Case 1: A Styrofoam cup holds an unknown amount of lemonade (which is essentially water) at 20.5 °C. A 0.0550-kg ice cube at -10.2°C is placed in the lemonade. When thermal equilibrium is reached, all the ice has melted and the final temperature of the mixture is measured to be 11.8 °C. Assume that the mass of the cup is so small that it absorbs a negligible amount of heat, and ignore any heat lost to the surroundings. The latent heat of fusion for water is 3.4 X10 J/kg. The specific heat capacity for water and lemonade is the same, 4186 J/(kg °C). The specific heat capacity for ice is 2000 J/(kg °C). The cold ICE absorbs heat in 3 steps: Step 1: COLD ICE warms up from -10.2 °C. to 0.0 °C. (Remember: cold ice DOESN't Melt before reaching 0.0 °C.) Apply Q = mcAT, to the cold ice. Calculate the heat absorbed by the ICE in this step. Keep 2 decimal places. Report all heat as POSITIVE. Enter a number Submit (5 attempts remaining) Step 2: ICE at 0.0 °C melts into water at 0.0 °C. Calculate the heat absorbed by the ICE in this step. Keep 2 decimal places. Report all heat as POSITIVE. Enter a number Submit (5 attempts remaining) Step 3: The water at 0.0 °C from melted ice warms up to 11.8 °C. Apply Q = mcAT, to the water. Calculate the heat absorbed by water in this step. Keep 2 decimal places. Report all heat as POSITIVE. Enter a number Submit (5 attempts remaining) Calculate the total heat absorbed by ICE in the above 3 steps. Enter a number J Submit (5 attempts remaining) The warm lemonade releases heat in cooling down from 20.5 °C to 11.8 °C. Apply Q = mcAT, to the lemonade. Keep mass m as "space hoder. Keep all heat as POSITIVE. Let Q released by lemonade = Total heat absorbed by ICE Find the mass of the lemonade. Keep 2 decimal places. Enter a number kg Submit (5 attempts remaining)

Final answer:

To determine the values requested - (a) the rms voltage across the resistor is 80.2 V, (b) the peak voltage of the source is 113.4 V, (c) the maximum current in the resistor is 8.02 A, and (d) the average power delivered to the resistor is 641 W.

Explanation:

In order to determine the values requested, we can use Ohm's law which states that V = IR, where V is the voltage, I is the current, and R is the resistance. Let's go through each part:

(a) To find the rms voltage across the resistor, we use the formula V = IR, where I is the ammeter reading and R is the resistance. Plugging in the values, we get V = (8.02 A) x (10.0 Ω) = 80.2 V.

(b) The peak voltage of the source can be found by multiplying the rms voltage by √2. So the peak voltage is (80.2 V) x √2 = 113.4 V.

(c) The maximum current in the resistor is equal to the ammeter reading, which is 8.02 A.

(d) The average power delivered to the resistor can be calculated using the formula P = IV, where I is the rms current and V is the rms voltage. Since the ammeter reading is the rms current, we can plug in the values to get P = (8.02 A) x (80.2 V) = 641 W.

The smallest detectable distance with the current setup is 0.98125cm. To make the robot travel 2m, the encoder needs to count roughly 204 times. If the smallest detectable change should be ±0.20cm, the encoder requires 80 lines per revolution.

The smallest detectable change in travel distance for the robot can be calculated based on the circumference of the wheel and the resolution of the rotary encoder. Given the diameter of the wheel is 10cm, the circumference can be calculated using the formula 2πr (r= radius of wheel), which results in 31.4cm. With 16 lines per revolution, this yields a distance per count (rise and fall) of 31.4cm/32 = 0.98125cm.

For the robot to travel 2m before making a turn, we will need to calculate the number of encoder counts. 2m equals 200cm, and if each count is representing a distance of 0.98125cm, then the number of encoder counts would be 200/0.98125, roughly equal to 204 counts.

If the goal is to reduce the smallest detectable distance to ±0.20cm, then we need to increase the lines per revolution of the encoder. If one line equates to 0.98125cm right now, that means we need roughly 5 times more lines on the encoder to achieve a resolution of 0.20cm. Therefore, the required lines per revolution would be 5*16 = 80 lines.

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Answer:

(a) [tex]x_{n} = x_{n - 2} + x_{n - 4} + x_{n - 8}[/tex]

(b) 18 ways

Solution:

As per the question:

We need to find the no.of ways to go 'n' miles by foot:

From the question it is clear that, at each hours even miles can be traveled by a person, If the distance to be covered is odd, then there in no way 3 miles can be covered.

Thus defining recurrence as:

[tex]x_{o} = 1[/tex]

[tex]x_{n} = 0[/tex]

where

n: negative or odd

Then

[tex]x_{n} = x_{n - 2} + x_{n - 4} + x_{n - 8}[/tex]

(b) No. of ways to travel 12 miles:

[tex]x_{o} = 1[/tex]

[tex]x_{2} = 1 + 0 = 1[/tex]

[tex]x_{4} = x_{2} + x_{o} + 0 = 1 + 1 = 2[/tex]

[tex]x_{6} = x_{4} + x_{2} + 0 = 2 + 1 + 0 = 3[/tex]

[tex]x_{8} = x_{6} + x_{4} + 0 = 3 + 2 = 6[/tex]

[tex]x_{10} = x_{8} + x_{6} + x_{2} = 6 + 3 + 1 = 10[/tex]

[tex]x_{12} = x_{10} + x_{8} + x_{4} = 10 + 6 + 2 = 18[/tex]

Thus clearly there are 18 ways to travel 12 miles

The recurrence relation for the number of ways to go n miles by foot at different speeds is An = An-2 + An-4 + An-8. To find the number of ways to go 12 miles, we can compute the values of An using the recurrence relation.

To find a recurrence relation for the number of ways to go n miles by foot walking at 2 miles per hour or jogging at 4 miles per hour or running at 8 miles per hour, we need to consider the choices made at each hour. Let's denote the number of ways to go n miles as An. At each hour, there are three choices: walk, jog, or run. If we choose to walk, we have An-2 ways to go n miles. If we choose to jog, we have An-4 ways. If we choose to run, we have An-8 ways. Therefore, the recurrence relation is:

An = An-2 + An-4 + An-8

To find the number of ways to go 12 miles, we need to find A12. Starting from the base cases, A0 = 1, A2 = 1, A4 = 2, and A8 = 4. Using the recurrence relation, we can compute the values of A12 as follows:

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To solve this problem it is necessary to apply the concepts given in the kinematic equations of movement description.

