Two wheels initially at rest roll the same distance without slipping down identical planes. Wheel B has twice the radius, but the same mass as wheel A. All the mass is concentrated in their rims so that the rotational inertias are I = mR2. Which has more translational kinetic energy when it gets to the bottom?

Answer:

a

  [tex]PE = 2.3 *10^{-16} \ J[/tex]

b

 [tex]T = 1.1 *10^{7} \ K[/tex]

Explanation:

From the question we are told that

      The distance of separation is  [tex]d = 1.00 *10^{-12} \ m[/tex]

Generally the electric potential energy can be mathematically represented as

            [tex]PE = \frac{k * q_1 q_2 }{d}[/tex]

Given that in a nuclei the only charged particle is the proton who charge is

     [tex]p = 1.60 *10^{-19} \ C[/tex]

Hence

     [tex]q_1 = q_2 = 1.60 *10 ^{-19} \ C[/tex]

And k is the coulomb constant with values   [tex]k = 9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.N/A2[/tex]

      So we have that

       [tex]PE = \frac{9*10^9 * (1.60 *10^{-19})^2}{ 1.00*10^{-12}}[/tex]

      [tex]PE = 2.3 *10^{-16} \ J[/tex]

The relationship between the electrical potential energy and the temperature is mathematically represented as

         [tex]PE = \frac{3}{2} kT[/tex]

Here  k is  the Boltzmann's constant with value  [tex]k = 1.38*10^{-23} JK^{-1}[/tex]

   making T the subject

       [tex]T = \frac{2}{3} * \frac{PE}{k}[/tex]

substituting values

      [tex]T = \frac{2}{3} * \frac{2.30 *10^{-16}}{ 1.38 *10^{-23}}[/tex]

     [tex]T = 1.1 *10^{7} \ K[/tex]

Answer: No, we can have a displacement equal to 0 while the distance traveled is different than zero.

Explanation:

Ok, let's write the definitions:

Displacement: The displacement is equal to the difference between the final position and the initial position.

Distance traveled: Total distance that you moved.

So, for example, if at t = 0s, you are in your house, then you go to the store, and then you return to your house, we have:

The displacement is equal to zero, because the initial position is your house and the final position is also your house, so the displacement is zero.

But the distance traveled is not zero, because you went from you traveled the distance from your house to the store two times.

So no, we can have a displacement equal to zero, but a distance traveled different than zero.

Answer:

t = 0.414s

Explanation:

In order to calculate the time that the ball takes to reach home plate, you assume that the speed of the ball is constant, and you use the following formula:

[tex]t=\frac{d}{v}[/tex]         (1)

d: distance to the plate = 18.4m

v: speed of the ball = 160.0km/h

You first convert the units of the sped of the ball to appropriate units (m/s)

[tex]160.0\frac{km}{h}*\frac{1h}{3600s}*\frac{1000m}{1km}=44.44\frac{m}{s}[/tex]

Then, you replace the values of the speed v and distance s in the equation (1):

[tex]t=\frac{18.4m}{44.44m/s}=0.414s[/tex]

THe ball takes 0.414s to reach the home plate

Answer:

Differences between regular and irregular objects are:

Hope this helps...

Good luck on your assignment..

Answer:

Explanation:

Given that

The electric fields of strengths E = 187,500 V/m and

and The magnetic  fields of strengths B = 0.1250 T

The diameter d is 25.05 cm which is converted to 0.2505m

The radius is (d/2)

= 0.2505m / 2 = 0.12525m

The given formula to find the magnetic force is [tex]F_{ma}=BqV---(i)[/tex]

The given formula to find the electric force is [tex]F_{el}=qE---(ii)[/tex]

The velocity of electric field and magnetic field is said to be perpendicular

Electric field is equal to magnectic field

Equate equation (i) and equation (ii)

[tex]Bqv=qE\\\\v=\frac{E}{B}[/tex]

[tex]v=\frac{187500}{0.125} \\\\v=15\times10^5m/s[/tex]

It is said that the particles moves in semi circle, so we are going to consider using centripetal force

[tex]F_{ce}=\frac{mv^2}{r}---(iii)[/tex]

magnectic field is equal to centripetal force

Lets equate equation (i) and (iii)

[tex]Bqr=\frac{mv^2}{r} \\\\\frac{q}{m}=\frac{v}{Br} \\\\\frac{q}{m} =\frac{15\times 10^5}{0.125\times0.12525} \\\\=\frac{15\times10^5}{0.015656} \\\\=95808383.23\\\\=958.1\times10^5C/kg[/tex]

Therefore,  the particle's charge-to-mass ratio is [tex]958.1\times10^5C/kg[/tex]

b)

To identify the particle

Then 1/ 958.1 × 10⁵ C/kg

The charge to mass ratio is very close to that of a proton, which is about 1*10^8 C/kg

Therefore the particle is proton.

Answer:

Explanation:

each grid corresponding  0.1s⁻¹.

0.2grid unit = 0.2×0.1 =0.02s⁻¹

distance of the star from telescope

d = 1/p

d= 1/0.02= 50 par sec

1par sec = 3.26 light year

1 light year = 9.5×10¹²km

3.26ly=3.084×10¹³km

d= 50×3.084×10¹³=1.55×10¹⁵km

Answer:

The magnitude of the frictional force between the car and the track is 367.763 N.

