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3 years ago

As you go up a food web is energy lost or gained? A/ lost b/ gained

Physics

2 answers:

Hatshy [7]3 years ago

8 0

I'm preeeety sure the answer is A, lost! Hope I helped, hon :)

Oliga [24]3 years ago

8 0

The answer is A hope it helps

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(a) According to Hooke's Law, the force required to hold any spring stretched x meters beyond its natural length is f(x)=kx. Sup

KengaRu [80]

Answer:

a) The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules, b) The area of the region enclosed by one loop of the curve $r(\theta) = 2\cdot \sin 5\theta$ is $4\pi$ .

Explanation:

a) The work, measured in joules, is a physical variable represented by the following integral:

$W = \int\limits^{x_{f}}_{x_{o}} {F(x)} \, dx$

Where

$x_{o}$ , $x_{f}$ - Initial and final position, respectively, measured in meters.

$F(x)$ - Force as a function of position, measured in newtons.

Given that $F = k\cdot x$ and the fact that $F = 25\,N$ when $x = 0.3\,m - 0.2\,m$ , the spring constant ( $k$ ), measured in newtons per meter, is:

$k = \frac{F}{x}$

$k = \frac{25\,N}{0.3\,m-0.2\,m}$

$k = 250\,\frac{N}{m}$

Now, the work function is obtained:

$W = \left(250\,\frac{N}{m} \right)\int\limits^{0.05\,m}_{0\,m} {x} \, dx$

$W = \frac{1}{2}\cdot \left(250\,\frac{N}{m} \right)\cdot [(0.05\,m)^{2}-(0.00\,m)^{2}]$

$W = 0.313\,J$

The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules.

b) Let be $r(\theta) = 2\cdot \sin 5\theta$ . The area of the region enclosed by one loop of the curve is given by the following integral:

$A = \int\limits^{2\pi}_0 {[r(\theta)]^{2}} \, d\theta$

$A = 4\int\limits^{2\pi}_{0} {\sin^{2}5\theta} \, d\theta$

By using trigonometrical identities, the integral is further simplified:

$A = 4\int\limits^{2\pi}_{0} {\frac{1-\cos 10\theta}{2} } \, d\theta$

$A = 2 \int\limits^{2\pi}_{0} {(1-\cos 10\theta)} \, d\theta$

$A = 2\int\limits^{2\pi}_{0}\, d\theta - 2\int\limits^{2\pi}_{0} {\cos10\theta} \, d\theta$

$A = 2\cdot (2\pi - 0) - \frac{1}{5}\cdot (\sin 20\pi-\sin 0)$

$A = 4\pi$

The area of the region enclosed by one loop of the curve $r(\theta) = 2\cdot \sin 5\theta$ is $4\pi$ .

5 0

3 years ago

(marking brainliest) pease help asap! both of the questions are in the pdf, and please let me know which is question one and whi

REY [17]

<h2>ANSWERS </h2>

1) The relationship in between the electrical energy carriesd by the transmission wires and the amount of the heat loss in it is due to the reason that when the electricity is flown through the wires there are some resistance found in these wires which creates a disturbance in the efficient flow of electricty.Also we know that current have an heating effect when it is in motion as due to if a large amount or magnitude of electricity is flown through the transmission wires it will carry a larger heat effected and also due to the resistance is provided by the wires and so the process of heat loss takes place.

2)It is important to minimize current in transmission wires due to minimize the heat loss and resistance on flowing electric current to make the system more efficient 

 3)Given Resistance = 250 ohms 

Electric potential = 150 volts 

so we know Power = 

volt^2/Resistance = 

=(150^2/250)(ohms/volts)

=(22500/250)watt = 90 watt 

4)Heat energy (H) = Power(P)×Time(t)

4)Heat energy (H) = Power(P)×Time(t)= (90×2)joules = 180 joules

Hope it helps

4 0

3 years ago

Four particles are in a 2-d plane with masses, x- and y- positions, and x- and y- velocities as given in the table below: what i

Arte-miy333 [17]

I attached the picture of the missing table.
Center of mass is the point such that if you apply force to it, the system would move without rotating.
We can use following formula to calculate the center of mass:
$R=\frac{1}{M}\sum_{i=1}^{n=i}m_ir_i$
Where M is the sum of the masses of all particles.
Part 1
To calculate the x coordinate of the center of mass we will use this formula:
$R_x=\frac{1}{M}\sum_{i=1}^{n=i}m_ix_i$
I will do all the calculations in the google sheet that I will share with you.
For the x coordinate of the center of mass we get:
$R_x=0.96m$
Part 2
To calculate the y coordinate of the center of mass we will use this formula:
$R_y=\frac{1}{M}\sum_{i=1}^{n=i}m_iy_i$
I will do all the calculations in the google sheet that I will share with you.
For the x coordinate of the center of mass we get:
$R_y=-0.84m$
Part 3
We will calculate speed along x and y-axis separately and then will add them together.
$v_x=\frac{\sum_{i=1}^{n=i}m_iv_x_i}{M}$
$v_y=\frac{\sum_{i=1}^{n=i}m_iv_y_i}{M}$
Total velocity is:
$v=\sqrt{v_x^2+v_y^2}$
Once we calculate velocities we get:
$v_x=-1.08\frac{m}{s}\\ v_y=-0.03\frac{m}{s}\\ v=\sqrt{(-1.08)^2+(-0.03)^2}=1.08\frac{m}{s}$
Part 4
Because origin is left to our center of mass(please see the attached picture) placing fifth mass in the origin would move the center of mass to the left along the x-axis.
Part 5
If you place fifth mass in the center of the mass nothing would change. The center of mass would stay in the same place.
Here is the link to the spreadsheet:
https://docs.google.com/spreadsheets/d/1SkQHbI1BxiJnwpWbLmP0XWgcNPrGquH1K2MfN6cznVo/edit?usp=sharing

3 0

3 years ago

(a) Calculate the absolute pressure at the bottom of a fresh- water lake at a depth of 27.5 m. Assume the density of the water i

ddd [48]

Answer:

a) $P = 370.993\,kPa$ , b) $F = 25.948\,kN$

Explanation:

a) The absolute pressure at a depth of 27.5 meters is:

$P = P_{atm} + P_{man}$

$P = 101.3\,kPa + \left(1000\,\frac{kg}{m^{3}}\right)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (27.5\,m)\cdot \left(\frac{1\,kPa}{1000\,Pa} \right)$