From the perspective of angular movement, we find the relationship with the tangential movement of velocity through

[tex]\omega = \frac{v}{R}[/tex]

Where,

[tex]\omega =[/tex]Angular velocity

v = Lineal Velocity

R = Radius

At the same time we know that the acceleration is given as the change of speed in a fraction of the time, that is

[tex]\alpha = \frac{\omega}{t}[/tex]

Where

[tex]\alpha =[/tex]Angular acceleration

[tex]\omega =[/tex] Angular velocity

t = Time

Our values are

[tex]v = 60\frac{km}{h} (\frac{1h}{3600s})(\frac{1000m}{1km})[/tex]

[tex]v = 16.67m/s[/tex]

[tex]r = 0.25m[/tex]

[tex]t=6s[/tex]

Replacing at the previous equation we have that the angular velocity is

[tex]\omega = \frac{v}{R}[/tex]

[tex]\omega = \frac{ 16.67}{0.25}[/tex]

[tex]\omega = 66.67rad/s[/tex]

Therefore the angular speed of a point on the outer edge of the tires is 66.67rad/s

At the same time the angular acceleration would be

[tex]\alpha = \frac{\omega}{t}[/tex]

[tex]\alpha = \frac{66.67}{6}[/tex]

[tex]\alpha = 11.11rad/s^2[/tex]

Therefore the angular acceleration of a point on the outer edge of the tires is [tex]11.11rad/s^2[/tex]

Answer:

Net filtration pressure is 2 mmHg

Solution:

As per the question:

Glomerular Pressure, [tex]P_{gh} = 46\ mmHg[/tex]

Colloidal Osmotic Pressure, [tex]P_{co} = 34\ mmHg[/tex]

Capsular Hydrostatic Pressure, [tex]P_{ch} = 10\ mmHg[/tex]

Now,

The net filtration Pressure is given by:

[tex]P_{net} = P_{gh} - (P_{co} + P_{ch})[/tex]

[tex]P_{net} = 46 - (34 + 10) = 2\ mmHg[/tex]

The net filtration pressure is 2 mmHg, which is lower than the normal value. This decrease would result in a decrease in the GFR.

The net filtration pressure is calculated by subtracting the colloid osmotic pressure and the capsular hydrostatic pressure from the glomerular hydrostatic pressure. In this case, the net filtration pressure would be 46 mmHg - 34 mmHg - 10 mmHg = 2 mmHg. This differs from the normal net filtration pressure of around 10 mmHg. A decrease in net filtration pressure would lead to a decrease in the glomerular filtration rate (GFR).

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Answer:

C. No work is done if the direction of motion is perpendicular to the force.

Explanation:

We know that work is the dot product of force and displacement.

Let's take the angle between the force and the displacement = θ

W= F . d cosθ

F=Force  , d=Displacement  

If θ = 0° then W= F.d

If θ = 90° then W= 0

So we can say that when force is perpendicular to the displacement then the work done by force will be zero.

Therefore the answer is C.

To determine the percentage of the human body above the surface in the Dead Sea, compare the body's density with the water's density. Use the formula to calculate the percentage.

To determine the percentage of the human body that would be above the surface in the Dead Sea, we need to compare the density of the human body with the density of the water. The average density of the human body is 945 kg/m3 after inhaling and 1020 kg/m3 after exhaling, whereas the density of the Dead Sea water is about 1230 kg/m3. Since the density of the human body is lower than the density of the Dead Sea water, a certain percentage of the body would be above the surface.

In the case of inhaling, the density of the human body is lower, so a larger percentage of the body would be above the surface compared to when exhaling. The exact percentage can be calculated using the formula:

Percentage above surface = (density of human body - density of water) / density of human body × 100

Once you plug in the values, you will get the percentage of the body above the Dead Sea surface.

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Answer:

v₀ = 13.9 10³ m / s

Explanation:

Let's analyze this exercise we can use the basic kinematics relationships to love the initial velocity and the acceleration we can look for from Newton's second law where force is gravitational attraction.

    F = m a

    G m M / x² = m dv / dt = m dv/dx  dx/dt

    G M / x² = dv/dx   v

    GM dx / x² = v dv

We integrate

    v² / 2 = GM (-1 / x)

We evaluate between the lower limits where x = Re = 6.37 10⁶m  and the velocity v = vo and the upper limit x = 2.50 10⁸m  with a velocity of v = 8.50 10³ m/s

    ½ ((8.5 10³)² - v₀²) = GM (-1 /(2.50 10⁸) + 1 / (6.37 10⁶))

    72.25 10⁶ - v₀² = 2 G M (+0.4 10⁻⁸ - 1.57 10⁻⁷)

    72.25 10⁶ - v₀² = 2 6.63 10⁻¹¹ 5.98 10²⁴ (-15.3  10⁻⁸)

    72.25 10⁶ - v₀² = -1.213  10⁸

    v₀² = 72.25 10⁶ + 1,213 10⁸

    v₀² = 193.6 10⁶

    v₀ = 13.9 10³ m / s

Answer:

2.79 m

Explanation:

Use the static equilibrium condition, net torque actin on the system is zero.

[tex]\sum \tau= 0[/tex]

[tex]T(Lsin\theta)- Mg\frac{L}{2}-mgx=0[/tex]

solve for the distance of the person from the wall x

[tex]x= \frac{TLsin\theta-Mg(L/2)}{Mg}[/tex]

now putting the values we get

[tex]x= \frac{1350\times4.25sin30-47\times9.81\times(4.25/2)}{69\times9.80}[/tex]

= 2.79 m

Answer:

[tex]T = m*g*cos\theta[/tex]

Explanation:

Since tarzan moves in a circular trajectory we can sum all forces on the centripetal-axis:

[tex]T - m*g*cos\theta = m*a_c[/tex]

[tex]T - m*g*cos\theta = m*V^2/R[/tex]   Since the speed is zero:

[tex]T - m*g*cos\theta = 0[/tex]

[tex]T = m*g*cos\theta[/tex] Having Tarzan's mass we could calculate the module of the tension in the vine.

The tension in Tarzan's vine can be determined at the highest point of his swing, when his velocity is zero, using Newton's laws of motion. At this point, the tension's vertical component equals his weight, while the horizontal component, providing centripetal force, is zero. The equation T=mg/cosθ can be used to calculate the tension.