Explanation:

The roller coster has an initial gravitational potential energy, which is partially dissipated by friction and final gravitational potential energy is less. According to the Principle of Energy Conservation and Work-Energy Theorem, the motion of roller coster is represented by the following expression:

[tex]U_{g,1} = U_{g,2} + W_{dis}[/tex]

Where:

[tex]U_{g,1}[/tex], [tex]U_{g,2}[/tex] - Initial and final gravitational potential energy, measured in joules.

[tex]W_{dis}[/tex] - Dissipated work due to friction, measured in joules.

Gravitational potential energy is described by the following formula:

[tex]U = m \cdot g \cdot y[/tex]

Where:

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

[tex]g[/tex] - Gravitational constant, measured in meters per square second.

[tex]y[/tex] - Height with respect to reference point, measured in meters.

In addition, dissipated work due to friction is:

[tex]W_{dis} = f \cdot \Delta s[/tex]

Where:

[tex]f[/tex] - Friction force, measured in newtons.

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

Now, the energy equation is expanded and frictional force is cleared:

[tex]m \cdot g \cdot (y_{1} - y_{2}) = f\cdot \Delta s[/tex]

[tex]f = \frac{m \cdot g \cdot (y_{1}-y_{2})}{\Delta s}[/tex]

If [tex]m = 1000\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{1} = 40\,m[/tex], [tex]y_{2} = 25\,m[/tex] and [tex]\Delta s = 400\,m[/tex], then:

[tex]f = \frac{(1000\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (40\,m-25\,m)}{400\,m}[/tex]

[tex]f = 367.763\,N[/tex]

The magnitude of the frictional force between the car and the track is 367.763 N.

Answer:

Hooking up the bulb to the battery in a series arrangement will draw the least amount of current.

Explanation:

In this case now, the bulb will serve as the load on the battery (resistance).

For the current to last longer, the least amount of energy must be drawn.

The least amount of energy will be drawn when the arrangement provides the maximum resistance possible.

Let us take the resistance of each bulb as 'R'

If we arrange the bulbs in series, then, the total resistance will be

Rt = R + R = 2R

at a EMF of V from the battery, current I through the battery will be

I = V/2R

If we arrange the bulbs in parallel, then , the total resistance will be

1/Rt = 1/R + 1/R

1/Rt = 2/R

therefore

Rt = R/2

at an EMF of V from the battery, the current I that will be drawn through the battery will be

I = 2V/R

we see that arranging the bulbs in parallel draws 4 times the current compared to arranging the bulb in series

From the above, we see that arranging the bulbs in series provides the maximum resistance, which means a lesser amount of current is drawn from the battery

Answer:

Explanation:

Hello,

Let's get the data for this question before proceeding to solve the problems.

Mass of flywheel = 40kg

Speed of flywheel = 590rpm

Diameter = 75cm , radius = diameter/ 2 = 75 / 2 = 37.5cm.

Time = 30s = 0.5 min

During the power off, the flywheel made 230 complete revolutions.

∇θ = [(ω₂ + ω₁) / 2] × t

∇θ = [(590 + ω₂) / 2] × 0.5

But ∇θ = 230 revolutions

∇θ/t = (530 + ω₂) / 2

230 / 0.5 = (530 + ω₂) / 2

Solve for ω₂

460 = 295 + 0.5ω₂

ω₂ = 330rpm

a)

ω₂ = ω₁ + αt

but α = ?

α = (ω₂ - ω₁) / t

α = (330 - 590) / 0.5

α = -260 / 0.5

α = -520rev/min

b)

ω₂ = ω₁ + αt

0 = 590 +(-520)t

520t = 590

solve for t

t = 590 / 520

t = 1.13min

60 seconds = 1min

X seconds = 1.13min

x = (60 × 1.13) / 1

x = 68seconds

∇θ = [(ω₂ + ω₁) / 2] × t

∇θ = [(590 + 0) / 2] × 1.13

∇θ = 333.35 rev/min

Answer:

Therefore, the new rotation rate of the satellite is 6.3 rev/s.

Explanation:

The expression for conservation of the angular momentum (L) is

[tex]L_{i} = L_{f} I_{i}\times\omega_{i} = I_{f}\times\omega_{f}[/tex]

Where

[tex]I_{i}\ and \ \omega_{i}[/tex] initial moment of inertia and angular velocity

[tex]I_{f}\ and \ \omega_{f}[/tex] is the final moment of inertia and angular velocity

The expression of moment of inertia of the satellite (a solid sphere) is

[tex]I_{i} = \frac{2}{5}m_{s}r^{2}[/tex]

Where [tex]m_{s}[/tex] is the satellite mass

r is the  radus of the sphere

Substititute 1900kg for m and 4.6m for r

[tex]I_{i} = \frac{2}{5}m_{s}r^{2}\\\\ = \frac{2}{5}\times1900 kg\times (4.6 m)^{2} \\\\= 1.61 \cdot 10^{4} kgm^{2}[/tex]

The final moment of inertia of the satellite about the centre of mass

[tex]I_{f} = I_{i} + 2\timesI_{x} \\\\= 1.61 \cdot 10^{4} kgm^{2} + 2\times\frac{1}{3}m_{x}l^{2}[/tex]

Where [tex]m_{x}[/tex] is the antenna's mass and

I is the length of the antenna

[tex]I_{f} = 1.61 \cdot 10^{4} kgm^{2} + 2\times\frac{1}{3}150.0 kg\times(6.6 m)^{2} \\\\= 2.05 \cdot 10^{4} kgm^{2}[/tex]

So, the Final rotation rate of the satellite is:

[tex]I_{i}\times\omega_{i} = I_{f}\times\omega_{f} \\\\\omega_{f} = \frac{I_{i}\times\omega_{i}}{I_{f}} \\\\= \frac{1.61 \cdot 10^{4} kgm^{2}\times8.0 \frac{rev}{s}}{2.05 \cdot 10^{4} kgm^{2}} \\\\= 6.3 rev/s[/tex]

Therefore, the new rotation rate of the satellite is 6.3 rev/s.

