An atom undergoes nuclear decay, but its atomic number is not changed.
What type of nuclear decay did the atom undergo?
A. Gamma decay
B. Beta decay
C. Nuclear fission
D. Alpha decay

Answer:

Workdone = 0.89 Joules

Explanation:

Given the following data;

Workdone = 6J

Extension = 39 - 26 = 13cm to meters = 13/100 = 0.13m

The workdone to stretch a string is given by the formula;

Workdone = ½ke²

Where;

k is the constant of elasticity.

e is the extension of the string.

We would solve for string constant, k;

6 = ½*k*0.13²

6 = ½*k*0.0169

Cross-multiplying, we have;

12 = 0.0169k

k = 12/0.0169

k = 710.06

a. To find the workdone when e = 30, 35.

Extension = 35 - 30 = 5 to meters = 5/100 = 0.05m

Workdone = ½*710.06*0.05²

Workdone = 355.03*0.0025

Workdone = 0.89 Joules

Therefore, the amount of work (in J) needed to stretch the spring from 30 cm to 35 cm is 0.89.

Answer:

- 21⁰C .

Explanation:

Speed of jet = 2.05 x 10³ km /h

= 2050 x 1000 / (60 x 60 ) m /s

= 569.44 m / s

Mach no represents times of speed of sound , the speed of jet

1.79 x speed of sound = 569.44

speed of sound = 318.12 m /s

speed of sound at 20⁰C = 343 m /s

Difference = 343 - 318.12 = 24.88⁰C

We know that 1 ⁰C change in temperature changes speed of sound

by .61 m /s

So a change in speed of 24.88 will be produced by a change in temperature of

24.88 / .61

= 41⁰C  

temperature = 20 - 41 = - 21⁰C .  

Answer and Explanation: To know how much tape he will need, we have to calculate the perimeter of each parallelogram-shaped stripe.

Perimeter is the sum of all the sides of a figure.

For a parallelogram:

P = 2*length + 2*width

So, we need to determine width and length of the stripe.

Width is 3 inches. Length is the hypotenuse of the right triangle, whose sides are 6 and 18 inches. Then, length is

[tex]h=\sqrt{18^{2}+6^{2}}[/tex]

[tex]h=\sqrt{360}[/tex]

h = 19 in

Perimeter of the first stripe is

P = (2*19) + (2*3)

P = 44 inches

The hazard sign has 3 stripes. So total perimeter is

[tex]P_{t}=[/tex] 44 + 44 + 44

[tex]P_{t}=[/tex] 132 inches

To outline the parallelogram-shaped stripes, Charles need a total of 132 inches of tape. Since one roll has 144 inches, he will have enough tape to finish the job.

Answer:

76.1N

Explanation:

Given parameters:

Mass of the ball  = 7.77kg

Unknow:

Weight of balloon  = ?

Solution:

Weight is the vertical force applied on a body.

  Weight = mass x acceleration due gravity

So;

  Weight  = mass x acceleration due to gravity

So;

 Weight  = 7.77 x 9.8  = 76.1N

Answer: F = 498.04 lbs

Explanation: The forces acting on the sled and Julie are show in the figure below. In it, we notice that, for the sled and Julie to go accross the field, they only need force of friction, because, force of friction is a force that resists the relative motion of surfaces.

Force of friction is given by the formula

[tex]F_{f}=\mu.F_{N}[/tex]

where

μ is coefficient of friction

[tex]F_{N}[/tex] is normal force

Normal force is the force the surface exerts on the object. It is always perpendicular and a force of contact.

In the case of the sled, since it is on a horizontal plane, Normal Force has the same magnitude of Gravitational Force. So

[tex]F_{N}=m.g[/tex]

Coefficient of friction is how much friction exists between two surfaces.

Rearraging friction force is

[tex]F_{f}=\mu.m.g[/tex]

Mass for this system is the sum of Julie and the sled, therefore

m = 109 + 12

m = 121 lb

Calculating Friction Force:

[tex]F_{f}=0.42.121.9.8[/tex]

[tex]F_{f}=[/tex] 498.04 lbs

LBS is a unit of measurement referred as pound by weight.

In conclusion, force Mary needs to start moving the sled is 498.04 lbs

Answer:

ummm hehe this is my time to shine

Explanation:

  MERICIA!!!!!!!!!!!!!!!!!!!!!!!

