how many grams of nitrogen gas are needed to produce 34 g of ammonia

Answer:

Choice A: approximately [tex]0.015\; \rm m^2[/tex], assuming that the two pistons are connected via some confined liquid to form a simple machine.

Explanation:

Assume that the two pistons are connected via some liquid that is confined. Pressure from the first piston:

[tex]\displaystyle P_1 = \frac{F_1}{A_1} = \frac{6.600\times 10^3\; \rm N}{1.0\times 10^{-2}\; \rm m^{2}} = 6.6\times 10^{5}\; \rm N \cdot m^{-2}[/tex].

By Pascal's Principle, because the first piston exerted a pressure of [tex]6.6\times 10^{5}\; \rm N \cdot m^{-2}[/tex] on the liquid, the liquid will now exert the same amount of pressure on the walls of the container.

Assume that the second piston is part of that wall. The pressure on the second piston will also be [tex]6.6\times 10^{5}\; \rm N \cdot m^{-2}[/tex]. In other words:

[tex]P_2 = P_1 = 6.6\times 10^{5}\; \rm N \cdot m^{-2}[/tex].

To achieve a force of [tex]9.900 \times 10^3\; \rm N[/tex], the surface area of the second piston should be:

[tex]\displaystyle A_2 = \frac{F_2}{P_2} = \frac{9.900\times 10^{3}\; \rm N}{6.6\times 10^5\; \rm N \cdot m^{-2}} \approx 0.015\; \rm m^{2}[/tex].

Answer: 0.3 m

So basically we want to think of this situation from the perspective of conservation of energy. As in, we can derive the velocity of both the swing and Stewie at the bottom of the swing via:

mgh = 0.5m*v^2

v = (2*g*h)^0.5 = (2*9.8*0.5)^0.5 = 3.13 m/s

This represents the velocity of Stewie and the swing at the bottom of the swing's path. Now, we will think of this from the perspective of conservation of momentum. Basically, the collective momentum of Stewie and the swing is equal to the sum of their subsequent momenta:

(5+2)*3.13 = 5*2+5*v(swing); v(swing) represents the velocity of the swing following Stewie jumping.

7*3.13 = 10 + 5v

v = 2.4 m/s

Now, we return to conservation of energy and find that the kinetic energy of the swing following Stewie jumping is equivalent to its final gravitational potential energy.

Based on the easily derivable formula: h=v^2/2g, we find that h = 2.4^2/19.6,

which is: 0.3 m.

If g is supposed to be 10 or you need a different degree of precision, then you can use my method. Hope this helps.

Answer:

no, because if he was pushing the box with constant force the box would have to move with constant speed

Explanation:

Complete Question

The diagram for this question is shown on the first uploaded image

Answer:

The magnitude of force is  [tex]F_R} = 5.33 \ m N[/tex]

Explanation:

 From the question we are told that

          The current in wire A , B and C are   [tex]I _a = I_b =I_c = I= 20 A[/tex]

          The distance is [tex]R = 15.0mm = \frac{15}{1000} = 0.015m[/tex]

           The length of wire C is [tex]L_c = 200cm = \frac{200}{100} = 2 m[/tex]

Generally the force exerted per unit length  that is acting in between two current carrying conductors can be mathematically represented as

             [tex]\frac{F}{L} = \frac{\mu_o }{2 \pi} \cdot \frac{I_1 I_2 }{R}[/tex]

   Where [tex]\mu_o[/tex] is the permeability  of free space with a constant value of  

              [tex]\mu_o = 4 \pi * 10^{-7} N/A^2[/tex]

When the current in the wire are of the same direction  the force is positive and when they are in opposite direction the force is negative

   Considering force between A and C

          [tex]F_{A -C} = \frac{\mu_o }{2 \pi} \cdot \frac{I_a I_c }{2R} * L[/tex]

   Considering force between B and C    

        [tex]F_{B -C} = \frac{\mu_o }{2 \pi} \cdot \frac{I_b I_c }{R} * L[/tex]

The resultant force is

        [tex]F_R = F_{B -C} - F_{A -C} = \frac{\mu_o }{2 \pi} \cdot \frac{I_a I_c }{2R} * L - \frac{\mu_o }{2 \pi} \cdot \frac{I_b I_c }{R} * L[/tex]

         [tex]F_R} = \frac{\mu_o }{2 \pi} \cdot \frac{I * I }{R} * L * ({1 - \frac{1}{2} })[/tex]

Substituting values

       [tex]F_R} = \frac{4 \pi * 10^{-7}}{2 * 3.142} \cdot \frac{ 20*20 }{0.015} * 2 ({\frac{1}{2} })[/tex]

       [tex]F_R} = 5.33 *10^{-3} N[/tex]

       [tex]F_R} = 5.33 \ m N[/tex]

Answer:

- Momentum is conserved

- Some kinetic energy is converted into thermal energy

Explanation:

Collision occurs when an object exert force on another. Collision can either be elastic or inelastic depending on whether the bodies stick together or separates after collision.

