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3 years ago

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Why is it so important to avoid pollution of freshwater ecosystems? A. Freshwater ecosystems contain more species than marine ec

osystems contain. B. Freshwater ecosystems are the source of most of our drinking water. C. There are more freshwater ecosystems than there are marine ecosystems. D. Freshwater organisms are more important than marine organisms are.

2 answers:

Mrrafil [7]3 years ago

5 0

The answer is B because it literally says it right there, and I don't wanna make it impossibly confusing for a middle school student, lol.

makkiz [27]3 years ago

3 0

it is b!!!!!!!!!!!!!!!!!!!!!!

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As a quality test of ball bearings, you drop bearings (small metal balls), with zero initial velocity, from a height of 1.94 m i

11Alexandr11 [23.1K]

Answer:

The average acceleration of the bearings is $0.77\times10^{3}\ m/s^2$

Explanation:

Given that,

Height = 1.94 m

Bounced height = 1.48 m

Time interval $t=14.86\times10^{-3}\ s$

Velocity of the ball bearing just before hitting the steel plate

We need to calculate the velocity

Using conservation of energy

$mgh=\dfrac{1}{2}mv^2$

Put the value into the formula

$9.8\times1.94=\dfrac{1}{2}\times v^2$

$v=\sqrt{2\times9.8\times1.94}$

$v=6.166\ m/s$

Negative as it is directed downwards

After bounce back,

We need to calculate the velocity

Using conservation of energy

$mgh=\dfrac{1}{2}mv^2'$

Put the value into the formula

$9.8\times1.48=\dfrac{1}{2}\times v^2'$

$v'=\sqrt{2\times9.8\times1.48}$

$v'=5.38\ m/s$

We need to calculate the average acceleration of the bearings while they are in contact with the plate

Using formula of acceleration

$a=\dfrac{v-v'}{t}$

Put the value into the formula

$a=\dfrac{5.38-(-6.166)}{14.86\times10^{-3}}$

$a=776.98\ m/s^2$

$a=0.77\times10^{3}\ m/s^2$

Hence,The average acceleration of the bearings is $0.77\times10^{3}\ m/s^2$

6 0

3 years ago

An astronaut is standing on the surface of a planetary satellite that has a radius of 1.74 × 10^6 m and a mass of 7.35 × 10^22 k

ExtremeBDS [4]

Answer:

2.87 km/s

Explanation:

radius of planet, R = 1.74 x 10^6 m

Mass of planet, M = 7.35 x 10^22 kg

height, h = 2.55 x 10^6 m

G = 6.67 x 106-11 Nm^2/kg^2

Use teh formula for acceleration due to gravity

$g=\frac{GM}{R^{2}}$

$g=\frac{6.67\times 10^{-11}\times 7.35\times 10^{22}}{1.74^{2}\times 10^{12}}$

g = 1.62 m/s^2

initial velocity, u = ?, h = 2.55 x 10^6 m , final velocity, v = 0

Use third equation of motion

$v^{2}=u^{2}-2gh$

0 = v² - 2 x 1.62 x 2.55 x 10^6

v² = 8262000

v = 2874.37 m/s

v = 2.87 km/s

Thus, the initial speed should be 2.87 km/s.

6 0

3 years ago

A closely wound search coil has an area of 3.21 cm2, 120 turns, and a resistance of 58.7 O. It is connected to a charge-measurin

Alexxx [7]

Answer:

The magnetic field in the System is 0.095T

Explanation:

To solve the exercise it is necessary to use the concepts related to Faraday's Law, magnetic flux and ohm's law.