Answer:

579600J

Explanation:

Given parameters:

Height of the building  = 828m

Weight of the man  = 700N

Unknown:

Work done by the man  = ?

Solution:

The work done by the man is the same as the potential energy expended.

 Work done:

             Work done  = Weight x height  = 700 x 828

        Work done  = 579600J

Answer:

[tex]m=1.63\times 10^{-27}\ kg[/tex]

Explanation:

The velocity of a particle is 90% of the speed of light.

The wavelength of the particle is [tex]1.5\times 10^{-15}\ m[/tex]

We need to find the mass of the particle.

The formula for the wavelength of a particle is given by :

[tex]\lambda=\dfrac{h}{mv}[/tex]

h is Planck's constant

v is 90% of speed of light

m is mass of the particle

[tex]m=\dfrac{h}{\lambda v}\\\\m=\dfrac{6.63\times 10^{-34}}{1.5\times 10^{-15}\times 0.9\times 3\times 10^8}\\\\m=1.63\times 10^{-27}\ kg[/tex]

So, the mass of the particle is [tex]1.63\times 10^{-27}\ kg[/tex].

Answer:

force = mass * acceleration

therefore the SI unit is kg*m/s2 or newton's

it's a vector quantity because it has both direction(acceleration) and size (mass)

Answer:

69.62 minutes

Explanation:

From the information we have here,

1 byte is = 8 bits

So 1 megabyte = 8 megabits

Then

783.216 x 8 megabits = 6265.728 megabits

This player has its reading capacity at 1.5 megabits / second

So 1 minute = 60x1.5

= 90 megabits / min

Then we have the entire reading time of this CD player to be =

6265.728/90

= 69.62 minutes.

This answers the question

Answer:

The specific heat of the unknown material is 131.750 joules per kilogram-degree Celsius.

Explanation:

Let suppose that sphere is cooled down at steady state, then we can estimate the rate of heat transfer ([tex]\dot Q[/tex]), measured in watts, that is, joules per second, by the following formula:

[tex]\dot Q = m\cdot c\cdot \frac{T_{f}-T_{o}}{\Delta t}[/tex] (1)

Where:

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

[tex]c[/tex] - Specific heat of the material, measured in joules per kilogram-degree Celsius.

[tex]T_{o}[/tex], [tex]T_{f}[/tex] - Initial and final temperatures of the sphere, measured in degrees Celsius.

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

In addition, we assume that both spheres experiment the same heat transfer rate, then we have the following identity:

[tex]\frac{m_{I}\cdot c_{I}}{\Delta t_{I}} = \frac{m_{X}\cdot c_{X}}{\Delta t_{X}}[/tex] (2)

Where:

[tex]m_{I}[/tex], [tex]m_{X}[/tex] - Masses of the iron and unknown spheres, measured in kilograms.

[tex]\Delta t_{I}[/tex], [tex]\Delta t_{X}[/tex] - Times of the iron and unknown spheres, measured in seconds.

[tex]c_{I}[/tex], [tex]c_{X}[/tex] - Specific heats of the iron and unknown materials, measured in joules per kilogram-degree Celsius.

[tex]c_{X} = \left(\frac{\Delta t_{X}}{\Delta t_{I}}\right)\cdot \left(\frac{m_{I}}{m_{X}} \right) \cdot c_{I}[/tex]

If we know that [tex]\Delta t_{I} = 6.35\,s[/tex], [tex]\Delta t_{X} = 4.59\,s[/tex], [tex]m_{I} = 0.515\,kg[/tex], [tex]m_{X} = 1.263\,kg[/tex] and [tex]c_{I} = 447\,\frac{J}{kg\cdot ^{\circ}C}[/tex], then the specific heat of the unknown material is:

[tex]c_{X} = \left(\frac{4.59\,s}{6.35\,s} \right)\cdot \left(\frac{0.515\,kg}{1.263\,kg} \right)\cdot \left(447\,\frac{J}{kg\cdot ^{\circ}C} \right)[/tex]

[tex]c_{X} = 131.750\,\frac{J}{kg\cdot ^{\circ}C}[/tex]

Then, the specific heat of the unknown material is 131.750 joules per kilogram-degree Celsius.