For elastic collision, the bodies separates after collision and due to this both their momentum and energy are conserved. Both separated object doesn't loose energy as such during collision and they possesses greater momentum.

In inelastic collision, the bodies stick together after collision. Their momentum is conserved but not their kinetic energy since they are not free to move independently.

Based on the question, the type of collision that occur is an inelastic collision since they stick together in air after collision. Some of the kinetic energy of the particles are turned into vibrational energy of the atoms leading to heating effect. Based on this, the following are true

- Their momentum before and after collision is conserved

- Some kinetic energy is converted into thermal energy(heat)

Using the lensmaker's equation with an index of refraction of 1.59 and radii of curvature of 1.15 m and -1.80 m, we find the focal length |f| to be approximately 1.19 m and the power |P| to be about 0.84 diopters.

To calculate the magnitude of the focal length |f| and the power |P| of the lens, we use the lensmaker's equation which is given by:

1/f = (n - 1) * (1/r1 - 1/r2)

where n is the index of refraction of the lens material, r1 and r2 are the radii of curvature of the lens surfaces. Given the index of refraction n = 1.59, and the radii of curvature r1 = 1.15 m (positive for convex surface) and r2 = -1.80 m (negative for concave surface), we can substitute these values into the equation.

Calculating |f|:

1/f = (1.59 - 1) * (1/1.15 + 1/1.80)
= 0.59(0.870 + 0.556)
= 0.59 * 1.426
= 0.841

Thus, |f| = 1/0.841 m ≈ 1.19 m.

To calculate the power of the lens P, which is given in diopters (D), we use: P = 1/f in meters. So, |P| = 1/1.19 m ≈ 0.84 D.

This calculation highlights the application of physics principles in analyzing optical systems.

Answer:

Energy used = 4KJ

Explanation:

Second law of thermodynamics states that as energy is transferred or transformed, more and more of it is wasted. The Second Law also states that there is a natural tendency of any isolated system to degenerate into a more disordered state.

Now when we apply that to heat engines, we'll see that;

Heat expelled = Heat removed + Work Done

We can write it as;

Q_h = Q_c + W

We are given that;

Heat removed; Q_c = 20KJ

Heat expelled into the room in each cycle; Q_h = 24KJ

Thus; plugging these 2 values into the equation, we obtain;

24 = 20 + W

W = 24 - 20

W = 4 KJ

Work done is energy used.

Thus, energy used = 4 KJ

Final answer:

The refrigerator uses 4 kJ of energy each cycle, calculated by the work done W, which is the difference between the heat ejected Qh (24 kJ) and the heat removed Qc (20 kJ).

Explanation:

The question is calculating the energy used by a refrigerator in each cycle. The refrigerator absorbs heat Qc from the inside and ejects a larger amount of heat Qh to the room. The extra energy being ejected comes from the work W done by the refrigerator, which is the energy used by it in each cycle.

If the refrigerator removes 20 kJ (Qc) from the freezing compartment and ejects 24 kJ (Qh) into the room each cycle, the work done (energy used) W can be found using the first law of thermodynamics which states that the conservation of energy principle - the change in internal energy of a system is equal to the heat added to the system minus the work done by the system:

W = Qh - Qc

Substituting in the given values:

W = 24 kJ - 20 kJ

W = 4 kJ

Thus, the work done (energy used) by the refrigerator in each cycle is 4 kJ.

Answer:[tex]y_1=1.9\ mm[/tex]

Explanation:

Given

slit width [tex]d=1.15\ mm[/tex]

Distance of screen [tex]D=3.53\ m[/tex]

wavelength [tex]\lambda =639\ nm[/tex]

Position of any bright fringe is given by

[tex]y_n=\dfrac{n\lambda D}{d}[/tex]

[tex]y_1=\frac{1\times 639\times 10^{-9}\times 3.53}{1.15\times 10^{-3}}[/tex]

[tex]y_1=0.0019\ m[/tex]

[tex]y_1=1.9\ mm[/tex]

Position of dark fringe is given by

[tex]y_D=\dfrac{(2n+1)\lambda D}{2d}[/tex]

for second dark fringe [tex]n=1[/tex]

[tex]y_D=\dfrac{1.5\times 639\times 10^{-9}\times 3.53}{1.15\times 10^{-3}}[/tex]

[tex]y_D=0.00294\ m\approx 2.94\ mm[/tex]

Answer:

0.026 meter

Explanation:

using distance time equation to determine the time for the puck to move 16.5 meters.

distance 'd' = velocity'v' x time't'

16.5 = 44 x t    

t =0.375 second

Here momentum is conserved. Since both objects are initially at rest, the initial momentum is 0.