The subject of your question involves physics, particularly mechanics. The tension in the vine at the highest point of Tarzan's swing, when his velocity is zero and the vine makes an angle of 40 degrees with the vertical, can be found using Newton's second law of motion, which can be written as ΣF=ma, where F represents force, m is mass, and a is acceleration.

At the highest point of the swing, the only forces acting on Tarzan are the tension in the vine (T) and his weight (mg), where m is his mass and g is the acceleration due to gravity. These two forces result in a net force that provides the centripetal acceleration necessary for Tarzan to change direction and swing back down. We have:

So we can get the tension in vine (T) from the equation Tcosθ=mg. Rearranging, we have T=mg/cosθ.

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Answer:

a) [tex] v = c \cdot 0.04 = 1.2\cdot 10^{7} m/s[/tex]

b) [tex] v = c \cdot 0.71 = 2.1\cdot 10^{8} m/s[/tex]

c) [tex] v = c \cdot 0.994 = 2.97\cdot 10^{8} m/s[/tex]

d) [tex] v = c \cdot 0.999 = 2.997\cdot 10^{8} m/s[/tex]

e) [tex] v = c \cdot 0.9999 = 2.999\cdot 10^{8} m/s[/tex]

Explanation:

At that energies, the speed of proton is in the relativistic theory field, so we need to use the relativistic kinetic energy equation.

[tex] KE=mc^{2}(\gamma -1) = mc^{2}(\frac{1}{\sqrt{1-\beta^{2}}} -1)[/tex]           (1)

Here β = v/c, when v is the speed of the particle and c is the speed of light in vacuum.

Let's solve (1) for β.

[tex] \beta = \sqrt{1-\frac{1}{\left (\frac{KE}{mc^{2}}+1 \right )^{2}}} [/tex]

We can write the mass of a proton in MeV/c².

[tex] m_{p}=938.28 MeV/c^{2} [/tex]

Now we can calculate the speed in each stage.

a) Cockcroft-Walton (750 keV)

[tex] \beta = \sqrt{1-\frac{1}{\left (\frac{0.75 MeV}{938.28 MeV}+1 \right )^{2}}} [/tex]

[tex] \beta = 0.04 [/tex]

[tex] v = c \cdot 0.04 = 1.2\cdot 10^{7} m/s[/tex]

b) Linac (400 MeV)

[tex] \beta = \sqrt{1-\frac{1}{\left (\frac{400 MeV}{938.28 MeV}+1 \right )^{2}}} [/tex]

[tex] \beta = 0.71 [/tex]

[tex] v = c \cdot 0.71 = 2.1\cdot 10^{8} m/s[/tex]

c) Booster (8 GeV)

[tex] \beta = \sqrt{1-\frac{1}{\left (\frac{8000 MeV}{938.28 MeV}+1 \right )^{2}}} [/tex]

[tex] \beta = 0.994 [/tex]

[tex] v = c \cdot 0.994 = 2.97\cdot 10^{8} m/s[/tex]

d) Main ring or injector (150 Gev)

[tex] \beta = \sqrt{1-\frac{1}{\left (\frac{150000 MeV}{938.28 MeV}+1 \right )^{2}}} [/tex]

[tex] \beta = 0.999 [/tex]

[tex] v = c \cdot 0.999 = 2.997\cdot 10^{8} m/s[/tex]

e) Tevatron (1 TeV)

[tex] \beta = \sqrt{1-\frac{1}{\left (\frac{1000000 MeV}{938.28 MeV}+1 \right )^{2}}} [/tex]

[tex] \beta = 0.9999 [/tex]

[tex] v = c \cdot 0.9999 = 2.999\cdot 10^{8} m/s[/tex]

Have a nice day!

Answer:

h = 15.34 m

Explanation:

given,

tuning fork vibration = 513 Hz

speed of sound = 343 m/s

frequency after deflection = 489 Hz

the source (the fork) moves away from the observer, its speed increases and hence the apparent frequency decreases

[tex]f_{apparent} = \dfrac{v}{v+u}f_0[/tex]

[tex]489 = \dfrac{343}{343+u}\times 513[/tex]

[tex]0.953 = \dfrac{343}{343+u}[/tex]

[tex]343+u = \dfrac{343}{0.953}[/tex]

[tex]343+u = 359.92[/tex]

u = 16.92 m/s

height of the building

v² = u² + 2 g s

16.92² = 2 x 9.8 x h

h = 14.61 m

time taken by sound to reach observer

[tex]t = \dfrac{14.61}{343}[/tex]

[tex]t =0.0426\ s[/tex]

in this time tuning fork has fallen one more now,

[tex]h' = u t + \dfrac{1}{2}gt^2[/tex]

[tex]h' = 16.92\times 0.0426 + \dfrac{1}{2}\times 9.8 \times 0.0426^2[/tex]

h' = 0.7296 m = 0.73 m

total distance

      h = 14.61 + 0.73

      h = 15.34 m

Final answer:

The tuning fork has fallen through a distance of 2,933.77 m.

Explanation:

To calculate the distance fallen by the vibrating tuning fork, we can use the equation:

f' = f(v/(v + vo))

where f' is the observed frequency, f is the original frequency of the tuning fork, v is the speed of sound, and vo is the velocity of the tuning fork as it falls.

Plugging in the given values: f = 513 Hz, v = 343 m/s, and f' = 489 Hz, we can rearrange the equation to solve for vo:

vo = v(f/f' - 1)

Substituting the values and solving for vo:

vo = 343((513/489) - 1) = 12.3 m/s

Since the equation for distance fallen is:

d = vo*t + (1/2)*g*t^2

and we are assuming the tuning fork is dropped from rest, we can simplify the equation to:

d = (1/2)*g*t^2

where g is the acceleration due to gravity and t is the time the tuning fork has been falling. Since we are solving for distance, we can rearrange the equation to solve for t:

t = sqrt(2*d/g)

Substituting the values and solving for t:

t = sqrt(2*d/9.8) = sqrt(2*d/10) = sqrt(d/5)

Now, we can substitute the value of vo and the known value of f' = 489 Hz into the equation:

vo*t = sqrt(d/5)*12.3 = d

Plugging in the known value of f' = 489 Hz and solving for d:

sqrt(d/5)*12.3 = d/489

Squaring both sides of the equation and solving for d:

d = (12.3^2 * 489^2)/(489^2 - 12.3^2*5) = 2933.77 m

The force on the wall can be calculated using the macroscopic balance for momentum. We need to calculate the mass flow rate of water and use the equation: mass flow rate = density × velocity × cross-sectional area of the nozzle. With the given values, we can find the force on the wall.

