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3 years ago

A particle moves according to the equation x = 11t^2, where x is in meters and t is in seconds.

1 answer:

4 0

We are given the equation:

x = 11t^2

We use that equation to calculate for the distance traveled.
For (a)

At t=2.20 sec,
x =53.24 meters

At t=2.95 sec,
x =95.73 meters

Velocity = (95.73 meters - 53.24 meters) / (2.95 s - 2.20 s ) = 56.65 m/s

For (b)

At t=2.20 sec,
x =53.24 meters

At t=2.40 sec,
x =63.36 meters

Velocity = (63.36 meters - 53.24 meters) / (2.40 s - 2.20 s ) = 50.6 m/s

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How are the variables speed and velocity different? how are they similar

victus00 [196]

Answer:

Speed, being a scalar quantity, is the rate at which an object covers distance. The average speed is the distance (a scalar quantity) per time ratio. ... Velocity is the rate at which the position changes. The average velocity is the displacement or position change (a vector quantity) per time ratio.

Explanation:

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2 years ago

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Which of the following is a chemical equation that accurately represents what happens when a sulfur and oxygen are produced from

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3 years ago

3. A large crane lifts a 25,000 kg mass in the air. The amount of work that must be done by the

andreev551 [17]

$\mathfrak{\huge{\orange{\underline{\underline{AnSwEr:-}}}}}$

Actually Welcome to the concept of Efficiency.

Here we can see that, the Input work is given as 2.2 x 10^7 J and the efficiency is given as 22%

The efficiency is => 22% => 22/100.

so we get as,

E = W(output) /W(input)

hence, W(output) = E x W(input)

so we get as,

W(output) = (22/100) x 2.2 x 10^7

=> W(output) = 0.22 x 2.2 x 10^7 => 0.484 x 10^7

hence, W(output) = 4.84 x 10^6 J

The useful work done on the mass is 4.84 x 10^6 J

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3 years ago

Select the correct answer. What is tan for the given triangle? 0.60 1.67 1.33 5.00 0.20

GuDViN [60]

The answer is 1.33 i hope this helps you

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3 years ago

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Air flows through an adiabatic turbine that is in steady operation. The air enters at 150 psia, 900oF, and 350 ft/s and leaves a

Nonamiya [84]

Answer:

$1486.5\frac{Btu}{s}$

Explanation:

The inlet specific volume of air is given by:

$v_1=\frac{RT_1}{P_1}\\\\v_1=\frac{(0.3704\frac{psia.ft^3}{lbm.R})(1360R)}{150psia}\\\\v_1=3.358\frac{ft^3}{lbm} \ \ \ \ \ \ \ \ \...i$

The mass flow rates is expressed as:

$\dot m=\frac{1}{v_1}A_1V_1\\\\\dot m=\frac{1}{3.358ft^3/psia}(0.1ft^2)(350ft/s)\\\\\dot m=10.42\frac{lbm}{s}$

The energy balance for the system can the be expresses in the rate form as:

$E_{in}-E_{out}=\bigtriangleup \dot E=0\\\\E_{in}=E_{out}\\\\\dot m(h_1+0.5V_1^2)=\dot W_{out}+\dot m(h_2+0.5V_2^2)+Q_{out}\\\\\dot W_{out}=\dot m(h_2-h_1+0.5(V_2^2-V_1^2))=-m({cp(T_2-t_1)+0.5(V_2^2-V_1^2)})\\\\\\\dot W_{out}=-(10.42lbm/s)[(0.25\frac{Btu}{lbm.\textdegree F})(300-900)\textdegree F+0.5((700ft/s)^2-(350ft/s)^2)(\frac{1\frac{Btu}{lbm}}{25037ft^2/s^2})]\\\\\\\\=1486.5\frac{Btu}{s}$

Hence, the mass flow rate of the air is 1486.5Btu/s

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3 years ago