Next is to determine the puck’s momentum.

Momemtum 'p' = m x v => 0.15 x 44 = 6.6kg⋅m/s

The player momentum is -6.6kg⋅m/s  .

In order to determine the player’s velocity, we'll use p=mv

-6.6 = 94v

v= -0.0702 m/s

The above negative sign represents that the player is moving in the opposite direction of the puck.

Lastly, how far does the player recoil in the time it takes the puck to reach the goal 16.5 m away?

d = v x t = 0.0702 x 0.375= 0.026 meter

Final answer:

The recoil velocity of the ice hockey player is 0.07021 m/s, and the time it takes for the puck to reach the goal is 0.375 s. Therefore, the player recoils a distance of 2.633 cm in the time the puck takes to reach the goal 16.5 m away.

Explanation:

When a hockey player hits a puck on frictionless ice, this scenario is an excellent example of the conservation of momentum where the total momentum before and after the event must be equal. Since the player and the puck are initially at rest, their combined momentum is zero. Thus, when the player imparts a velocity to the puck, the player must recoil with a momentum equal in magnitude and opposite in direction to preserve the momentum balance.

To find the recoil speed of the player, we set the momentum of the puck equal to the momentum of the recoiling player:

So, \\(0.150 kg\\) \\( imes\\) 44.0 m/s = \\(94.0 kg\\) \\( imes\\) recoil velocity of player

Recoil velocity of player = \\((0.150 kg \\( imes\\) 44.0 m/s) / 94.0 kg)

Recoil velocity of player = 0.07021 m/s

To find the distance the player recoils, you first calculate the time it takes for the puck to reach the goal. Since velocity = distance / time, we rearrange to find time = distance / velocity.

Time for puck to reach the goal = Distance to goal / Speed of puck

Time for puck to reach the goal = 16.5 m / 44.0 m/s = 0.375 s

Then, we find the distance the player recoils by using the recoil velocity we found earlier:

Distance player recoils = Recoil velocity of player \\( imes\\) Time for puck to reach the goal

Distance player recoils = 0.07021 m/s \\( imes\\) 0.375 s

Distance player recoils = 0.02633 m, or 2.633 cm

Answer:

[tex]f_{police}=1268.7 Hz[/tex]    

Explanation:

We can use Doppler equation to find the frequency of the siren.

First of all we have the police car moving behind the car. Hence, the frequency detected by the car will be:  

[tex]f_{car1}=f_{police}(\frac{v_{s}-v_{car}}{v_{s}-v_{police}})[/tex]      (1)

Now, when the police car is moving in front of the car, the frequency detected by the car will be:

[tex]f_{car2}=f_{police}(\frac{v_{s}+v_{car}}{v_{s}+v_{police}})[/tex]        (2)            

By solving equation (1) and equation (2) for [tex]v_{police}[/tex] we have:

[tex]v_{police} = 44.67 m/s[/tex]

Knowing that:

Finally, we just need to put this value into the first equation to find frequency of the police car.

[tex]f_{police}=f_{car}(\frac{v_{s}-v_{police}}{v_{s}-v_{car}})[/tex]    

[tex]f_{police}=1310(\frac{343-44.7}{343-35})[/tex]  

[tex]f_{police}=1268.7 Hz[/tex]    

I hope it helps you!

Answer:

See explaination

Explanation:

Magnetic flux density definition, a vector quantity used as a measure of a magnetic field.

It can be defined as the amount of magnetic flux in an area taken perpendicular to the magnetic flux's direction.

See attached file for detailed solution of the given problem.

The magnetic flux density inside a solenoid can be derived using the formula B = μ₀(NI)/L. As L approaches infinity, this simplifies to B = μ₀nI, aligning with the standard magnetic field expression for a long solenoid.