By Faraday's law we know that

$\epsilon = \frac{NBA}{t}$

Where,

$\epsilon =$ electromotive force

N = Number of loops

B = Magnetic field

A = Area

t= Time

For Ohm's law we now that,

V = IR

Where,

I = Current

R = Resistance

V = Voltage (Same that the electromotive force at this case)

In this system we have that the resistance in series of coil and charge measuring device is given by,

$R = R_c + R_d$

And that the current can be expressed as function of charge and time, then

$I = \frac{q}{t}$

Equation Faraday's law and Ohm's law we have,

$V = \epsilon$

$IR = \frac{NBA}{t}$

$(\frac{q}{t})(R_c+R_d) = \frac{NBA}{t}$

Re-arrange for Magnetic Field B, we have

$B = \frac{q(R_c+R_d)}{NA}$

Our values are given as,

$R_c = 58.7\Omega$

$R_d = 45.5\Omega$

$N = 120$

$q = 3.53*10^{-5}C$

$A = 3.21cm^2 = 3.21*10^{-4}m^2$

Replacing,

$B = \frac{(3.53*10^{-5})(58.7+45.5)}{120*3.21*10^{-4}}$

$B = 0.095T$

Therefore the magnetic field in the System is 0.095T

3 0

3 years ago

A block (mass = 61.2 kg) is hanging from a massless cord that is wrapped around a pulley (moment of inertia = 1/2MR2 kg · m2, wh

kolezko [41]

Answer:

The angular velocity is $w = 53.35 \ rounds /minute$

Explanation:

From the question we are told that

The mass of the block is $m = 61.2kg$

The of the pulley is $M = 14.2 kg$

The radius of the pulley is $R = 1.5m$

The radius of the cord around the pulley is $r = 1.5 m$

The distance of the block to the floor is $d = 8.0 m$

From the question we are told that the moment of inertia of the pulley is

$I = \frac{1}{2} MR^2 kg \cdot m^2$

Substituting value

$I = \frac{1}{2} * 14.2 * (1.5)^2$

$I = 15.975 kg \cdot m^2$

Using the Newtons law we can express the force acting on the vertical axis as

$ma = mg -T$

=> $T = mg -ma$

Now when the pulley is rotated that torque generated on the massless cord as a r result of the tension T and the radius of the cord around the pulley is mathematically represented as

$\tau = I \alpha$

Here $\alpha$ is the angular acceleration

Here $\tau$ is the torque which can be equivalent to

$\tau = T r$

Substituting this above

$Tr = I \alpha$

Substituting for T

$(mg - ma ) r = I\ r \alpha$

Here $a$ is the linear acceleration which is mathematically represented as

$a = r\alpha$

$(mg - m(r\alpha ) ) r = I\ r \alpha$

$mgr = I\alpha + m(r\alpha ) r$

$mgr = \alpha [ I + mr^2]$

making $\alpha$ the subject

$\alpha = \frac{mgr}{I -mr ^2}$

Substituting values

$\alpha = \frac{61.2 * 1.5 * 9.8}{15.975 + (61.2 ) * (1.5)^2}$

$\alpha =5.854 rad /s^2$

Now substituting into the equation above to obtain the acceleration

$a = 5.854 * 1.5$

$a=8.78 m/s^2$

This acceleration is $a = \frac{v}{t}$

and v is the linear velocity with is mathematically represented as

$v = \frac{d}{t}$

Substituting this into the formula acceleration

$a = \frac{d}{t^2}$

making t the subject

$t = \sqrt{\frac{d}{a} }$

substituting value

$t = \sqrt{\frac{8}{8.78}}$

$t = 0.9545 \ s$

Now the linear velocity is

$v = \frac{8}{0.9545}$

$v = 8.38 m/s$

The angular velocity is

$w = \frac{v}{r}$

So

$w = \frac{8.38}{1.5}$

$w = 5.59 rad/s$

Generally 1 radian is equal to 0.159155 rounds or turns

So 5.59 radian is equal to x

Now x is mathematically obtained as

$x = \frac{5.59 * 0.159155}{1}$

$= 0.8892 \ rounds$

Also

60 second = 1 minute

So 1 second = z

Now z is mathematically obtained as

$z = \frac{ 1}{60}$

z $= 0.01667 \ minute$

Therefore

$w = \frac{0.8892}{0.01667}$

$w = 53.35 \ rounds /minute$

8 0

3 years ago

Can someone explain how the weight of the block is 10.26N, with reference to an appropriate law of motion?

spayn [35]

this process is called parellelogram method of resolving vectors.

5 0

3 years ago