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1 month ago

The δe of a system that releases 12.4 j of heat and does 4.2 j of work on the surroundings is __________ j.

1 answer:

5 0

For this issue, the answer is clarified as the system takes in energy (+). The surroundings contribute 84 KJ of work. Whenever a system is receiving work from its surroundings, the value will be positive. Therefore, it sums to 12.4 KJ + 4.2 = 16.6 KJ.

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A running mountain lion can make a leap 10.0 mlong, reaching a maximum height of 3.0 m. What is the speed of the mountain lion j

ValentinkaMS [3465]

To tackle this question we will apply the kinematic equations that describe the motion of a projectile, where both maximum height and distance traveled are defined. This scenario demonstrates a lion achieving a height (H) of 3m and covering a horizontal distance (R) of 10m. The equations governing this kind of motion are expressed as follows:

$H = \frac{v_0^2sin^2\theta}{2g}$

$R = \frac{v_0^2 sin 2\theta}{g}$

By dividing the two equations, we determine:

$\frac{H}{R}=\frac{\frac{v_0^2sin^2\theta}{2g}}{\frac{v_0^2 sin 2\theta}{g}}$

$\frac{H}{R}= \frac{sin^2\theta}{2}*\frac{1}{sin2\theta}$

$\frac{H}{R}= \frac{sin^2\theta}{2}*\frac{1}{2sin\theta cos\theta}$

$\frac{H}{R}= \frac{1}{4} \frac{sin\theta}{cos\theta}$

$\frac{H}{R}= \frac{1}{4} tan\theta$

Plugging in the values for H and R yields:

$\frac{3}{10} = \frac{1}{4} tan\theta$

$\theta = tan^{-1} \frac{12}{10}$

$\theta = 50.2\°$

After substituting \theta into the relevant equation, we find:

$H = \frac{v_0^2sin^2\theta}{2g}$

$v_0^2 = \frac{H 2g}{sin^2\theta}$

$v_0^2 = \frac{3*2*9.8}{sin^2(50.2)}$

$v_0^2 = 99.62$

$v_0 = \sqrt{99.62}$

$v_0 = 9.98m/s$

In conclusion, the mountain lion's launch speed upon takeoff is approximately 9.98m/s at an angle of 50.2°.

5 0

2 months ago

The parasailing system shown uses a winch to pull the rider in towards the boat, which is traveling with a constant velocity. Du

ValentinkaMS [3465]

Answer:

The force magnitude is $F_{net}= 1.837 *10^4N$

and it is directed at 57.98° from the horizontal in a counterclockwise manner.

Explanation:

The problem states that

At t = 0, $\theta = 20^o$

The angular rate of increase is $w = 2 \ ^o/s$

Converting to revolutions per second gives us $\theta ' = 2 \ ^o/s * \frac{\pi}{180} =0.0349\ rps$

The rope length is defined by

$r = 125- \frac{1}{3}t^{\frac{3}{2} }$

At $\theta =30^o$ , Tension T of the rope is 18 kN.

The weight of the para-sailor is $M_p = 75kg$

In analyzing the question, we observe that the equation for length can be represented as a linear displacement equation.

The derivative of displacement results in velocity.

Hence,

$r' = -\frac{1}{3} [\frac{3}{2} ] t^{\frac{1}{2} }$

signifies the velocity, and further differentiation yields acceleration.

Therefore,

$r'' = -\frac{1}{4} t^{-\frac{1}{2} }$

Now considering the moment when the rope forms a 30° angle with the water,

typically angular velocity is expressed as

$w = \frac{\Delta \theta}{\Delta t}$

where $\theta$ represents the angular displacement.

Next, evaluating the interval from $20^o \ to \ 30^o$ gives us

$2 = \frac{30 -20 }{t -0}$

making t the focal point.

$t = \frac{10}{2}$

$= 5s$

At this time, the displacement measures

$r = 125- \frac{1}{3}(5)^{\frac{3}{2} }$

$= 121.273 m$

The linear velocity computes to

$r' = -\frac{1}{3} [\frac{3}{2} ] (5)^{\frac{1}{2} }$

$= -1.118 m/s$

Whereas linear acceleration calculates as

$r'' = -\frac{1}{4} (5)^{-\frac{1}{2} }$

$= -0.112m/s^2$

Generally, radial acceleration is given by

$\alpha _R = r'' -r \theta'^2$

$= -0.112 - (121.273)[0.0349]^2$

$= 0.271 m/s^2$

Simultaneously, angular acceleration can be represented as

$\alpha_t = r \theta'' + 2 r' \theta '$

Then $\theta '' = \frac{d (0.0349)}{dt} = 0$

Thus,

$\alpha _t = 121.273 * 0 + 2 * (-1.118)(0.0349)$

$= -0.07805 m/s^2$

The resultant acceleration is mathematically denoted as

$a = \sqrt{\alpha_R^2 + \alpha_t^2 }$

$= \sqrt{(-0.07805)^2 +(-0.027)^2}$

$= 0.272 m/s^2$

Now the acceleration's direction is mathematically expressed as

$tan \theta_a = \frac{\alpha_R }{\alpha_t }$

$\theta_a = tan^{-1} \frac{-0.271}{-0.07805}$

$= 73.26^o$

The y-axis force acting on the para-sailor is mathematically shown as

$F_y = mg + Tsin 30^o + ma sin(90- \theta )$

$= (75 * 9.8) + (18 *10^3) sin 30 + (75 * 0.272)sin(90-73.26)$

$= 9.74*10^3 N$

The x-axis force acting on the para-sailor is represented as

$F_x = mg + Tcos 30^o + ma cos(90- \theta )$

$= (75 * 9.8) + (18 *10^3) cos 30 + (75 * 0.272)cos(90-73.26)$

$= 1.557 *10^4 N$

The overall force is calculated as

$F_{net} = \sqrt{F_x^2 + F_y^2}$

$=\sqrt{(1.557 *10^4)^2 + (9.74*10^3)^2}$

$F_{net}= 1.837 *10^4N$

The directional force is evaluated as

$tan \theta_f = \frac{F_y}{F_x}$

$\theta_f = tan^{-1} [\frac{1.557*10^4}{9.74*10^3} ]$

$= tan^{-1} (1.599)$

$= 57.98^o$

7 0

2 months ago

How many electrons does it take to make 80 μc (microcoulombs) of charge?

serg [3582]

The charge for a single electron is 1.602*10^ -19 C

80 µC can be expressed as 8*10^ - 5 C

This is basic arithmetic

Total Charge divided by the charge of one electron = Number of electrons

(8*10^ -5 C / 1.602*10^ -19 C) equals 4.99 * 10^14 electrons.

6 0

2 months ago

Describe several uses of plastic, and explain why plastic is a good choice for these products

Maru [3345]

Numerous items, including bags, toys, and various goods, incorporate plastic. Its remarkable durability and flexibility are key characteristics.:) I hope this information is beneficial to you! c;

5 0

1 month ago

In an amusement park rocket ride, cars are suspended from 3.40-m cables attached to rotating arms at a distance of 5.90 m from t

ValentinkaMS [3465]

Answer:

The rotational angular speed is measured at 1.34 rad/s.

Explanation:

Considering the following parameters,

Length = 3.40 m

Distance = 5.90 m

Angle = 45.0°

We are tasked with finding the angular speed of rotation