The force on the wall can be computed using the macroscopic balance for momentum. Since water discharges from a nozzle and travels horizontally, the initial momentum of the water is zero. The change in momentum of the water is equal to the force exerted on the wall. To calculate the force, we need to find the mass flow rate of the water. Assuming the water is an ideal fluid with a flat velocity profile, we can use the equation: mass flow rate = density × velocity × cross-sectional area of the nozzle. Plugging in the given values, we can then calculate the force on the wall.

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Final answer:

To calculate the force exerted on the wall by the water discharging from the nozzle, we use the mass flow rate and velocity of the water to find the momentum change. This calculation applies the macroscopic momentum balance principle and yields the force in Newtons.

Explanation:

The student is asking to calculate the force on a wall when water from a nozzle hits it horizontally. To find this force, we can use the macroscopic momentum balance principle which states that the rate of change of momentum of the fluid is equal to the sum of the external forces applied to the fluid.

As gravity and frictional effects are negligible, the only external force is the force on the wall exerted by the water. Key information given includes the diameter of the nozzle (12 mm) and the velocity of the water (6.0 m/s).

To calculate the force, we need to find the mass flow rate of the water, which can be determined by multiplying the density of water (ρ = 1000 kg/m³ at standard conditions) by the cross-sectional area of the nozzle (A) and the velocity (v) of the water.

The cross-sectional area, A, of the nozzle is A = π * (d/2)², where d is the diameter of the nozzle. Substituting the given values, the area A is π * (0.012/2)² m². The mass flow rate (ṭ) then is ρ * A * v.

With the mass flow rate known, the force (F) exerted on the wall can be calculated by taking the rate of change of momentum as ṭ * v, because the water's velocity is reduced to zero upon hitting the wall, thus the change in momentum is just the initial momentum. Therefore, F = ṭ * v.

After calculating the mass flow rate and applying the equation for force, the student will get the desired force in Newtons (N) on the wall.

Answer:

 W = 1.8 J

Explanation:

given,

mass of square plate = 0.6 Kg

length of side of square plate = 0.15 m

speed is change from   0 to 40.0 rad/s

work done = ?

moment of inertia through the plane perpendicular to plane

[tex]I = \dfrac{Mr^2}{6}[/tex]

r will be equal to 0.15 m

[tex]I = \dfrac{0.6 \times 0.15^2}{6}[/tex]

[tex]I =0.00225\ kgm^2[/tex]

work done is equal to change in kinetic energy

 [tex]W = \dfrac{1}{2}I(\omega_f^2-\omega_i^2)[/tex]

 [tex]W = \dfrac{1}{2}\times 0.00225(40^2-0^2)[/tex]

   W = 1.8 J

work done when speed changes from 0 to 40 rad/s is equal to  W = 1.8 J

The speed of the car is 60 m/s

Explanation:

The speed of an object is a scalar quantity telling "how fast" the object is moving, regardless of its direction. It can be calculated as follows:

[tex]speed=\frac{d}{t}[/tex]

where

d is the distance covered

t is the time taken

For the car in this problem, we have

[tex]d=108 km =1.08\cdot 10^5 m[/tex] is the distance covered

[tex]t=30 min \cdot 60 =1800 s[/tex] is the time taken

Subsituting into the equation, we find the speed:

[tex]speed=\frac{1.08\cdot 10^5}{1800}=60 m/s[/tex]

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Answer:

a. 1.12m/s

b. 0.34m/s

Explanation:

Let Vgi = the velocity of the girl relative to the ice

Let Vgp = the velocity of the girl relative to the plank

Let Vpi = the velocity of the plank relative to the ice

Vgi = Vgp + Vpi

From the question, Vgp = 1.46/s

So, Vgi = 1.46 + Vpi

Conservation of Momentum (Relative to the ice), we have

Initial Momentum = Final Momentum

Because the girl and the plank are at rest initially, the initial Momentum = 0

So, 0 = Mg * Vgi + Mp * Vpi

Where Mg = Mass of the girl = 45.8kg

Mo = Mass of the Plank = 151kg

Make Vpi the subject of the formula, we then have

-Mg*Vgi = Mp*Vpi ---------- Divide through by Mp

Vpi = -Mg * Vgi/Mp

Vpi =( -45.8 * Vgi)/151

Vpi = -45.8Vgi/151

Remember that, we have (Vgi = 1.46 + Vpi)

We then substitute the expression of Vpi in the above equation

That is;

Vgi = 1.46 + (-45.8Vgi/151) ------- Open the bracket

Vgi = 1.46 - 45.8Vgi/151 ----------- Multiply through by 151

151 * Vgi = 151 * 1.46 - 45.8Vgj

151Vgi = 220.46 - 45.8Vgi --------- Collect like terms

151Vgi + 45.8Vgi = 220.46

196.8Vgi = 220.46 --------------- Divide through by 196.8

Vgi = 220.46/196.6

Vgi = 1.1213631739572736

Vgi = 1.12 m/s (Approximated)

So, the velocity of the girl relative to the ice is 1.12m/s

b. Velocity of the plank, relative to the ice

We can solve this using (Vgi = 1.46 + Vpi)

All we need to do is substitute 1.12 for Vgi in the equation

So, we have

Vgi = 1.46 + Vpi becomes

1.12 = 1.46 + Vpi -------- Collect like terms

1.12 - 1.46 = Vpi

-0.34 = Vpi

So, the Velocity of the plank is 0.34m/s to the left

Answer: (a) 2.095 m/s (b) 0.635 m/s

Explanation: The initial momentum is = 0. Also, the final momentum is = 0. At the initial momentum she is at rest hence it is = 0. Whereas, the final momentum is = 0 because it has to equal the initial momentum.

This helps us to understand that when final momentum is = 0 then the girl's  direction would cancel the affect of the plank's momentum in the opposite direction.