Answer:

240 meters

Explanation:

The distance traveled by the vehicle can be calculated using the following equation:

[tex] v_{f}^{2} = v_{0}^{2} + 2ax [/tex]   (1)

Where:

x: is the displacement

[tex]v_{f}[/tex]: is the final speed = 0 (reduces its velocity back to zero)                    

[tex]v_{0}[/tex]: is the initial speed = 60 m/s

a: is the acceleration = -7.5 m/s²

By solving equation (1) for x we have:

[tex] x = \frac{v_{f}^{2} - v_{0}^{2}}{2a} = \frac{0 - (60 m/s)^{2}}{2*(-7.5 m/s^{2})} = 240 m [/tex]

Therefore, the vehicle undergoes 240 meters of displacement during the deceleration period.

I hope it helps you!

Answer:

v₁ =√2gh,      v₂ = v₁ /√2

Explanation:

Let's use the concepts of energy and work to analyze each case

hill without rubbing. Energy is conserved

starting point. Highest part

         Em₀ = U = mg h

final point. Lower part

         [tex]Em_{f}[/tex] = K = ½ m v²

         Em₀ = Em_{f}

         m g h = ½ m v²

         v₁ =√2gh

rubbing hill

in this case the energy is not conserved because it is converted into work of the friction force, therefore the variation of the energy is the work of the friction

        W = Em_{f} - Em₀

they indicate half of the initial mechanical energy is lost due to friction

          W = ½ Em₀

we substitute

          - ½ Em₀ = Em_{f} - Em₀

The negative sign is because the friction work always opposes the movement

         Em_{f} = ½ Em₀

        ½ m v₂² = ½ m g h

         v₂ = √½    √2gh

         v₂ = v₁ /√2

Answer:

[tex]Q = 4.9216 * 10^{-7}C[/tex]

Explanation:

Given

[tex]E = 12596.37 N/C[/tex]

[tex]r = 0.593m[/tex]

Required

Determine the magnitude point charge (Q)

This question will be solved using [tex]the\ magnitude[/tex] of the electric field formula

[tex]E = \frac{kQ}{r^2}[/tex]

Where

[tex]k = 9 * 10^9\ Nm^2 / C^2[/tex]

Make Q the subject in [tex]E = \frac{kQ}{r^2}[/tex]

[tex]E * r^2 = kQ[/tex]

[tex]Q = \frac{E * r^2}{k}[/tex]

Substitute values for E, r and k

[tex]Q = \frac{12596.37 * 0.593^2}{9 * 10^9}[/tex]

[tex]Q = \frac{4429.50}{9 * 10^9}[/tex]

[tex]Q = \frac{492.16}{10^9}[/tex]

[tex]Q = 492.16 * 10^{-9}[/tex]

Express in standard form

[tex]Q = 4.9216 * 10^2 * 10^{-9}[/tex]

[tex]Q = 4.9216 * 10^{2-9}[/tex]

[tex]Q = 4.9216 * 10^{-7}C[/tex]

Answer:

The torsion constant for the wire is [tex]2.856\times 10^{-4}\,N\cdot m[/tex].

Explanation:

The angular frequency of the torsional pendulum ([tex]\omega[/tex]), measured in radians per second, is defined by the following expression:

[tex]\omega = \sqrt{\frac{\kappa}{I} }[/tex] (1)

Where:

[tex]\kappa[/tex] - Torsional constant, measured in newton-meters.

[tex]I[/tex] - Moment of inertia, measured in kilogram-square meters.

The angular frequency and the moment of inertia are represented by the following formulas:

[tex]\omega = \frac{2\pi}{T}[/tex] (2)

[tex]I = \frac{m\cdot L^{2}}{12}[/tex] (3)

Where:

[tex]T[/tex] - Period, measured in seconds.

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

[tex]L[/tex] - Length of the stick, measured in meters.

