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3 years ago

14

1When you look through a magnifying glass, the objects you are looking at

1 answer:

Vadim26 [7]3 years ago

8 0

convex lenses.

eyeglasses with convex lenses increase refraction, and accordingly reduce the focal lenght

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A tennis ball is dropped from 1.13 m above the

Alex73 [517]

Answer:

-4.71 m/s

Explanation:

Given:

y₀ = 1.13 m

y = 0 m

v₀ = 0 m/s

a = -9.8 m/s²

Find: v

v² = v₀² + 2a (y − y₀)

v² = (0 m/s)² + 2(-9.8 m/s²) (0 m − 1.13 m)

v = -4.71 m/s

7 0

3 years ago

an electrically charged particle with mass m, velocity v, and charge q moves in a circle in magnetic field b. what is the formul

kolbaska11 [484]

Answer:

$r=\frac {mv}{qb}$

Explanation:

In this case, since the charged particle moves in circular motion, the centripetal force is equivalent to the magnetic force.

$\frac {mv^{2}}{r}=qvb\\\frac {mv}{r}=qb\\r=\frac {mv}{qb}$

5 0

3 years ago

Which of the following is formed by constructive erosion? 1. delta 2. gully 3. rill 4. glacier

lana [24]

1. Delta, is formed by constructive erosion.

7 0

3 years ago

What is kelvin scale?

Darina [25.2K]

<span>a scale of temperature with absolute zero as zero, and the triple point of water as exactly 273.16 degrees.</span>

4 0

4 years ago

Read 2 more answers

1. Bone has a Young’s modulus of about

Blababa [14]

#1

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

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

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

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

$\Delta L = 5.04 mm$

#2

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

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

#3

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

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

$\rho = 1185.3 kg/m^3$

#4

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

#5

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