An aircraft carrier is sailing at a constant speed of 15.0m/s. it is suddenly spotted by an enemy bomber aircraft flying in the same direction as the ship is moving. the bomber is flying at a constant speed of 100 m/s at an altitude of 1.00 km above the carrier when it decides to attack the carrier. if the bomber drops its bomb at a horizontal distance pf 1.3 km from the aircraft carrier, will it score a direct hit?

Answer:

84.44N

Explanation:

Hi!

The force F between two charges q₁ and q₂ at a distance r from each other is given by Coulomb's law:

[tex]F = k_c \frac{q_1 q_2}{r^2}[/tex]

The force on the negative charge q₁ is the sum of the forces from the other two charges. This forces have equal magnitude as both distances are 48cm. The magnitud is:

[tex]F_{1,2} =F_{1,3} = -k_c\frac{(26\mu C)^2}{(48cm)^2}=-9*10^9Nm^2C^{-2}*0.54*\frac{10^{-12}C^2}{10^{-4}m^2}=-48.75N\\[/tex]

(negative means attractive)

The sum of the forces, because of symmetry reasons actos along line L (see the figure), and its magnitud is:

[tex]F = 2*48.75*\cos(30\º)N = 84.44N[/tex]

Answer:

285 seconds

Explanation:

Jenny speed is 3.8 m/s

Alyssa speed in 4.0 m/s

Alyssa starts after 15 seconds

Find the distance covered by Jenny, when Alyssa starts

Distance=Speed*time

Distance covered by Jenny in 15 seconds= 3.8×15=57m

Relative speed of the two members heading same direction will be;

4.0m/s-3.8m/s=0.2m/s

To find the time Alyssa catch up with Jenny you divide the distance to be covered by Alyssa by the relative speed of the two

Distance=57m, relative speed=0.2m/s  t=57/0.2 =285 seconds

=4.75 minutes

Answer:

Explanation:

[tex]v_{Jenny} =3.8m/s[/tex]

[tex]t_{Jenny}=t[/tex]

[tex]d_{Jenny}=d[/tex]

[tex]v_{Alyssa}=4.0m/s[/tex]

[tex]t_{Alyssa}=t-15[/tex], because she has a difference of 15 seconds.

[tex]d_{Alyssa}=d[/tex]

Both move at a constant speed, that means there's no acceleration, their speed is always the same.

Now, the equation of each movement is

[tex]d=3.8t[/tex] and [tex]d=4(t-15)[/tex], then we solve this two.

We replace the first equation into the second one

[tex]3.8t=4(t-15)\\3.8t=4t-20\\20=4t-3.8t\\0.2t=20\\t=\frac{20}{0.2}\\ t=100[/tex]

That means after 100 seconds Alyssa catch up with Jenny.

Answer:

The continental drift speed is   6.18 cm/yr

Explanation:

In order to calculate the drift speed of the continents, we can assume that it is constant for what the relationship meets

       v = d / t

Where v is the speed, t the time and d the distance

To find the distance we use the closest points that are in the South Atlantic, after reviewing a world map, these points are Brazil and Namibia that has an approximate distance of 7415 km

To start the calculation let's reduce the magnitude

    d = 7415 km (1000m/1km) (100cm/1m)

    d = 7.415 10⁸ 8 cm

    t = 120000000 years  

    t = 1.2 10⁸  year

With these values ​​we calculate the average speed

   v = 7.415 10 3 / 1.2 108 [km / yr.]

   v = 6.18 10-5 Km / yr

   v = 7.415 10 8 / 1.2 10 8 [cm / yr]

   v = 6.18 cm / yr.

The continental drift speed is   6.18 cm/yr

The direction of vector R is [tex]138.47^o[/tex]counterclockwise from the +x-direction and the angle is measured in counterclockwise direction.

For the direction of vector R, which is the angle measured counterclockwise from the +x-direction, we utilize trigonometry.

Given:

x-component of vector R [tex](R_x) = -23.2\ units[/tex]

y-component of vector R [tex](R_y) = 21.4\ units[/tex]

The angle is determined using the arctangent function:

[tex]\theta = tan^{-1}(R_y / R_x)[/tex]

Substituting the given values in the equation:

[tex]\theta = tan^{-1}(21.4 / -23.2).[/tex]

Calculating using a calculator:

[tex]\theta=-41.53^o[/tex]

Since angles are measured counterclockwise from the positive x-direction, adding 180° to the calculated angle gives the direction in that context. The final angle is calculated as:

[tex]\theta= -41.53^o + 180^o \\\theta= 138.47^o\\[/tex]

Therefore, the direction of vector R is [tex]138.47^o[/tex]counterclockwise from the +x-direction.