Complete Question:

The elastic energy stored in your tendons can contribute up to 35 % of your energy needs when running. Sports scientists have studied the change in length of the knee extensor tendon in sprinters and nonathletes. They find (on average) that the sprinters' tendons stretch 43 mm , while nonathletes' stretch only 32 mm . The spring constant for the tendon is the same for both groups, [tex]31 {\rm {N}/{mm}}[/tex]. What is the difference in maximum stored energy between the sprinters and the nonathlethes?

Answer:

[tex]\triangle E = 12.79 J[/tex]

Explanation:

Sprinters' tendons stretch, [tex]x_s = 43 mm = 0.043 m[/tex]

Non athletes' stretch, [tex]x_n = 32 mm = 0.032 m[/tex]

Spring constant for the two groups, k = 31 N/mm = 3100 N/m

Maximum Energy stored in the sprinter, [tex]E_s = 0.5kx_s^2[/tex]

Maximum energy stored in the non athletes, [tex]E_m = 0.5kx_n^2[/tex]

Difference in maximum stored energy between the sprinters and the non-athlethes:

[tex]\triangle E = E_s - E_n = 0.5k(x_s^2 - x_n^2)\\\triangle E = 0.5*3100* (0.043^2 - 0.032^2)\\\triangle E = 0.5*31000*0.000825\\\triangle E = 12.79 J[/tex]

Answer:

The necessary separation between  the two parallel plates is 0.104 mm

Explanation:

Given;

length of each side of the square plate, L = 6.5 cm = 0.065 m

charge on each plate, Q = 12.5 nC

potential difference across the plates, V = 34.8 V

Potential difference across parallel plates is given as;

[tex]V = \frac{Qd}{L^2 \epsilon_o} \\\\d = \frac{V L^2 \epsilon_o}{Q}[/tex]

Where;

d is the separation or distance between the two parallel plates;

[tex]d = \frac{VL^2 \epsilon_o}{Q} \\\\d = \frac{34.8*(0.065)^2 *8.854*10^{-12}}{12.5*10^{-9}} \\\\d = 0.000104 \ m\\\\d = 0.104 \ mm[/tex]

Therefore, the necessary separation between  the two parallel plates is 0.104 mm

Answer:

E = (1 / 4π ε₀ )  q r / R³

Explanation:

Thomson's stable model that the negative charge is mobile within the atom and the positive charge is uniformly distributed, to calculate the force we can use Coulomb's law

       F = K q₁ q₂ / r²

we used law Gauss

Ф = ∫ E .dA = q_{int} /ε₀

E 4π r² = q_{int} /ε₀  

E = q_{int} / 4π ε₀ r²

we replace the charge inside  

E = (1 / 4π ε₀ r²) ρ 4/3 π r³  

E = ρ r / 3 ε₀

the density for the entire atom is  

ρ = Q / V  

V = 4/3 π R³  

we substitute  

E = (r / 3ε₀ ) Q 3/4π R³  

E = (1 / 4π ε₀ ) q r / R³

Answer: Acceleration of the car at time = 10 sec is 108 [tex]m/s^{2}[/tex] and velocity of the car at time t = 10 sec is 918.34 m/s.

Explanation:

The expression used will be as follows.

[tex]M\frac{dv}{dt} = u\frac{dM}{dt}[/tex]

[tex]\int_{t_{o}}^{t_{f}} \frac{dv}{dt} dt = u\int_{t_{o}}^{t_{f}} \frac{1}{M} \frac{dM}{dt} dt[/tex]

       = [tex]u\int_{M_{o}}^{M_{f}} \frac{dM}{M}[/tex]

[tex]v_{f} - v_{o} = u ln \frac{M_{f}}{M_{o}}[/tex]

[tex]v_{o} = 0[/tex]

As, [tex]v_{f} = u ln (\frac{M_{f}}{M_{o}})[/tex]

u = -2900 m/s

[tex]M_{f} = M_{o} - m \times t_{f}[/tex]

           = [tex]2500 kg + 1000 kg - 95 kg \times t_{f}s[/tex]

           = [tex](3500 - 95t_{f})s[/tex]

[tex]v_{f} = -2900 ln(\frac{3500 - 95 t_{f}}{3500}) m/s[/tex]

Also, we know that

     a = [tex]\frac{dv_{f}}{dt_{f}} = \frac{u}{M} \frac{dM}{dt}[/tex]

        = [tex]\frac{u}{3500 - 95 t} \times (-95) m/s^{2}[/tex]

        = [tex]\frac{95 \times 2900}{3500 - 95t} m/s^{2}[/tex]

At t = 10 sec,

[tex]v_{f}[/tex] = 918.34 m/s

and,   a = 108 [tex]m/s^{2}[/tex]

Answer:

497.6 N

Explanation:

From the question,

The net force on the skydiver = weight of the skydiver- the resistive force of air

F' = W-F...................... Equation 1

Where W = weight of the skydiver, F = resistive force of air.

