A spherical tank with radius 3 m is half full of a liquid that has a density of 900 kg/m3. The tank has a 1 m spout at the top. Find the work W required to pump the liquid out of the spout. (Use 9.8 m/s2 for g.)

Answer:

Explanation:

component of force along the ramp

= 370 cos θ where θ is the slope of the ramp with respect to ground.

sinθ = 1.4 / 5

θ = 16 degree

370 cos16

= 355.67 N

Work done

= component of force along the ramp x length of the ramp

= 355.67 x 5

= 1778.35 J

Answer:

The spring compressed is 0.44 m.

Explanation:

Given that,

Number of person = 25

Average Mass of each person [tex]m_{p}= 25\times69 = 1725\ kg[/tex]

Mass of goat [tex]m_{g}= 3\times15 = 45\ kg[/tex]

Mass of chicken [tex]m_{c}= 5\times3 = 15\ kg[/tex]

Mass of bananas = 25 kg

If the spring constant is [tex]4\times10^{4}\ N/m[/tex]

We need to calculate the total mass

[tex]M=m_{p}+m_{g}+m_{c}+m_{b}[/tex]

Put the value in the formula

[tex]M=1725+45+15+25[/tex]

[tex]M=1810\ kg[/tex]

The weight of all these things is

[tex]W=Mg[/tex]

[tex]W=1810\times9.8[/tex]

[tex]W=17738\ N[/tex]

We need to calculate the distance

Using formula of restoring force

[tex]F=kx[/tex]

[tex]x=\dfrac{F}{k}[/tex]

Put the value into the formula

[tex]x=\dfrac{17738}{4\times10^{4}}[/tex]

[tex]x=0.44\ m[/tex]

Hence, The spring compressed is 0.44 m.

Answer: the total velocity of the air is 67.69km/h to the north and 35.4km/h to the east.

Explanation: The initial velocity of the plane is 200km/h south (supose that south is our positive x-axis here and east is the positive y-axis)

In one hour, the plane is located 137km away from the initial position, and the position in X is equal to 137km*cos(15°) = 132.33, this means that the velocity in the x axis is equal to 132.33 km/h, knowing that the initial velocity of the plane was 200km in the x-axis, this means that the velocity of the air must be:

132.33km/h - 200km/h = -67.69km/h

km and the position in "y" is equal to 137km*sin(15°) = 35.4km

This means that the velocity of the air in the y-axis is 35.4km/h

So the total velocity of the air is 67.69km/h to the north and 35.4km/h to the east.

Answer:

V = 5.24m/s

Explanation:

See attachment below.

Answer with Explanation:

We are given that

Mass of ball=m=0.145 kg

Initially horizontal  velocity of ball,ux=50 m/s

[tex]\theta=30^{\circ}[/tex]

Final velocity of ball,v=38m/s

Time ,t=1.75 ms=[tex]1.75\times 10^{-3} s[/tex]

[tex]1 ms=10^{-3} s[/tex]

Horizontal component of average force, [tex]F_x=\frac{m(vcos\theta-u_x)}{t}[/tex]

Using the formula

Horizontal component of average force, [tex]F_x=\frac{0.145(-38cos30-50)}{1.75\times 10^{-3}}=-6.9\times 10^3[/tex]N

Vertical component of average force, [tex]F_y=\frac{m(vsin\theta-u_y)}{t}[/tex]

Vertical component of average force,[tex]F_y=\frac{0.145(38sin30-0}{1.75\times 10^{-3}}=1.6\times 10^3 N[/tex]

Answer: The work done in J is 324

Explanation:

To calculate the amount of work done for an isothermal process is given by the equation:

[tex]W=-P\Delta V=-P(V_2-V_1)[/tex]

W = amount of work done = ?

P = pressure = 732 torr = 0.96 atm    (760torr =1atm)

[tex]V_1[/tex] = initial volume = 5.68 L

[tex]V_2[/tex] = final volume = 2.35  L

Putting values in above equation, we get:

[tex]W=-0.96atm\times (2.35-5.68)L=3.20L.atm[/tex]

To convert this into joules, we use the conversion factor:

[tex]1L.atm=101.33J[/tex]

So, [tex]3.20L.atm=3.20\times 101.3=324J[/tex]

The positive sign indicates the work is done on the system

Hence, the work done for the given process is 324 J

Given values:

We know the equation,

→ [tex]W = - P\Delta V[/tex]

       [tex]= -P(V_2-V_1)[/tex]

By substituting the values,

       [tex]= -0.96\times (2.35-5.68)[/tex]

       [tex]= 3.20 \ L.atm[/tex]

By converting it in "J",

       [tex]= 3.20\times 101.3[/tex]

       [tex]= 324 \ J[/tex]

Thus the above answer is right.

