the highest theoretical efficiency of a certain engine is 30% if the engine uses the atmosphere which has a temperature of 300k as its cold reservoir what is temperature of its hot reservoir

Answer:

Mass of block = 6kg

wright = 6 × g = 60 N using g = 10 m/s^2 .

compression= 0.20m

spring constant =( weight / compression) =( 60/0.20) = 300 N/m .

Work done in compressing 0.20 m =( 1/2 × 300 × 0.20 × 0.20)= 6 joule.

Additional compression = 0.10 m

total compression = 0.20+ 0.10= 0.30 m.

Total work done = ( 1/2 × 300 × 0.30× 0.30) = 13.5 joule.

Hence work done in compressing from 0.20 m to 0.30 m will be = ( 13.5 - 6) = 7.5 joule.

Note :- work done in compressing a spring having spring constant K N/m by x m is equal to 1/2× (K) × X^ 2 .

Answer:

No because a pot plant system must interact with its surroundings (air, water, etc) and these are not closed systems.

Only if the pot plant were completely enclosed, say in a large glass jar would the system be closed.

(1) If the ship moves upward with a constant velocity, the period increases, false.

(2) If the length of the pendulum is doubled, the new period will be: square root of 2 times T₀. true.

(3) If the ship accelerates upward, the period increases, true.

(4) If the ship accelerates downward at 9.81 m/s² , the pendulum will no longer oscillate, false.

(5) If the mass of the pendulum is halved, the period decreases, false.

The period of a simple pendulum is given as;

T = 2π√(L/g)    ------- (1)

where;

When the ship moves upward with a constant velocity, the period will not change.

T ∝√L

when, L = 2L₀

T = T₀√2

Thus, if the length of the pendulum is doubled, the new period will be: square root of 2 times T₀.

t = √2h/g

T = √I/mgd

where;

when the mass is halved, the period increases.

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Answer:

X component= 46 N

Y component= 80 N

Magnitude =( 46 ^2 + 80^2 )^0.5 = 92.3 N

The minimum necessary height of the IV bag above the position of the needle is 0.37 m.

The minimum necessary height of the IV bag above the position of the needle is calculated as follows;

P = ρgh

where;

Substitute the given parameters and solve for minimum height

h = P/ρg

h = (4759.609) / (9.8 x 1308)

h = 0.37 m

Thus, the minimum necessary height of the IV bag above the position of the needle is 0.37 m.

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Answer:

because he's fast and speedy

Explanation:

Answer:

Explanation:

yes

Answer:

Astrobiologists have put in tremendous efforts till now in the search for extra-terrestrial life. They are searching not only for tremendously developed life forms like humans, but also for microbes that are found to inhabit even the harshest of environments on Earth. Due to the tremendous development in technology, the odds of finding life beyond Earth are increasing day by day. Nasa's Kepler mission was launched to hunt for Earth like planets beyond our Solar System, and till now this mission has established the presence of 2,337 exoplanets, along with 1,284 new planets. This hugely increases the chances of finding another Earth like planet outside the limits of our Solar system. NASA's Astrobiology researchers are presently studying the most severe environments on Earth for improved understanding of the kind of conditions that would be able to support life in every part of the universe. Moreover, programs for instance SETI are keeping an eye on the galaxy for electromagnetic communications occurring or situated between stars from the societies on new worlds.

There is still a tremendous possibility to find life in our solar system. By the discovery of life forms in the extreme environments here on Earth, scientists have realized that the presence of solar energy is not an absolute necessity for the subsistence of life. Energy from chemical reactions can also trigger the organic elements to interact and start the formation of life. In the Solar system, Jupiter's moons Io and Europa and Saturn's moons Enceladus and Titan have been realized to the very strong candidates for the presence of microscopic life forms.

Explanation:

I hold a very optimistic view on the presence of life beyond Earth. It is highly probable that life on Earth is not an exceptional case but it should be a general one in the Universe. While we are always excited to imagine that we might encounter species as much developed or more highly developed than us, there is very good chance that the Universe is teeming with microscopic life forms. Also, to discover another intelligent life form would also be highly welcome.

