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3 years ago

How to use producer in a sentence science terms

Physics

1 answer:

Lunna [17]3 years ago

5 0

A producer makes its own food unlike herbivores who eat the plants.

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Magnetism is a type of A. force B. energy C. gravity D. electricity

erastovalidia [21]

Answer:

A. force

Explanation:

Magnetism is a type of force.

5 0

3 years ago

Read 2 more answers

Integrated Science- Study Guide: Water

postnew [5]

Answer:

Not all the pictures are popping up. So I can't answer them all. :(

Explanation:

The fist one is B. And the second is D. That's all I can see. I hope this helps a little bit :(

7 0

3 years ago

A 0.50 mm-wide slit is illuminated by light of wavelength 500 nm.What is the width of the central maximum on a screen 2.0m behin

Novay_Z [31]

Answer:

0.004 m

Explanation:

For light passing through a single slit, the position of the nth-minimum in the diffraction pattern is given by

$y=\frac{n\lambda D}{d}$

where

$\lambda$ is the wavelength

D is the distance of the screen from the slit

d is the width of the slit

Therefore, the width of the central maximum is equal to twice the value of y for n=1 (first minimum):

$w=2\frac{\lambda D}{d}$

where we have

$\lambda=500 nm = 5\cdot 10^{-7}m$ is the wavelength

D = 2.0 m is the distance of the screen

$d=0.50 mm=5\cdot 10^{-4}m$ is the width of the slit

Substituting, we find

$w=2\frac{(5\cdot 10^{-7} m)(2.0 m)}{5\cdot 10^{-4} m}=0.004 m$

5 0

3 years ago

A 1-meter-long wire consists of an inner copper core with a radius of 1.0 mm and an outer aluminum sheathe, which is 1.0 mm thic

Mazyrski [523]

Answer:

The total resistance of the wire is = $1.917\times10^{-3}$

Explanation:

Since the wires will both be in contact with the voltage source at the same time and the current flows along in their length-wise direction, the two wires will be considered to be in parallel.

Hence, for resistances in parallel, the total resistance, $R_{Total}$

$\frac{1}{R_{Total}} =\frac{1}{R_{cu} }+\frac{1}{R_{al}}$

Parameters given:

Length of wire = 1 m

Cross sectional area of copper $A_{cu}= \pi r^{2}= \pi \times (1\times 10^{-3} )^{2} =3.142\times10^{-6} m^{2}$

Cross sectional area of aluminium wire

$A_{al}= \pi( R^{2}-r^{2})\\\\ = \pi \times [ (2\times 10^{-3} )^{2}-(1\times 10^{-3} )^{2}] =9.42\times10^{-6} m^{2}\\$

Resistivity of copper $\rho _{cu}=1.7\times 10^{-8} \Omega .m$

Resistivity of Aluminium $\rho _{al}=2.8\times 10^{-8} \Omega .m$

Resistance of copper $R_{cu}= \frac{\rho_{cu} \times l}{A_{cu} } =\frac{1.7\times 10^{-8} \times 1}{3.142\times10^{-6} } =5.41\times 10^{-3}\Omega$

Resistance of aluminium $R_{al}= \frac{\rho_{al} \times l}{A_{al} } =\frac{2.8\times 10^{-8} \times 1}{9.42\times10^{-6} } =2.97\times 10^{-3}\Omega$

The total resistance of the wire can be obtained as follows;

$\frac{1}{R_{Total}} =\frac{1}{5.41\times10^{-3} }+\frac{1}{2.97\times10^{-3}}=521.52\frac{1}{\Omega}$

$R_{Total}= 1.917\times 10^{-3}\Omega$

∴ The total resistance of the wire = $1.917\times 10^{-3}\Omega$

4 0

3 years ago

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the electrostatic force between two objects is 40N. if the charge of one object is cut in half, and the distance is doubled, wha

n200080 [17]

Answer:

F1 = K Q1 Q2 / R1^2

F2 = K Q1 / 2 * Q2 / (2 R1)^2

F2 / F1 = 1/2 / 4 = 1/8

The new force is 5N (1/2 due to charge and 1/4 due to distance)

8 0

3 years ago