Find out more information about Work done here:

https://brainly.com/question/25573309

Answer:

minimum speed v=[tex]\sqrt({Fr}/m)}[/tex]

Explanation:

Recall the formula for centripetal force;

Centripetal force is the force that is required to keep an object moving in circular part

     [tex]F=mv^{2} /r[/tex]

where;

F=centripetal force

m=mass of object

r=radius of curvature

v= minimum speed

To find minimum speed make v the subject of formula;

   v=[tex]\sqrt({Fr}/m)}[/tex]

The question is incomplete.

The complete question is:

A student with a mass of m rides a roller coaster with a loop with a radius of curvature of r. What is the minimum speed the rollercoaster can maintain and still make it all the way around the loop? g= 9.8m/s

r=5.5 m= 55kg

Answer: 7.3m/s

Explanation:

Centripetal force is the force acting on a body causing circular motion towards the center which allows to keep the body on track or right path. Any combination of forces in nature can cause centripetal acceleration in various systems.

Where:

V is the tangential velocity

W is the angular velocity

M is the Mass in kg

g is the gravitational force

r is the circular radius

Apply Newton's second law of motion.

The minimum speed is also known as the critical speed is the point where the tension, frictional or normal force is zero and the only thing keeping the object in circular motion is the force of gravity.

This is explained as follows:

N + mg = mw^2r

mg = mv^2/r

Therefore,the minimum speed is:

v^2 = gr

v = square root(gr)

= sqrt(9.8N/kg)(5.5m)

=(9.8N/kg)(5.5m) ^1/2

v = 7.3m/s

Answer:

Δy=70.3885 downwards

Explanation:

First we get the acceleration on moon by substituting the values we have(v=0m/s at t=0s and v=-3.40m/s at t=2.0 s) into equation of simple motion:

[tex]v_{f}-v_{i}=at\\-3.40m/s-0=(2s)a\\a=-1.7m/s^{2}[/tex]

As we know that

Δy=vit+(1/2)at²

Substitute the values of acceleration and time t=9.10s

So

Δy=vit+(1/2)at²

[tex]=(0)(9.10s)+(1/2)(-1.7m/s^{2})(9.10s)^{2}\\ =-70.3885m[/tex]  

Δy=70.3885 downwards

The camera has fallen approximately 402.5 meters after 9.10 seconds.

To find the distance the camera has fallen after 9.10 s, we can use the equation of motion:

s = ut + (1/2)at^2

Where:
s = distance fallen
u = initial velocity (0 m/s)
t = time (9.10 s)
a = acceleration (due to gravity, -9.8 m/s^2)

Plugging the values into the equation, we get:

s = 0 + (1/2)(-9.8)(9.10)^2

Simplifying, we find that the camera has fallen approximately 402.5 meters after 9.10 seconds.

https://brainly.com/question/33015658

#SPJ3

Answer:

(A) t = 4s

(B) H = 40m

(C) g = 5m/s²

(D) V = -20m/s

(E) t = 8s

The detailed solution to this problem requires the knowledge of costant linear acceleration motion.xplanation:

The detailed solution to this problem requires the knowledge of costant linear acceleration motion.

Explanation:

The full solution can be found in the attachment below.

To answer part A, we have been given some values of velocities bounding the time jnterval of 1s from which we can calculate the acceleration due to gravity and then the time.

Part B

Requires just using the acceleration due to gravity and the time taken in the equation

H = ut - 1/2gt²

Part C

Has already been calculated in part A

Part D

V = -20m/s because the tennis ball is coming down as the upward direction was assumed positive.

Part E

When the ball retirns to its original position it is the same as it never left and so H = 0m

The calculation can be found in the attachment below.

Final answer:

Ball reaches peak at 4s, rises 80m, experiences -5m/s² gravity. Returns at -20m/s and spends 8s in the air (4s up + 4s down).

Explanation:

Here are the answers to your questions about the tennis ball on the atmosphere-free planet:

A) Time to reach maximum height: Between the beginning and one second later, the ball's velocity decreases by 5.0 m/s. Since acceleration is constant (no atmosphere), this decrease is due to gravity acting against the initial velocity. We can use the equation:

v = u + at

where:

v = final velocity (15.0 m/s)

u = initial velocity (20.0 m/s)

a = acceleration due to gravity (unknown, negative direction)

t = time (1 second)

Solving for a:

a = (v - u) / t = (15.0 m/s - 20.0 m/s) / 1 s = -5.0 m/s²

Now, to find the time to reach maximum height, we know that at the peak, the velocity is 0 m/s. We can again use the same equation:

0 = u + at

Solving for t:

t = -u / a = -20.0 m/s / (-5.0 m/s²) = 4.0 seconds (positive result since time cannot be negative)

Therefore, it takes 4.0 seconds for the ball to reach its maximum height.