By (2) and (3), (1) is now expanded:

[tex]\frac{2\pi}{T} = \sqrt{\frac{12\cdot \kappa}{m\cdot L^{2}} }[/tex]

[tex]\frac{2\pi}{T} = \frac{2}{L}\cdot \sqrt{\frac{3\cdot \kappa}{m} }[/tex]

[tex]\frac{\pi\cdot L}{T} = \sqrt{\frac{3\cdot \kappa}{m} }[/tex]

[tex]\frac{\pi^{2}\cdot L^{2}}{T^{2}} = \frac{3\cdot \kappa}{m}[/tex]

[tex]\kappa = \frac{\pi^{2}\cdot m\cdot L^{2}}{3\cdot T^{2}}[/tex]

If we know that [tex]m = 5\,kg[/tex], [tex]L = 1\,m[/tex] and [tex]T = 240\,s[/tex], then the torsion constant for the wire is:

[tex]\kappa = \frac{\pi^{2}\cdot (5\,kg)\cdot (1\,m)^{2}}{3\cdot (240\,s)^{2}}[/tex]

[tex]\kappa = 2.856\times 10^{-4}\,N\cdot m[/tex]

The torsion constant for the wire is [tex]2.856\times 10^{-4}\,N\cdot m[/tex].

Answer:

Mediums in which the speed of sound is different generally have differing acoustic impedances, so that, when a sound wave strikes an interface between

Explanation:The propagation of a wave through a medium will depend on the properties of the medium. For example, waves of different frequencies may travel

Answer:

[tex]B=5.2*10^-^8T[/tex]

Explanation:

From the question we are told that

Capacitor [tex]c=1000nf[/tex]

Radius [tex]r=1cm =>0.001m[/tex]

Charge rate [tex]Q/t=52mC/s[/tex]

Distance [tex]d=20cm =0.2m[/tex]

Generally the rate of charge I is mathematically given as

[tex]I=\frac{dQ}{dt}[/tex]

[tex]I=52Cm/s[/tex]

[tex]I=52*10^-^3C[/tex]

Generally the the magnetic field intensity at distance d is mathematically given as

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

[tex]B=\frac{(4 * \pi *10^-^7)(52*10^-^3C)}{2\pi (0.2)}[/tex]

[tex]B=5.2*10^-^8T[/tex]

Answer:

a)[tex]V=\pi *r^2 * \sqrt{2gd}[/tex]

b)[tex]dh / dt = 0.2658 mm / min[/tex]

Explanation:

From the question we are told that

Diameter of hole [tex]d_h=4mm=>0.004m[/tex]

Depth of hole [tex]D=0mm=>0.001m[/tex]

Diameter of tank [tex]d_t=2mm=>0.002m[/tex]

Generally the equation for pressure is mathematically given as

[tex]Pressure P= \rho*g*d[/tex]

[tex]P= 1/2*\rho *v^2[/tex]

Where

[tex]v = \sqrt {2gd}[/tex]

[tex]V = Area*v[/tex]

[tex]V=\pi *r^2 * \sqrt{2gd}[/tex]

Generally the level at which the water level will initially drop if the water is not replenished is mathematically given by

[tex]dh / dt = (r/R)^2 *sqrt{2gd}\\dh / dt = (2/2000)^2 *sqrt(2*9.81*1) \\dh / dt = 4.429*10^-3 mm/s \\[/tex]

Therefore the level at which the water level will initially drop if the water is not replenished

[tex]dh / dt = 0.2658 mm / min[/tex]

The rate, in mm/min, at which the water level will initially drop will be 1.0625 mm/min.

Given data:

The diameter of hole is, d = 4.0 mm = 0.004 m.

The depth of hole is, h = 1.0 m.

The diameter of tank is, d' = 2.0 m.

The given problem is based on the flow rate, which is defined as the flow of liquid through a given section per unit time.

Let us first obtain the equation of pressure as,

[tex]P=\dfrac{1}{2} \times \rho \times v^{2}[/tex]

Here, v is the velocity of efflux and its value is,

[tex]v=\sqrt{2gh} \\\\v^{2}=2gh[/tex]

And the level at which the water level will initially drop if the water is not replenished is mathematically given by,

[tex]\dfrac{dH}{dt}=(r/R)^{2} \times v[/tex]

Here,

r is the radius of hole.

R is the radius of tank.

Solving as,

[tex]\dfrac{dH}{dt}=((d/2) /(d'/2))^{2} \times \sqrt{2gh} \\\\\dfrac{dH}{dt}=((0.004/2) /(2/2))^{2} \times \sqrt{2 \times 9.8 \times 1}\\\\\dfrac{dH}{dt}=1.77 \times 10^{-5} \;\rm m/s\\\\\dfrac{dH}{dt}=1.77 \times 10^{-5} \times 6 \times 10^{4} \;\rm mm/min\\\\\dfrac{dH}{dt}=1.0625 \;\rm mm/min[/tex]

Thus, we can conclude that the rate, in mm/min, at which the water level will initially drop will be 1.0625 mm/min.