To find the magnetic flux density at a point on the axis of a solenoid with radius b, length L, a current I, and N turns of closely wound coil, use the following steps:

Step 1 : Consider an infinitesimal element dz of the solenoid. The element is located at a distance z from the center of the solenoid. The magnetic field contribution ( dB ) from this element at point ( P ) is given by the Biot-Savart law as -

[tex]\[ dB = \frac{\mu_0 I d\ell}{4\pi} \frac{1}{r^2} \][/tex]

Step 2 : For a solenoid, considering a differential element, the distance r from the element to the point P on the axis is -

[tex]\( \sqrt{b^2 + (z - x)^2} \).[/tex]

Step 3 : For a small segment dz, the current element is -

[tex]\[ dB = \frac{\mu_0 (n I dz)}{4\pi} \frac{1}{(b^2 + (z - x)^2)^{3/2}} \][/tex]

Step 4 :  Integrate dB from -L/2 to L/2 :

[tex]\[ B = \int_{-L/2}^{L/2} \frac{\mu_0 (n I dz)}{4\pi} \frac{1}{(b^2 + (z - x)^2)^{3/2}} \][/tex]

Substitute, [tex]\( n = \frac{N}{L}[/tex]

[tex]\[ B = \frac{\mu_0 I N}{4\pi L} \int_{-L/2}^{L/2} \frac{dz}{(b^2 + (z - x)^2)^{3/2}} \][/tex]

Solving the integral part -

[tex]\[ \int_{-L/2}^{L/2} \frac{dz}{(b^2 + (z - x)^2)^{3/2}} = \frac{z}{b^2 \sqrt{b^2 + (z - x)^2}} \Bigg|_{-L/2}^{L/2} \][/tex]

[tex]\[ \left[ \frac{L/2}{b^2 \sqrt{b^2 + (L/2 - x)^2}} - \frac{-L/2}{b^2 \sqrt{b^2 + (-L/2 - x)^2}} \right] \][/tex]

Step 5: When  L  is very large, the ends of the solenoid are far away, and we can approximate:

[tex]\[ \frac{L/2 - x}{b^2 \sqrt{b^2 + (L/2 - x)^2}} \approx \frac{1}{b^2} \quad \text{and} \quad \frac{-L/2 - x}{b^2 \sqrt{b^2 + (-L/2 - x)^2}} \approx -\frac{1}{b^2} \][/tex]

This simplifies the result to -

[tex]\[ \left[ \frac{L/2 - x}{b^2 (L/2 - x)} - \frac{-L/2 - x}{b^2 (L/2 + x)} \right] = \left[ \frac{1}{b^2} - \left(-\frac{1}{b^2}\right) \right] = \frac{2}{b^2} \][/tex]

Conclusion, For a very long solenoid:

[tex]\[B = \mu_0 \frac{N}{L} I\][/tex]

Which simplifies to,

[tex]\[B = \mu_0 n I\][/tex]

Thus, for a very long soleniod the magnetic flux density can be approximated to [tex]\[B = \mu_0 n I\][/tex].

The intensity I2 after light passes through polarizer #2, oriented at 60° to polarizer #1, is 25% of the intermediate intensity I1.

To determine the intensity [tex]I_2[/tex] after light passes through the second polarizer, we use Malus's Law. Malus's Law states that when polarized light passes through a polarizing filter, the intensity of the transmitted light is given by:

[tex]I = I_0 * cos^2(\theta),[/tex]

where I0 is the initial intensity and θ is the angle between the light’s polarization direction and the axis of the polarizer.

In this case:

Since [tex]cos(60^o) = 0.5:[/tex]

[tex]I_2 = I_1 * (0.5)^2= I_1 * 0.25[/tex]

Hence, the intensity [tex]I_2[/tex] is 25% of [tex]I_1[/tex].

When a star moves towards or away from Earth, its light is shifted to longer or shorter wavelengths, respectively. The red dwarf moving away from Earth and the blue dwarf moving away from Earth would have their light shifted towards longer wavelengths, resulting in a redshift.

The Doppler effect is the change in the observed frequency of sound or light waves due to the relative motion between the source of the waves and the observer.

When a star moves towards or away from Earth, its light is shifted to a longer or shorter wavelength, respectively.

In this case, the red dwarf moving away from Earth at 39.1 km/s would have its light shifted towards longer wavelengths, resulting in a redshift.

Similarly, the blue dwarf moving away from Earth at 25.9 km/s would also experience a redshift.

On the other hand, the red giant moving towards Earth at 23.3 km/s would have its light shifted towards shorter wavelengths, resulting in a blueshift.

The yellow dwarf moving transversely at 15.1 km/s would not exhibit any shift in its light because its motion is perpendicular to the line of sight.

The red dwarf moving transversely at 14.1 km/s would also not show any shift in its light.