To know more about the vector:

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The angle of vector R relative to the positive x-axis can be calculated using the arctangent function with the given x and y components, modified to reflect the vector's presence in quadrant II of the Cartesian plane.

To determine the direction of vector R given its x and y components, we can use the arctangent function to find the angle of the vector relative to the positive x-axis. Since the x-component (Rx) is -23.2 units and the y-component (Ry) is 21.4 units, we calculate the angle θ using the formula θ = tan⁻¹(Δy/Δx). Plugging in our values, we get θ = tan⁻¹(21.4 / -23.2). Remember to adjust the angle based on the signs of Rx and Ry since tan⁻¹ only provides results for quadrants I and IV, and our vector lies in quadrant II. This angle will be measured counterclockwise from the positive x-direction.

Answer:

V=37.5miles/h

Explanation:

For convenience, let's convert the average speed to miles per minute:

V=7miles/h * 1h/60min = 0.1167 miles/min.

The distance traveled during the first 15 min was:

D = V*t = 1.75 miles   So, the remaining distance is 6.25miles.

Since you only have 10min left:

Vr = Dr / tr = 6.25 / 10 = 0.625 miles / min. If we take that to miles per hour we get final answer:

Vr = 0.625 miles/min * 60min/1h = 37.5 miles/h

Answer:

The impulse is (10.88 i^ + 7.04 j^) N s

maximum force on the ball is  (4.53 10 2 i^ + 2.93 102 j ^) N  

Explanation:

In a problem of impulse and shocks we must use the impulse equation

       I = dp = pf-p₀         (1)

       p = m V

With we have vector quantities, let's decompose the velocities on the x and y axes

      V₀ = -19 m / s

      θ₀ = 40.0º  

      Vf = 46.0 m / s

      θf = 30.0º

Note that since the positive direction of the x-axis is from the batter to the pitcher, the initial velocity is negative and the angle of 40º is measured from the axis so it is in the third quadrant

      Vcx = Vo cos θ

      Voy = Vo sin θ

      Vox= -19 cos (40) = -14.6 m/s

      Voy = -19 sin (40) =  -12.2 m/s

      Vfx = 46 cos 30 = 39.8 m/s

      Vfy = 46 sin 30 =  23.0 m/s

   a) We already have all the data, substitute and calculate the impulse for each axis

      Ix = pfx -pfy

      Ix = m ( vfx -Vox)

      Ix = 0.200 ( 39.8 – (-14.6))

      Ix = 10.88 N s

      Iy = m (Vfy -Voy)

      Iy = 0.200 ( 23.0- (-12.2))

      Iy=  7.04 N s

In vector form it remains

       I =  (10.88 i^ + 7.04 j^) N s

   b) As we have the value of the impulse in each axis we can use the expression that relates the impulse to the average force and your application time, so we must calculate the average force in each interval.

         I = Fpro Δt

In the first interval

        Fpro = (Fm + Fo) / 2

With the Fpro the average value of the force, Fm the maximum value and Fo the minimum value, which in this case is zero

         Fpro = (Fm +0) / 2

In the second interval the force is constant

          Fpro = Fm

In the third interval

         Fpro = (0 + Fm) / 2

Let's replace and calculate

         I =  Fpro1 t1 +Fpro2 t2  +Fpro3 t3

         I = Fm/2 4 10⁻³ + Fm 20 10⁻³+ Fm/2 4 10⁻³  

         I = Fm  24 10⁻³ N s

         Fm = I / 24 10⁻³

         Fm = (10.88 i^ + 7.04 j^) / 24 10⁻³

         Fm = (4.53 10² i^ + 2.93 10² j ^) N

maximum force on the ball is  (4.53 10 2 i^ + 2.93 102 j ^) N  

The impulse delivered to the ball is (-10.87 î + 7.04 ĵ) N·s. The maximum force on the ball is -452.92 N, based on the given force-time relationship.

Solution:

(a) To find the impulse delivered to the ball, we can use vector components of velocity and the formula for impulse.