B) Maximum height:

We can use the kinematic equation to find the maximum height (h):

h = ut + 1/2 × at²

Substituting the values:

h = 20.0 m/s × 4.0 s + 1/2 × (-5.0 m/s²) × (4.0 s)² = 40.0 m + 40.0 m = 80.0 meters

C) Acceleration due to gravity:

We already calculated the acceleration due to gravity in part A: -5.0 m/s² (the negative sign indicates downward direction).

D) Velocity upon returning to the original position:

Since the motion is symmetrical (same initial and final positions), the velocity when the ball returns will be the negative of its initial velocity: -20.0 m/s.

E) Time spent in the air:

The total time in the air is the sum of the time to reach the maximum height and the time to fall back to the original position. Since the motion is symmetrical, the time to fall back is also 4.0 seconds. Therefore, the total time in the air is:

4.0 seconds (upward) + 4.0 seconds (downward) = 8.0 seconds

Answer:

Explanation:

[tex]A=\left[\begin{array}{ccc}3&-9&-3\\-1&2&0\\-2&3&-1\end{array}\right] \\\\R_2\rightarrow 3R_2+R_1,R_3\rightarrow 3R_3+2R_1\\\\=\left[\begin{array}{ccc}3&-9&-3\\0&-3&-3\\0&-9&-9\end{array}\right] \\\\R_3\rightarrow 3R_3-9R_2\\\\=\left[\begin{array}{ccc}3&-9&-3\\0&-3&-3\\0&0&0\end{array}\right][/tex]

This is the row echelon form of A. This means that only two of the vectors in our set are linearly independent. In other words, the first two vectors alone will span the same subspace of [tex]R^4[/tex] as all three vectors.

Therefore, the linearly independent spanning set for the subspace is

[tex]\left[\begin{array}{ccc}3\\-1\\-2\end{array}\right] \left[\begin{array}{ccc}-9\\2\\3\end{array}\right] \left[\begin{array}{ccc}3\\0\\-1\end{array}\right][/tex]

Answer

Newton's third law is used for the impact test on the vehicle.                    

In the frontal collision of the vehicle, the impact of the vehicle is made against the concrete wall then the impact on the passengers and driver is calculated.  

In the impact test time of the opening of the airbag is also calculated.  

Similar tests also take place when the impact is sideways.

Answer:

(A) The ratio between the restoring force and the density of the medium remain constant

Explanation:

The ratio between the restoring force and the density of the medium is equal to the square of the velocity of the wave.

[tex]v = \sqrt{\frac{F}{\mu}}[/tex]

The general formula that relates the displacement and velocity (x = vt) can be written in wave mechanics such that

[tex]v = \lambda f[/tex]

where f is the frequency, λ is the wavelength, and v is the velocity of the wave.

According to this equation, in order to halve the wavelength by doubling the frequency, the velocity should be constant. Therefore, the correct answer is (A).

The wavelength of a mechanical wave can be decreased by half by doubling the frequency ONLY if; Choice (A) The ratio between the restoring force and the density of the medium remain constant and Choice D: The same medium is used at the same temperature.

Discussion:

The speed of the mechanical wave is dependent on the ratio of the restoring force and the density of the medium.

Additionally, when the same medium is used at the same temperature; the density of the medium remains constant.

Read more;

https://brainly.com/question/17969340

Answer:

[tex]9.7\times 10^{-5} T[/tex]

Explanation:

Velocity =[tex]5\times 10^4i-2\times 10^4j[/tex]

r=[tex]0.03i+0.05j[/tex]

r=[tex]\mid r\mid=\sqrt{(0.03)^2+(0.05)^2}=0.058[/tex]

v=[tex]\mid V\mid=\sqrt{(5\times 10^4)^2+(-2\times 10^{4})^2}=5.39\times 10^{2}[/tex]

We know that

[tex]B=\frac{mv}{qr}[/tex]

Where q=[tex]1.6\times 10^{-19} C[/tex]

Mass of proton=[tex]1.67\times 10^{-27} kg[/tex]

Using the formula

[tex]B=\frac{1.67\times 10^{-27}\times 5.39\times 10^2}{1.6\times 10^{-19}\times 0.058}[/tex]

[tex]B=9.7\times 10^{-5} T[/tex]

The vector r represents the position at which the magnetic field is being calculated. The formula used to calculate the magnetic field is the Biot-Savart law. The given values can be plugged into the formula to find the magnitude and direction of the magnetic field.