Answer:

0.99 T/s

Explanation:

Solution

From the example given, we recall that,

The emf induced in the loop is V = 7.0

Closed loop conductor with r = 1.5 m

∅B = BA

The emf =  d∅B/ dt

which is

d/dt  (BA)

so,

emf = A dB/ dt

emf =πr² dB/ dt

Now,

dB/dt = emf /πr²

=  7/π * (1.5)²

Therefore dB/dt = 0.99 T/s

Answer:

E = -118 KN / C

Explanation:

In this exercise we are given the expression of the electric field for a wire

       E = 1 /4πε₀ 2λ / r

they also indicate the linear density of charge

     λ = 6,6 10⁻⁷ C / m

ask to calculate the electric field at a position at r = 10 cm from the wire

    k = 1 /4πε₀  = 8,988 10⁹ N m² / C²

    E = 8,988 10⁹ 2 (-6.6 10⁻⁷ / 0.10)

     E = -1.18 105 N / C

we reduce to KN / C

    1 KN / C = 10³ N / C

   E = -1.18 10² KN / C

   E = -118 KN / C

the negative sign indicates that the field is directed to the charged

Answer:

the aquifer’s hydraulic conductivity  is K = 10.039 m/day

transmissivity T = 331.287 m/day

the drawdown 100 m away from the well is s = 1.452 m

Explanation:

Given that :

The  constant pumps rate Q = 2000 m³/day

R₁ =160 m     →    H₁  = 249 m

R₂ = 453 m    →    H₂ = 250 m

The  confined aquifer B is 33 m thick

The hydraulic conductivity K = [tex]\frac{Q*In (\frac{R_1}{R_2}) }{2 \pi B(H_2-H_1)}[/tex]

K = [tex]\frac{2000*In (\frac{160}{453}) }{2 \pi *33(250-249)}[/tex]

K = [tex]\frac{2081.43662}{207.3451151}[/tex]

K = 10.039 m/day

Transmissivity T = K × B

T = 10.039×33

T = 331.287 m/day

TO find the drawdown 100 m away from the well; we have:

K = [tex]\frac{Q* In(\frac{R_2}{R_1} )}{2 \pi B (H_2-H_1)} =\frac{Q* In(\frac{R_2}{R_3} )}{2 \pi B (H_2-H_3)}[/tex]

[tex]\frac{ In(\frac{453}{160} )}{(250-249)} =\frac{ In(\frac{453}{100} )}{ (250-H_3)}[/tex]

H₃ = 248.548 m

Drawdown (s) = H₂ - H₃

s = (250 -  248.548)m

s = 1.452 m

Answer:

Explanation:

a ) If x be the position of n the bright fringe on the screen , following formula holds .

x = n (λD / d) ; λ is wave length , D is screen distance and d is slit separation .

If we increase the value of λ or  wave length, x will increase so central fringe along with all the fringes will shift away from the centre .

If we increase the value of D or screen distance , it will also increase x ,  so fringes along with central fringe  will shift away from the center.

b ) Fringes ,whether bright or dark , both shift together , either towards or away from the center .

So to move the dark spots towards the center , we need to do the opposite to what we did in the first case , ie decrease the wavelength or decrease the screen distance .

Answer:

12.3

Explanation:

Answer:

12.3

Explanation:

got it right

Answer:

a) n2 = 1.55

b) 408.25 nm

c) 4.74*10^14 Hz

d) 1.93*10^8 m/s

Explanation:

a) To find the index of refraction of the syrup solution you use the Snell's law:

[tex]n_1sin\theta_1=n_2sin\theta_2[/tex]   (1)

n1: index of refraction of air

n2: index of syrup solution

angle1: incidence angle

angle2: refraction angle

You replace the values of the parameter in (1) and calculate n2:

[tex]n_2=\frac{n_1sin\theta_1}{sin\theta_2}=\frac{(1)(sin30.2\°)}{sin18.82\°}=1.55[/tex]

b) To fond the wavelength in the solution you use:

[tex]\frac{\lambda_2}{\lambda_1}=\frac{n_1}{n_2}\\\\\lambda_2=\lambda_1\frac{n_1}{n_2}=(632.8nm)\frac{1.00}{1.55}=408.25nm[/tex]

c) The frequency of the wave in the solution is:

[tex]v=\lambda_2 f_2\\\\f_2=\frac{v}{\lambda_2}=\frac{c}{n_2\lambda_2}=\frac{3*10^8m/s}{(1.55)(408.25*10^{-9}m)}=4.74*10^{14}\ Hz[/tex]

d) The speed in the solution is given by:

[tex]v=\frac{c}{n_2}=\frac{3*10^8m/s}{1.55}=1.93*10^8m/s[/tex]

Answer:

A) 1010430 pa

B) 1111755 pa

C) 2020860 N

Explanation:

Guage pressure = pgh

= 1030 kg/m3 x 9.81 m/s2 x 100 m

= 1010430 pa

Absolute pressure is Guage pressure + atmospheric pressure.