But,

W = mg................ Equation 2

Where m = mass of the skydiver, g = acceleration due to gravity.

Substitute equation 2 into equation 1

F' = mg-F............ Equation 3

Given: m = 87 kg, F = 355 N, g = 9.8 m/s²

Substitute these values into equation 3

F' = 87(9.8)-355

F' = 852.6-355

F' = 487.6 N

Answer:

498 N down

Explanation:

Explanation:

F = ma, and a = Δv / Δt.

F = m Δv / Δt

Given: m = 60 kg and Δv = -30 m/s.

a) Δt = 5.0 s

F = (60 kg) (-30 m/s) / (5.0 s)

F = -360 N

b) Δt = 0.50 s

F = (60 kg) (-30 m/s) / (0.50 s)

F = -3600 N

c) Δt = 0.05 s

F = (60 kg) (-30 m/s) / (0.05 s)

F = -36000 N

360N, 3600N and 36000N forces are required to stop a 60 kg person traveling at 30 m/s during a time of a)5.0 seconds, b) 0.50 seconds, c)0.05 seconds respectively.

To find the force, we need to know about the mathematical formulation of force.

Mathematically, it is written as

F= m×a= m×(∆V/∆t)

Here, initially the velocity of the person is 30m/s. But after applying the force, he came to rest. So his final velocity is 0 m/s. ∆V= 30m/s

When ∆t=5 seconds, F= 60×(30/5)=360N

When ∆t=0.5 seconds, F= 60×(30/0.5)=3600N

When ∆t=0.05 seconds, F= 60×(30/0.05)=36000N

Thus, we can conclude that 360N, 3600N and 36000N forces are required to stop a 60 kg person traveling at 30 m/s during a time of a)5.0 seconds, b) 0.50 seconds, c)0.05 seconds respectively.

Learn more about force here:

brainly.com/question/12785175

#SPJ2

Answer:

-17.8 V

Explanation:

The induced emf in a coil is given as:

[tex]E = \frac{-NdB\pi r^2}{dt}[/tex]

where N = number of loops

dB = change in magnetic field

r = radius of coil

dt = elapsed time

From the question:

N = 50

dB = final magnetic field - initial magnetic field

dB = 0.35 - 0.10 = 0.25 T

r = 3 cm

dt = 2 ms = 0.002 secs

Therefore, the induced emf is:

[tex]E = \frac{-50 * 0.25 * \pi * 0.03^2}{0.002} \\E = -17.8 V[/tex]

Note: The negative sign implies that the EMf acts in an opposite direction to the change in magnetic flux.

There is only one possible state: constant uniform motion. That means constant speed in a straight line.

(If the constant speed happens to be zero, this description also covers the case where the object isn't moving. That special case is called "at rest".)

Answer:

Hope this helps

Answer:

The final angular speed of the merry-go-round is [tex]3.118\,\frac{rad}{s}[/tex] [tex]\left(0.496\,\frac{rev}{s} \right)[/tex].

Explanation:

Given the absence of external forces, the final angular speed of the merry-go-round can be determined with the resource of the Principle of Angular Momentum Conservation, which is described in this case as:

[tex]I_{g, m} \cdot \omega_{o,m} + I_{g, p}\cdot \omega_{o,p} = (I_{g,m} + I_{g, p})\cdot \omega_{f}[/tex]

Where:

[tex]I_{g,m}[/tex] - Moment of inertia of the merry-go-round with respect to its axis of rotation, measured in [tex]kg\cdot m^{2}[/tex].

[tex]I_{g,p}[/tex] - Moment of inertia of the person with respect to the axis of rotation of the merry-go-round, measured in [tex]kg\cdot m^{2}[/tex].

[tex]\omega_{o, m}[/tex] - Initial angular speed of the merry-go-round with respect to its axis of rotation, measured in radians per second.

[tex]\omega_{o,p}[/tex] - Initial angular speed of the merry-go-round with respect to the axis of rotation of the merry-go-round, measured in radians per second.

[tex]\omega_{f}[/tex] - Final angular speed of the merry-go-round-person system, measured in radians per second.

The final angular speed is cleared:

[tex]\omega_{f} = \frac{I_{g,m}\cdot \omega_{o,m}+I_{g,p}\cdot \omega_{o,p}}{I_{g,m}+I_{g,p}}[/tex]

Merry-go-round is modelled as uniform disk-like rigid body, whereas the person can be modelled as a particle. The expressions for their moments of inertia are, respectively:

Merry-go-round

[tex]I_{g,m} = \frac{1}{2}\cdot M \cdot R^{2}[/tex]

Where:

[tex]M[/tex] - The mass of the merry-go-round, measured in kilograms.

[tex]R[/tex] - Radius of the merry-go-round, measured in meters.

Person

[tex]I_{g,p} = m\cdot r^{2}[/tex]

Where:

[tex]m[/tex] - The mass of the person, measured in kilograms.

[tex]r[/tex] - Distance of the person with respect to the axis of rotation of the merry-go-round, measured in meters.