The vector r represents the position of the point where we want to calculate the magnetic field. In this case, r = < 0.03, 0.05, 0 > m.

To calculate the magnetic field at this point, we can use the Biot-Savart law. The Biot-Savart law states that the magnetic field at a point due to a moving charge is given by B = (μ₀/4π)  imes ((qv)×r)/r³, where B is the magnetic field, μ₀ is the permeability of free space, q is the charge, v is the velocity of the charge, and r is the displacement vector from the charge to the point where the magnetic field is being calculated.

Plugging in the values given, we have q = +e, v = <5e4, -2e4, 0> m/s, and r = <0.03, 0.05, 0> m. The magnitude of the magnetic field can be calculated using the formula B = (μ₀/4π)  imes ((qv)×r)/r³ and the direction can be determined using the right-hand rule.

https://brainly.com/question/36936308

#SPJ3

Answer:

Explanation:

Let T be the tension in the string .

For hanging mass

net force = m₁g - T

m₁g - T = m₁a

For mass placed on horizontal plane

T = m₂a

m₁g - m₂a = m₁a

m₁g = a ( m₂ +m₁)

a = m₁g  / ( m₂ +m₁)

= 5 x 9.8 / 19

= 2.579 m /s²

m₁g - T = m₁a

T = m₁g - m₁a

= m₁ ( g - a )

= 5 ( 9.8 - 2.579)

= 36.10 N

Final answer:

To determine how far below the bridge Kate will eventually be hanging, we need to consider the forces acting on her. When she stops oscillating and comes to rest, the gravitational force pulling her downwards will be balanced by the spring force exerted by the bungee cord. To determine the spring constant k, we need to use the equation Fs = -kx and substitute the values of the gravitational force and the displacement x.

Explanation:

(a) To determine how far below the bridge Kate will eventually be hanging, we need to consider the forces acting on her. When she stops oscillating and comes to rest, the gravitational force pulling her downwards will be balanced by the spring force exerted by the bungee cord. At this point, her net force will be zero. The gravitational force can be calculated as mg, where m is her mass and g is the acceleration due to gravity. The spring force can be calculated using Hooke's Law: Fs = -kx, where k is the spring constant and x is the displacement of the cord from its equilibrium position. Equating the gravitational force and the spring force and solving for x will give us the distance below the bridge where Kate will be hanging.

(b) To determine the spring constant k, we need to use the equation Fs = -kx and substitute the values of the gravitational force and the displacement x. Solving for k will give us the spring constant of the bungee cord.

Answer:

In the explanation the answers for the two questions are analyzed.

Explanation:

1. When a collision occurs, it can be said that the total energy is not conserved, but the momentum is conserved. Kinetic energy is converted to heat in the case of an inelastic collision, therefore the sum of all energies is the same before and after the collision. Regarding the net moment before the collision is equal to the net moment after said collision.

2. Regarding the impulse, this is not conserved in the case of an oscillating pendulum because the resulting force acting on the pendulum is not equal to zero. While the total energy would be equal to the sum of the potential energy plus the kinetic energy.

Find the speed with which the disk slides across the surface

Answer:

[tex]\boxed{2.85 m/s}[/tex]

Explanation:

The Potential energy of spring is transformed to kinetic energy hence

[tex]0.5kx^{2}=0.5mv^{2}\\kx^{2}=mv^{2}\\v=\sqrt{\frac {kx^{2}}{m}}[/tex]

Here k is the spring constant, x is the extension of spring, v is the velocity of disk and m is the mass of the disk.

Substituting 0.117 Kg for m, 194 N/m for k and 0.07 m for x then

[tex]v=\sqrt{\frac {194\times 0.07^{2}}{0.117}}=2.850401081 m/s\approx \boxed{2.85 m/s}[/tex]

To solve this problem we will apply the concepts related to the potential energy in the molecules and obtain its electric field through the relationship given by the dipole moment. The change in potential energy from one state to another is given by,

[tex]U_2-U_1 = 1.4*10^{-21} J[/tex]

From this difference we can identify that [tex]U_1[/tex] is equivalent to the potential energy when it is perpendicular to the electric field. At the same time, the potential energy [tex]U_2[/tex] would be equivalent when it is aligned with the electric field.