Learn more about the flow rate here:

https://brainly.com/question/11816739

Answer:

The acceleration experienced by the occupants of the spaceship during launch is 282652.782 meters per square second.

Explanation:

Let suppose that spaceship is accelerated uniformly. A yard equals 0.914 meters. A feet equals 0.304 meters. If air viscosity and friction can be neglected, then acceleration ([tex]a[/tex]), measured in meters per square second, is estimated by this kinematic formula:

[tex]a = \frac{v^{2}-v_{o}^{2}}{2\cdot \Delta s }[/tex] (1)

Where:

[tex]\Delta s[/tex] - Travelled distance, measured in meters.

[tex]v_{o}[/tex], [tex]v[/tex] - Initial and final speeds of the spaceship, measured in meters.

If we know that [tex]v_{o} = 0\,\frac{m}{s}[/tex], [tex]v = 10968\,\frac{m}{s}[/tex] and [tex]\Delta s = 212.8\,m[/tex], then the acceleration experimented by the spaceship is:

[tex]a = \frac{\left(10968\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{2\cdot (212.8\,m)}[/tex]

[tex]a = 282652.782\,\frac{m}{s^{2}}[/tex]

The acceleration experienced by the occupants of the spaceship during launch is 282652.782 meters per square second.

Answer:

MINORS,    SYSTEMATIC,   STATISTICAL,  BOTOM LINE, ZERO MATCHES

Explanation:

In general the sources of error or uncertainty can be classified

* Statistics. Which are those that describe the statistical formulas, for example: average, standard deviation, absolute error, etc.

* Systematic. That they occur due to an inappropriate measurement or to an interaction between the system and the instrument that cannot be quantified, in general this error shifts the measurements towards an explicit side

* Random, so errors that sometimes occur in the measurement and sometimes not, for example temperature changes during the medical process

In this case, you are asked to complete the sentences with the appropriate word

the measured diameters were slightly ___ MINORS________ compared to what a non-contact method of measuring would provide

. This represents an ____SYSTEMATIC_______ error in ________________.

This additional source of error ________STATISTICAL________ significant.

When the caliper jaws closed, the zero mark on the sliding Vernier scale, BOTOM LINE AND THE __________ line up with the zero mark on the measuring scale.

This means the caliper ___ZERO MATCHES_____________ calibrated correctly.

Answer: 109.4 mm

Explanation: Distance is a scalar quantity and it is the measure of how much path there are between two locations. It can be calculated as the product of velocity and time:  d = vt

The separation between the two steamrollers is 105 mm or 0.105 m. They collide to each other at the middle of the separation:

location of collision = [tex]\frac{0.105}{2}[/tex] = 0.0525 m

To reach that point, both steamrollers will have spent

[tex]v=\frac{\Delta x}{t}[/tex]

[tex]t=\frac{\Delta x}{v}[/tex]

[tex]t=\frac{0.0525}{1.2}[/tex]

t = 0.04375 s

The fly is travelling with speed of 2.5 m/s. So, at t = 0.04375 s:

d = 2.5*0.04375

d = 0.109375 m

Until it is crushed, the fly will have traveled 109.4 mm.

Answer:

32000 N

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 40 m/s

Distance (s) = 10 m

Final velocity (v) = 0 m/s

Mass (m) of car = 400 Kg

Force (F) =?

Next, we shall determine the acceleration of the the car. This can be obtained as follow:

Initial velocity (u) = 40 m/s

Distance (s) = 10 m

Final velocity (v) = 0 m/s

Acceleration (a) =?

v² = u² + 2as

0² = 40² + (2 × a × 10)

0 = 1600 + 20a

Collect like terms

0 – 1600 = 20a

–1600 = 20a

Divide both side by –1600

a = –1600 / 20

a = –80 m/s²

The negative sign indicate that the car is decelerating i.e coming to rest.

Finally, we shall determine the force needed to stop the car. This can be obtained as follow:

Mass (m) of car = 400 Kg

Acceleration (a) = –80 m/s²

Force (F) =?

