

y = Asin(kx - ωt)

= (6.44 m)sin[(2.34 m-1)(1.21 m) - (2.88 rad/s)(0.71 s)]

Y=6.44sin(0.7866 rad)

0.7866rad*(180 degrees/pi rad) =45.1

Y=6.44sin(45.1)

Y=4.55m

when the same amount of heat is added to equal masses of water and copper at the same temperature the copper is heated to a higher final temperature than water. on a molecular level what explains this difference

a. the average kinetic energy of water molecules is greater than the average kinetic energy of the copper
b.more of the heat is transferred to the potential energy of the water molecules than the potential energy of the copper atoms
c.the intermolecular forces between copper atoms are stronger than those between water molecules
d.more of the heat is transferred to the kinetic energy of the water molecules than to the kinetic energy of the copper atoms​

Answers

Answer:

C

Explanation:

The intermolecular forces between the water molecule is less binding than that of the copper molecule. Hence the water would take a shorter time to be converted to vapour where the temperature of boiling is constant however the temperature of that of the copper molecule keeps increasing.

A 0.150 kg lump of clay is dropped from a height of 1.45 m onto the floor. It sticks to the floor and does not bounce.

What is the magnitude of the impulse imparted to the clay by the floor during the impact? Assume that the acceleration due to gravity is =9.81 m/s2.

Answers

Answer:

J = 0.800 kg m/s

Fmax = 291 N

Explanation:

During the fall, energy is conserved.

PE = KE

mgh = ½ mv²

v = √(2gh)

v = √(2 × 9.81 m/s² × 1.45 m)

v = 5.33 m/s

Alternatively, you can use kinematics to find the velocity.

Impulse = change in momentum

J = Δp

J = mΔv

J = (0.150 kg) (5.33 m/s)

J = 0.800 kg m/s

Impulse = area under F vs t graph

J = ∫ F dt

J = ½ Fmax Δt

(0.800 kg m/s) = ½ Fmax (0.0055 ms)

Fmax = 291 N

A block is supported on a compressed spring, which projects the block straight up in the air at velocity . The spring and ledge it sits on then retract. You can win a prize by hitting the block with a ball. When should you throw the ball and in what direction to be sure the ball hits the block? (Assume the ball can reach the block before the blochk reaches the ground and that the ball is thrown from a height equal to the release position of the block.)
A. At the instant when the block is at the highest point, directed at the spring.
B. At the instant when the block is at the highest point, directed at the block.
C. At the instant when the block leaves the spring, directed at the spring.
D. At the instant when the block leaves the spring, directed at the block.
E. When the block is back at the spring's original position, directed at that position.

Answers

Answer:

B. At the instant when the block is at the highest point, directed at the block.

Explanation:

Motion of an object is the change in the position of the object with respect to time. On the earth, gravity has a great influence on the motion of an object (especially in a vertical direction).

When the block is projected up in the air, it moves with a varying velocity until the velocity becomes zero due to gravity. Which make the object to rest a little in the air (when velocity = gravity) and starts to fall freely.

To ensure hitting the block by the ball, it is thrown at the block when the block is at its highest point in the air. Since the block would be at rest at this instant before it start to fall at a constant acceleration under gravity.

Help with this answer please

Answers

Answer:

Everytime you do an experiment you need something that is regular. For example if you try and measure how much germs spread in bread. you need 1 bread thats clean and 3 different breads for different molds. So thats called a CONTROL

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A uniform thin spherical shell of mass M=2kg and radius R=0.23m is given an initial angular speed w=18.3rad/s when it is at the bottom of an inclined plane of height h=3.5m, as shown in the figure. The spherical shell rolls without slipping. Find wif the shell comes to rest at the top of the inclined plane. (Take g-9.81 m/s2, Ispherical shell = 2/3 MR2 ).Express your answer using one decimal place.

Answers

Answer:

47.8rad/s

Explanation:

For energy to be conserved.

The potential energy sustain by the object would be equal to K.E

P.E = m× g× h = 2 × 9.81× 3.5= 68.67J

Now K.E = 1/2 × I × (w1^2 - w0^2)

I = 2/3 × M × R2

= 2/3 × 2 × (0.23)^2= 0.0705

Hence

W1 = final angular velocity

Wo = initial angular velocity

From P.E = K.E we have;

68.67J = 1/2 × 0.0705 × (w1^2 - w0^2)

(w1^2 - w0^2) = 1948.09

W1^2 = 1948.09 + (18.3^2)

W1^2=2282.98

W1 = √2282.98

=47.78rad/s

= 47.8rad/s to 1 decimal place.