From there the relationship between energy, the dipole moment and the electric field would be subject to

[tex]U_2-U_1 = pE[/tex]

Here,

[tex]p = \text{Dipole moment} = 6.2*10^{-30} C \cdot m[/tex]

Rearranging to find the electric field,

[tex]E = \frac{(U_2-U_1)}{p}[/tex]

[tex]E = \frac{(1.4*10^{-21})}{(6.2*10^{-30})}[/tex]

[tex]E =2.25*10^8 N/C[/tex]

Therefore the electric field is [tex]2.25*10^8N/C[/tex]

A mass weighing 32 pounds stretches a spring 2 feet.

(a) Determine the amplitude and period of motion if the mass is initially released from a point 1 foot above the equilibrium position with an upward velocity of 6 ft/s.

(b) How many complete cycles will the mass have completed at the end of 4 seconds?

Answer:

[tex]A = 1.803 ft[/tex]

Period = [tex]\frac{\pi}{2}[/tex] seconds

8 cycles

Explanation:

A mass weighing 32 pounds stretches a spring 2 feet;

it implies that the mass (m) = [tex]\frac{w}{g}[/tex]

m= [tex]\frac{32}{32}[/tex]

= 1 slug

Also from Hooke's Law

2 k = 32

k = [tex]\frac{32}{2}[/tex]

k = 16 lb/ft

Using the function:

[tex]\frac{d^2x}{dt} = - 16x\\\frac{d^2x}{dt} + 16x =0[/tex]

[tex]x(0) = -1[/tex]        (because of the initial position being above the equilibrium position)

[tex]x(0) = -6[/tex]          ( as a result of upward velocity)

NOW, we have:

[tex]x(t)=c_1cos4t+c_2sin4t\\x^{'}(t) = 4(-c_1sin4t+c_2cos4t)[/tex]

However;

[tex]x(0) = -1[/tex] means

[tex]-1 =c_1\\c_1 = -1[/tex]

[tex]x(0) =-6[/tex] also implies that:

[tex]-6 =4(c_2)\\c_2 = - \frac{6}{4}[/tex]

[tex]c_2 = -\frac{3}{2}[/tex]

Hence, [tex]x(t) =-cos4t-\frac{3}{2} sin 4t[/tex]

[tex]A = \sqrt{C_1^2+C_2^2}[/tex]

[tex]A = \sqrt{(-1)^2+(\frac{3}{2})^2 }[/tex]

[tex]A=\sqrt{\frac{13}{4} }[/tex]

[tex]A= \frac{1}{2}\sqrt{13}[/tex]

[tex]A = 1.803 ft[/tex]

Period can be calculated as follows:

= [tex]\frac{2 \pi}{4}[/tex]

= [tex]\frac{\pi}{2}[/tex] seconds

How many complete cycles will the mass have completed at the end of 4 seconds?

At the end of 4 seconds, we have:

[tex]x* \frac{\pi}{2} = 4 \pi[/tex]

[tex]x \pi = 8 \pi[/tex]

[tex]x=8[/tex] cycles

Answer:

  The mean free time is given as z  [tex]= 2.3*10^{-14}sec[/tex]

Explanation:

Generally the formula for resistivity

                 [tex]\rho = \frac{m_e}{e^2 n_ez}[/tex]

Where

              [tex]\rho[/tex] is he resistivity of metal

              [tex]m_e[/tex]  is the mass of the electron

             [tex]e[/tex]  is the charge of the electron

             [tex]n_e[/tex] is  electron density

            [tex]z[/tex] is the mean free time

  Now making z the subject  of the formula

              =>  [tex]z = \frac{m_e}{e^2n_e\rho}[/tex]

    Substituting the given values

           [tex]z = \frac{9*10^{-31}}{(1.6*10^{-19}(16.1*10^{28}(2.5*10^{-8})))}[/tex]

                [tex]= 2.3*10^{-14}sec[/tex]

Answer:

[tex]30\Omega, 10\Omega[/tex]

Explanation:

Let two resistors R1 and R2 are wired in series.

Potential difference, V=12 V

Current=I=0.3 A

We have to find the value of two resistors.