If [tex]M = 185\,kg[/tex], [tex]m = 63.4\,kg[/tex], [tex]R = r = 2.83\,m[/tex], the moments of inertia are, respectively:

[tex]I_{g,m} = \frac{1}{2}\cdot (185\,kg)\cdot (2.83\,m)^{2}[/tex]

[tex]I_{g,m} = 740.823\,kg\cdot m^{2}[/tex]

[tex]I_{g,p} = (63.4\,kg)\cdot (2.83\,m)^{2}[/tex]

[tex]I_{g,p} = 507.764\,kg\cdot m^{2}[/tex]

The angular speed experimented by the person with respect to the axis of rotation of the merry-go-round is:

[tex]\omega_{o,p} = \frac{v_{p}}{r}[/tex]

[tex]\omega_{o,p} = \frac{3.51\,\frac{m}{s} }{2.83\,m}[/tex]

[tex]\omega_{o,p} = 1.240\,\frac{rad}{s}[/tex]

Given that [tex]I_{g,m} = 740.823\,kg\cdot m^{2}[/tex], [tex]I_{g,p} = 507.764\,kg\cdot m^{2}[/tex], [tex]\omega_{o,m} = 4.405\,\frac{rad}{s}[/tex] and [tex]\omega_{o,p} = 1.240\,\frac{rad}{s}[/tex], the final angular speed of the merry-go-round is:

[tex]\omega_{f} = \frac{(740.823\,kg\cdot m^{2})\cdot \left(4.405\,\frac{rad}{s} \right)+(507.764\,kg\cdot m^{2})\cdot \left(1.240\,\frac{rad}{s} \right)}{740.823\,kg\cdot m^{2}+507.764\,kg\cdot m^{2}}[/tex]

[tex]\omega_{f} = 3.118\,\frac{rad}{s}[/tex]

[tex]\omega_{f} = 0.496\,\frac{rad}{s}[/tex]

The final angular speed of the merry-go-round is [tex]3.118\,\frac{rad}{s}[/tex] [tex]\left(0.496\,\frac{rev}{s} \right)[/tex].

Answer:

(a) σ = 3.41*10⁻7C/m^2

(b) E = 38,530.1 N/C

Explanation:

(a) In order to calculate the resulting surface charge density, you use the following formula:

[tex]\sigma=\frac{Q}{S}[/tex]     (1)

σ: surface charge density

Q: charge of the satellite = 3.1 µC = 3.1*10^-6C

S: surface area of the satellite

The satellite has a spherical form, then, the area of the surface is given by:

[tex]S=4\pi r^2[/tex]     (2)

r: radius of the satellite = d/2 = 1.7m/2 = 0.85m

You replace the equation (2) into the equation (1) and solve for the surface charge density:

[tex]\sigma=\frac{3.1*10^{-6}C}{4\pi (0.85m)^2}=3.41*10^{-7}\frac{C}{m^2}[/tex]

The surface charge density acquired by the satellite on one orbit is 3.41*10⁻7C/m^2

(b) The electric field just outside the surface is calculate d by using the following formula:

[tex]E=k\frac{Q}{R^2}[/tex]      (3)

k: Coulomb's constant = 8.98*10^9 Nm^2/C^2

R: radius of the satellite = 0.85m

[tex]E=(8.98*10^9Nm^2/C^2)\frac{3.1*10^{-6}C}{(0.85m)^2}=38530.1\frac{N}{C}[/tex]

The magnitude of the electric field just outside the sphere is 38,530.1 N/C

Answer:

The found acceleration in terms of h and t is:

[tex]a=\frac{h}{5(t_1)^2}[/tex]

Explanation:

(The complete question is given in the attached picture. We need to find the acceleration in terms of h and t in this question)

We are given 3 stages of movement of elevator. We'll first model them each of the stage one by one to find the height covered in each stage. After that we'll find the total height covered by adding heights covered in each stage, and equate it to Total height h. From that we can find the formula for acceleration.

Constant acceleration, starts from rest.

Distance = [tex]y = \frac{1}{2}a(t_1)^2[/tex]

Velocity = [tex]v_1=at_1[/tex]

Constant velocity where

Velocity = [tex]v_o=v_1=at_1[/tex]

Distance =

Constant deceleration where

Velocity = [tex]v_0=v_1=at_1[/tex]

Distance =

[tex]y_3=v_1t_3-\frac{1}{2}a(t_3)^2\\\text{Where}~t_3=t_1\\y_3=v_1t_1-\frac{1}{2}a(t_1)^2\\\text{Where}~ v_1t_1=a(t_1)^2\\y_3=a(t_1)^2-\frac{1}{2}a(t_1)^2\\\text{Subtracting both terms:}\\y_3=\frac{1}{2}a(t_1)^2[/tex]

Total height = y₁ + y₂ + y₃

Total height = [tex]\frac{1}{2}a(t_1)^2+4a(t_1)^2+\frac{1}{2}a(t_1)^2 = 5a(t_1)^2[/tex]

Find acceleration by rearranging the found equation of total height.

Total Height = h

h = 5a(t₁)²

[tex]a=\frac{h}{5(t_1)^2}[/tex]

Answer:

force on larger piston = [tex]\frac{fA}{a}[/tex]

Explanation:

we label the pistons as piston A and piston B

small piston A:

area = a

force = f

large piston B:

area = A

force  = ?

Pascal's law of pressure state that the pressure delivered to a liquid is transmitted undiminished to every portion of the fluid.

we know that pressure = force ÷ area

pressure of piston A = [tex]\frac{f}{a}[/tex]

pressure of piston B = [tex]\frac{(force on piston B)}{A}[/tex]

obeying Pascal's law, the system pressures must be equal. Therefore

[tex]\frac{f}{a} = \frac{(force on piston B)}{A}[/tex]

force on large piston (B) = F = [tex]\frac{fA}{a}[/tex]

Answer:

The height of the building is 158.140 meters.

