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3 years ago

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A ball is thrown down vertically with an initial speed of 20 m/s from a height of 60 m. Find (a) its speed just before it strike

s the ground and (b) how long it takes for the ball to reach the ground. Repeat (a) and (b) for the ball thrown directly up from the same height and with the same initial speed.

1 answer:

vivado [14]3 years ago

3 0

Answer:

Explanation:

Ball is thrown downward:

initial velocity, u = - 20 m/s (downward)

height, h = - 60 m

Acceleration due to gravity, g = - 9.8 m/s^2 (downward)

(a) Let the speed of the ball as it hits the ground is v.

Use third equation of motion

$v^{2}=u^{2}+2as$

$v^{2}=(-20)^{2}+2\times 9.8 \times 60$

v = 39.69 m/s

(b) Let t be the time taken

Use First equation of motion

v = u + a t

- 39.69 = - 20 - 9.8 t

t = 2 second

Now the ball is thrown upwards:

initial velocity, u = 20 m/s (upward)

height, h = - 60 m

Acceleration due to gravity, g = - 9.8 m/s^2 (downward)

(c) Let the speed of the ball as it hits the ground is v.

Use third equation of motion

$v^{2}=u^{2}+2as$

$v^{2}=(-20)^{2}+2\times 9.8 \times 60$

v = 39.69 m/s

(d) Let t be the time taken

Use First equation of motion

v = u + a t

- 39.69 = + 20 - 9.8 t

t = 6.09 second

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Explanation:

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1) <em>convert your given </em>

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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

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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

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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

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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

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3 years ago

Find the potential energy associated with a 61-kg hiker atop New Hampshire's Mount Washington, 1900 m above sea level. Take the

Natali [406]

Answer:

The potential energy of the hiker is $1.13\times 10^6\ J$ .

Explanation:

Given that,

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We need to find the potential energy associated with a 61-kg hiker atop New Hampshire's Mount Washington. The potential energy is given by :

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g is the acceleration due to gravity

$E=61\ kg\times 9.8\ m/s^2\times 1900\ m\\\\E=1.13\times 10^6\ J$

So, the potential energy of the hiker is $1.13\times 10^6\ J$ . Hence, this is the required solution.

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3 years ago

Based on the replacement reaction, what would the products of the reaction be?

cricket20 [7]

Answer:

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Explanation:

In a double replacement reaction, two ionic compounds exchange their ions to produce two different ionic compounds.

In this question, the two ionic compounds are:

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In particular,

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In a binary ionic compound, cations (positive ions) can only bond to anions (negative ions.)

$\rm Be^{2+}$ is a cation. In $\rm BeSO_4$ , $\rm Be^{2+}$ was bounded $\rm {SO_4}^{2-}$ anions. During the reaction, it bonds with $\rm OH^{-}$ anions to produce $\rm Be(OH)_2$ .

$\rm {NH_4}^{+}$ is also a cation. In $\rm NH_4 OH$ , $\rm {NH_4}^{+}$ was bounded to $\rm OH^{-}$ ions. During the reaction, it bonds with $\rm {SO_4}^{2-}$ anions to produce $\rm (NH_4)_2 SO_4$ .

Hence, the two products will be $\rm Be(OH)_2$ and $\rm (NH_4)_2 SO_4$ .

Note that charges on the ions must balance. For example, a $\rm Be^{2+}$ ion carries twice as much charge as an $\rm {NH_4}^{+}$ ion. As a result, each $\rm Be^{2+}$ ion would bond with twice as many $\rm OH^{-}$ ions as $\rm {NH_4}^{+}$ would in $\rm NH_4 OH$ .

4 0

3 years ago

Read 2 more answers

The uniform 100-kg I-beam is supported initially by its end rollers on the horizontal surface at A and B. by means of the cable

Ann [662]

I attached a free body diagram for a better understanding of this problem.

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3 years ago