When two resistors are connected in series

[tex]R=R_1+R_2[/tex]

[tex]V=IR=I(R_1+R_2)[/tex]

Substitute the values

[tex]12=0.3(R_1+R_2)[/tex]

[tex]R_1+R_2=\frac{12}{0.3}=40[/tex]

[tex]R_1+R_2=40[/tex]..(1)

In parallel

[tex]\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}[/tex]

[tex]\frac{1}{R}=\frac{R_2+R_1}{R_1R_2}[/tex]

[tex]R=\frac{R_1R_2}{R_1+R_2}[/tex]

Current in parallel, I=1.6 A

[tex]V=IR[/tex]

[tex]V=1.6(\frac{R_1R_2}{R_1+R_2})[/tex]

[tex]\frac{12}{1.6}=\frac{R_1R_2}{40}[/tex]

[tex]R_1R_2=\frac{12\times 40}{1.6}=300[/tex]

[tex]R_1-R_2=\sqrt{(R_1+R_2)^2-4R_1R_2}[/tex]

[tex]R_1-R_2=\sqrt{(40)^2-4(300)}=20[/tex]....(2)

Adding equation (1) and (2)

[tex]2R_1=60[/tex]

[tex]R_1=\frac{60}{2}=30\Omega[/tex]

Substitute the value in equation (1)

[tex]30+R_2=40[/tex]

[tex]R_2=40-30=10\Omega[/tex]

Final answer:

To find the values of two resistors based on their behavior in series and parallel circuits, one must calculate the total equivalent resistances for each configuration with Ohm's law and then solve the equations relating the individual resistances.

Explanation:

When looking to determine the values of two resistors R1 and R2 based on current measurements in both series and parallel configurations, the approach involves some electrical principles and algebra. Let's break down the steps needed to solve for the resistors' values.

Determination of resistors in series

Firstly, we calculate the equivalent resistance for the series circuit using Ohm's law (V = I × R), where V is the voltage, I is the current, and R is the resistance. With a 12 V battery and a current of 0.30 A, the equivalent resistance, Req-series, is 12 V / 0.30 A = 40 Ω.

Determination of resistors in parallel

For the parallel configuration, the current through the battery increases to 1.6 A with the same voltage of 12 V. Again, using Ohm's law, we find the equivalent parallel resistance, Req-parallel, which is 12 V / 1.6 A = 7.5 Ω.

In parallel circuits, the reciprocal of the total resistance is the sum of the reciprocals of each individual resistance. This can be expressed as 1/Req-parallel = 1/R1 + 1/R2. To find R1 and R2, we need to relate the resistances in parallel to the sum of their series counterpart (R1 + R2 = Req-series).

Finding Individual Resistances

Given the equations R1 + R2 = 40 Ω and 1/R1 + 1/R2 = 1/7.5 Ω, we can simplify the latter to R1×R2/(R1 + R2) = 7.5 Ω. Substituting R1 + R2 = 40 Ω into this equation, we finally solve for the individual resistance values, R1 and R2.

Answer:

Explanation:

Let x be the horizontal distance of airplane . angle of elevation of airplane from observer = θ , altitude of airplane = 2000 ft ,

from the construction  θ = 45 degree. , x = 2000 ft .

Tanθ = 2000 / x

sec²θ dθ / dt = (2000 / x²)  dx / dt

dθ / dt = 2000 /(sec²θ x²) x dx / dt

dx / dt  = 150 ft /s

dθ / dt = 2000x 150 /(sec²θ x²)

= 300000 / sec²45 x 2000²

= .15 degree/ s

The rate of change of the angle of elevation when an airplane (traveling at 150 ft/s) is over a point 2000 ft from the observer is approximately -2.15° per second.

This problem falls under the department of mathematics known as trigonometry specifically, its applications in real-world scenarios. In this case, we're looking to find the rate of change of the angle of elevation when the airplane is at a certain point.

When the airplane is directly over a point 2000 ft from the observer, it forms a right triangle with the observers' location and the point over which it is flying. The hypotenuse of this triangle is the line between observer and airplane.

The angle of elevation, θ, from the observer to the plane changes over time as the airplane moves overhead so its rate of change(dθ/dt) is what we need to find. We use the trigonometric relation tangent (tan), which in this case equals to the altitude (opposite side, 2000 ft) over the horizontal distance between the observer and the plane (adjacent side, 2000 ft).

The relation is tan θ = opposite/adjacent = 2000/2000 = 1, thus θ = 45 degrees. With the plane's speed (150 ft/s), this changes the horizontal distance over time, so we differentiate tan θ giving us sec² θ*dθ/dt = -150/2000² = -0.0375 radians/sec, by applying the chain rule and remembering sec² θ is 1/cos² θ.

In degrees per second, this is approximately -2.15°/s, so that is the rate of change of the angle of elevation when the airplane is directly over a point 2000 ft from the observer.

https://brainly.com/question/30323081

#SPJ11

Answer:

The amplitude, in meters, of the resulting simple harmonic motion of the block 1/40 m

Explanation:

given information:

the block mass, m = 0.1 kg

spring of force constant, k = 40 N/m

release at t = 0

the formula for simple harmonic motion is

F = kx

where

F = force (N)

k = spring of force constant (N/m)

x = amplitude (m)

thus,

F = kx

x = F/k

  = m g/k

  = 0.1 x 10/40

  = 1/40 m

Explanation:

Below is an attachment containing the solution.

