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FinnZ [79.3K]
2 years ago
8

What is the efficiency of a machine if your work on the machine is 1200 j and the machines output work is 300 j?

Physics
2 answers:
inysia [295]2 years ago
7 0

Answer:

The efificiency is 0,25 of the machine  (25%). See the explanation below

Explanation:

We calculate the efficiency with this formula:

Efficiency = energy obtained/energy supplied= 300 Joule/1200 Joule

Efficiency= 0,25

Efficiency(%) = 0,25 x100 = 25%

julsineya [31]2 years ago
7 0

Answer:  

The efficiency of the machine is 25%.  

<u>Explanation:  </u>

Efficiency is defined as the ratio of work done to the total energy supplied. Here, we know that the energy obtained or work done by a machine is 300J and the energy supplied to the machine is 1200J

We know that, \text { Efficiency }=\frac{\text { work done }}{\text { energy supplied }}=\frac{300}{1200}=0.25(\%)=0.25 \times 100=25 \%

Hence efficiency of the machine is 25%.

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A sample of a gas occupies a volume of 90 mL at 298 K and a pressure of 702 mm Hg. What is the correct expression for calculatin
aleksandr82 [10.1K]

Answer:

Explanation:

Given

volume of sample V_1=90\ mL

Temperature T_1=298\ K

Pressure P_1=702\ mm\ of\ Hg

for different conditions

Temperature T_2=273\ K

Pressure P_2=760\ mm\ of\ Hg

suppose V_2 is the volume of sample

Using ideal gas equation

PV=nRT

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

\frac{702\times 90}{298}=\frac{760\times V_2}{273}

V_2=76.15\ mL

             

8 0
2 years ago
An ocean liner is cruising at 10 meters/second and is about to approach a stationary ferryboat. A parcel is released from the oc
Afina-wow [57]
The parcel will undergo projectile motion, which means that it will have motion in both the horizontal and vertical direction.

First, we determine how long the parcel will fall using:

s = ut + 1/2 at²

where s will be the height, u is the initial vertical velocity of the parcel (0), t is the time of fall and a is the acceleration due to gravity. 

5.5 = (0)(t) + 1/2 (9.81)(t)²
t = 1.06 seconds

Now, we may use this time to determine the horizontal distance covered by the parcel by using:
distance = velocity * time

The horizontal velocity of the parcel will be equal to the horizontal velocity of the cruise liner.

Distance = 10 * 1.06
Distance = 10.6 meters

The boat should be 10.6 meters away horizontally from the point of release.
4 0
2 years ago
Keisha looks out the window from a tall building at her friend Monique standing on the ground, 8.3 m away from the side of the b
Salsk061 [2.6K]

Answer:

Explanation:

GIVEN DATA:

Distance between keisha and her friend 8.3 m

angle made by keisha toside building 30 degree

height of her friend monique is 1.5 m

from the figure

\Delta ACB

tan 30 = \frac{8.3}{h}

h= \frac{8.3}{tan 30} = 14.376 m

therefore

height of keisha is = h  + 1.5 m

                               = 14.376 + 1.5

= 15.876 \simeq 16 m

therefore option c is correct

5 0
2 years ago
A 4.0-m-diameter playground merry-go-round, with a moment of inertia of 350 kg⋅m2 is freely rotating with an angular velocity of
Flauer [41]

Answer:

v = 4.375\,\frac{m}{s}

Explanation:

The situation of the system Ryan - merry-go-round is modelled after the Principle of the Angular Momentum Conservation:

(350\,kg\cdot m^{2})\cdot (1.5\,\frac{rad}{s} ) - (2\,m)\cdot (60\,kg)\cdot v = 0\,kg\cdot \frac{m^{2}}{s}

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v = 4.375\,\frac{m}{s}

5 0
2 years ago
Read 2 more answers
When you urinate, you increase pressure in your bladder to produce the flow. For an elephant, gravity does the work. An elephant
weqwewe [10]

Answer:

a) v =  1.19 m / s , b)   P₁ = 0.922 10⁵ Pa

Explanation:

1) Let's use the fluid continuity equation

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      A = π r2 = π d²/4

     

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2) write Bernoulli's equation, where point 1 is the bladder and point 2 is the urine exit point

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The exercise tell us

P₂ = 1.0013 105 Pa

v₁ = 0

y₁ = 1 m

y₂=0  

Rho (water) = 1000 kg / m³

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      P₁ = P₂ + ½ rho v₂² - rho g y₁

      P₁ = 1.013 10⁵ + ½ 1000 (1.19)² - 1000 9.8 1

      P₁ = 1.013 10⁵ +708.5  - 9800

      P₁ =  92208.5Pa

      P₁ = 0.922 10⁵ Pa

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2 years ago
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