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2 years ago

Find the efficiency of a machine that does 800 J of work if the input work is 2,000 J.

Physics

1 answer:

torisob [31]2 years ago

3 0

Answer:

Efficiency: 0.4 (40%)

Explanation:

The efficiency of a simple machine is given by the ratio

$\eta = \frac{W_{out}}{W_{in}}$

where

$W_{in}$ is the input work of the machine

$W_{out}$ is the output work of the machine

For the machine in this problem, we have:

$W_{out}=800 J$ is the output work

$W_{in}=2000 J$ is the input work

Therefore, the efficiency of the machine is

$\eta=\frac{800}{2000}=0.4$

Which can be also written as percentage:

$\eta=0.4\cdot 100 = 40\%$

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Romeo lanza suavemente guijarros a la ventana de julieta y quiere que los guijarros golpeen la ventana solo con con un component

Yuki888 [10]

Answer:

$5.219\,\frac{m}{s}$

Explanation:

Las condiciones del problema requieren el cálculo de la rapidez inicial de los guijarros. Se sabe que el componente vertical de la rapidez final es cero. Por tanto, el tiempo se determina a continuación: (The conditions of this problems require the calculation of the initial speed of the peebles. It is known that vertical component of the final speed is zero. Therefore, the time is determined herein:).

$(0\,\frac{m}{s})^{2} = v_{o,y}^{2} - 2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (4.5\,m)$

$v_{o,y} = 9.395\,\frac{m}{s}$

$0\,\frac{m}{s} = 9.395\,\frac{m}{s} - \left(9.807\,\frac{m}{s^{2}} \right)\cdot \Delta t$

$\Delta t = 0.958\,s$

Además, se determina el componente horizontal de la rapidez inicial (Likewise, the horizontal component of the initial speed is determined):

$v_{o,x} = \frac{5\,m}{0.958\,s}$

$v_{o,x} = 5.219\,\frac{m}{s}$

El guijarro tiene una rapidez de $5.219\,\frac{m}{s}$ cuando golpea la ventana (The peeble has a speed of $5.219\,\frac{m}{s}$ when it hits the window).

6 0

2 years ago

A punted football is observed to have velocity components vhorizontal = 15 m/s to the right and vvertical = 1.25 m/s directed do

enot [183]

Answer:

v₀ₓ = 15 m / s, $v_{oy}$ = 5.2 m / s

v = 15.87 m / s , θ = 19.1

Explanation:

This is a projectile launch problem. The horizontal speed that is constant throughout the entire path is worth 15 m / s, instead the vertical speed changes in value due to the acceleration of gravity, let's look for the initial vertical speed

Vy² = $v_{oy}$ ² - 2 g y

$v_{oy}$ ² = $v_{y}$ ² + 2 g y

$v_{oy}$ = √ ( $v_{y}$ ² + 2 gy

Let's calculate

$v_{oy}$ = √ (1.25² + 2 9.8 1.3)

$v_{oy}$ = √ (27.04)

$v_{oy}$ = 5.2 m / s

The initial speed can be calculated by the initial speed

v = √ v₀ₓ² + $v_{oy}$ ²

v = RA (15² + 5.2²)

v = 15.87 m / s

We look for the angle with trigonometry

tan θ = voy / vox

θ = tan⁻¹ I'm going / vox

θ = tan⁻¹ 5.2 / 15

θ = 19.1

The answer is

v₀ₓ = 15 m / s

$v_{oy}$ = 5.2 m / s

5 0

2 years ago

The human ear canal is, on average, 2.5cm long and aids in hearing by acting like a resonant cavity that is closed on one end an

Troyanec [42]

Answer:

$3400$ Hz

Explanation:

We know that

1 cm = 0.01 m

$L$ = Length of the human ear canal = 2.5 cm = 0.025 m

$V$ = Speed of sound = 340 ms⁻¹

$f$ = First resonant frequency

The human ear canal behaves as a closed pipe and for a closed pipe, nth resonant frequency is given as

$f = \frac{(2n - 1)V}{4L}$

for first resonant frequency, we have n = 1

Inserting the values

$f = \frac{(2(1) - 1) 340}{4(0.025)}$

$f = \frac{340}{4(0.025)}$

$f = 3400$ Hz

4 0

2 years ago

A car travels at a constant rate for 25 miles, going due east for one hour. Then it travels at a constant rate another 60 miles

egoroff_w [7]

60 mph east...........

6 0

2 years ago

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Joule’s law is a linear relationship, that is, the more heat you provide, the greater the temperature change. However, in the pr

Nitella [24]

Answer:

hydrogen bridge

Explanation:

Joule's relationship to heat and temperature is true for all materials where we assume that interatomic forces are linear, when atoms separate these forces decrease. There is a point where the separation between atoms is enough that thermal agitation can separate the molecules and there is a change of state, generally from solid to liquid and from liquid to vapor. When these changes of state are occurring all the energy supplied is used to break the links, so the temperature does not change.

In the specific case of water, there is a bond called a hydrogen bridge that breaks around 4ºC, therefore, at this temperature there is a deviation from the curve since this link is being broken, this does not lead to a change of macroscopic state.

For the other temperatures the water behaves like the other bodies.

7 0

2 years ago