Initial velocity vector of the ball:

vix = 19.0 m/s × cos(40.0°) = 14.55 m/s

viy = 19.0 m/s × sin(-40.0°) = -12.22 m/s

Final velocity vector of the ball:

vfx = -46.0 m/s × cos(30.0°) = -39.78 m/s

vfy = 46.0 m/s × sin(30.0°) = 23.00 m/s

Change in velocity vector:

Δvx = vfx - vix = -39.78 m/s - 14.55 m/s = -54.33 m/s

Δvy = vfy - viy = 23.00 m/s - (-12.22 m/s) = 35.22 m/s

Impulse vector (I = m × Δv):

Ix = 0.200 kg × (-54.33 m/s) = -10.87 N·s

Iy = 0.200 kg × (35.22 m/s) = 7.04 N·s

Total impulse vector:

I = (-10.87 î + 7.04 ĵ) N·s

(b) To determine the maximum force (Fmax) on the ball, consider the force-time relationship given:

Force increases linearly for 4.00 ms

Force holds constant for 20.0 ms

Force decreases linearly to zero in another 4.00 ms

The total duration of the force application is 28 ms (4 + 20 + 4 = 28 ms). The impulse is the area under the force-time graph, which is:

Impulse (I) = (1/2 × Fmax × 4.00 ms) + (Fmax × 20.0 ms) + (1/2 × Fmax × 4.00 ms)

-10.87 N·s = (0.002 × Fmax + 0.020 × Fmax + 0.002 × Fmax)

-10.87 N·s = 0.024 × Fmax

Fmax = -10.87 N·s / 0.024 s = -452.92 N

The negative sign indicates the direction opposite to the assumed positive direction.

Answer:81.24 m/s

Explanation:

Given

Horizontal displacement([tex]R_x[/tex])=500

Angle of projection[tex]=24 ^{\circ}[/tex]

Let u be the launching velocity

and horizontal range is given by

[tex]R_x=\frac{u^2sin2\theta }{g}[/tex]

[tex]500=\frac{u^2sin48}{9.81}[/tex]

[tex]u^2=\frac{500\times 9.81}{0.7431}[/tex]

[tex]u^2=6600.32854[/tex]

[tex]u=\sqrt{6600.32854}=81.24 m/s[/tex]

Answer:

4.18 μC

Explanation:

given,

charge place at the center of conducting spherical shell = -3.07 μC                    

total charge place in the shell itself = +7.25 μC                        

to calculate charge on the outer surface = ?                          

total charge on the outer surface = +7.25 μC - 3.07 μC

                                                    = 4.18 μC

hence, the charge on the outer surface of the shell is 4.18 μC

Answer:

43.3 m

Explanation:

d1 = 25.1 m in 15.4° west of north

d2 = 38.8 m in 23.1° south of west

Write the displacements in vector form

[tex]\overrightarrow{d_{1}}=25.1\left ( -Sin15.4\widehat{i}+Cos15.4\widehat{j} \right )=-6.67\widehat{i}+24.2\widehat{j}[/tex]

[tex]\overrightarrow{d_{2}}=38.8\left ( -Cos23.1\widehat{i}-Sin23.1\widehat{j} \right )=-35.69\widehat{i}-15.22\widehat{j}[/tex]

The resultant displacement is given by

[tex]\overrightarrow{d}=\overrightarrow{d_{1}}+\overrightarrow{d_{2}}[/tex]

[tex]\overrightarrow{d}}=\left ( -6.67-35.69 \right )\widehat{i}+\left ( 24.2-15.22 \right )\widehat{j}[/tex]

[tex]\overrightarrow{d}}=\left ( -42.36 \right )\widehat{i}+\left ( 8.98 \right )\widehat{j}[/tex]

The magnitude of the resultant displacement is given by

[tex]d=\sqrt{8.98^{2}+\left ( -42.36 \right )^{2}}=43.3 m[/tex]

Thus, you are 43.3 m far from your starting point.

Answer:

chapter book

Explanation:

To answer this question, you substitute t = 40 s into the given parametric equations to find the horizontal distance from the airport and the altitude. Then, you take the derivative of both equations to find the speed, and the second derivative to find the acceleration.

To solve this problems, you will need to use the given parametric equations. The horizontal distance from the airport (a) is given by x = 1.25 t^2. At t = 40 s, you can simply substitute the value of t into the equation to find x. (b) The altitude of the jet is represented by y = 0.03 t^3.


Similarly, substitute t = 40 s into this equation to find y. (c) The speed of the jet can be found by calculating the derivative of both x and y with respect to t and then using these to find the magnitude of the velocity vector. (d) The acceleration of the jet can be found by taking the second derivative of both x and y with respect to t and again using these to find the magnitude of the acceleration vector.

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Answer:

The time is 5.71 sec.