Therefore,

Vg = Velocity of girl

Vp= Velocity of plank

(45.8)(Vg) + (151)(Vp) = 0

Girl's velocity relative to the plank is 1.46 m/s

Therefore, Vg - Vp = 1.46

Vg = 1.46 + Vp

(45.8)(1.46 + Vp) = 151Vp

66.868 + 45.8Vp = 151Vp

66.868 = 105.2Vp

Vp = 0.635 m/s

Vg = 1.46 + 0.635

Vg = 2.095

Answer:

D) 50%

Explanation:

According to conservation of momentum:

[tex]m_{1} v_{1i} + m_{2} v_{2i} = m_{1} v_{1f} + m_{2} v_{2f}[/tex]

Assuming that the second taffy started at rest, both pieces have the same mass and that they combined after the collision, their final velocity is:

[tex]m v_{1i} = (m+m) v_{f}\\v_{f} = 0.5 v_{1i}[/tex]

The initial kinetic energy of the system is:

[tex]E_{ki} = \frac{m*v_{1i}^2}{2}[/tex]

Since the second taffy was not moving, it had no kinetic energy at first.

The initial kinetic energy of the system is:

[tex]E_{kf} = \frac{2m*v_{f}^2}{2}\\E_{kf} = \frac{2m*(0.5v_{1i})^2}{2}\\E_{kf} = \frac{0.5m*v_{1i}^2}{2}[/tex]

The percentage of kinetic energy that becomes heat is given by:

[tex]H=1 - \frac{E_{kf}}{E_{ki}}\\H=1 - \frac{\frac{0.5m*v_{1i}^2}{2}}{\frac{m*v_{1i}^2}{2}}\\\\H=1- 0.5 = 0.5[/tex]

THerefore, 50% of the kinetic energy becomes heat

In inelastic collisions, such as the described scenario where two objects stick together, 75% of the initial kinetic energy is lost to other forms, including heat.

The question involves a type of collision known as an inelastic collision, where two objects stick together post-collision. In these scenarios, while momentum is conserved, kinetic energy is not. The initial kinetic energy before the collision gets partly converted into other forms, such as heat.

Considering a scenario where a moving object collides with a stationary one of identical mass and they stick together, the resulting velocity of the combined mass is half of the original moving object's velocity. This result emerges from the conservation of momentum. Since kinetic energy is proportional to the square of the velocity, the kinetic energy of the combined object will be one-fourth of the initial kinetic energy. Consequently, 75% of the initial kinetic energy is converted into other forms of energy like heat, which answers the original question.

Answer:

a) J = 13067 kg*m/s

b) J = 9240 kg*m/s

c) F = 2666.73 N

d) F = 21488.37 N

e) 135°

Explanation:

We know that the impulse could be calculated by:

J = ΔP

where ΔP is the change in the linear momentum:

ΔP = [tex]P_f -P_i[/tex]

Also:

P = MV

Where M is the mass and V is the velocity.

so:

J= [tex]MV_f-MV_i[/tex]

where [tex]V_f[/tex] is the final velocity and [tex]V_i[/tex] is the inicial velocity

a)The impluse from turn is:

[tex]J_x=MV_{fx}-MV_{ix}[/tex]

[tex]J_y=MV_{fy}-MV_{iy}[/tex]

On the turn, [tex]V_{ix}=0[/tex] and [tex]V_{fy}=0[/tex], the magnitude of the impulse on direction x and y are:

[tex]J_x=9240 kg*m/s[/tex]

[tex]J_y=-9240 kg*m/s[/tex]

So, using pythagoras theorem the magnitude of the impulse is:

J = [tex]\sqrt{9240^2+9240^2}[/tex]

J = 13067 kg*m/s

b) The impluse from the collision is:

[tex]J=MV_{f}-MV_{i}[/tex]

J = 0 - (1400)(6.6)

J = 9240 kg*m/s

c) Using the next equation:

FΔt = J

where F is the force, Δt is the time and J is the impulse.

Replacing J by the impulse due to the turn, Δt by 4.9s and solving for F we have that:

F = J / Δt

F = 13067 / 4.9 s

F = 2666.73 N

d) At the same way, replacing J by the impulse during the collision, Δt by 0.43s and solving for F we have that:

F = J / Δt

F = 9240 / 0.43

F = 21488.37 N

e) The force have the same direction than the impulse due to the turn, Then, if the impulse have a direction of -45°, the force have -45° or 135°

Answer:

22.02°C

Explanation:

P = Power output = 30 kW

Volume flow rate of water = [tex]1.49\times 10^{-4}\ m^3/s[/tex]

S = Specific heat of water = 4.186 kJ/mol°C

[tex]T_i[/tex] = Initial temperature = 10°C

[tex]T_f[/tex] = Final temperature

Total volume of water is

[tex]v=4\times 1.49\times 10^{-4}\\\Rightarrow v=5.96\times 10^{-4}\ m^3/s[/tex]

Mass flow rate is

[tex]m=\rho\times v\\\Rightarrow m=1000\times 5.96\times 10^{-4}\\\Rightarrow m=0.596\ kg/s[/tex]

Heat is given by

[tex]\frac{q}{t}=\frac{m}{t}S\Delta T\\\Rightarrow 30=0.596\times 4.186(T_f-10)\\\Rightarrow T_f=\frac{30}{0.596\times 4.186}+10\\\Rightarrow T_f=22.02\ ^{\circ}C[/tex]

The maximum possible temperature of the hot water that each showering resident receives is 22.02°C

To estimate the speed of the swinging orangutan, we can use the principles of rotational motion and trigonometry. By considering the angle of swing and the length of the orangutan's arm, we can calculate the horizontal distance traveled in one swing. However, without the time taken for one swing, we cannot provide an exact value for the speed.

To estimate the speed at which the orangutan is moving while swinging from branch to branch, we can use the principles of rotational motion and trigonometry. When the orangutan swings to a 20° angle, it forms a right triangle with the vertical height being 0.90 m and the hypotenuse being the length of the arm. By using the sine function, we can find the horizontal distance traveled by the orangutan in one swing.

Using the formula sin(20°) = horizontal distance / 0.90 m, we can rearrange the equation to find the horizontal distance traveled:

horizontal distance = 0.90 m * sin(20°)

Since the orangutan takes one swing immediately after another, the time taken for one swing is negligible. Therefore, the speed at which the orangutan is moving can be calculated by dividing the horizontal distance by the time taken for one swing. As the time taken is not given in the question, we cannot provide an exact value for the speed.

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To estimate the speed of an orangutan swinging, we use the physics formula for the speed of an object in pendulum motion. Under the given conditions of a 20° swing and arm length of 0.9 m, the orangutan's speed is approximately 2.54 m/s.

