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2 years ago

14

A hot-air balloon is 11.0 m above the ground and rising at a speed of 7.00 m/s. A ball is thrown horizontally from the balloon b

asket at a speed of 9.00 m/s. Ignore friction and air resistance and find the speed of the ball when it strikes the ground.

2 answers:

Nesterboy [21]2 years ago

7 0

Answer:

18.6 m/s

Explanation:

$h$ = Initial height of the balloon = 11 m

$v_{o}$ = initial speed of the ball

$v_{oy}$ = initial vertical speed of the ball = 7 m/s

$v_{ox}$ = initial horizontal speed of the ball = 9 m/s

initial speed of the ball is given as

$v_{o} = \sqrt{v_{ox}^{2} + v_{oy}^{2}} = \sqrt{9^{2} + 7^{2}} = 11.4 m/s$

$v_{f}$ = final speed of the ball as it strikes the ground

$m$ = mass of the ball

Using conservation of energy

Final kinetic energy before striking the ground = Initial potential energy + Initial kinetic energy

$(0.5) m v_{f}^{2} = (0.5) m v_{o}^{2} + mgh \\(0.5) v_{f}^{2} = (0.5) v_{o}^{2} + gh\\(0.5) v_{f}^{2} = (0.5) (11.4)^{2} + (9.8)(11)\\(0.5) v_{f}^{2} = 172.78\\v_{f} = 18.6 m/s$

vampirchik [111]2 years ago

6 0

Answer:

0.95 second

Explanation:

height, h = 11 m

ux = 9 m/s

uy = 7 m/s

Let it takes time t to strike the ground.

Use second equation of motion

$h = u_{y}t + 0.5 gt^{2}$

- 11 = 7 t - 0.5 x 9.8 t²

-11 = 7t - 4.9t²

4.9t² - 7t - 11 = 0

$t=\frac{-7\pm \sqrt{49+4\times4.9\times11}}{9.8}$

$t=\frac{-7\pm 16.27}{9.8}$

take positive sign

t = 0.95 second

Thus, the time taken to reach the ground is 0.95 second.

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An airplane propeller is 2.68 m in length (from tip to tip) and has a mass of 107 kg. When the airplane's engine is first starte

telo118 [61]

Answer:

a)30.14 rad/s2

b)43.5 rad/s

c)60633 J

d)42 kW

e)84 kW

Explanation:

If we treat the propeller is a slender rod, then its moments of inertia is

$I =\frac{mL^2}{12} = \frac{107*2.68^2}{12} = 64.04 kgm^2$

a. The angular acceleration is Torque divided by moments of inertia:

$\alpha = \frac{T}{I} = \frac{1930}{64.04} = 30.14 rad/s^2$

b. 5 revolution would be equals to $10\pi$ rad, or 31.4 rad. Since the engine just got started

$\omega^2 = 2\alpha\theta = 2*30.14*31.4 = 1893.5$

$\omega = \sqrt{1893.5} = 43.5 rad/s$

c. Work done during the first 5 revolution would be torque times angular displacement:

$W = T*\theta = 1930 * 31.4 = 60633 J$

d. The time it takes to spin the first 5 revolutions is

$t = \frac{\omega}{\alpha} = \frac{43.5}{30.14} = 1.44 s$

The average power output is work per unit time

$P = \frac{W}{t} = \frac{60633}{1.44} = 41991 W$ or 42 kW

e.The instantaneous power at the instant of 5 rev would be Torque times angular speed at that time:

$P_i = T*\omega = 1930*43.5=83983 W$ or 84 kW

7 0

2 years ago

Air enters a turbine operating at steady state at 8 bar, 1600 K and expands to 0.8 bar. The turbine is well insulated, and kinet

kobusy [5.1K]

Answer:

the maximum theoretical work that could be developed by the turbine is 775.140kJ/kg

Explanation:

To solve this problem it is necessary to apply the concepts related to the adiabatic process that relate the temperature and pressure variables