Explanation:

Given that,

Acceleration [tex]a= -4.20 m/s^2[/tex]

Initial velocity = 24.0 m/s

We need to calculate the time

Using equation of motion

v = u+at[/tex]

Where, v = final velocity

u = inital velocity

t = time

a = acceleration

Put the value into the formula

[tex]0 =24.0 +(-4.20)\times t[/tex]

[tex]t = \dfrac{-24.0}{-4.20}[/tex]

[tex]t=5.71\ sec[/tex]

Hence, The time is 5.71 sec.

The heat transferred to the cold end of the square steel bar as a function of time can be calculated using a derivation of the heat conduction formula, factoring in the steel bar's thermal conductivity, cross-sectional area, temperature differential, time, and length.

This question relates to the transfer of heat through a square steel bar using conduction. The heat transfer can be calculated using the formula Q = k·A·(T1 - T2)·t / L, where Q is the heat transferred, k is the thermal conductivity of the material, A is the cross-sectional area, T1 and T2 are the temperatures at both ends of the material respectively, t is the time of heat transfer, and L is the length of the material.

In this case, the steel bar is square, so its cross-sectional area A is calculated by squaring the side length w, so A = w². So, the exact formulation to calculate the heat transferred to the cold end of the bar as a function of time will be Q= 14.6 J/(s⋅m⋅°C)⋅(0.18 m)²⋅(88°C - T2)·t / 1.7 m.

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The heat transferred to the cold end of the steel bar is calculated by finding the rate of heat transfer and multiplying it by the time. The rate of heat transfer is approximately 23.5729 W.

To find the heat transferred to the cold end of the bar as a function of time, we use the formula for the rate of heat transfer through a material.

Given data:

The rate of heat transfer Q/t can be calculated using the formula:

[tex]Q/t = k * A * (T_1 - T_2) / L[/tex]

Plugging in the values:

[tex]Q/t = 14.6 J/(s.m.^oC) * 0.0324 m^2 * (88^oC - 0^oC) / 1.7 m[/tex]

[tex]Q/t = 14.6 * 0.0324 * 88 / 1.7[/tex]

[tex]Q/t = 23.5729 W[/tex]

Therefore, the heat transferred to the cold end of the bar as a function of time is:

[tex]Q = (23.5729 W) * t[/tex]

The heat transferred to the cold end of a square steel bar over time can be calculated by finding the rate of heat transfer and multiplying it by the time. The rate of heat transfer for the given data is approximately 23.5729 W.

Answer:

Option c

Explanation:

Magnetic field lines form loops starting from north pole to south pole outside the magnet and from south pole to north pole inside the magnet.

Thus the field is such that it is directed outwards from the North pole and directed inwards to the South pole of the magnet.

A compass in a magnetic field will will comply with the magnet's North pole directing towards the magnetic field.

Answer:

The charge q₃ must be placed at X = +2.5 cm

Explanation:

Conceptual analysis

The electric field at a point P due to a point charge is calculated as follows:

E = k*q/d²

E: Electric field in N/C

q: charge in Newtons (N)

k: electric constant in N*m²/C²

d: distance from charge q to point P in meters (m)

The electric field at a point P due to several point charges is the vector sum of the electric field due to individual charges.

Equivalences

1µC= 10⁻6 C

1cm= 10⁻² m

Data

k = 8.99*10⁹ N×m²/C²

q₁ =+3 µC =3*10⁻⁶ C

q₂ = -10 µC =-10*10⁻⁶ C

q₃= +6µC =+6*10⁻⁶ C

d₁ = 3cm =3×10⁻² m

d₂ = 4cm = 4×10⁻² m

Graphic attached

The attached graph shows the field due to the charges:

E₁:Field at point P due to charge q₁. As the charge is positive ,the field leaves the charge. The direction of E1 is (+ x).

E₂: Field at point P due to charge q₂. As the charge is positive ,the field leaves the charge. The direction of E1 is (+ x).

Problem development

E₃: Field at point P due to charge q₃. As the charge q₃ is positive, the field leaves the charge.

The direction of E₃ must be (- x) so that the electric field can be equal to zero at point P since E₁ and E₂ are positive, then, q₃must be located to the right of point P.