Explanation:

A barometer is system that helps measuring atmospheric pressure. Manometric pressure is the difference between total and atmospheric pressures. Manometric pressure difference is directly proportional to fluid density and height difference. That is:

[tex]\Delta P \propto \rho \cdot \Delta h[/tex]

[tex]\Delta P = k \cdot \rho \cdot \Delta h[/tex]

Where:

[tex]\Delta P[/tex] - Manometric pressure difference, measured in kilopascals.

[tex]\rho[/tex] - Fluid density, measured in kilograms per cubic meter.

[tex]\Delta h[/tex] - Height difference, measured in meters.

Now, an equivalent height difference with a different fluid can be found by eliminating manometric pressure and proportionality constant:

[tex]\rho_{air} \cdot \Delta h_{air} = \rho_{Hg} \cdot \Delta h_{Hg}[/tex]

[tex]\Delta h_{air} = \frac{\rho_{Hg}}{\rho_{air}} \cdot \Delta h_{Hg}[/tex]

Where:

[tex]\Delta h_{air}[/tex] - Height difference of the air column, measured in meters.

[tex]\Delta h_{Hg}[/tex] - Height difference of the mercury column, measured in meters.

[tex]\rho_{air}[/tex] - Density of air, measured in kilograms per cubic meter.

[tex]\rho_{Hg}[/tex] - Density of mercury, measured in kilograms per cubic meter.

If [tex]\Delta h_{Hg} = 0.015\,m[/tex], [tex]\rho_{air} = 1.29\,\frac{kg}{m^{3}}[/tex] and [tex]\rho_{Hg} = 13600\,\frac{kg}{m^{3}}[/tex], the height difference of the air column is:

[tex]\Delta h_{air} = \frac{13600\,\frac{kg}{m^{3}} }{1.29\,\frac{kg}{m^{3}} }\times (0.015\,m)[/tex]

[tex]\Delta h_{air} = 158.140\,m[/tex]

The height of the building is 158.140 meters.

158.13m

Force exerted over a unit area is called Pressure. Also, in a given column of air, the pressure(P) is given as the product of the density(ρ) of the air, the height(h) of the column of air and the acceleration due to gravity(g). i.e

P = ρhg

Let;

Pressure measured at the roof top =  ([tex]P_{R}[/tex])

Pressure measured at the ground level =  ([tex]P_{G}[/tex])

Pressure at the ground level = Pressure at the roof + Pressure at the column height of air.

[tex]P_{G}[/tex] = [tex]P_{R}[/tex] + P               ---------------(i)

(a) P = ρhg             -----------(***)

But;

ρ = density of air = 1.29kg/m³  

h = height of column of air = height of building

g = acceleration due to gravity = 10m/s²

Substitute these values into equation (***)

P = 1.29 x h x 10

P = 12.9h Pa

(b) [tex]P_{G}[/tex] =  ρ[tex]_{mercury}[/tex] x h[tex]_{(mercury)_{ground} }[/tex] x g ------------(*)

But;

ρ[tex]_{mercury}[/tex] = density of mercury = 13600kg/m³  

h[tex]_{(mercury)_{ground} }[/tex] = height of mercury on the ground = 760.0mm = 0.76m

g = acceleration due to gravity = 10m/s²

Substitute these values into equation (*)

[tex]P_{G}[/tex] =  13600 x 0.76 x 10

[tex]P_{G}[/tex] = 103360 Pa

(c) [tex]P_{R}[/tex] = ρ[tex]_{mercury}[/tex] x h[tex]_{(mercury)_{roof} }[/tex] x g       --------------(**)

But;

ρ[tex]_{mercury}[/tex] = density of mercury = 13600kg/m³  

h[tex]_{(mercury)_{roof} }[/tex] = height of mercury on the roof = 745.0mm = 0.745m

g = acceleration due to gravity = 10m/s²

Substitute these values into equation (**)

[tex]P_{R}[/tex]  =  13600 x 0.745 x 10

[tex]P_{R}[/tex]  = 101320 Pa

(d) Now that we know the values of P, [tex]P_{G}[/tex] and [tex]P_{R}[/tex] , let's substitute them into equation (i) as follows;

[tex]P_{G}[/tex] = [tex]P_{R}[/tex] + P  

103360 = 101320 + 12.9h

Solve for h;

12.9h = 103360 - 101320

12.9h = 2040

h = [tex]\frac{2040}{12.9}[/tex]

h = 158.13m

Therefore, the height of the building is 158.13m

Answer:

a.     F = 2.32*10^-18 N

b.     The force F is 2.59*10^11 times the weight of the electron

Explanation:

a. In order to calculate the magnitude of the force exerted on the electron you first calculate the acceleration of the electron, by using the following formula:

[tex]v^2=v_o^2+2ax[/tex]         (1)

v: final speed of the electron = 6.60*10^5 m/s

vo: initial speed of the electron = 4.00*10^5 m/s

a: acceleration of the electron = ?

x: distance traveled by the electron = 5.40cm = 0.054m

you solve the equation (2) for a and replace the values of the parameters:

[tex]a=\frac{v^2-v_o^2}{2x}=\frac{(6.60*10^5m/s)^2-(4.00*10^5m/s)^2}{2(0.054m)}\\\\a=2.55*10^{12}\frac{m}{s^2}[/tex]

Next, you use the second Newton law to calculate the force:

[tex]F=ma[/tex]

m: mass of the electron = 9.11*10^-31kg

[tex]F=(9.11*10^{-31}kg)(2.55*10^{12}m/s^2)=2.32*10^{-18}N[/tex]

