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2 years ago

7. A perfect cube has a width of 2 cm. What is the cube's volume? Show your work!

Physics

1 answer:

Mariana [72]2 years ago

8 0

Answer:

8cm^3

Explanation:

length*width*height=volume

2*2*2=8

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What are (a) the kinetic energy, (b) the rest energy, and (c) the total energy of a 1.50 g particle with a speed of 0.600 c ?

MAVERICK [17]

Kinetic energy = 1/2 m v^2 = 1/2 x1.5 x10^-3 x 0.36

5 0

3 years ago

Calculate the electric field at the center of a square

pantera1 [17]

Answer:

$E_y=1175510.2\ N.C^{-1}$

The Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

Explanation:

Given:

side of a square, $a=52.5\ cm$

charge on one corner of the square, $q_1=+45\times 10^{-6}\ C$

charge on the remaining 3 corners of the square, $q_2=q_3=q_4=-27\times 10^{-6}\ C$

Distance of the center from each corners $=\frac{1}{2} \times diagonals$

$diagonal=\sqrt{52.5^2+52.5^2}$

$diagonal=74.25\cm=0.7425\ m$

∴Distance of center from corners, $b=0.3712\ m$

Now, electric field due to charges is given as:

$E=\frac{1}{4\pi\epsilon_0}\times \frac{q}{b^2}$

For charge $q_1$ we have the field lines emerging out of the charge since it is positively charged:

$E_1=9\times 10^9\times \frac{45\times 10^{-6}}{0.3712^2}$

$E_1=2938775.5\ N.C^{-1}$

Force by each of the charges at the remaining corners:

$E_2=E_3=E_4=9\times 10^9\times \frac{27\times 10^{-6}}{0.3712^2}$

$E_2=E_3=E_4=1763265.3\ N.C^{-1}$

Now, net electric field in the vertical direction:

$E_y=E_1-E_4$

$E_y=1175510.2\ N.C^{-1}$

Now, net electric field in the horizontal direction:

$E_y=E_2-E_3$

$E_y=0\ N.C^{-1}$

So the Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

8 0

3 years ago

Which activity is the best example of cardiovascular and strength training exercises work together

Nataly [62]

it got to be jumping jacks that all i got

4 0

3 years ago

Read 2 more answers

1. Bone has a Young’s modulus of about

Blababa [14]

As we know that

$Y = \frac{stress}{strain}$

now plug in all data into this

$1.8\times 10^{10} = \frac{1.68 \times 10^8}{strain}$

$strain = 9.33 \times 10^{-3}$

now from the formula of strain

$strain = \frac{\Delta L}{L}$

$9.33 \times 10^{-3} = \frac{\Delta L}{0.54}$

$\Delta L = 5.04 \times 10^{-3} m$

$\Delta L = 5.04 mm$

As we know that

pressure * area = Force

here we know that

$Area = 3.53 \times 11.6 = 40.95 m^2$

$P = 0.2 atm = 0.2 \times 1.01 \times 10^5 = 2.02\times 10^4 Pa$

now force is given as

$F = 40.95 \times (2.02\times 10^4) = 8.27 \times 10^5 N$

As we know that density of water will vary with the height as given below

$\rho = \frac{\rho_0}{1 - \frac{\Delta P}{B}}$

here we know that

$\Delta P = 2600 atm = 2600 \times 1.01 \times 10^5 = 2.63\times 10^8 Pa$

$B = 2.3 \times 10^9 N/m^2$

now density is given as

$\rho = \frac{1050}{1 - \frac{2.63\times 10^8}{2.3 \times 10^9}}$

$\rho = 1185.3 kg/m^3$

as we know that pressure changes with depth as per following equation

$P = P_o + \rho g h$

here we know that

$P = 3 P_0$

now we will have

$3P_0 = P_0 + \rho g h$

$2P_0 = \rho g h$

$2(1.01 \times 10^5) = 1025 (9.81)(h)$

here we will have

$h = 20.1 m$

so it is 20.1 m below the surface

Here net buoyancy force due to water and oil will balance the weight of the block

so here we will have

$mg = \rho_1V_1g + \rho_2V_2g$

$A(0.0476)979 = 922(A)(0.0476 - x) + 1000(A)(x)$

$46.6 = 43.89 - 922x + 1000x$

$x = 3.48 cm$

so it is 3.48 cm below the interface

5 0

3 years ago

How much time is required for reflected sunlight to travel from the Moon to Earth if the distance between Earth and the Moon is

mafiozo [28]

1.3 second of time will be required for reflected sunlight to travel from the Moon to Earth if the distance between Earth and the Moon is 3.85 × 105 km

<h3>What is Speed ?</h3>

Speed is the distance travelled per time taken. It is a scalar quantity. And the S.I unit is meter per second. That is, m/s

In the given question, we want to find how much time is required for reflected sunlight to travel from the Moon to Earth if the distance between Earth and the Moon is 3.85 × 10^5 km.

What are the parameters to consider ?

The parameters are;

The distance S = 3.85 × $10^{5}$ km

The Speed of Light C = 3 × $10^{8}$ m/s

The time taken t = ?

Speed = distance S ÷ Time t

Convert kilometer to meter by multiplying it by 1000

C = S/t

3 × $10^{8}$ = 3.85 × $10^{8}$ / t

Make t the subject of formula

t = 3.85 × $10^{8}$ / 3 × $10^{8}$

t = 1.2833

t = 1.3 s

Therefore, 1.3 second of time will be required for reflected sunlight to travel from the Moon to Earth if the distance between Earth and the Moon is 3.85 × 105 km

Learn more about Speed here: brainly.com/question/4931057

#SPJ1

3 0

2 years ago