We make the algebraic sum of fields at point P due to the charges q1, q2, and q3:

E₁+E₂-E₃=0

[tex]\frac{k*q_{1} }{d_{1}^{2}  } +\frac{k*q_{2} }{d_{2}^{2}  } -\frac{k*q_{3} }{d_{3}^{2}  } =0[/tex]

We eliminate k

[tex]\frac{q_{1} }{d_{1} ^{2} } +\frac{q_{2} }{d_{2} ^{2} }+\frac{q_{3} }{d_{3} ^{2} }=0[/tex]

We replace data

[tex]\frac{3*10^{-6} }{(3*10^{-2})^{2} } +\frac{10*10^{-6} }{(4*10^{-2})^{2} } +\frac{6*10^{-6} }{d_{3} ^{2} } =0[/tex]

we eliminate 10⁻⁶

[tex]\frac{3}{9*10^{-4} } +\frac{10}{16*10^{-4} } =\frac{6}{d_{3}^{2}  }[/tex]

[tex](\frac{1}{10^{-4} }) *(\frac{1}{3} +\frac{5}{8}) =\frac{6}{d_{3}^{2}  }[/tex]

[tex]\frac{23*10^{4} }{24} =\frac{6}{d_{3} ^{2} }[/tex]

[tex]d_{3} =\sqrt{\frac{6*24}{23*10^{4} } }[/tex]

[tex]d_{3} =2.5*10^{-2} m\\d_{3} =2.5 cm[/tex]

The charge q₃ must be placed at X = +2.5 cm

Answer:

The uncertainty in the volume of the sphere is [tex]1.059\times 10^{4} m^{3}[/tex]

Solution:

As per the question:

Measured radius of the sphere, R = 135.4 m

Uncertainty in the radius, [tex]\Delta R = 4.6 cm = 4.6\times 10^{- 2} = 0.046 m[/tex]

We know the volume of the sphere is:

[tex]V_{s} = \frac{4}{3}\pi R^{3}[/tex]

We know that the fractional error for the given sphere is given by:

[tex]\frac{\Delta V_{s}}{V_{s}} = \frac{4}{3}\pi.\frac{|Delta R}{R}[/tex]

where

[tex]\Delta V_{s}[/tex] = uncertainty in volume of sphere

Now,

[tex]\Delta V_{s} = \frac{4}{3}\pi 3R^{2}\Delta R[/tex]

Now, substituting  the suitable values:

[tex]\Delta V_{s} = 4\pi (135.4)^{2}\times 0.046 = 1.059\times 10^{4} m^{3}[/tex]

Answer:

1.61 second

Explanation:

Angle of projection, θ = 53°

maximum height, H = 7.8 m

Let T be the time taken by the ball to travel into air. It is called time of flight.

Let u be the velocity of projection.

The formula for maximum height is given by

[tex]H = \frac{u^{2}Sin^{2}\theta }{2g}[/tex]

By substituting the values, we get

[tex]7.8= \frac{u^{2}Sin^{2}53 }{2\times 9.8}[/tex]

u = 9.88 m/s

Use the formula for time of flight

[tex]T = \frac{2uSin\theta }{g}[/tex]

[tex]T = \frac{2\times 9.88\times Sin53 }{9.8}[/tex]

T = 1.61 second

Final answer:

The acceleration due to gravity on this planet is approximately [tex]g = 9.82 m/s^2.[/tex]

Explanation:

To determine the acceleration due to gravity on an unknown planet using a pendulum, you can utilize the formula for the period of a simple pendulum: T = 2π[tex]\sqrt{(L/g}[/tex], where T is the period, L is the length, and g is the acceleration due to gravity.

With the given length of the pendulum being 0.81 m and the period being 1.138 s, you can rearrange the formula to solve for  [tex]g = (4*22/7)^2[/tex][tex](L/T^2)[/tex]. Plugging in the known values, [tex]g = (4*22/7)^2[/tex][tex](0.81 m / 1.138^2 s^2)[/tex].

Computing the value, we find that the acceleration due to gravity on this planet is approximately [tex]g = 9.82 m/s^2.[/tex]

Answer:

(A) only I

Explanation:

According to Newton's first law of motion, a particle in motion will continue moving in a straight line at constant speed or will stay at rest, if at rest, as long as there's no external force acting on it.

ALL of the statements except #2 COULD be true. (Choice-D)

Answer:

[tex]a_{c}=v^{2}/R[/tex]

The radius of curvature changes so that centripetal acceleration is similar along the entire roller coaster.

Explanation:

We know that the centripetal acceleration is directly proportional to the tangential velocity and inversely proportional to the radius of curvature:

[tex]a_{c}=v^{2}/R[/tex]

By energy conservation (and common sense), we know that the speed at the top of the roller coaster is smaller. Therefore if the roller coaster has similar accelerations (therefore also similar normal forces) at the top and at the bottom, it is necessary that the difference in speed be compensated with the radius of curvature, i.e. smaller radius at the top than at the bottom.