The magnitude of the force exerted on the electron is 2.32*10^-18 N

b. The weight of the electron is given by:

[tex]F_g=mg=(9.11*10^{-31}kg)(9.8m/s^2)=8.92*10^{-30}N[/tex]

The quotient between the weight of the electron and the force F is:

[tex]\frac{F}{F_g}=\frac{2.32*10^{-18}N}{8.92*10^{-30}N}=2.59*10^{11}[/tex]

The force F is 2.59*10^11 times the weight of the electron

Given that,

Sun exposure = 30%, 45%, 60%, 75%, 90%

Stem mass (g) = 275, 415, 563, 815, 954

Stem volume (ml) = 1100, 1215, 1425, 1610, 1742

(a). We need to convert the mass measurements to kilograms (kg) and the volume measurements to cubic meters

Using conversion of mass

[tex]1\ g=0.001\ kg[/tex]

Conservation of volume

[tex]1\ Lt=0.001\ m^3[/tex]

[tex]1\ mL=1\times10^{-6}\ m^3[/tex]

So, mass in kg

Stem mass (kg) = 0.275, 0.415, 0.563, 0.815, 0.954

Volume in m³,

Stem volume (m³) = 0.0011, 0.001215, 0.001425, 0.001610, 0.001742

(b). We need to calculate the density of the samples

Using formula of density

[tex]\rho=\dfrac{m}{V}[/tex]

Where, m = mass

V = volume

If the m = 0.275 kg and V = 0.0011 m³

Put the value into the formula

[tex]\rho=\dfrac{0.275}{0.0011}[/tex]

[tex]\rho=250\ kg/m^3[/tex]

If the m = 0.415 kg and V = 0.001215 m³

Put the value into the formula

[tex]\rho=\dfrac{0.415}{0.001215}[/tex]

[tex]\rho=341.56\ kg/m^3[/tex]

[tex]\rho=342\ kg/m^3[/tex]

If the m = 0.563 kg and V = 0.001425 m³

Put the value into the formula

[tex]\rho=\dfrac{0.563}{0.001425}[/tex]

[tex]\rho=395.08\ kg/m^3[/tex]

If the m = 0.815 kg and V = 0.001610 m³

Put the value into the formula

[tex]\rho=\dfrac{0.815}{0.001610}[/tex]

[tex]\rho=506.21\ kg/m^3[/tex]

If the m = 0.954 kg and V = 0.001742 m³

Put the value into the formula

[tex]\rho=\dfrac{0.954}{0.001742}[/tex]

[tex]\rho=547.6\ kg/m^3[/tex]

[tex]\rho=548\ kg/m^3[/tex]

(c). We need to convert the density values to scientific notation

In scientific notation

The densities are

[tex]\rho\ (kg/m^3)= 2.50\times10^{2}, 3.42\times10^{2}, 3.95\times10^{2}, 5.06\times10^{2}, 5.48\times10^{2}[/tex]

Hence, This is required solution.

Answer:

c. The astronaut does not need to worry: the charge will remain on the outside surface.

Explanation:

The astronaut need not worry because according to Gauss's law of electrostatic, a hollow charged surface will have a net zero charge on the inside. This is the case of a Gauss surface, and all the charges stay on the surface of the metal chamber. This same principle explains why passengers are safe from electrostatic charges, in an enclosed aircraft, high up in the atmosphere; all the charges stay on the surface of the aircraft.

Answer:

The total internal energy change for the entire process is  -0.94 kJ

Explanation:

Process A to B is an isothermal process, therefore, [tex]u_A[/tex] - [tex]u_B[/tex] = 0

Process B to C

P = -mV + C

When P = 12, V = 0.12

When P = 4, V = 0.135

Therefore, we have;

12 = -m·0.12 + C

4 = -m·0.135 + C

Solving gives

m = 533.33

C = 76

[tex]T = \dfrac{1}{nR} \times (-533.33 \times V^2 + 76 \times V)[/tex]

p₂ = p₁V₁/V₂ = 12*0.1/0.12 = 10 kPa

The work done = 0.5*(0.135 - 0.12)*(4 - 10.0) = -0.045 kJ = -45 J

For heat supplied

Assuming an approximate polytropic process, we have;

Work done = (p₃×v₃ - p₂×v₂)/(n - 1)

Which gives;

-45 = (4*0.135 - 10*0.12)/(n -1)

∴ n -1 = (4*0.135 - 10*0.12)/-45 =   14.67

n = 15.67

Q = W×(n - γ)/(γ - 1)

Q = -45*(15.67 - 1.4)/(1.4 - 1) = -1,605.375 J

u₃ - u₂ = Q + W = -1,605.375 J - 45 J = -1650 J = -1.65 kJ

For the constant pressure process D to C, we have;

[tex]Q = c_p \times \dfrac{p}{R} \times (V_4 -V_3) = \dfrac{5}{2} \times p \times (V_4 -V_3)[/tex]

Q₄₋₃ = (0.1 - 0.135) * 4*5/2 = -0.35 kJ

W₄₋₃ = 4*(0.1 - 0.135) = -0.14 kJ

u₄ - u₃ = Q₄₋₃ + W₄₋₃ = -0.14 kJ + -0.35 kJ = -0.49 kJ

For the process D to A, we have a constant volume process

[tex]Q_{1-4} = \dfrac{c_v}{R} \times V \times (p_1 - p_4) = \dfrac{3}{2} \times 0.1 \times (12 - 4) = 1.2 \ kJ[/tex]

W₁₋₄ = 0 for constant volume process, therefore, u₁ - u₄ = 1.2 kJ

The total internal energy change Δ[tex]u_{process}[/tex] for the entire process is therefore;

Δ[tex]u_{process}[/tex] = u₂ - u₁ + u₃ - u₂ + u₄ - u₃ + u₁ - u₄ = 0  - 1.65 - 0.49 + 1.2 = -0.94 kJ.

