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3 years ago

Because the current in the equivalent resistor of Sample Problem A

Physics

1 answer:

Serhud [2]3 years ago

4 0

V1= IR1= 0.50×2 = 1 V

V2= IR2= 0.50×4= 2 V

V3= IR3= 0.50×5= 2.5 V

V4= IR4= 0.50×7 = 3.5 V

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If you place 48 cm object 15 m in front of a convex mirror of focal length 5 m, what is the magnification of the image?

Arisa [49]

Answer:

m = 0.25

Explanation:

Given that,

Object distance, u = -15cm

Height of the object, h = 48

Focal length, f = cm

We need to find the magnification of the image.

Let v is the image distance. Using mirror's equation.

$\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}\\\\\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}\\\\=\dfrac{1}{5}-\dfrac{1}{(-15)}\\\\=3.75\ cm$

Magnification,

$m=\dfrac{-v}{u}\\\\=\dfrac{-3.75}{-15}\\\\=0.25$

Hence, the magnification of the image is 0.25.

7 0

3 years ago

The volume of a ball was measured at 500.0 cm3, and its mass was measured to be 404.2 g.

KonstantinChe [14]

Answer:

0.8084g/cm³

Explanation:

For calculating density, mass is divided by volume.

Its units is commonly in grams per cubic centimeters.

Therefore,

density= mass/volume

= 404.2/500

=0.8084g/cm³

4 0

3 years ago

Read 2 more answers

Which are functions of the endoplasmic reticulum (ER)?

Oksana_A [137]

Answer:

a. DNA storage

Explanation:

6 0

3 years ago

Isotope A contains 56 protons and 80 neutrons. Isotope B contains 55 protons and 81 neutrons. Isotope C contains 57 protons and

ioda

mass number is defined as sum of number of protons and neutrons

so here for Isotope A

Isotope A contains 56 protons and 80 neutrons.

Mass number will be given as

$A = 56 + 80 = 136$

For Isotope B

Isotope B contains 55 protons and 81 neutrons.

$A = 55 + 81 = 136$

For Isotope C

Isotope C contains 57 protons and 80 neutrons.

$A = 57 + 80 = 137$

For Isotope D

Isotope D contains 56 protons and 74 neutrons.

$A = 56 + 74 = 130$

<em>So here Isotope A and Isotope B has same mass number</em>

3 0

3 years ago

The beautiful Multnomah Falls in Oregon are approximately 206m high. If the Columbia river is flowing horizontally at 2.90m/s ju

zhenek [66]

Answer:

The overall velocity of the water when it hits the bottom is:

$v_f=63.61\ \frac{m}{s}$

Explanation:

Use the law of conservation of energy.

Call it instant [1] to the moment when the water is just before reaching the falls.

At this moment its height h is 206 meters and its velocity horizontally $v_i$ is $v_i = 2.90$ m/s.

At the instant [1] the water has gravitational power energy $E_g$

$E_g = mgh$

The water also has kinetic energy Ek.

$E_k = 0.5mv_i ^ 2$

Then the Total E1 energy is:

$E_1 = mgh + 0.5mv_i ^ 2$

In the instant [2] the water is within an instant of touching the ground. At this point it only has kinetic energy, since the height h = 0. However at time [2] the water has maximum final velocity $v_f$

So:

$E_2 = 0.5mv_f ^ 2$

As the energy is conserved then $E_1 = E_2$

$mgh + 0.5mv_i ^ 2 = 0.5mv_f ^ 2$

Now we solve for $v_f$ .

$gh + 0.5v_i ^ 2 = 0.5v_f ^ 2\\\\9.8(206) + 0.5(2.90) ^ 2 = 0.5v_f 2\\\\v_f^2 = \frac{9.8(206) + 0.5(2.90) ^ 2}{0.5}\\\\v_f = \sqrt{\frac{9.8(206) + 0.5(2.90) ^ 2}{0.5}}\\\\v_f=63.61\ \frac{m}{s}$

7 0

3 years ago