F = ma

F = 400 × –80

F = – 32000 N

NOTE: The negative sign indicate that the force is in opposite direction to the motion of the car.

Answer:

it can melt orcan put them past their boiling point

Explanation:

Answer:

[tex]d=8.657mm[/tex]

Explanation:

From the question we are told that

Pressurized air pressure is [tex]P_{air}=350 kPa,[/tex]

Atmospheric pressure is [tex]P_a=100 kPa[/tex]

Initial acceleration of the water rocket is [tex]a_i=0.5g.[/tex]

Acceleration of the water rocket is  [tex]a_r=0.5g[/tex]

Mass of water is [tex]M_w=0.5 kg[/tex]

Generally total mass is given mathematically given as

[tex]T_M=0.5+0.5=>1kg[/tex]

Generally the tension on the rocket is given mathematically given as

[tex]T=(P_{air}-P_a)A[/tex]

[tex]T=(350-100) \frac{\pi d^2}{4}[/tex]

T is also

[tex]T=\frac{3Mg}{2}[/tex]

Therefore

[tex]T=>(350-100) \frac{\pi d^2}{4}= \frac{3Mg}{2}[/tex]

[tex]T=>(350-100) \frac{\pi d^2}{4}= \frac{3*1*9.81}{2}[/tex]

[tex]d^2= \frac{3*1*9.81*4}{2(350-100) \pi}[/tex]

[tex]d=\sqrt{\frac{3*1*9.81*4}{2(350-100) \pi}}[/tex]

[tex]d=8.657mm[/tex]

therefore diameter of nozzle is mathematically given as

[tex]d=8.657mm[/tex]

Answer:

The order = [tex]\mathbf{\sqrt{\dfrac{h}{2 \pi m}} \sqrt[4]{\dfrac{H}{g}} \sqrt[4]{\dfrac{1}{2}} }}[/tex]

Explanation:

To miss the crack at a given distance is apparently not the same as the uncertainty that occurred in the distance while falling from the tower. However, it is believed that the uncertainties in both cases appear to be the same.

So, let's work it out together

According to Heisenberg's uncertainty principle:

[tex]\Delta s. \Delta p =\dfrac{h}{2} =\dfrac{h}{4 \pi}[/tex]

Also; if we recall from the equation of motion that:

[tex]v = u + at ---(1) \\ \\ v^2 - u^2 = 2as --- (2) \\ \\ s = ut + \dfrac{1}{2}at^2 --- (3)[/tex]

So, if u = 0 and a = g

Then;

[tex]v = gt --- (1) \\ \\ v^2 = 2gs - - - ( 2) \\ \\ s = \dfrac{1}{2}gt^2 --- (3)[/tex]

From (2)

Making (s) the subject, we have:

[tex]s = \dfrac{v^2}{2g}[/tex]

[tex]s = \dfrac{p^2}{2gm^2}[/tex]

By differentiation;

[tex]ds = d (\dfrac{p^2}{2gm^2})[/tex]

[tex]ds = \dfrac{2pdp}{2gm^2}[/tex]

[tex]\Delta \ s = \dfrac{p \Delta p}{gm^2 }[/tex]

where;

[tex]\Delta p = \dfrac{h}{4 \pi \Delta \ s}[/tex] from uncertainty principle

This implies that:

[tex]\Delta s = \dfrac{p(\dfrac{h}{4 \pi \Delta s }) }{gm^2}[/tex]

[tex]\Delta s = p(\dfrac{h}{4 \pi gm^2 }) \times \dfrac{1}{ \Delta s}}[/tex]

[tex](\Delta s)^2 = \dfrac{hmv} {4 \pi gm^2 }[/tex]

here;

v = 2gH

So;

[tex](\Delta s)^2 = \dfrac {h \sqrt{2gH} }{4 \pi gm }[/tex]

[tex]\mathbf{(\Delta s)^2 = \sqrt{\dfrac{h}{2 \pi m}} \sqrt[4]{\dfrac{H}{g}} \sqrt[4]{\dfrac{1}{2}} }[/tex]

Thus, the order = [tex]\mathbf{\sqrt{\dfrac{h}{2 \pi m}} \sqrt[4]{\dfrac{H}{g}} \sqrt[4]{\dfrac{1}{2}} }}[/tex]

Answer:

change in kinetic energy of the trolley is 53.91 J.