In each pair, select a substance that is a better heat conductor.

1. copper wire / wood 3. water / iron
2. water / air 4. iron / glass

Answers

Answer:

1)copper wire

Explanation:

it is the best electric conductor

Cooper wire / wood
Cooper is a good conductor of heat and electricity. Cooper is used in making wires

The current through an inductor of inductance L is given by I(t) = Imax sin(ωt).
(a) Derive an expression for the induced emf in the inductor as a function of time.
(b) At t = 0, is the current through the inductor increasing or decreasing?
(c) At t = 0, is the induced emf opposing or aiding the flow of the charge carriers? (Remember that the direction of a positive induced emf is the same as the current direction and the direction of a negative induced emf is opposite the current direction.)
(d) How are the answers to parts b and c consistent with the behavior of inductors discussed in the text?

Answers

Answer:

(a) [tex]emf_L=-LI_{max}\omega cos(\omega t)[/tex]

(b) neither increasing or decreasing

(c) opposite to the flow of charge carriers

Explanation:

The current through an inductor of inductance L is given by:

[tex]I(t)=I_{max}sin(\omega t)[/tex]   (1)

(a) The induced emf is given by the following formula

[tex]emf_L=-L\frac{dI}{dt}[/tex]    (2)

You derivative the expression (1) in the expression (2):

[tex]emf_L=-L\frac{d}{dt}(I_{max}sin(\omega t))\\\\emf_L=-LI_{max}\omega cos(\omega t)[/tex]

(b) At t=0 the current is zero

(c) At t = 0 the emf is:

[tex]emf_L=-\omega LI_{max}[/tex]

w, L and Imax have positive values, then the emf is negative. Hence, the induced emf is opposite to the flow of the charge carriers.

(d) read the text carefully

At t zero,  the current through the inductor neither increasing nor decreasing because current is zero.

The current through an inductor of inductance L can be calculated by

[tex]\bold {I_t = I_m_a_x sin (\omega t)}[/tex].........1  

(a) The induced emf can be calculated by  

[tex]\bold {emf_L = - L \dfrac {dI}{dt}}[/tex]............2  

Derivative the equation (1) in the equation (2)

[tex]\bold {emf _L= -L \dfrac {d (I _m_a_x sin (\omega t)} {dt}}\\\\\bold {emf _L= -L (I _m_a_x \omega cos( \omega t) }[/tex]

(b) At t=0 the current is zero,

 

(c) At t = 0 the emf is:

[tex]\bold {emf_L = -\omega LI _m_a_x}[/tex]  

Therefore, at t zero,  the current through the inductor neither increasing nor decreasing because current is zero.

To know more about inductance,

https://brainly.com/question/17431388

You have a suction cup that creates a circular region of low pressure with a 30 mm diameter. It holds the pressure to 85 % of atmospheric pressure. What "holding force" does the suction cup generate in N

Answers

Answer:

Force = 60.08 N

Explanation:

Given that

Diameter d = 30 mm  

Holding pressure = 85 %  of Atmospherics pressure

Solution

As we know that  here 1 atm = 10⁵ N/m²

and pressure is known as force per unit area

pressure = [tex]\frac{F}{A}[/tex]   ................1

put here value and we will get

F = [tex]0.85\times 10^5\times \frac{\pi}{4}\times 0.03^2\ N[/tex]

solve it we get

Force = 60.08 N

This problem explores the behavior of charge on conductors. We take as an example a long conducting rod suspended by insulating strings. Assume that the rod is initially electrically neutral. For convenience we will refer to the left end of the rod as end A, and the right end of the rod as end B. In the answer options for this problem, "strongly attracted/repelled" means "attracted/repelled with a force of magnitude similar to that which would exist between two charged balls.A small metal ball is given a negative charge, then brought near (i.e., within about 1/10 the length of the rod) to end A of the rod. What happens to end A of the rod when the ball approaches it closely this first time?

Answers

Answer:

rod end A is strongly attracted towards the balls

rod end B is weakly repelled by the ball as it is at a greater distance

Explanation:

When the ball with a negative charge approaches the A end of the neutral bar, the charge of the same sign will repel and as they move they move to the left end, leaving the rod with a positive charge at the A end and a negative charge of equal value at end B.

Therefore rod end A is strongly attracted towards the balls and

rod end B is weakly repelled by the ball as it is at a greater distance

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