The behavior you're describing, where an orangutan swings like a pendulum beneath handholds, involves the study of pendulum motion, which is a concept in physics. Under simplified physics assumptions, we can model the swinging motion using the formula for the speed (v) of an object in pendulum motion: v = √(2*g*L(1-cos(θ))), where g is the acceleration due to gravity (9.8 m/s^2), L is the length of the pendulum (which we can approximate as the length of an orangutan's arm, 0.9 m), and θ is the angle swung through (20° or 0.349 radians after conversion).

Plugging the given values into the formula gives us: v = √(2*9.8*0.9(1-cos(0.349))), so the speed of an orangutan swinging would be approximately 2.54 m/s.

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so it is not bad not good. pefectly balanced just as all things should be

The radius of the orbits and the rocket's mass, The changes would involve transitioning from the elliptical to the circular orbit, taking into account Earth's gravitational pull.

This question delves into detail regarding the physics of celestial movements, in particular, satellites and their orbits. The problem pertains to the transition from an elliptical trajectory to a circular one.

The change in speed at point A would need to equal the difference between the speed in the elliptical orbit at the apogee and the speed necessary to maintain the desired circular orbit, factoring in the gravitational pull of Earth. The change in speed at point B would need to address any residual speed from the elliptical orbit to ensure the rocket's speed aligns with the necessary speed to maintain the circular orbit around the Earth.

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Answer:

V = 8.26 m/s

Explanation:

The sum of forces on the centripetal-axis is:

[tex]Ff = m*a_c[/tex]

Maximum speed happens when friction force is maximum, so:

[tex]\mu*N = m*V^2/R[/tex]

[tex]\mu*m*g = m*V^2/R[/tex]

[tex]V=\sqrt{R*\mu*g}[/tex]

V = 8.26 m/s

The maximum speed that he can travel through the arc without slipping is mathematically given as

v = 9.93m/s

Question Parameters:

Generally the equation for the Maximum velocity   is mathematically given as

[tex]\frac{mv^2}{r} = \mu_s mg[/tex]

Therefore

[tex]v = \sqrt{(0.63)(9.8)(16)}[/tex]

v = 9.93m/s

Therefore, the maximum speed that he can travel through the arc without slipping is

v = 9.93m/s

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To develop this problem it is necessary to apply the Rayleigh Criterion (Angular resolution)criterion. This conceptos describes the ability of any image-forming device such as an optical or radio telescope, a microscope, a camera, or an eye, to distinguish small details of an object, thereby making it a major determinant of image resolution.  By definition is defined as:

[tex]\theta = 1.22\frac{\lambda}{d}[/tex]

Where,

[tex]\lambda[/tex]= Wavelength

d = Width of the slit

[tex]\theta[/tex]= Angular resolution

Through the arc length we can find the radius, which would be given according to the length and angle previously described.

The radius of the beam on the moon is

[tex]r = l\theta[/tex]

Relacing [tex]\theta[/tex]

[tex]r = l(\frac{1.22\lambda}{d})[/tex]

[tex]r = 1.22\frac{l\lambda}{d}[/tex]

Replacing with our values we have that,

[tex]r = 1.22*(\frac{(384*10^3km)(\frac{1000m}{1km})(550*10^{-9}m)}{7*10^{{-2}}})[/tex]

[tex]r = 3680.91m[/tex]

Therefore the diameter of the beam on the moon is

[tex]d = 2r[/tex]

[tex]d = 2 * (3690.91)[/tex]

[tex]d = 7361.8285m[/tex]

Hence, the diameter of the beam when it reaches the moon is 7361.82m

Final answer:

Using diffraction formulas, it is estimated that a flashlight beam with an aperture of 7.0cm and an average wavelength of 550nm would have a diameter of approximately 3.7 km when it reaches the Moon.

Explanation:

To estimate the diameter of a flashlight beam when it reaches the Moon, we'll use the concept of diffraction spreading, as the beam is assumed to spread due to diffraction only. According to the diffraction formula for a circular aperture, θ = 1.22 λ/D. The angle θ is the angle of the spread, λ is the average wavelength of the light, and D is the diameter of the aperture through which the light passes.

Using the provided average wavelength of λ = 550nm and an aperture diameter of D = 7.0cm for the flashlight, we can calculate the minimum angular spread of the beam, θ.

θ = 1.22 × (550 x 10^-9 m) / (7.0 x 10^-2 m) = 0.0000095714 radians.

Then, to find out how broad the beam would be when it reaches the Moon, we use the angular spread to calculate the diameter of the beam at the distance to the Moon, which is 384,000 km (or 384 x 10^6 m):

Beam Diameter on the Moon = θ × Distance to the Moon = 0.0000095714 radians × 384 x 10^6 m = 3677.388 m.

Therefore, the flashlight's beam would have an estimated diameter of approximately 3.7 km when it reaches the Moon, considering only the diffraction spreading.

The question involves using the Conservation of Energy to find the speed of an object in simple harmonic motion. The concept relies on the fact that the total energy in the harmonic system remains constant, allowing us to determine the kinetic energy of the object and consequently its speed.

This question can be answered using the Conservation of Energy principle.

In the given problem, the object attached to the spring performs simple harmonic motion, meaning that the total energy of the system (kinetic energy + potential energy) remains constant.

At any instant in time, the total mechanical energy of the object will be:

E = 1/2 k A²

Where.

E is total energy, k is the spring constant, and A is the system's amplitude.

When the displacement of the object is at x = 0.040 m, the potential energy of the object (U) will be:

U = 1/2  k x²

At this point, the kinetic energy (K) can be found by subtracting the potential energy (U) from the total energy (E).

K = E - U

Knowing that the kinetic energy (K) is also given by 1/2 m  v² (where m is mass and v is speed), we can rearrange this equation to find the speed of the object:

v = sqrt((2K)/m)

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The speed of the object at a displacement of 0.040 m is 0.49 m/s.

Calculating Speed in Simple Harmonic Motion Using Conservation of Energy

To find the speed of a 0.20 kg object attached to a spring (k = 10 N/m) at a displacement of 0.040 m, we can use the principle of conservation of energy.

Calculate the total mechanical energy (E) in the system using the amplitude (A):

E = 1/2 x k x [tex]A^2[/tex]

E = 1/2 x 10 N/m x [tex](0.080 m)^2[/tex]

E = 0.032 J

Determine the potential energy (U) at the displacement (x = 0.040 m):

U = [tex]1/2 \times k \times x^2[/tex]

U = [tex]1/2 \times 10 N/m \times (0.040 m)^2[/tex]

U = 0.008 J

Find the kinetic energy (K) at that displacement:

K = E - U

K = 0.032 J - 0.008 J

K = 0.024 J

Calculate the speed (v) using the kinetic energy:

K = 1/2 x m x [tex]v^2[/tex]

[tex]0.024 J = 1/2 \times 0.20 kg \times v^2[/tex]

[tex]v^2 = 0.24 m^2/s^2[/tex]

v = 0.49 m/s

Therefore, the speed of the object at a displacement of 0.040 m is 0.49 m/s.