Answer:

[tex]\frac{W}{F} = 8.8\times10^7[/tex]

Explanation:

According to newton's law of gravitation

[tex]F=G\frac{m_1\times m_2}{r^2}[/tex]

here m_1=m_2=15000 kg

r= 3 meters and G= 6.67[tex]6.67\times10^{-11}[/tex]

putting values we get

[tex]F= 6.67\times10^{-11}\frac{15000^2}{3^2}[/tex]

solving the above equation we get

Force of gravitation F= [tex]1.6675\times 10^{-3}[/tex] newton

weight of the one of the truck W = mg= 15000×9.81 N

=147000 N

therefore [tex]\frac{W}{F} = \frac{147000}{1.66\times10^{-3}}[/tex]

=8.8×10^{7)

Answer:

1) Speed in m/s equals 22.22 m/s.

2) Speed in miles per hour equals 49.712 mph.

Explanation:

Since we know that in 1 kilometer there are 1000 meters and in 1 hour there are 3600 seconds hence we can write

[tex]80km/h=\frac{80\times 1000m}{3600s}=22.22m/s[/tex]

Now we know that 1 mile equals 1.609 kilometer  hence we conclude that 1 kilometer equals [tex]\frac{1}{1.609}=0.6214[/tex]mile

Hence

[tex]80km/h=\frac{80\times 0.6214miles}{1h}=49.712mph[/tex]

The speed limit of 80 km/h is approximately 22.2 meters per second and 49.7 miles per hour.

The speed limit on some interstate highways is roughly 80 km/h. To convert this speed to meters per second, you divide by 3.6 (since 1 km/h is equal to about 0.27778 m/s). So, 80 km/h divided by 3.6 gives us approximately 22.2 m/s.

To convert the speed to miles per hour, you would use the conversion factor that 1 kilometer is approximately 0.621371 miles. Therefore, 80 km/h multiplied by 0.621371 gives us approximately 49.7 mi/h.

Answer:

The compression is [tex] \sqrt{2} \  d [/tex].

Explanation:

A Hooke's law spring compressed has a potential energy

[tex]E_{potential} = \frac{1}{2} k (\Delta x)^2[/tex]

where k is the spring constant and [tex]\Delta x[/tex] the distance to the equilibrium position.

A mass m moving at speed v has a kinetic energy

[tex]E_{kinetic} = \frac{1}{2} m v^2[/tex].

So, in the first part of the problem, the spring is compressed a distance d, and then launch the mass at velocity [tex]v_1[/tex]. Knowing that the energy is constant.

[tex]\frac{1}{2} m v_1^2 = \frac{1}{2} k d^2[/tex]

If we want to double the kinetic energy, then, the knew kinetic energy for a obtained by compressing the spring a distance D, implies:

[tex] 2 * (\frac{1}{2} m v_1^2) = \frac{1}{2} k D^2[/tex]

But, in the left side we can use the previous equation to obtain:

[tex] 2 * (\frac{1}{2} k d^2) = \frac{1}{2} k D^2[/tex]

[tex]  D^2 =  \frac{2 \ (\frac{1}{2} k d^2)}{\frac{1}{2} k} [/tex]

[tex]  D^2 =  2 \  d^2 [/tex]

[tex]  D =  \sqrt{2 \  d^2} [/tex]

[tex]  D =  \sqrt{2} \  d [/tex]

And this is the compression we are looking for

Answer:

[tex]d'=\sqrt{2} d[/tex]

Explanation:

By hooke's law we have that the potential energy can be defined as:

[tex]U=\frac{kd^{2} }{2}[/tex]

Where k is the spring constant and d is the compression distance, the kinetic energy can be written as

[tex]K=\frac{mv^{2} }{2}[/tex]

By conservation of energy we have:

[tex]\frac{mv^{2} }{2}=\frac{kd^{2} }{2}[/tex] (1)

If we double the kinetic energy

[tex]2(\frac{mv^{2} }{2})=\frac{kd'^{2} }{2}[/tex] (2)

where d' is the new compression, now if we input (1) in (2) we have

[tex]2(\frac{kd^{2} }{2})=\frac{kd'^{2} }{2}[/tex]

[tex]2(\frac{d^{2} }{2})=\frac{d'^{2} }{2}[/tex]

[tex]d'=\sqrt{2} d[/tex]

a. There is nothing suggesting that the balloon is accelerating horizontally, so we can assume that its horizontal speed is constant. The time before the balloon crash into the hill is simply the distance between the balloon and the hill divided by its velocity. Remember that velocity is simply the amount of distance that a object travels in a certain amount of time:

[tex]t = \frac{130m}{2.8 m/s} = 46.43 s[/tex]

b. Know that you know the maximum amount of time that the balloon can take to gain 28m of altitude, the minimum acceleration can be found using the  equations constant acceleration motion:

[tex]x = \frac{1}{2}at^2 + v_ot +x_0[/tex]

where a is the acceleration, v_o is the initial vertical velocity, 0 as the balloon is not moving vertically before starting to ascend. xo is the initial position, which we will give a value of 0m.