Answer:

See below

Explanation:

A) bearing y reaches the FLOOR in the same amount of time   .5 seconds  since there is no initial vertical velocity component

B.  in .5 seconds y will travel     5 m/s * .5 s = 2.5 meters horizontally

C.  d = 1/2 a t^2

         = 1/2 (9.81)(.5^2) = 1.23 m table height

(a) Bearing y takes 0.2 seconds to reach the table. (b) Bearing y travels 1 meter horizontally before hitting the floor. (c)  The height of the table top above the floor level is 1.225 meters.

a) Time is taken for bearing y to reach the table:

For a horizontal motion, the equation to calculate time is:

Time (t) = Distance (d) ÷ Velocity (v)

Time (t) = Width of the table / Velocity of y

t = 1  / 5

t = 0.2 sec

So, bearing y takes 0.2 seconds to reach the table.

b) Horizontal distance traveled by bearing y before hitting the floor:

Distance (d) = Velocity (v) × Time (t)

d = 5 × 0.2

d = 1 m

So, Bearing y travels 1 meter horizontally before hitting the floor.

c) The height of the table top above the floor level:

h = (1/2) × g × t²

h = (1/2) × 9.8 × (0.5)²

h = (1/2) × 9.8 × 0.25

h = 1.225 m

So, the height of the table top above the floor level is 1.225 meters.

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Answer:

.12 N

Explanation:

Velocity goes from 2 m/s    to   0  m/s   so average velocity = 1 m/s

.250 m / 1 m/s  =  .25 second to stop

change in velocity / change in time = accel

  2 m/s / .25 sec =  8 m/s^2

F = ma

  = .015 kg *  8 m/s^2  = .12 N

The oxygen–hemoglobin dissociation curve depicts the relationship between the partial pressure of oxygen (pao2) and the arterial oxygen.

The oxygen-hemoglobin dissociation curve (OHDC) indicates the relationship between the oxygen saturation of hemoglobin and the partial pressure of arterial oxygen. This curve can change its position or shifted to another level depending on various factors.

So we can conclude that the oxygen–hemoglobin dissociation curve depicts the relationship between the partial pressure of oxygen (pao2) and the arterial oxygen.

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B. As the resistance increases, the slope on the current vs voltage graph gets flatter.

Resistance is the opposition to the follow of current in electrical material.

The mathematical relationship between resistance, current and voltage is given by Ohm's law.

R = V/I

where;

From the equation above, increase in resistance means an increase in voltage or decrease in current.

In Current vs voltage graph, the following will occur;

Thus, as the resistance increases, the slope on the current vs voltage graph gets flatter.

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Answer:

an open fracture contains some sort of water which if it comes in contact with electrons could result to a shock

Answer:

[tex]V = 900\; {\rm V}[/tex] for a resistor where [tex]I = 30\; {\rm A}[/tex] and [tex]R = 30\; {\rm \Omega}[/tex].

Explanation:

By Ohm's Law, if the electric resistance of a resistor is [tex]R[/tex] and the current going through that resistor is [tex]I[/tex], the voltage across that resistor would be [tex]V = I\, R[/tex].

Note that electric resistance is typically measured in Ohms, where [tex]1\; {\rm \Omega} = 1\; {\rm V \cdot A^{-1}}[/tex].

Given that [tex]I = 30\; {\rm A}[/tex] and [tex]R = 30\; {\rm \Omega}[/tex] for the resistor in this question, apply the formula [tex]V = I\, R[/tex] to find the voltage [tex]V[/tex] across this resistor:

[tex]\begin{aligned} V &= I\, R \\ &= 30\; {\rm A} \times 30\; {\rm \Omega} \\ &= 30\; {\rm A} \times 30\; {\rm V \cdot A^{-1}} \\ &= 900\; {\rm V} \end{aligned}[/tex].

The average density of the material from which the coin is made is 9.67 g/cm³.

The volume of the coin at the given diameter is calculated as follows;

V = Ah

where;

V = πd²/4 x h

V = π(2.8)²/4  x (0.21 cm)

V = 1.293 cm³

The average density of the material from which the coin is made is calculated as follows;

density = mass/volume

density = 12.5 g / (1.293 cm³)

density = 9.67 g/cm³

Thus, the average density of the material from which the coin is made is 9.67 g/cm³.

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Answer:

Warm weather may be found in the tropical zone since the sun's heat is concentrated in regions close to the equator. At 23.5 degrees north of the equator and 23.5 degrees south of the equator, this area is nearly in the middle of the world. It has a monthly average temperature of 18 degrees Celsius or higher.

Explanation:

The time it takes for the ball to hit the ground from the platform is determined as 1.56 seconds.

The time of motion of the ball is calculated as follows;

h = vt + ¹/₂gt²

where;

h = 0 + ¹/₂gt²

h = ¹/₂gt²

t = √(2h/g)

t = √(2 x 12 / 9.8)

t = 1.56 seconds

Thus, the time it takes for the ball to hit the ground from the platform is determined as 1.56 seconds.