Answer:

Explanation:

F(r) = Kr/|r|3

= k / r²

Work done

= ∫F(r) dr

= ∫ k / r²  dr

limit from

r₁ = 2 to r₂ = √ ( 2² + 4² + 5² ) = 6.7

Work done

∫ k / r² dr

[ - k / r ]

Taking limit from 2 to 6.7

Work done  = K ( - 1 / 6.7 + 1 / 2 )

= 0.35 K

The work done as a particle moves along a straight line from one point to another in the presence of an electric field is 0.

To find the work done as a particle moves along a straight line from one point to another in the presence of an electric field, you can use the following formula for the work done:

[tex]\[W = \int_{\mathbf{r_1}}^{\mathbf{r_2}} \mathbf{F} \cdot d\mathbf{r}\][/tex]

Where:

- [tex]\(W\)[/tex] is the work done.

- [tex]\(\mathbf{F}\)[/tex] is the force acting on the particle.

- [tex]\(d\mathbf{r}\)[/tex] is the infinitesimal displacement vector along the path of the particle.

- [tex]\(\mathbf{r_1}\)[/tex] and [tex]\(\mathbf{r_2}\)[/tex] are the initial and final positions of the particle.

In this case, the force [tex]\(\mathbf{F}\)[/tex] is given as [tex]\(F(\mathbf{r})[/tex] = [tex]\frac{K\mathbf{r}}{|\mathbf{r}|^3}\)[/tex], and the particle moves along a straight line from [tex]\(\mathbf{r_1} = (2, 0, 0)\)[/tex] to [tex]\(\mathbf{r_2} = (2, 4, 5)[/tex].

We can rewrite [tex]\(\mathbf{F}[/tex] as a vector:

[tex]\[\mathbf{F} = \frac{K}{|\mathbf{r}|^3}\mathbf{r}\][/tex]

Now, let's calculate the differential displacement [tex]\(d\mathbf{r}\)[/tex] along the path from [tex]\(\mathbf{r_1}\)[/tex] to [tex]r_2[/tex]:

[tex]\[d\mathbf{r} = (dx, dy, dz)\][/tex]

Now, let's calculate the dot product [tex]\(\mathbf{F} \cdot d\mathbf{r}\)[/tex]:

[tex]\[\mathbf{F} \cdot d\mathbf{r} = \frac{K}{|\mathbf{r}|^3}\mathbf{r} \cdot (dx, dy, dz)\][/tex]

Since the path is a straight line, [tex]\(\mathbf{r}\) and \(d\mathbf{r}\)\\[/tex] are parallel, so[tex]\(\mathbf{r} \cdot d\mathbf{r} = |\mathbf{r}||d\mathbf{r}|\cos(0^\circ) = |\mathbf{r}||d\mathbf{r}|\)[/tex].

Therefore, we have:

[tex]\[\mathbf{F} \cdot d\mathbf{r} = \frac{K}{|\mathbf{r}|^3}|\mathbf{r}||d\mathbf{r}| = \frac{K}{|\mathbf{r}|^2}|d\mathbf{r}|\][/tex]

Now, we can integrate [tex]\(\mathbf{F} \cdot d\mathbf{r}\)[/tex] from [tex]\(\mathbf{r_1}\)[/tex] to [tex]\(\mathbf{r_2}\)[/tex] along the straight line path:

[tex]\[W = \int_{\mathbf{r_1}}^{\mathbf{r_2}} \frac{K}{|\mathbf{r}|^2}|d\mathbf{r}|\][/tex]

The integration limits and the path are straight, so this simplifies to:

[tex]\[W = K \int_{2}^{2} \frac{1}{(2^2+4^2+5^2)^{3/2}}|d\mathbf{r}|\][/tex]

Since the particle moves from [tex]\(\mathbf{r_1}\) to \(\mathbf{r_2}\)[/tex], the displacement [tex]\(|d\mathbf{r}| = |\mathbf{r_2} - \mathbf{r_1}| = |(0, 4, 5)| = \sqrt{0^2 + 4^2 + 5^2} = \sqrt{41}\)[/tex].

So, the work done is:

[tex]\[W = K \int_{2}^{2} \frac{1}{(2^2+4^2+5^2)^{3/2}}\sqrt{41} dr\][/tex]

Since the limits of integration are the same, the integral evaluates to zero. Therefore, the work done as the particle moves along a straight line from (2, 0, 0) to (2, 4, 5) is zero.

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To find the translational speed of the bowling ball at the top of the vertical rise, we can use the principle of conservation of mechanical energy.

To find the translational speed of the bowling ball at the top of the vertical rise, we can use the principle of conservation of mechanical energy. Initially, the ball has kinetic energy due to its translational speed. As it moves up the vertical rise, its gravitational potential energy increases. Since there is no friction, we can assume there is no loss of energy.

Using the equation for conservation of mechanical energy:

Initial kinetic energy = final potential energy

0.5 * m * v^2 = m * g * h

Where m is the mass of the ball, v is the translational speed at the bottom, g is the acceleration due to gravity, and h is the height of the vertical rise.

Plugging in the given values:

0.5 * m * (8.57 m/s)^2 = m * 9.8 m/s^2 * 0.760 m

Simplifying and solving for the mass:

m = (9.8 m/s^2 * 0.760 m) / (0.5 * (8.57 m/s)^2)

Finally, we can plug the mass back into the initial equation to find the translational speed at the top of the rise.

Answer:

ωf = 4.53 rad/s

Explanation:

By conservation of the angular momentum:

Ib*ωb = (Ib + Ic)*ωf

Where

Ib is the inertia of the ball

ωb is the initial angular velocity of the ball

Ic is the inertia of the catcher

ωf is the final angular velocity of the system

We need to calculate first Ib, Ic, ωb:

[tex]Ib = mb*(L/2)^2=0.15*(1.2/2)^2=0.054 kg.m^2[/tex]

[tex]Ic = mc/12*L^2=2.2/12*1.2^2=0.264 kg.m^2[/tex]

ωb = Vb / (L/2) = 16 / (1.2/2) = 26.67 m/s

Now, ωf will be:

[tex]\omega f = \frac{Ib*\omega b}{Ib + Ic}  = 4.53rad/s[/tex]

The change in magnetic flux when the coil is rotated is equivalent to the initial flux through the coil. The magnitude of the average induced emf is found by dividing the change in magnetic flux by the time over which it occurs, multiplied by the number of turns in the coil, following Faraday's law.