[tex]x = \frac{1}{2}at^2\\ 28m = \frac{1}{2}(46.43s)^2a\\ a = \frac{2*28m}{(46.43s)^2} = 0.026 m/s^2[/tex]

c. As we said before, there isn't any kind of force that accelerates the balloon horizontally, therefore, we can consider that its horizontal velocity is constant and equal to 2.8m/s

d. Acceleration is the amount of change in velocity after a given amount of time. So, with the acceleration and the time we can fin the velocity:

[tex]v_y = a_y*t = 0.026m/s^2*46.43s=1.206 m/s[/tex]

The pilot has approximately 46.43 seconds to gain 28 m in altitude to clear the hill. The minimum constant upward acceleration needed is approximately 0.026 m/s². The horizontal component of the balloon's velocity at clearance will be 2.8 m/s, while the vertical component will be approximately 1.21 m/s.

The time the pilot has to make the altitude change without crashing into the hill can be found using the horizontal velocity and the distance to the hill. Since the balloon drifts horizontally at 2.8 m/s and needs to cover 130 m, the time (t) it will take can be calculated as:

t = distance / horizontal velocity = 130 m / 2.8 m/s = 46.43 seconds.

To find the minimum constant upward acceleration (a) needed to clear the hill, we use the kinematic equation:

s = ut + (1/2)at2

Where s is the vertical displacement (28 m), u is the initial vertical velocity (0 m/s), and t is the time calculated above. Rearranging for a gives:

a = 2s / t2 ≈ 2(28 m) / (46.43 s)2 ≈ 0.026 m/s2.

As the horizontal velocity is not affected by the vertical motion in the absence of air resistance, the horizontal component of the balloon's velocity when it clears the top of the hill will remain 2.8 m/s.

To find the vertical component of the velocity at the instant it clears the top of the hill, we can use the equation:

vf = u + at

Where vf is the final vertical velocity, u is the initial vertical velocity, a is the acceleration, and t is the time taken. Substituting the known values gives:

vf = 0 m/s + (0.026 m/s2)(46.43 s) ≈ 1.21 m/s.

Answer:

Answered

Explanation:

The girl whirling the ball should let go off ball when the ball is at a position such that tangent to the circle is in the direction of the target.

the tangent at any point in a circular path indicates the direction of velocity at that point. And the moment when the centripetal force is removed the ball will follow the tangential path at that moment.

Final answer:

The girl should release the ball when it aligns with the target in a straight line from her, following principles of inertia and circular motion. In a scenario where a ball winds around a post, it would speed up due to the conservation of angular momentum, aligning with Michelle's prediction.

Explanation:

The question about when a girl should release a ball while whirling it around her head in a horizontal plane to hit a target involves understanding circular motion and projectile motion principles in physics. According to the laws of physics, the ball will continue to move tangentially to the circle at the point of release because of inertia. Thus, she should let go of the string when the ball is directly in line with the target, assuming no air resistance and that the path follows a straight line in the horizontal direction.

The conservation of angular momentum and the conservation of energy are relevant in situations where objects are in circular motion, like a ball tied to a string being whirled around. For an object in circular motion, when the radius of the motion decreases (as in the ball winding around a post), it will speed up because angular momentum is conserved. This implies Michelle’s view on the ball having to speed up as it approaches the post is correct, contrasting Astrid’s expectation of constant speed due to energy conservation.

Answer:

v = √rg.

Explanation:

The Minimum speed of the stone that can have to the stone when it is rotated in a vertical circle is √rg.

Mathematical Proof ⇒

at the top point on the circle we have

T + mg = m v²/r

We know that minimum speed will be at the place when its tension will be zero.

∴ v² = rg

⇒ v = √rg.

So, the minimum speed or the critical speed is given as  v = √rg.

Answer:

[tex]v=\sqrt{rg}[/tex]

Explanation:

radius of circle = R

Let T be the tension in the string.

At highest point A, the tension is equal to or more than zero, so that it completes the vertical circle. tension and weight is balanced by the centripetal force.

According to diagram,

[tex]T + mg = \frac{mv^{2}}{R}[/tex]

T ≥ 0

So, [tex]mg = \frac{mv^{2}}{R}[/tex]

Where, v be the speed at the highest point, which is called the critical speed.