Mathematically this can be determined as

$\frac{T_2}{T_1} = (\frac{P_2}{P_1})^{(\frac{\gamma-1}{\gamma})}$

Where

Temperature at inlet of turbine

Temperature at exit of turbine

Pressure at exit of turbine

Pressure at exit of turbine

The steady flow Energy equation for an open system is given as follows:

$m_i = m_0 = mm(h_i+\frac{V_i^2}{2}+gZ_i)+Q = m(h_0+\frac{V_0^2}{2}+gZ_0)+W$

Where,

m = mass

m(i) = mass at inlet

m(o)= Mass at outlet

h(i)= Enthalpy at inlet

h(o)= Enthalpy at outlet

W = Work done

Q = Heat transferred

v(i) = Velocity at inlet

v(o)= Velocity at outlet

Z(i)= Height at inlet

Z(o)= Height at outlet

For the insulated system with neglecting kinetic and potential energy effects

$h_i = h_0 + WW = h_i -h_0$

Using the relation T-P we can find the final temperature:

$\frac{T_2}{T_1} = (\frac{P_2}{P_1})^{(\frac{\gamma-1}{\gamma})}\\$

$\frac{T_2}{1600K} = (\frac{0.8bar}{8nar})^{(\frac{1.4-1}{1.4})}\\ = 828.716K$

From this point we can find the work done using the value of the specific heat of the air that is 1,005kJ / kgK

$W = h_i -h_0W = C_p (T_1-T_2)W = 1.005(1600 - 828.716)W = 775.140kJ/Kg$

the maximum theoretical work that could be developed by the turbine is 775.140kJ/kg

4 0

2 years ago

A spring with spring constant 33N/m is attached to the ceiling, and a 4.8-cm-diameter, 1.5kg metal cylinder is attached to its l

mylen [45]

Answer:

0.423m

Explanation:

Conversion to metric unit

d = 4.8 cm = 0.048m

Let water density be $\who_w = 1000 kg/m^3$

Let gravitational acceleration g = 9.8 m/s2

Let x (m) be the length that the spring is stretched in equilibrium, x is also the length of the cylinder that is submerged in water since originally at a non-stretching position, the cylinder barely touches the water surface.

Now that the system is in equilibrium, the spring force and buoyancy force must equal to the gravity force of the cylinder. We have the following force equation:

$F_s + F_b = W$

Where $F_s = kx$ N is the spring force, $F_b = W_w = m_wg = \rho_w V_s g$ is the buoyancy force, which equals to the weight $W_w$ of the water displaced by the submerged portion of the cylinder, which is the product of water density $\rho_w$ , submerged volume $V_s$ and gravitational constant g. W = mg is the weight of the metal cylinder.

$kx + \rho_w V_s g = mg$

The submerged volume would be the product of cross-section area and the submerged length x

$V_s = Ax = \pi(d/2)^2x$

Plug that into our force equation and we have

$kx + \rho_w \pi(d/2)^2x g = mg$

$x(k + \rho_w g \pi d^2/4) = mg$

$x = \frac{m}{(k/g) + (\rho_w\pi d^2/4)} = \frac{1.5}{(33/9.8) + (100*\pi * 0.048^2/4)} = 0.423 m$

6 0

2 years ago

Suppose our apparatus will only allow us to distinguish the first order bright fringe from the central bright spot if they are s

ozzi

Answer:

λ = 8.716 mm

Explanation:

Given:

- d = 10 cm

- Q >= 5 degrees

Find:

- Find the shortest wavelength of light for which this apparatus is useful

Solution:

- The formula that relates the split difference and angle of separation between successive fringes is given by:

d*sin(Q) = n*λ

Where,

λ: wavelength

d: split separation

Q: angle of separation between successive fringes

m: order number.

- Since this apparatus only shows the first order light so m =1

- the shortest possible wavelength corresponds to:

d*sin(Q) = λ

λ = 0.1*sin(5)

λ = 8.716 mm

3 0

3 years ago

If a scale of the solar system was built where 1 mm equaled 1 mile, could the model be practically built in a city?

Y_Kistochka [10]

No, it couldn't be.
On that scale, Neptune would be almost 1,740 MILES from the sun.
ON THAT SCALE !

7 0

3 years ago

Read 2 more answers