= 1010430 + 101325 = 1111755 pa

If the pressure inside submarine is 1 atm, then net pressure will be

1111755 - 101325 = 1010430 pa

Force required to open hatch against this pressure will be,

F = pghA

pgh = 1010430 pa

F = 1010430 pa x 2 m^2

F = 2020860 N

Based on the data provided:

The gauge pressure is calculating using the formula:

Guage pressure = pgh

where:

p is density of seawater = 1030 kg/m^3

g is acceleration due to gravity = 9.81 m/s^2

h is depth = 100 m

Substituting:

Gauge pressure = 1030 kg/m3 x 9.81 m/s2 x 100 m

Gauge pressure = 1010430 pa

Therefore, the gauge pressure is 1010420 pa

The absolute gauge pressure is calculated using the formula:

Absolute pressure = Guage pressure + atmospheric pressure.

Atmospheric pressure = 1 atm = 101325 pa

Thus:

Absolute gauge pressure = 1010430 + 101325

Absolute gauge pressure = 1111755 pa

Therefore, the absolute gauge pressure is 1111755 pa

First determine the net pressure.

Pressure inside submarine = 1 atm

Net pressure = 1111755 - 101325

Net pressure = 1010430 pa

Force required to open hatch against this pressure is then calculated from the formula:

Force = net pressure × area

Force = 1010430 pa x 2 m^2

Force = 2020860 N

Therefore, the force required to open the hatch is 2020860 N

Learn more about about guage pressure and force at: https://brainly.com/question/9376763

Answer:

positive, positive

You throw a rock upward. The rock is moving upward, but it is slowing down. If we define the ground as the origin, the position of the rock is positive and the velocity of the rock is positive

Explanation:

Given that the ground is defined as the origin.

The position of the rock is positive since the rock is thrown upward, the position also increases with time until it reaches the maximum height. Also, since the rock is thrown upward with the ground as the origin, the velocity of the rock is positive but the velocity reduces with time (change in height per unit time as the rock moves up is positive)

The correct option for the position of the rock is positive, and for the velocity of the rock, it is positive.

When the rock is thrown upward, it moves away from the ground, which we have defined as the origin. Since the rock is above the ground, its position is positive. Although the rock is slowing down as it moves upward, it still has an upward velocity until it reaches its peak height. Therefore, the velocity of the rock is also positive because it is still moving in the upward direction, even though it is decelerating.

To summarize:

- The position of the rock is positive because it is above the origin (ground).

- The velocity of the rock is positive because it is moving upward, despite the fact that it is slowing down.

The correct answer is: positive, positive.

Answer:

Acceleration, [tex]a=5\ m/s^2[/tex]

Explanation:

We have,

Mass of the object is 20 kg

Force acting on the object is 100 N

It is required to find the resulting acceleration of the  object. Let a is the acceleration. The force acting on object is given by :

[tex]F=ma[/tex]

a is acceleration

[tex]a=\dfrac{F}{m}\\\\a=\dfrac{100\ N}{20\ kg}\\\\a=5\ m/s^2[/tex]

So, the resulting acceleration of the  object is [tex]5\ m/s^2[/tex].

Answer:

4.04m

Explanation:

First you calculate the velocity of the block when it leaves the spring. You calculate this velocity by taking into account that the potential energy of the spring equals the kinetic energy of the block, that is:

[tex]U=K\\\\\frac{1}{2}kx^2=\frac{1}{2}mv^2\\\\v=\sqrt{\frac{kx^2}{m}}=\sqrt{\frac{(10000N/m)(0.08m)^2}{12kg}}=2.3\frac{m}{s}[/tex]

To find the distance in which the block stops you use the following expression (net work done by the friction force is equal to the difference in the kinetic energy of the block):

[tex]W_{T}=\Delta K\\\\F_f d=\frac{1}{2}m[v^2-v_o^2]\\\\d=\frac{mv^2}{2F_f}[/tex]

where Ff is the friction force. By replacing the values of the parameters you obtain:

[tex]d=\frac{(12kg)(2.3m/s)^2}{2(8N)}=3.96m[/tex]

hence, the distance to the original position is 3.96m+0.08m=4.04m

The 12 kg block travels a distance of 4 meters before coming to a stop. This is calculated using the principles of conservation of energy and work done by a force, in this case, the friction force.

The question involves a situation related to Physics, specifically kinematics and mechanics. In order to calculate how far the block travels, we first need to understand how the forces in this situation work. When the block is released, the potential energy stored in compressed spring converts into kinetic energy which propels the block forward. However, due to the presence of friction, this kinetic energy gradually diminishes causing the block to eventually come to a stop.