The average power supplied by the force of kinetic friction acting on the block if the block moves 4 m before coming to stop is  -31.25 W.

In physics, power is the amount of energy that is transferred or transformed in a given amount of time. The International System of Units uses the watt, or one joule per second, as the unit of power. Ancient literature frequently used the term "activity" to describe power. Powers are scalar variables.

According to the question, the given values are :

Block mass, m = 4 kg,

The initial speed of the block at the horizontal surface, v₁ = 5 m/s,

The final speed of the block will be, v₂ = 0 m/s.

Distance covered, d = 4 m

Let t be the time when the box is going to stop.

d = [(v₁+v₂)/2] × t

4 = (5/2) t

t = 1.6 sec.

The change in kinetic energy of the block will be ΔE.

ΔE = mv₂²/2 - mv₁²/2

ΔE = 0 - (4)(5)²/2

ΔE = -50 J

Average power supplied by the force of kinetic friction acting on the block = P(avg)

P(avg) = ΔE / t

P(avg) = -50 / 1.6

P(avg) = -31.26 W.

So, the average power supplied by the force of kinetic friction on the block will be -31.26 W.

To get more information about Power :

https://brainly.com/question/2289248

#SPJ2

Answer:

The temperature is 749.8 K

Explanation:

Final temperature (T2) = (distance apart/thermal coefficient of expansion×length) + initial temperature

distance apart = 1.5 mm = 1.5/1000 = 0.0015 m

thermal coefficient of expansion for copper = 16.6×10^-6/K

Length of copper = 20 cm = 20/100 = 0.2 m

Initial temperature = 25 °C = 25 + 273 = 298 K

T2 = (0.0015/16.6×10^-6×0.2) + 298 = 451.8 + 298 = 749.8 K

Answer with Explanation:

We are given that

Force=F=112 N

[tex]\theta=45^{\circ}[/tex]

Distance,[tex]s=84 m[/tex]

Time, t=3.33 minutes

We have to find the work done by Johnson on the bag and the power generated by Johnson.

Work done, W=[tex]Fscos\theta[/tex]

Using the formula

Work done, W=[tex]112\times 84cos45=6652.46 J[/tex]

Power, P=[tex]\frac{W}{t}=\frac{6652.46}{3.33\times 60}=33.3 watt[/tex]

Using 1 minute=60 s

Hence, the power generated by Johnson=33.3 watt

Answer:

0.67 m/s² or 2/3 m/s²

3 m

Explanation:

Using

F-F' = ma............ Equation 1

Where F = Horizontal force applied to the box, F' = Frictional force, m = mass of the box, a = acceleration of the box.

make a the subject of the equation

a = (F-F')/m............ Equation 2

Given: F = 230 N, F' = 210 N, m = 30 kg.

substitute into equation 2

a = (230-210)/30

a = 20/30

a = 0.67 m/s² or 2/3 m/s²

The acceleration of the box = 2/3 m/s²

Using,

s = ut+1/2at²............ Equation 3

Where u = initial velocity, t = time, a = acceleration, s = distance.

Given: u = 0 m/s (from rest), t = 3 s, a = 2/3 m/s²

substitute into equation 3

s = 0(3)+1/2(2/3)(3²)

s = 3 m.

Hence the box moves 3 m

The acceleration of the 30.0-kg box is approximately 0.67 m/s². If it starts from rest, it would move around 3.02 meters in 3 seconds.

In order to solve the problem, we'll apply the principle of Newton's second law, which states that the acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.

First, we identify the net force on the box. The box is being pulled by a force of 230 N, and there is friction acting against this pull with a force of 210 N. The net force (F) is therefore 230 N (pulling force) - 210 N (friction) = 20 N.

To find the acceleration (a), we use the formula:
a = F/m
Substituting the given values, we get:
a = 20 N / 30.0 kg = 0.67 m/s².

Now, the second part of the problem asks for the distance the box would move in 3 seconds if it starts from rest. We use the formula for distance (d) traveled under constant acceleration:
d = 0.5 * a * t²
Substituting the calculated acceleration and time (3 seconds), we find:
d = 0.5 * 0.67 m/s² * (3 s)² = 3.02 m.

So, under the given conditions, the box's acceleration is 0.67 m/s², and it would move approximately 3.02 meters in 3 seconds.

https://brainly.com/question/13447525

#SPJ3

Answer:

magnetic field strength increases.