Answer:

20 J

Explanation:

From the question, since there is a lost in kinetic energy, Then the collision is an inelastic collision.

m'u'+mu = V(m+m')........... Equation 1

Where m' = mass of the moving object, m = mass of the stick, u' = initial velocity of the moving object, initial velocity of the stick, V = common velocity after collision.

make V the subject of the equation above

V = (m'u'+mu)/(m+m')............. Equation 2

Given: m' = 2 kg, m = 8 kg, u' = 5 m/s, u = 0 m/s (at rest).

Substitute into equation 2

V = [(2×5)+(8×0)]/(2+8)

V = 10/10

V = 1 m/s.

Lost in kinetic energy = Total kinetic energy before collision- total kinetic energy after collision

Total kinetic energy before collision = 1/2(2)(5²) = 25 J

Total kinetic energy after collision = 1/2(2)(1²) +1/2(8)(1²) = 1+4 = 5 J

Lost in kinetic energy = 25-5 = 20 J

The collision is inelastic collision. As a result of collision the kinetic energy lost by the given system is 20 J.

Since there is a lost in kinetic energy, the collision is inelastic collision.  

m'u'+mu = V(m+m')

[tex]\bold {V =\dfrac { (m'u'+mu)}{(m+m')} }[/tex]  

Where

m' = mass of the moving object = 2 kg

m = mass of the stick = 8 kg,

u' = initial velocity of the moving object = 5 m/s

V = common velocity after collision= ?    

u = 0 m/s (at rest).

put the values in the formula,  

[tex]\bold {V = \dfrac {(2\times 5)+(8\times 0)}{(2+8)}}\\\\\bold {V = \dfrac {10}{10}}\\\\\bold {V = 1\\ m/s.}[/tex]

  kinetic energy before collision

[tex]\bold { = \dfrac 1{2} (2)(5^2) = 25 J}[/tex]  

kinetic energy after collision

[tex]\bold { = \dfrac 12(2)(1^2) + \dfrac 12(8)(1^2) = 5\ J}[/tex]  

Lost in kinetic energy = 25-5 = 20 J

Therefore, As a result of collision the kinetic energy lost by the given system is 20 J.

To know more about Kinetic energy,

https://brainly.com/question/12669551

Answer:

Φ = 361872 N.m^2 / C

Explanation:

Given:-

- The area of the two plates, [tex]A_p = 180 cm^2[/tex]

- The charge on each plate, [tex]q = 17 * 10^-^6 C[/tex]

- Permittivity of free space, [tex]e_o = 8.85 * 10^-^1^2 \frac{C^2}{N.m^2}[/tex]

- The radius for the flux region, [tex]r = 3.3 cm[/tex]

- The angle between normal to region and perpendicular to plates, θ = 4°

Find:-

Find the flux (in N · m2/C) through a circle of radius 3.3 cm between the plates.

Solution:-

- First we will determine the area of the region ( Ar ) by using the formula for the area of a circle as follows. The region has a radius of r = 3.3 cm:

                             [tex]A_r = \pi *r^2\\\\A_r = \pi *(0.033)^2\\\\A_r = 0.00342 m^2[/tex]

- The charge density ( σ ) would be considered to be uniform for both plates. It is expressed as the ratio of the charge ( q ) on each plate and its area ( A_p ):

                           σ = [tex]\frac{q}{A_p} = \frac{17*10^-^6}{0.018} \\[/tex]

                           σ = 0.00094 C / m^2

- We will assume the electric field due to the positive charged plate ( E+ ) / negative charged plate ( E- ) to be equivalent to the electric field ( E ) of an infinitely large charged plate with uniform charge density.

                         [tex]E+ = E- = \frac{sigma}{2*e_o} \\\\[/tex]

- The electric field experienced by a region between two infinitely long charged plates with uniform charge density is the resultant effect of both plates. So from the principle of super-position we have the following net uniform electric field ( E_net ) between the two plates:

                        [tex]E_n_e_t = (E+) + ( E-)\\\\E_n_e_t = \frac{0.00094}{8.85*10^-^1^2} \\\\E_n_e_t = 106214689.26553 \frac{N}{C} \\[/tex]

- From the Gauss-Law the flux ( Φ ) through a region under uniform electric field ( E_net ) at an angle of ( θ ) is:

                        Φ = E_net * Ar * cos ( θ )

                        Φ = (106214689.26553) * (0.00342) * cos ( 5 )

                        Φ = 361872 N.m^2 / C

Answer:

1.35 kJ  

Explanation:

KE = ½mv² = ½ × 0.030  kg × (300 m·s⁻¹)² = 1350 J = 1.35 kJ

Given:-

To Find: Kinetic energy of the bullet.

We know,

Eₖ = ½mv²

where,

thus,

Eₖ = ½(30 g)(300 m/s)²

= (15 g)(90000 m²/s²)

= 1350000 g m²/s²

= 1350 kg m²/s²

= 1350 J

= 1.35 kJ (Ans.)