[tex]v=\sqrt{rg}[/tex]

Thus, the critical speed at the highest point to complete the vertical circle is  [tex]v=\sqrt{rg}[/tex].

Final answer:

The bead rests at a height of 9.03 x 10^7 meters above the fixed charge.

Explanation:

To find the height above the fixed charge where the bead rests in equilibrium, we need to consider the electric forces acting on the bead. The electric force is given by the equation:

F = k * (q1 * q2) / r^2

Where F is the force between the two charges, q1 and q2 are the charges, r is the distance between them, and k is the electrostatic constant. In this case, the two charges are the fixed charge at the base of the rod and the charge on the bead. Setting the gravitational force equal to the electric force, we can solve for the height.

First, we need to convert the given charge of 20 nC to coulombs by dividing it by 10^9:

q2 = 20 nC / 10^9 = 20 * 10^-9 C

Next, we calculate the gravitational force and the electric force:

F_gravity = m * g

F_electric = k * q1 * q2 / r^2

Since the bead is in equilibrium, the two forces must be equal:

m * g = k * q1 * q2 / r^2

Now, we can solve for the height:

h = sqrt(k * q1 * q2 / (m * g))

Plugging in the given values:

h = sqrt((9 * 10^9 N * m^2 / C^2) * (20 * 10^-9 C) / (0.050 x 10^-3 kg * 9.8 m/s^2))

Simplifying:

h = sqrt(4 * 10^11 / (0.049 x 10^-3))

h = sqrt(8.16 x 10^14)

h = 9.03 x 10^7 m

Therefore, the bead rests at a height of 9.03 x 10^7 meters above the fixed charge.

Answer:

The rock will rise 2.3 m above the top of the window

Explanation:

The equations for the position and velocity of the rock are the following:

y = y0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

y = height of the rock at time t

v0 = initial velocity

y0 = initial height

g = acceleration due to gravity

t = time

v = velocity at time t

If we place the center of the frame of reference at the bottom of the window, then, y0 = 0 and at t = 0.22 s, y = 1.7 m. With this data, we can calculate v0:

1.7 m = 0.22 s · v0 - 1/2 · 9.8 m/s² · (0.22 s)²

Solving for v0:

v0 = 8.8 m/s

Now that we have the initial velocity, we can calculate the time at which the rock reaches its maximum height, knowing that at that point its velocity is 0.

Then:

v = v0 + g · t

0 = 8.8 m/s - 9.8 m/s² · t

-8.8 m/s / -9.8 m/s² = t

t = 0.90 s

Now, we can calculate the max height of the rock:

y = y0 + v0 · t + 1/2 · g · t²

y = 8.8 m/s · 0.90 s - 1/2 · 9.8 m/s² · (0.90 s)²

y = 4.0 m

Then the rock will rise (4.0 m - 1.7 m) 2.3 m above the top of the window

Answer:

(A) Torque required is 21.205 N-m

(b) Wok done will be equal to 1199.1286 j

Explanation:

We have given moment of inertia [tex]I=12kgm^2[/tex]

Wheel deaccelerate from 135 rpm to 0 rpm

135 rpm = [tex]135\times \frac{2\pi }{60}=14.1371rad/sec[/tex]

Time t = 8 sec

So angular speed [tex]\omega _i=135rpm[/tex] and [tex]\omega _f=0rpm[/tex]

Angular acceleration is given by [tex]\alpha =\frac{\omega _f-\omega _i}{t}=\frac{0-14.1371}{8}=--1.7671rad/sec^2[/tex]

Torque is given by torque [tex]\tau =I\alpha[/tex]

[tex]=12\times 1.7671=21.205N-m[/tex]

Work done to accelerate the vehicle is

[tex]\Delta w=K_I-K_F[/tex]

[tex]\Delta W=\frac{1}{2}\times 12\times 14.137^2-\frac{1}{2}\times 12\times0^2=1199.1286J[/tex]

Answer:

The refractive index of the sphere is 2.66

Solution:

The refractive index, [tex]n_{w} = 1.33[/tex] and since the sphere is surrounded by water.

Therefore, according to the question, the parallel rays that affect one of the faces of the sphere converges on the second vortex:

Thus the image distance from the pole  of surface 1, v' = 2R

where

R = Radius of the sphere

Now, using the eqn:

[tex]\frac{n_{w}}{v} + \frac{n}{v'} = \frac{n - n_{w}}{R}[/tex]

[tex]0 + \frac{n}{2R} = \frac{n - 1.33}{R}[/tex]

Since, v is taken as infinite

n = 2.66