The question is about applying Faraday's law of electromagnetic induction to calculate the change in magnetic flux and the induced electromotive force (emf).

Part A: Change in Magnetic Flux

The magnetic flux \\(\Phi\\) through a coil is given by \\(\Phi = B \cdot A \cdot \cos(\theta)\\), where \\(B\\) is the magnetic field strength, \\(A\\) is the area of the coil, and \\(\theta\\) is the angle between the field and the normal to the coil's plane. Initially, with the coil parallel to the field, \\(\theta = 0\\) degrees, so \\(\cos(\theta) = 1\\), and the initial flux is \\(\Phi_{initial} = B \cdot A\\). After rotation, \\(\theta = 90\\) degrees, so \\(\cos(\theta) = 0\\), and the final flux is \\(\Phi_{final} = 0\\). The change in flux is the final minus the initial. Therefore, the change in magnetic flux is equal to the initial flux \\(\Phi_{initial}\\).

Part B: Magnitude of the Average Induced Emf

To find the magnitude of the average induced emf, Faraday's law is used, which states that the induced emf is proportional to the rate of change of magnetic flux. It is given by the equation \\(emf = -N \cdot \Delta\Phi / \Delta t\\), where \\(N\\) is the number of turns in the coil, \\(\Delta\Phi\\) is the change in flux, and \\(\Delta t\\) is the time over which the change occurs. Plugging in the values given, including the rate of change of flux from part A, we can calculate \\(emf\\).

Answer:

a) 1.18 seconds

b) 8.6 m

c) 5.19 revolutions

d) 6.07 m/s

Explanation:

Step 1: Data given

radius of the ball = 11.0 cm

Initial speed of the ball = 8.50 m/s

The coefficient of kinetic friction between the ball and the lane is 0.210.

(a) For what length of time does the ball skid?

The velocity at time t can be written as v(t) = v0 + at

 ⇒ with v(t) = the velocity at time t

⇒ with v0 : the initial velocity = 8.50 m/s

⇒ with a = the acceleration (in m/s²)

   ⇒The acceleration (negative) due to friction: a = -µg

           ⇒ with µ = 0.210

          ⇒ with g = 9.81 m/s²

v(t) =8.5m/s - 0.21*9.81m/s² * t = 8.5 - 2.06t

Torque τ = Iα = (2m(0.11m)²/5)α = 0.00484m*α

τ = F * r = µm*g*R = 0.21 * M * 9.81m/s² * 0.11m = 0.227m

so α = 0.227m / 0.00484m = 46.9 rad/s²

angular velocity ω(t) = ωo + αt = 0 + 46.9 rad/s² * t

The ball stops sliding when v(t) = ω(t) * r

8.5 - 2.06t  = 46.9*0.11*t = 5.159t

7.219t = 8.5

t = 1.18 seconds

b) How far down the lane does it skid?

s = Vo*t + ½at² = 8.5m/s * 1.18s - ½* 2.06 m/s² * (1.18s)² = 8.6 m

c) How many revolutions does it make before it starts to roll?

The angular acceleration of the ball is:

α =  τ/I

 ⇒ with  τ = the torque experienced by the ball due the frictional force

   ⇒  τ = fk*R

α = fk*R /I

 ⇒ I = 2/5 m*R²

 ⇒ fk = µk*m*g

α = (µk*m*g*R)/(2/5mR²)

α = 5µk*g /2R

The angular displacement of the ball is:

∅ = 1/2αt²

⇒ The ball does not have an initial angular velocity

∅ =1/2*(5µk*g/2)*t²

∅ = 5µkgt²/4R

∅ = (5*0.21*9.81*1.18²)/(4*11.0 *10^-2)

∅ = 32.6 rad

Number of revolutions = 32.6 rad /2π

Number of revolutions = 5.19

(d) How fast is it moving when it starts to roll?

v = Vo + at = 8.5m/s - 2.06m/s² * 1.18s = 6.07 m/s

To find the time, distance, number of revolutions, and speed at which the bowling ball starts to roll, we can use equations of motion and the relationship between linear and angular speed.

To find the answers to the given questions, we need to use the equations of linear and rotational motion as well as the relationship between linear and angular speed for the bowling ball.

(a) To find the time for the ball to skid, we can use the equation v = u + at, where v = 0 (since the ball stops skidding), u = 8.50 m/s (initial speed), and a = μg, where μ = 0.210 (coefficient of kinetic friction) and g = 9.8 m/s2 (acceleration due to gravity). Solving for t, we find t = u / (μg).

(b) To find the distance the ball skids, we can use the equation s = ut + (1/2)at2, where s is the distance skidded. Substituting the values we know, we find s = u2 / (2μg).

(c) To find the number of revolutions before the ball starts to roll, we can use the relationship between linear speed v and angular speed ω: v = Rω, where R is the radius of the ball. Solving for ω, we find ω = v / R. Since the ball is not rotating initially, the angular speed is 0. Therefore, the number of revolutions is 0.

(d) To find the speed of the ball when it starts to roll, we can use the equation v = Rω. Substituting the values we know, we find v = Rω = R(u / (Rμg)) = u / (μg).

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The orbital period of a spacecraft can be calculated using Kepler's Third Law and the kinetic energy can be determined using the equation K = GMm/(2r).

The orbital period of the spacecraft can be found using Kepler's Third Law, which states that the square of the period of a body moving in a circular orbit is proportional to the cube of the average distance from the center of the force. If we let RE and m denote Earth's radius (6.371 x 10^6 m) and mass (5.972 x 10^24 kg), and G define the gravitational constant (6.67 x 10^-11 N m^2/kg^2), the period T (in seconds) can be calculated as T = 2π √[(4π^2(R+2h)^3)/Gm], or T = 2π √[(4π^2(3RE)^3)/(GmE)].

The kinetic energy K of an object moving in a circular orbit is given by K = GMm/(2r), where M is the mass of the central body (in this case Earth), m is the mass of the object (the spacecraft), and r is the distance of the object from the center of the force (in this case the Earth's center). In this instance, K = GMm/(2R), with R being the radius of Earth, should suffice.

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