Explanation:

No, the solenoid will not attract any magnetic substance if it does not carries a direct electric current.

The formula for the magnetic field strength of the solenoid is given as:

[tex]B=\mu.n.I[/tex]

where:

[tex]\mu=[/tex] permeability of free space

[tex]n=[/tex] no. of coil turns in the solenoid

[tex]I=[/tex] current in the coil

So, when the current & no. of coil turns in the coil are increased then the magnetic field strength also increases.

Answer:

595391.482946 m/s

[tex]3.21875\times 10^{6}[/tex]

Explanation:

E = Energy = 1.85 keV

I = Current = 5.15 mA

e = Charge of electron = [tex]1.6\times 10^{-19}\ C[/tex]

t = Time taken = 1 second

m = Mass of proton = [tex]1.67\times 10^{-27}\ kg[/tex]

Velocity of proton is given by

[tex]v=\sqrt{\dfrac{2E}{m}}\\\Rightarrow v=\sqrt{\dfrac{2\times 1.85\times 10^3\times 1.6\times 10^{-19}}{1.67\times 10^{-27}}}\\\Rightarrow v=595391.482946\ m/s[/tex]

The speed of the proton is 595391.482946 m/s

Current is given by

[tex]I=\dfrac{\Delta Q}{t}\\\Rightarrow \Delta Q=It\\\Rightarrow \Delta Q=5.15\times 10^{-3}\times (1\ sec)\\\Rightarrow Q=5.15\times 10^{-3}\ C[/tex]

Number of protons is

[tex]n=\dfrac{Q}{e}\\\Rightarrow n=\dfrac{5.15\times 10^{-3}}{1.6\times 10^{-19}}\\\Rightarrow n=3.21875\times 10^{6}\ protons[/tex]

The number of protons is [tex]3.21875\times 10^{6}[/tex]

The speed of the protons accelerated by the Van de Graaff generator is approximately 3.19 million meters per second. Additionally, the generator produces roughly 3.21 x 10^16 protons every second.

The Van de Graaff generator is a particle accelerator that can be used to accelerate charged particles. In the given scenario, a beam of protons with an energy of 1.85 keV and a beam current of 5.15 mA is being generated.

(a) The energy of a proton (kinetic energy = 1/2 mv²) is given by the equation E = mv²/2, where m is the mass of the proton (1.67262192369 × 10⁻²⁷ kg) and v is the speed of the proton. Solving the equation v = sqrt((2*E)/m), where E is the energy in joules (1.85 keV = 1.85 * 10^-16 joules), we find that v ≈ 3.19 x 10^6 m/s. This implies that the speed of the protons is approximately 3.19 x 10^6 meters per second.

(b) The number of protons produced each second, i.e., the beam current, can be calculated using the formula I = qN/t, where I is the current, q is the charge of a proton (1.602 x 10^-19 Coulombs), N is the number of protons, and t is time. Rearranging the formula, we find that N = It/q. Substituting the values I=5.15 mA = 5.15 x 10^-3 A and t=1s, we get N ≈ 3.21 x 10^16 protons. Therefore, approximately 3.21 x 10^16 protons are produced every second under the mentioned conditions.

https://brainly.com/question/31871736

#SPJ3

Answer:

a) 607.5 J

b) 160.531875 J

c)  0 J

d)  0 J

e) 2.925 m\s

Explanation:

The given data :-

Solution:-

a) The work done by applied force ( W )

W = force applied × displacement = 150 × 4.05 = 607.5 J

b)  The increase in internal energy in the box-floor system due to friction.

Frictional force ( f ) = μ × N = 0.3 × 367.875 = 110.3625 N

Change in internal energy = change in kinetic energy.

ΔU = ( K.E )₂ - ( K.E )₁

Since the initial velocity is zero so the  ( K.E )₁ = 0  

ΔU = ( K.E )₂ = ( F - f ) × ( x ) = ( 150 - 110.3625 ) × 4.05 = 160.531875 J

c) The work done by the normal force .

Displacement of box vertically = 0

W = force applied × displacement = 367.875 × 0 = 0 J

d)  The work done by the gravitational force.

Displacement of box vertically = 0

W = force applied × displacement = 367.875 × 0 = 0 J

e) The change in kinetic energy of the box

( K.E )₂ - ( K.E )₁ = ( K.E )₂ - 0 = ( F - f ) × ( x ) = ( 150 - 110.3625 ) × 4.05 = 160.531875 J

f) The final speed of the box

( K.E )₂ = 160.531875 J = 0.5 × 37.5 × v²

v² = 8.56

v = 2.925 m\s.