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2 years ago

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A weight of 1400 pounds is suspended from two cables as shown in the figure. What is the tension in the left cable? _________ po

unds (Round to one decimal place as needed).

1 answer:

juin [17]2 years ago

4 0

Answer:

Following are the solution to this question:

Explanation:

Law:

$\to \theta= 180^{\circ}- 50^{\circ}- 25^{\circ}$

$= 180^{\circ}- 75^{\circ}\\\\= 105^{\circ}$

$\to \frac{T_{L}}{\sin (90+50)}= \frac{T_{R}}{\sin (25+90)}=\frac{1400}{\sin (105)}$

$\to T_L=931.65 \ pounds \\\\ \to T_R=1313.59 \ pounds \\\\$

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According to the graph, what is the motorcycle’s Average velocity ?

borishaifa [10]

According to the position vs time graph, the <em>average</em> <em>velocity</em> of the motorcycle is the change in position divided by the change in time. Also, note that the slope is linear and positive throughout the 5 hours, it doesn't change direction.

Therefore, we have
Avg velocity = change in direction/change in time
Avg velocity = (150km - 30km)/(5h - 0h)
Avg velocity = 24km/hr south.

3 0

2 years ago

A stone is thrown horizontally from the top of a tower at the same instant a ball is droppedvertically. Which object is travelin

Igoryamba

The ball is travelling faster when the two objects hits the level ground below.

<h3>Time of motion of the objects</h3>

The time of motion of the objects depends on height and initial velocity of projection of the objects.

The stone has no initial vertical velocity while the ball has initial vertical velocity.

Thus, the ball is travelling faster when the two objects hits the level ground below.

Learn more about time of motion here: brainly.com/question/2364404

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2 years ago

I need help on number 5. We need to use one of the four kinematics equations.

Gnom [1K]

ANSWER

$35.02m$

EXPLANATION

Parameters given:

Initial velocity, u = 26.2 m/s

When the vase reaches its maximum height, its velocity becomes 0 m/s. That is the final velocity.

We can now apply one of Newton's equations of motion to find the height:

$v^2=u^2-2as$

where a = g = acceleration due to gravity = 9.8 m/s²

Therefore, we have that:

$\begin{gathered} 0=26.2^2-2(9.8)s \\ \Rightarrow19.6s=686.44 \\ s=\frac{686.44}{19.6} \\ s=35.02m \end{gathered}$

That is the height that the vase will reach.

5 0

8 months ago

Air enters a turbine operating at steady state with a pressure of 75 Ibf/in.^2, a temperature of 800º R and velocity of 400 ft/s

Arturiano [62]

Answer:

(a) W/m = 49.334 Btu/lb

(b) $\frac{E_{d} }{m}$ = 22.12 Btu/lb

Explanation:

For the given problem, it can be assumed that the system is operating at steady state and the effects of potential energy can be neglected.

(a) Using the thermodynamic table for air.

At the temperature ( $T_{1}$ )of 800 ºR and pressure ( $P_{1}$ ) of 75 Ibf/in.^2, we can deduce that:

Specific enthalpy ( $h_{1}$ ) = 191.81 BTu/lb

Specific entropy ( $s_{1}$ ) = 0.6956 Btu/(lb.ºR)

At the temperature ( $T_{2}$ )of 600 ºR and pressure ( $P_{2}$ ) of 15 Ibf/in.^2, we can deduce that:

Specific enthalpy ( $h_{2}$ ) = 143.47 BTu/lb

Specific entropy ( $s_{2}$ ) = 0.6261 Btu/(lb.ºR)

The work done can be calculated using energy rate equation:

$\frac{W}{m} = \frac{Q}{m} + (h_{1} - h_{2}) + \frac{V_{1}^{2} - V_{2}^{2}}{2}$

Q/m = heat transfer = -2 Btu/lb

$V_{1}$ = 400 ft/s

$V_{2}$ = 100 ft/s

$\frac{W}{m} = -2 + (191.81 - 143.47) + \frac{400^{2} - 100^{2}}{2}*[tex]\frac{1}{2*32.2*778}$ [/tex] = -2 + 48.34 + 29.938 = 49.334 Btu/lb

(b) To calculate the exergy destruction, we will use the equation for exergy rate:

$\frac{E_{d} }{m} = [1-\frac{T_{o} }{T_{b} }](\frac{Q}{m}) - \frac{W}{m} + [(h_{1} - h_{2}) -T_{o}(s_{1} - s_{2}) + \frac{V^{2} _{1} - V_{2} ^{2}}{2}]$

The equation above is further simplified to:

$\frac{Ed}{m} = T_{o}[(s_{2} -s_{1}) - Rln\frac{P_{2} }{P_{1} } - \frac{Q/m}{T_{b} }]$

Using a reference temperature (To) = 500 °R

Average surface temperature (Tb = 620°R

$\frac{Ed}{m} = 500*[(0.6261 -0.6956) - (1.986/28.97)ln\frac{15 }{75 } - \frac{-2}{620}}]$

$\frac{E_{d} }{m}$ = 500*[-0.0695 +0.068688*1.609 +0.003225] = 22.12 Btu/lb

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3 years ago

With two identical light bulbs and two identical batteries how would you arrange in order to get the maximum total power to the

Norma-Jean [14]

Answer:

The batteries would be connected in series while the bulbs would be connected in parallel

Explanation:

Power (W) = VI

where V = voltage, I = current and R = resistance

from V = IR , I = V/R

Power (W) now becomes = V (V/R) = $\frac{V^{2} }{R}$

Power (W) = $\frac{V^{2} }{R}$

from the above equation, power is directly dependent on voltage, hence the voltage has to be high for the power to be high and the power is also inversely dependent on the resistance (in this case the bulbs which act as the load)

We have to batteries, when batteries are connected in series the total voltage becomes the summation of the two voltages hence giving a higher voltage and when they are connected in parallel their voltage remains the same. Since we want to get higher voltage we will connect the two batteries in series.
we have two bulbs which are the resistance here, from the equation above the power is inversely dependent on the resistance so we would need its value to be minimal. When resistance is connected in series the resistance individual will be added to get the total resistance, hence the total resistance will be high but when the resistors are arranged in parallel you get the total resistance by applying the formula $\frac{R1R2}{R1+R2}$ which will give us a lower resistance. Hence we would connect the bulbs in parallel.

Take note that the power from this connection should not exceed the bulbs power rating so as to avoid damage of the bulbs.

4 0

3 years ago