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2 years ago

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Three professional wrestlers are fighting over a champion's belt, and each exerts a force on the belt. Wrestler 1 exerts a force

F1 = 40.0 N in the +x-direction, wrestler 2 exerts a force F2 = 80.0 N in the -y-direction, and wrestler 3 exerts a force F3 = 60.0 N at an angle of 36.9∘ counterclockwise from the +x-direction. Find the x- and y-components of the net external force F⃗ on the belt.

1 answer:

seropon [69]2 years ago

6 0

Answer:

The x-component of the net external force Fx = 88.0 N î

The y-component of the net external force Fy = - 44 N ĵ

Explanation:

Net force acting on an object is the sum of all forces acting upon it.

F = F1 + F2 + F3

Forces in terms of x,y-components:

F₁ = 40 N î (where î is the unit vector in x-direction)

F₂ = -80 N ĵ (where ĵ is the unit vector in y-direction)

F₃ = 60 N

= F₃cosθ î + F₃sinθ ĵ

= 60*cos(36.9°) î + 60*sin(36.9°) ĵ

= 48î + 36ĵ

Net External Force = F1 + F2 + F3

= 40 î - 80 ĵ + 48 î + 36 ĵ

= 40 î + 48 î - 80 ĵ + 36 ĵ

= 88 î - 44 ĵ

Here, only like components(same direction) can be added into each other.

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Explanation & answer:

Given:

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Specific gravity = 0.8

output power, P = 55 kW

heating value, H = 44,000 kJ/kg

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E = C*P*H

= (22 L/h) / (3600 s/h) * (1000 mL/L) * (0.8 g/mL) * (44000 kJ/kg)

= (22/3600)*1000*0.8*44000 j/s

= 215111.1 j/s

Calculate output power

P = 55 kW

= 55000 j/s

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= output / input

= P/E

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2 years ago

Which vector has a y-component with a length of 1?

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In this 2-dimensional graph, the x-component of each vector is the horizontal distance from the origin, while the y-component of each vector is the vertical distance from the origin. It can be seen that the c vector is 1 vertical unit away from the origin, which means that it has a y-component of 1.

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Water evaporating from a pond does so as if it were diffusing across an air film 0.15 cm thick. The diffusion coefficient of wat

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Answer:

The water level will drop by about 1.24 cm in 1 day.

Explanation:

Here Mass flux of water vapour is given as

$j_{H_2O}=\frac{D}{l} \bigtriangleup c$

where

$j_{H_2O}$ is the mass flux of the water which is to be calculated.

D is diffusion coefficient which is given as $0.25 cm^2/s$

l is the thickness of the film which is 0.15 cm thick.
$\bigtriangleup c$ is given as

$\bigtriangleup c= \frac{P_{sat}-P_a}{RT}$

In this

$P_{sat}$ is the saturated water pressure, which is look up from the saturated water property at 20°C and 0.5 saturation given as 2.34 Pa

$P_a$ is the air pressure which is given as 0.5 times of $P_{sat}$

R is the universal gas constant as $8.314 kJ/kmol-K$

T is the temperature in Kelvin scale which is $20+273= 293K$

By substituting values in the equation

$\bigtriangleup c= \frac{P_{sat}-P_a}{RT} \\ \bigtriangleup c= \frac{P_{sat}-0.5P_{sat}}{RT} \\ \bigtriangleup c= \frac{0.5P_{sat}}{RT} \\ \bigtriangleup c= \frac{0.5 \times 2.34}{8.314 \times 293} \\\bigtriangleup c= 0.48 mol/m^3$

Converting $\bigtriangleup c$ into $cm^3/cm^3$

As 1 mole of water 18 $cm^3$ so

$\bigtriangleup c= 0.48 mol/m^3 \\ \bigtriangleup c= 0.48 \times 18 \times 10^{-6} cm^3/cm^3 \\ \bigtriangleup c= 8.64 \times 10^{-6} cm^3/cm^3$

Putting this in the equation of mass flux equation gives

$j_{H_2O}=\frac{D}{l} \bigtriangleup c \\ j_{H_2O}=\frac{0.25}{0.15} \times 8.64 \times 10^{-6} \\ j_{H_2O}=14.4 \times 10^{-6} cm/s$

For calculation of water level drop in a day, converting mass flux as

$j_{H_2O}=14.4 \times 10^{-6} \times 24 \times 3600 cm/day\\ j_{H_2O}=1.24 cm/day$

So the water level will drop by about 1.24 cm in 1 day.

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Anne is working on a research project that involves the use of a centrifuge. Her samples must first experience an acceleration o

oee [108]

Answer:

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Explanation:

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a=rω²

where ω is angular speed and r radius of circle

Rearranging equation

ω= $\sqrt{\frac{a}{r} }$

The initial acceleration is given by

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The final acceleration of particle is

a₂ = 6 (100g)

=600g

The final and initial rotation speed are related as

$\frac{w1}{w2} =\sqrt{\frac{a2}{a1 }$

ω2= $\sqrt{\frac{600g}{100g} } *w1$

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2 years ago

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A.

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Explanation:

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They are held together by very weak attractive forces
Gases are random and they frequently collide with one another and the walls of the containers they are put into.
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learn more:

Kinetic theory of matter brainly.com/question/12362857

#learnwithBrainly

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