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Answer:

an instrument measuring atmospheric pressure, used especially in forecasting the weather and determining altitude.

Explanation:

hope it helps ya

Answer: 1.348 meters

Explanation: Although the sign is missing from the location of the 4.00 kg object, it is assumed to be positive. The net moment of all the objects about the center of mass must be zero. Let the center of mass be on the y axis at a point  c . Adding the four moments together, we get:

(2.00)(3.00−c)+(3.00)(2.50−c)+(2.50)(0−c)+(4.00)(0.500−c)=0

6.00−2.00c+7.50−3.00c+0−2.50c+2.00−4.00c=0

11.5c=15.50

c=  1.348 metres

The center of mass is on the y axis at  y  = 1.348 metres.

Answer:

B. 2.5 Ω

Explanation:

[tex]\frac{1}{5} +\frac{1}{5}=0.4[/tex]

Put a one on top of the 0.4

[tex]\frac{1}{0.4} =2.5[/tex]

Answer:

v

f=90km/h=25m/s

v_{_{i}}=0v

i=0

t=10\;\mathrm{s}t=10s

m=1000\;\mathrm{kg}m=1000kg

The acceleration of the car is 2.5 m/s². The net force acting on the car is 2500 Newtons (N). The correct answers are  2.5 m/s², 2500 Newtons (N).

Given:

Mass of the car (m) = 1000 kg

Acceleration (a) = 2.5 m/s²

a) The acceleration of the car is:

Acceleration (a) = Change in velocity / Time taken

Final velocity (v(f)) = 90 × 0.27778  = 25 m/s

Acceleration (a) = (Final velocity - Initial velocity) / Time taken

Acceleration (a) = (25 - 0 ) / 10

Acceleration (a) = 2.5 m/s²

Therefore, The acceleration of the car is 2.5 m/s².

b) Newton's second law of motion:

F = m × a

F = 1000 × 2.5

F = 2500 N

The net force acting on the car is 2500 Newtons (N).

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The final vertical velocity of the skydiver at 50.8 m of fall is 31.56 m/s.

The time of motion of the girl is calculated as follows;

h = vt + ¹/₂gt²

where;

Substitute the given parameters and solve for time of motion;

50.8 = 0 + ¹/₂(9.8)t²

2(50.8) = 9.8t²

101.6 = 9.8t²

t² = 101.6/9.8

t² = 10.367

t = √10.367

t = 3.22 seconds

vf = vi + gt

where;

vi is the initial vertical velocity = 0

vf = 0 + 9.8(3.22)

vf = 31.56 m/s

Thus, the final vertical velocity of the skydiver at 50.8 m of fall is 31.56 m/s.

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The period of the oscillations.T = 1.2042s

Opposition is the process of any quantity or measure fluctuating repeatedly about its equilibrium value throughout time. This process is referred to as oscillation. Oscillation, a periodic fluctuation of a substance, can also be described as alternating between two values or rotating around a central value.

Typically, the mathematical formula for the moment of inertia is

T = 2 π √(I / mgd)

Therefore, a moment of inertia

I = 9.00×10-3 + md^2 ;

I=9.00*10^{-3}+ 0.5 * 0.3^2

I=0.054

T=2[tex]\pi \sqrt{0.5*9.8*0.3}[/tex]

T=1.2042s

The period of the oscillations.T = 1.2042s

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The maximum rate at which energy can be added to the circuit element mathematically given as

[tex]MER=5.044 \times 10^{-4} \mathrm{~J} / \mathrm{sec}[/tex]

Generally, the equation for P is  mathematically given as

[tex]P=\ln s \frac{\Delta T}{\Delta t}[/tex]

Therefore

[tex]Rate\ of\ Change\ of\ Temp =\frac{p}{lnS}[/tex]

[tex]\frac{p}{lnS}=\frac{7.4 \times 10^{-3}}{23 \times 10^{-6} \times 705}[/tex]

[tex]\frac{p}{lnS}=0.456^{\circ \mathrm{c}} / \mathrm{sec}[/tex]

Max temp Change

[tex]MaxT=5.6^{\circ} \mathrm{C}[/tex]

[tex]\text { time }=3 \times 60[/tex]

t=180s

In conclusion, Max Energy Rate

[tex]MER =23 \times 10^{-6} \times \frac{301 \times 5.6}{180}[/tex]

[tex]MER=5.044 \times 10^{-4} \mathrm{~J} / \mathrm{sec}[/tex]

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Answer:

water at 4°c that's abnormal stage of water