The first step is to find the initial speed of the block just when it is released. We can use the principle of conservation of energy for this: the spring potential energy is equal to the initial kinetic energy. For potential energy in a spring, we use the formula PE = 0.5 * k * x^2, where k is the spring constant and x is the amount of compression (so PE = 0.5 * 10000 * 0.08^2 = 32 Joules). Kinetic energy is given by KE = 0.5 * m * v^2, where m is the mass and v is the velocity. Equating these (since initial PE = initial KE), we get v = sqrt((2 * PE) / m) = sqrt((2 * 32) / 12) = approx 4.08 m/s.

Now, knowing the initial speed, we can calculate how far the block travels before friction brings it to stop. The work done by the friction force will equal the initial kinetic energy of the block: Work = Friction force * distance = KE. Solving this equation for distance gives us distance = KE / Friction force = 32 Joules / 8 N = 4 meters.

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Answer: Scientists found evidence of Earth's magnetic field reversal in rocks on the ocean floor at plate boundaries. These rocks have alternating polarity due to magnetization that were during their cooling period. Using radiometric dating, scientists estimate that reversals occur approximately every several hundred thousand years.

Explanation: The earth’s magnetic field impacts into the alignment of elements like iron in rocks due to Ferro-magnetization. This causes the elements to align in a north-south direction. Accordingly scientists have studied this alignment in ancient rocks, such as in the layers of sedimentary rocks, and realized that the earth magnetic field has flipped around 200,000 to 300,000 in the last 20 million years.

Answer:

Scientists found evidence of Earth's magnetic field reversal in rocks on the ocean floor at plate boundaries. These rocks have alternating polarity due to magnetization that occurred during their cooling period. Using radiometric dating, scientists estimate that reversals occur approximately every several hundred thousand years.

Explanation:

Answer:

A. The resultant force in the same direction as the satellite’s acceleration.

Explanation:

Launching a satellite in the space and then placing it in orbit around the Earth is a complicated process but at the very basic level it works on simple principles. Gravitational force pulls the satellite towards Earth whereas it acceleration pushes it in straight line.

The resultant force of gravity and acceleration makes the satellite remain in orbit around the Earth. It is condition of free fall where the gravity is making the satellite fall towards Earth but the acceleration doesn't allow it and keeps it in orbit.

In a circular orbit around the Earth, the resultant force acting on a satellite is in the same direction as its acceleration.

In a satellite orbiting the Earth in a circular orbit, there are several forces at play. The gravitational force between the satellite and the Earth provides the centripetal force that keeps the satellite in its orbit. The centripetal force acts towards the center of the circular orbit, while the satellite's acceleration is directed towards the center as well. Therefore, option A is correct: the resultant force is in the same direction as the satellite's acceleration.

The gravitational force acting on the satellite is not negligible; in fact, it is crucial in providing the necessary centripetal force. Therefore, option B is incorrect.

Option C is incorrect as well. There is a resultant force acting on the satellite relative to the Earth, which is responsible for keeping the satellite in its circular orbit.

Lastly, option D is also incorrect. According to Newton's third law of motion, the satellite exerts an equal and opposite force on the Earth, keeping the Earth and the satellite in orbit around their common center of mass.

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Complete Question

The diagram for this question is shown on the first uploaded image

Answer:

The distance traveled in horizontal direction is [tex]D = 1.38 m[/tex]

Explanation:

From the question we are told that

      The length of the string is  [tex]L = 1.6 \ m[/tex]

      The mass of the ball is  [tex]m = 200 g = \frac{200}{1000} = 0.2 \ kg[/tex]

       The height of ball is  [tex]h = 1.5 \ m[/tex]

Generally the work energy theorem can be mathematically represented as

               [tex]PE = KE[/tex]

   Where PE is the loss in potential energy which is mathematically represented as

                   [tex]PE =mgh[/tex]

Where h is the difference height of ball at A and at B  which is mathematically represented as

                 [tex]h = y_A - y_B[/tex]

So        [tex]PE =mg(y_A - y_B)[/tex]              

While KE is the gain in kinetic energy which is mathematically represented as

               [tex]KE = \frac{1}{2 } (v_b ^2 - 0)[/tex]

Where [tex]v_b[/tex] is the velocity of the of the ball

  Therefore we have from above that

                    [tex]PE =KE \equiv mg (y_A - y_B) = \frac{1}{2} m (v_b ^2 - 0)[/tex]

