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3 years ago

7

FREE BRAINLIEST! if you can answer this correctly ill give you brainliest and answer some of the questions you have posted :) th

ank you very much!!! (22pts)

1 answer:

Gala2k [10]3 years ago

4 0

b) accelerate to the left as much more pressure is pulling it in that direction and on the right however , there is less force .

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In an engine, an almost ideal gas is compressed adiabatically to half its volume. In doing so, 2820 J of work is done on the gas

Dmitry_Shevchenko [17]

Answer:

$Q=0$
$U=2820$
Energy increases

Explanation:

From the question we are told that

Work done $W=2820$

a)Generally the heat flow for an adiabatic process is 0 (zero)

$Q= U + W =>0$

$Q=0$

b)Generally Change in internal energy of gas is mathematically given by

Since $W=-2820J$

Therefore

$U=2820$

Giving

$Q= 2820 -2820$

$Q=O$

c)With increases in internal energy brings increase in temperature

Therefore

Energy increases

3 0

2 years ago

At a certain harbor, the tides cause the ocean surface to rise and fall a distance d (from highest level to lowest level) in sim

OleMash [197]

Answer:

1.869 hours

Explanation:

$T$ = Time period = 11.1 h

Angular frequency is given by

$\omega=\frac{2\pi}{T}\\\Rightarrow \omega=\frac{2\pi}{11.1}\\\Rightarrow \omega=0.566\ rad/h$

The distance moved from highest to lowest level is given by

$d=2x_{m}\\\Rightarrow x_m=\frac{d}{2}\\\Rightarrow x_m=0.5d$

At the ocean surface $x_m=0.25d$

$x=x_mcos(\omega t+\phi)$

$\phi$ = Phase constant = 0 as clock is started at $x_0=x_m$

$0.25d=0.5dcos(0.56t)\\\Rightarrow cos(0.56t)=\frac{0.25}{0.5}\\\Rightarrow 0.56t=cos^{-1}0.5\\\Rightarrow t=\frac{1.047}{0.56}\\\Rightarrow t=1.869\ h$

The time taken for the water to fall the distance is 1.869 hours

7 0

3 years ago

The engine of a locomotive exerts a constant force of 6.8 105 N to accelerate a train to 80 km/h. Determine the time (in min) ta

Setler [38]

Answer:

t = 6 minutes

Explanation:

Given that,

Force, $F=6.8\times 10^5\ N$

Initial speed of the train, u = 0

Final speed of the train, v = 80 km/h = 22.22 m/s

The mass of the train, $m=1.1\times 10^7\ kg$

We need to find the time taken by the train to come to rest. We know that,

F = ma

$F=\dfrac{m(v-u)}{t}\\\\t=\dfrac{m(v-u)}{F}\\\\t=\dfrac{1.1\times10^7\times (22.22-0)}{6.8\times 10^5}\\\\t=359.44\ s$

or

t = 6 minutes (approx)

So, the required time is equal to 6 minutes.

3 0

2 years ago

A) The current theory of the structure of the Earth, called plate tectonics, tells us that the continents are in constant motion

suter [353]

A) The mass of the continent is $2.5\cdot 10^{21} kg$

B) The kinetic energy is 2016 J

C) The speed of the jogger should be 7.1 m/s

Explanation:

A)

The mass of the continent can be calculated as

$m = \rho V$

where

$\rho = 2800 kg/m^3$ is its density

V is its volume

We have to calculate its volume. We know that the continent is represented as a slab of side 5900 km (so its surface is 5900 x 5900, assuming it is a square) and depth of 26 km, so its volume is:

$V=(5900 km)^2 (26 km)=9.05\cdot 10^8 km^3 =9.05 \cdot 10^8 \cdot (10^9 m^3/k^3)=9.05\cdot 10^7 m^3$

So, the mass of the continent is

$m=\rho V = (2800)(9.05\cdot 10^{17})=2.5\cdot 10^{21} kg$

B)

The kinetic energy of a body is given by

$K=\frac{1}{2}mv^2$

where

m is the mass of the body

v is its speed

For the continent, we have:

$m=2.5\cdot 10^{21} kg$ is the mass

$v=4 cm/year$ is the speed

We have to convert the speed into SI units. we have:

1 cm = 0.01 m

$1 year = (365)(24)(60)(60) s = 3.15\cdot 10^7 s$

So, the speed is

$v=4 cm/year = 0.04 m/year \cdot \frac{1}{3.15\cdot 10^7}=1.27\cdot 10^{-9} m/s$

Therefore, the kinetic energy is

$K=\frac{1}{2}(2.5\cdot 10^{21} kg)(1.27\cdot 10^{-9} m/s)^2=2016 J$

C)

Again, the kinetic energy of an object is

$K=\frac{1}{2}mv^2$

For the jogger in this problem, his mass is

m = 80 kg

And we want its kinetic energy to be equal to that of the continent, so

K = 2016 J

Re-arranging the equation for v, we find what speed the jogger needs to have this kinetic energy:

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(2016)}{80}}=7.1 m/s$

Learn more about kinetic energy here:

brainly.com/question/6536722

#LearnwithBrainly

8 0

3 years ago

A javelin is thrown at 28.5 m/s from flat

zavuch27 [327]

The time for the javelin in the air will be 3.977 second

<h3>What is the period?</h3>

The value of time needed to complete the upward and the downward motion is the total period. Its unit is the second and is denoted by t. T is the time at which the ball first hits the ground.

Given data;

Θ is the angle of throw = 43.2°

u is the thrown velocity = 28.5 m/s

t is the time for the javelin in the air

The time period is found as;

$\rm t = \frac{2u sin \theta }{g} \\\\ t = \frac{ 2 \times 28.5 \ sin 430.2^0}{9.81 } \\\\ t = 3.977 \ sec.$

Hence the time for the javelin in the air will be 3.977 second

To learn more about the period, refer to the link;

brainly.com/question/569003

#SPJ2

4 0

2 years ago

Read 2 more answers