Explanation:

mass of the trolley, m = 5.0 kg

initial velocity of the trolley, u = 1.0 m/s

external force on the trolley, F = 25 N

time of force action, t = 0.75 s

The final velocity of the trolley at the end of 0.75 s is calculated as follows;

[tex]F = \frac{m(v-u)}{t} \\\\25 = \frac{5(v-1)}{0.75}\\\\5(v-1) = 18.75\\\\v-1 = \frac{18.75}{5} \\\\v-1 = 3.75\\\\v = 4.75 \ m/s \ in \ the \ direction \ of \ the \ applied \ force[/tex]

The change in kinetic energy of the trolley is calculated as;

Δ K.E = ¹/₂m(v² - u²)

Δ K.E = ¹/₂ x 5(4.75² - 1²)

Δ K.E = 53.91 J.

Therefore, change in kinetic energy of the trolley is 53.91 J.

Answer:

x₂ / x₁ = √2

Explanation:

To solve this exercise we can use the projectile launch ratios, let's find the time it takes for the second book to reach the ground

             y = y₀ + [tex]v_{oy}[/tex] t - ½ g t²

as the book is thrown horizontally v_{oy} = 0, when it reaches the ground its height is zero y= 0

            0 = y₀ - ½ g t²

            t = [tex]\sqrt{ \frac{2y_o}{ g} }[/tex]

            t = \sqrt{    \frac{2 \ 2h}{ g} }

with this time we calculate the horizontal distance traveled

            x = v₀ t

            x₂ = v₀ [tex]\sqrt{ \frac{4h}{g} }[/tex]

now let's calculate the time it takes him to get to the floor when he leaves from the first floor

           t =\sqrt{    \frac{2y_o}{ g} }

the horizontal distance traveled is

           x₁ = v₀ [tex]\sqrt{ \frac{2h}{g} }[/tex]

therefore the difference in distance between the two runs is

           Δx = x₂-x₁

           Δx = v₀ \sqrt{ \frac{4h}{g} } - v₀ \sqrt{ \frac{2h}{g} }

            Δx = v₀ \sqrt{ \frac{2h}{g} }    √2

            Δx =√2    x₁

the relationship between the two distances is

             x₂ / x₁ = √2

Answer:

The importance of the sediments permeability is that if it is permeable, water will flow easily through the sediment and thereby produce a very good supply of water for the well.

Explanation:

When digging a well into saturated sediments, the possibility of the sediment with either little saturation or full saturation being able to provide steady water supply will be limited by how permeable it is. Now, the importance of the sediments permeability is that if it is permeable, water will flow easily through the sediment and thereby produce a very good supply of water for the well.

Answer:

400 Ns

Explanation:

Impulse = Change in momentum

i.e I = ΔP

So that,

Impulse experienced by Max = Change in Max's momentum

Change in Max's momentum = m(v - u)

Where m is the mass, v is the velocity after collision, and u is the velocity before collision.

m = 100.0 kg, v = 2.0 m/s, u = 6.0 m/s

Change in Max's momentum = 100 x (2 -6)

                             = -400 kg m/s

The negative sign shows that the change in momentum was against his direction of motion.

Impulse experienced by Max = 400 Ns.

Thus,

Max experienced an impulse of 400 Ns as a result of the collision.

Answer:

the charge on the object is 71.043×10^-20 C and the number of electron is 4.44

Explanation:

from coulumbs law, The force that is acting over both charge can be computed as

F=( kq1q2)/r^2..............eqn(1)

Where

F=electrostatic force= 3.89 x 10-21 N

k= column constant= 9 x 10^9 Nm^2/C^2

q1 and q2= magnitude of the charges

r= distance between the charges= 1.08 x 10-3 m.

Since both charges are experiencing the same force, eqn(1) can be written as

F=( kq^2)/r^2.

We can make q subject of the formula

q= √(Fr^2)/k

= √[(3.89 x 10^-21× (1.08 x 10^-3)^2]/8.99 x 10^9

q= 71.043×10^-20 C

Hence, the charge is 71.043×10^-20 C

From quantization law, the number of electron can be computed as

N=q/e

Where

N= number of electron

q= charges

=1.6×10^-19C

N=71.043×10^-20/1.6×10^-19

=4.44

Hence, the charge on the object is 71.043×10^-20 C and the number of electron is 4.44