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3 years ago

Why are some machines less than 100% efficient?

Physics

2 answers:

evablogger [386]3 years ago

7 0

Answer:

The output work is always less than the input work because some of the input work is used to overcome friction. Therefore, efficiency is always less than 100 percent. The closer to 100 percent a machine's efficiency is, the better it is at reducing friction

Explanation:

Contact [7]3 years ago

3 0

Answer:

Not all machines work effectevly.

Explanation:

A disrepare objest may not work as vigorus as a new object.

You might be interested in

Calculate the force of gravity between a 2.50 kg newborn baby and a 80.0 kg doctor standing 0.250 m away. G = 6.67 E -11 N*m2/kg

Alex787 [66]

Answer: 2.13 × 10⁻⁷ N

Explanation:

Gravitational force exists between any two bodies having mass.

Force of gravity is given by:

$F =G\frac{Mm}{r^2}$

It is given that, mass of newborn baby is M = 2.50 kg

Mass of the doctor, m = 80.0 kg

Distance between the two, r = 0.250 m

Gravitational constant, G = 6.67 × 10⁻¹¹ N m²/kg²

⇒F = (6.67 × 10⁻¹¹ N m²/kg² × 2.50 kg × 80.0 kg )÷ (0.250 m)² = 2.13 × 10⁻⁷ N

Thus, the force of gravity between new born baby and doctor is 2.13 × 10⁻⁷ N.

7 0

3 years ago

Salmon often jump waterfalls to reach their breeding grounds starting downstream, 2.9 meters away from a waterfall .436 meters i

nikklg [1K]

The minimum speed must a Salmon jumping with to leave the water

to continue upstream is 5.79 m/s

Explanation:

At first let us find the two component of the jumping velocity of the fish

1. Horizontal component $u_{x}$ = u cosФ

2. Vertical component $u_{y}$ = u sinФ

where u is the initial velocity and Ф is the angle between the horizontal

and the initial velocity u

→ Ф = 44.7°

→ $u_{x}$ = u cos(44.7)

→ $u_{y}$ = u sin(44.7)

The horizontal distance x is 2.9 meters away from a waterfall

The vertical distance y is 0.436 meters

3. The horizontal distance x = $u_{x}$ t

4. The vertical distance y = $u_{y}$ t + $\frac{1}{2}$ gt²

where g is the acceleration of gravity

→ x = u cos(44.7) t

→ x = 2.9 meters

→ 2.9 = u cos(44.7) t

Divided both sides by u cos(44.7)

→ t = $\frac{2.9}{ucos(44.7)}$ ⇒ (1)

→ y = u sin(44.7) t + $\frac{1}{2}$ gt²

→ y = 0.436 meters , g = -9.81 m/s²

→ 0.436 = u sin(44.7) t - 4.905 t² ⇒ (t)

Substitute (1) in (2) to make the equation of u only

→ 0.436 = u sin(44.7)( $\frac{2.9}{ucos(44.7)}$ ) - 4.905 ( $\frac{2.9}{ucos(44.7)}$ )²

→ 0.436 = 2.9 ( $\frac{sin(44.7)}{cos(44.7)}$ - $\frac{41.25105}{u^{2}[cos(44.7)]^{2}}$

→ 0.436 = 2.8698 - $\frac{81.4671}{u^{2} }$

Subtract 2.8698 from both sides

→ -2.4338 = - $\frac{81.4671}{u^{2} }$

Multiply both sides by -1

→ 2.4338 = $\frac{81.4671}{u^{2} }$

By using cross multiplication

∴ 2.4338 u² = 81.4671

Divide both sides by 2.4338

→ u² = 33.4732

Take √ for both sides

→ u = 5.79 m/s

The minimum speed must a Salmon jumping with to leave the water

to continue upstream is 5.79 m/s

Learn more:

You can learn more about the equation of trajectory of the projectile in brainly.com/question/2814900

brainly.com/question/5531630

#LearnwithBrainly

4 0

3 years ago

High-altitude mountain climbers do not eat snow, but always melt it first with a stove. To see why, calculate the energy absorbe

Vikki [24]

Answer:

a) $Q=467443.8\ J$

b) $Q_m=299700\ J$

c) $Q_2=131859\ J$

Explanation:

Given:

specific heat of ice, $c_i=2100\ J.kg^{-1}.^{\circ}C^{-1}$

latent heat of fusion of ice, $L=333000\ J.kg^{-1}$

specific heat of water, $c_w=4186\ J.kg^{-1}.^{\circ}C^{-1}$

(a)

mass of snow, $m_s=0.9\ kg$

initial temperature of snow, $T_{is}=-15^{\circ}C$

Final temperature of the consumed mass, $T_f=37^{\circ}C$

Now the energy absorbed from the body after eating this snow:

$Q=m_s.c_i.\Delta T_i+m_s.L+m_s.c_w.\Delta T_w$

$Q=0.9\times 2100\times (0-(-15))+0.9\times 333000+0.9\times 4186\times (37-0)$

$Q=467443.8\ J$

(b)

Energy absorbed from the body in melting the ice is the total latent heat:

$Q_m=m_s.L$

$Q_m=0.9\times 333000$

$Q_m=299700\ J$

(c)

initial temperature of water, $T_{iw}=2^{\circ}C$

final temperature of water, $T_{iw}=37^{\circ}C$

Now, the amount of energy invested by body for the water at this condition:

$Q_2=m_s.c_w.\Delta T_2$

$Q_2=0.9\times 4186\times (37-2)$

$Q_2=131859\ J$

3 0

3 years ago

An ideal Otto cycle has a compression ratio of 8. At the beginning of the compression process, air is at 95 kPa and 27°C, and 75

katen-ka-za [31]

Answer:

Part A) 3899 kPa

Part B) 392.33 kJ/kg

Part C) 0.523

Part D) 495 kPa

Explanation:

Part A

First from the temperature at state 1 the relative specific volume and the internal energy at that state are determined from:

$u_{1}$ = 214.07 kJ/kg

$\alpha$ $r_{1}$ = 621.2

The relative specific volume at state 2 is obtained from the compression ratio:

$\alpha$ $r_{2}$ = $\frac{\alpha r_{1} }{r}$

=621.2/ 8

= 77.65

From this the temperature and internal energy at state 2 can be determined using interpolation with data from A-17(table):

$T_{2}$ = 673 K

$u_{2}$ = 491.2 kJ/kg

The pressure at state 2 can be determined by manipulating the ideal gas relations at state 1 and 2:

$P_{2}$ = $P_{1} r\frac{T_{2} }{T_{1} }$

= 95*8*673/300

= 1705 kPa

Now from the energy balance for stage 2-3 the internal energy at state 3 can be obtained:

$deltau_{2-3} =q_{in}\\ u_{3} -u_{2} =q_{in}\\u_{3}=u_{2}+q_{in}$

= 1241.2 kJ/kg

From this the temperature and relative specific volume at state 3 can be determined by interpolation with data from A-17(table):

$T_{3}$ = 1539 K

$\alpha r_{3}$ = 6.588

The pressure at state 3 can be obtained by manipulating the ideal gas relations for state 2 and 3:

$P_{3} =P_{2} \frac{T_{3} }{T_{2} }$

= 3899 kPa

Part B

The relative specific volume at state 4 is obtained from the compression ratio:

$\alpha r_{4}= r\alpha r_{3}$

= 52.7

From this the temperature and internal energy at state 4 can be determined by interpolation with data from A-17:

$T_{4}$ =775 K

$u_{4}$ = 571.74 kJ/kg

The net work output is the difference of the heat input and heat rejection where the heat rejection is determined from the decrease in internal energy in stage 4-1:

$w=q_{in}-q_{out}\\q_{in}-(u_{4} -u_{1} )\\=392.33 kJ/kg$