               Making [tex]v_b[/tex] the  subject we have

      [tex]v_b = \sqrt{2g (y_A - y_B)}[/tex]

substituting values

      [tex]v_b = \sqrt{2g (1.5 - 0.40)}[/tex]

     [tex]v_b = 4.6 \ m/s[/tex]

Considering velocity of the ball when it hits the  floor in terms of its vertical and horizontal component we have

         [tex]v_x = 4.6 m/s \ while \ v_y = 0 m/s[/tex]

The time taken to travel  vertically from the point the ball broke loose  can be obtained using the equation of motion

            [tex]s = v_y t - \frac{1}{2} g t^2[/tex]

Where s is distance traveled vertically which given in the diagram as [tex]s = -0.4[/tex]

The negative sign is because it is moving downward

     Substituting values

              [tex]-0.4 = 0 -\frac{1}{2} * 9.8 * t^2[/tex]

         solving for t we have  

               [tex]t = 0.3 \ sec[/tex]

Now the distance traveled on the horizontal is mathematically evaluated as

           [tex]D = v_b * t[/tex]

           [tex]D = 4.6 * 0.3[/tex]

           [tex]D = 1.38 m[/tex]

Answer:

bjknbjk;jln

Explanation:

Answer:

Check the explanation

Explanation:

given

charge of [tex]1.11 * 10^{-10}[/tex] C

radius 250m

The potential at a point 1m

and 1 V

mass 2 micrograms

charge of +2 Coulombs

the potential is V = K Q / r

V = 9 X 109 X 1.11 X 10-10 / 1

V = 9 X 109 X 1.11 X 10-10

V = 0.999 volt

nearly it is V = 1 v

the energy stored is U = q X V

U = 2 X 1

U = 2 J

the energy is stored in this configuration is U = 2 J

Answer:

the coefficient of speed fluctuation is 0.08

the thickness of the rim is [tex]1.423*10^{-4}\ \ m[/tex]

Explanation:

a.Estimate the coefficient of speed fluctuation

Let first determine the average speed of the flywheel by using the expression:

[tex]n = \frac{n_1+n_2}{2}[/tex]

where;

[tex]n_1 =[/tex] minimum speed  = 240 rev/min

[tex]n_2 =[/tex] maximum speed = 260 rev/min

∴ [tex]n = \frac{240 +260}{2}[/tex]

n = 250 rev/ min

To find the coefficient of speed fluctuation; we have:

[tex]C_s = \frac{n_2-n_1}{n}[/tex]

[tex]C_s = \frac{260-240}{250}[/tex]

[tex]C_s = 0.08[/tex]

Hence; the coefficient of speed fluctuation is 0.08

b . If the weight of the spokes is neglected, determine the thickness of the rim

Let's start solving this process by finding the moment of inertia of the flywheel.

The moment of inertia of the fly wheel is given by the equation:

[tex]I = \frac {E_2-E_1}{C_s \omega^2}[/tex]

where ;

[tex]\omega = \frac{2 \pi *n}{60}[/tex]  (since we are converting to rad/s)

[tex]\omega = \frac{2 \pi *250}{60}[/tex]

= 26.18 rad/s

[tex]E_2-E_1[/tex] = the energy fluctuation = 6.75 J

[tex]I = \frac{6.75}{0.08*26.18^2}[/tex]

= 0.123 kg.m²

To determine the weight of the flywheel ; we have the following expression;

[tex]I = \frac{W}{8g}(D^2+d^2)[/tex]

[tex]W = \frac{8gl}{D^2_o+d_i^2}[/tex]

where;

[tex]D_0[/tex] = outer diameter = 1.5 m

[tex]d_i =[/tex] inner diameter = 1.4 m

[tex]W = \frac{8*9.81*0.123}{1.5^2+1.4^2}[/tex]

W = 2.29 N

Let employ the weight density of the cast-iron flywheel [tex]w \rho[/tex] = 70575.5N/m²

Then the volume of the flywheel:

[tex]V = \frac{W}{ w \ rho}[/tex]

[tex]V = \frac{2.29}{70575.5}[/tex]

[tex]V = 3.24*10^{-5} m^3[/tex]

Let t be the thickness of the rim;

the thickness of the rim can be calculate by using the formula;

[tex]V = \frac{\pi t}{4}(D^2-d^2)[/tex]

[tex]t = \frac{4V}{\pi(D^2-d^2)}[/tex]

[tex]t = \frac{4*3.24*10^{-5}}{\pi(1.5^2-1.4^2)}[/tex]

[tex]t = 1.423*10^{-4}m[/tex]

Hence, the thickness of the rim is [tex]1.423*10^{-4}\ \ m[/tex]